14_Petrucci10e_CSM K 6.85 M All 1B CSM
User Manual: K 6.85 M
Open the PDF directly: View PDF 
.
Page Count: 486
| Download | |
| Open PDF In Browser | View PDF |
CHAPTER 14
CHEMICAL KINETICS
PRACTICE EXAMPLES
1A
(E) The rate of consumption for a reactant is expressed as the negative of the change in molarity
divided by the time interval. The rate of reaction is expressed as the rate of consumption of a
reactant or production of a product divided by its stoichiometric coefficient.
A 0.3187 M 0.3629 M 1min
rate of consumption of A =
=
= 8.93 105 M s 1
t
8.25 min
60 s
rate of reaction = rate of consumption of A2 =
8.93 105 M s 1
4.46 105 M s 1
2
1B
(E) We use the rate of reaction of A to determine the rate of formation of B, noting from the
balanced equation that 3 moles of B form (+3 moles B) when 2 moles of A react (–2 moles A).
(Recall that “M” means “moles per liter.”)
0.5522 M A 0.5684 M A 3moles B
rate of B formation=
1.62 104 M s 1
60s
2 moles A
2.50 min
1min
2A
(M)
(a) The 2400-s tangent line intersects the 1200-s vertical line at 0.75 M and reaches 0 M at
3500 s. The slope of that tangent line is thus
0 M 0.75 M
slope =
= 3.3 104 M s 1 = instantaneous rate of reaction
3500 s 1200 s
The instantaneous rate of reaction = 3.3 104 M s 1 .
(b)
At 2400 s, H 2 O 2 = 0.39 M. At 2450 s, H 2 O 2 = 0.39 M + rate t
At 2450 s, H 2 O 2
= 0.39 M + 3.3 10 4 mol H 2 O 2 L1s 1 50s
= 0.39 M 0.017 M = 0.37 M
2B
(M) With only the data of Table 14.2 we can use only the reaction rate during the first 400 s,
H 2 O 2 /t = 15.0 104 M s 1 , and the initial concentration, H 2 O 2 0 = 2.32 M.
We calculate the change in H 2 O 2 and add it to H 2 O 2 0 to determine H 2 O 2 100 .
H 2 O 2 = rate of reaction of H 2 O 2 t = 15.0 104 M s 1 100 s = 0.15 M
H 2O2 100 = H 2O 2 0 + H 2O2 = 2.32 M + 0.15 M = 2.17 M
This value differs from the value of 2.15 M determined in text Example 14-2b because
the text used the initial rate of reaction 17.1104 M s 1 , which is a bit faster than the average
rate over the first 400 seconds.
611
Chapter 14: Chemical Kinetics
3A
(M) We write the equation for each rate, divide them into each other, and solve for n.
R1 = k N 2 O5 1 = 5.45 105 M s 1 = k 3.15 M
n
n
R2 = k N 2 O5 2 = 1.35 105 M s 1 = k 0.78 M
n
n
n
k N 2 O5 1
k 3.15 M
R1 5.45 105 M s 1
n
3.15
=
=
4.04
=
=
=
= 4.04
n
n
5
1
R2 1.35 10 M s
0.78
k N 2 O5 2
k 0.78 M
n
n
We kept an extra significant figure (4) to emphasize that the value of n = 1. Thus, the reaction is
first-order in N 2 O5 .
3B
1
2
(E) For the reaction, we know that rate = k HgCl2 C 2O 4 . Here we will compare Expt. 4 to
Expt. 1 to find the rate.
2
2
1
2
2
0.025 M 0.045 M
rate 4 k HgCl2 C 2 O 4
rate4
=
=
= 0.0214 =
2
2
1
2
rate1 k HgCl C O
1.8 105 M min 1
0.105 M 0.150 M
2 2 4
The desired rate is rate4 = 0.0214 1.8 105 M min 1 = 3.9 107 M min 1 .
4A
(E) We place the initial concentrations and the initial rates into the rate law and solve for k .
2
2
rate = k A B = 4.78 102 M s 1 = k 1.12 M 0.87 M
4.78 102 M s 1
k=
4B
5A
1.12 M
= 4.4 102 M 2 s 1
0.87 M
2
1
2
(E) We know that rate = k HgCl2 C2 O 4 and k = 7.6 103 M 2 min 1 .
Thus, insertion of the starting concentrations and the k value into the rate law yields:
1
2
Rate = 7.6 103 M 2 min 1 0.050 M 0.025 M = 2.4 107 M min 1
(E) Here we substitute directly into the integrated rate law equation.
ln A t = kt + ln A 0 = 3.02 103 s1 325 s + ln 2.80 = 0.982 +1.030 = 0.048
b g
At= e
5B
2
0.048
= 1.0 M
(M) This time we substitute the provided values into text Equation 14.13.
HO
0.443
1.49 M
2 2 t
k=
ln
= kt = k 600 s = ln
= 0.443
= 7.38 104 s 1
600 s
HO
2.32 M
2 2 0
Now we choose H 2 O2 0 = 1.49 M, H 2 O 2 t = 0.62,
ln
H 2 O 2 t
2 0
H 2O
= kt = k 1200 s = ln
0.62 M
= 0.88
1.49 M
t = 1800 s 600 s = 1200 s
k=
0.88
= 7.3 104 s 1
1200 s
These two values agree within the limits of the experimental error and thus, the reaction is firstorder in [H2O2].
612
Chapter 14: Chemical Kinetics
6A
(M) We can use the integrated rate equation to find the ratio of the final and initial
concentrations. This ratio equals the fraction of the initial concentration that remains at time t.
A t
= kt = 2.95 103 s 1 150 s = 0.443
A
0
A t
= e 0.443 = 0.642; 64.2% of A 0 remains.
A
0
ln
6B
(M) After two-thirds of the sample has decomposed, one-third of the sample remains.
Thus H 2 O 2 t = H 2 O 2 0 3 , and we have
H 2 O 2 t
2 0
ln
H2O
t=
7A
= kt = ln
H 2 O 2 0 3
H 2O
2 0
= ln 1/ 3 = 1.099 = 7.30 104 s 1t
1.099
1 min
= 1.51103 s
= 25.1 min
4 1
7.30 10 s
60 s
(M) At the end of one half-life the pressure of DTBP will have been halved, to 400 mmHg. At the
end of another half-life, at 160 min, the pressure of DTBP will have halved again, to 200 mmHg.
Thus, the pressure of DTBP at 125 min will be intermediate between the pressure at
80.0 min (400 mmHg) and that at 160 min (200 mmHg). To obtain an exact answer, first we
determine the value of the rate constant from the half-life.
k=
ln
0.693
0.693
=
= 0.00866 min 1
80.0 min
t1/2
PDTBP t
PDTBP 0
= kt = 0.00866 min 1 125 min = 1.08
PDTBP t
= e 1.08 = 0.340
PDTBP 0
PDTBP t = 0.340 PDTBP 0 = 0.340 800 mmHg = 272 mmHg
7B
(M)
(a) We use partial pressures in place of concentrations in the integrated first-order rate
equation. Notice first that more than 30 half-lives have elapsed, and thus the ethylene oxide
30
pressure has declined to at most 0.5 = 9 1010 of its initial value.
b g
ln
P30
P
3600 s
= kt = 2.05 104 s 1 30.0 h
= 22.1 30 = e 22.1 = 2.4 1010
P0
P0
1h
P30 = 2.4 1010 P0 = 2.4 1010 782 mmHg = 1.9 107 mmHg
613
Chapter 14: Chemical Kinetics
Pethylene oxide initially 782 mmHg 1.9 107 mmHg (~ 0). Essentially all of the ethylene
oxide is converted to CH4 and CO. Since pressure is proportional to moles, the final
pressure will be twice the initial pressure (1 mole gas 2 moles gas;
782 mmHg 1564 mmHg). The final pressure will be 1.56 103 mmHg.
(b)
(D) We first begin by looking for a constant rate, indicative of a zero-order reaction. If the rate is
constant, the concentration will decrease by the same quantity during the same time period. If we
choose a 25-s time period, we note that the concentration decreases 0.88 M 0.74 M = 0.14 M
during the first 25 s, 0.74 M 0.62 M = 0.12 M during the second 25 s,
0.62 M 0.52 M =
0.10 M during the third 25 s, and 0.52 M 0.44 M = 0.08 M during the
fourth 25-s period. This is hardly a constant rate and we thus conclude that the reaction is not
zero-order.
We next look for a constant half-life, indicative of a first-order reaction. The initial concentration
of 0.88 M decreases to one half of that value, 0.44 M, during the first 100 s, indicating a 100-s
half-life. The concentration halves again to 0.22 M in the second 100 s, another 100-s half-life.
Finally, we note that the concentration halves also from 0.62 M at 50 s to 0.31 M at 150 s, yet
another 100-s half-life. The rate is established as first-order. The rate constant is
0.693 0.693
k=
=
= 6.93 103 s1 .
t1/2
100 s
That the reaction is first-order is made apparent by the fact that the ln[B] vs time plot is a straight
line with slope = -k (k = 6.85 103 s1).
Plot of 1/[B] versus
Time
Plot of ln([B]) versus
Time
Plot of [B]
versus Time
7
0
-0.2
0.75
-0.4
0.65
-0.6
0
200
6
5
-0.8
ln([B])
0.55
0.45
-1
-1.2
0.35
400
1/[B] (M -1)
0.85
[B] (M)
8A
4
3
-1.4
2
-1.6
0.25
-1.8
0.15
0
100
200
Time(s)
300
1
-2
0
Time(s)
614
100
200
Tim e (s)
300
Chapter 14: Chemical Kinetics
8B
(D) We plot the data in three ways to determine the order. (1) A plot of [A] vs. time is linear if
the reaction is zero-order. (2) A plot of ln [A] vs. time will be linear if the reaction is first-order.
(3) A plot of 1/[A] vs. time will be linear if the reaction is second-order. It is obvious from the
plots below that the reaction is zero-order. The negative of the slope of the line equals
k = 0.083 M 0.250 M 18.00 min = 9.28 103 M/min (k = 9.30 103 M/min using a
graphical approach).
Plot of [A]
versus Time
0.25
0
10
Plot of ln([A])
versus Time
-1.5
0.23
0.21
20
12
Plot of 1/[A]
versus Time
11
10
-1.7
ln([A])
0.17
0.15
1/[A] (M-1)
[A] (M)
0.19
-1.9
9
8
7
-2.1
0.13
6
0.11
-2.3
5
0.09
y = -0.00930x + 0.2494
4
0.07
0
9A
10
Time (min)
-2.5
20
0
Time (min)
10
Time (min)
20
(M) First we compute the value of the rate constant at 75.0 C with the Arrhenius equation. We
know that the activation energy is Ea = 1.06 105 J/mol, and that k = 3.46 105 s1 at 298 K.
The temperature of 75.0 o C = 348.2 K.
ln
FG
H
IJ
K
FG
H
IJ
K
k2
k2
Ea 1 1
1.06 105 J / mol
1
1
= ln
=
=
= 6.14
5 1
1
1
k1
3.46 10 s
R T1 T2
8.3145 J mol K 298.2 K 348.2 K
k2 = 3.46 105 s 1 e +6.14 = 3.46 105 s 1 4.6 102 = 0.016 s 1
t1/2 =
9B
0.693
0.693
=
= 43 s at 75 C
k
0.016 s 1
(M) We use the integrated rate equation to determine the rate constant, realizing that one-third
remains when two-thirds have decomposed.
N 2 O5 t
5 0
ln
k=
N 2O
= ln
N 2 O5 0 3
N 2O
5 0
1
= ln = kt = k 1.50 h = 1.099
3
1.099
1h
= 2.03 104 s 1
1.50 h 3600 s
615
Chapter 14: Chemical Kinetics
Now use the Arrhenius equation to determine the temperature at which the rate constant is 2.04
104 s1.
FG
H
FG
H
IJ
K
k2
2.04 104 s1
Ea 1 1
1.06 105 J / mol
1
1
ln = ln
= 1.77 =
=
5 1
1
1
k1
3.46 10 s
R T1 T2
8.3145 J mol K 298 K T2
1
1
1.77 8.3145 K 1
=
= 3.22 103 K 1
5
1.06 10
T2 298 K
IJ
K
T2 = 311 K
10A (M) The two steps of the mechanism must add, in a Hess's law fashion, to produce the overall
reaction.
Overall reaction: CO + NO 2
CO 2 + NO
or
CO + NO 2
CO 2 + NO
Second step: NO3 + CO
NO 2 + CO 2 or + NO 2 + CO 2
NO3 + CO
2 NO 2
NO + NO 3
First step:
If the first step is the slow step, then it will be the rate-determining step, and the rate of that step
2
will be the rate of the reaction, namely, rate of reaction = k1 NO2 .
10B (M)
(1) The steps of the mechanism must add, in a Hess's law fashion, to produce the overall reaction.
This is done below. The two intermediates, NO 2 F2 g and F(g), are each produced in one step
and consumed in the next one.
bg
(2)
Fast:
NO 2 g + F2 g
NO2 F2 g
Slow:
Fast:
Net:
NO 2 F2 g NO 2 F g + F g
F g + NO 2 g NO 2 F g
2 NO 2 g + F2 g 2 NO 2 F g
bg
bg bg
bg bg
bg
bg bg
bg
The proposed mechanism must agree with the rate law. We expect the rate-determining step
to determine the reaction rate: Rate = k3 NO 2 F2 . To eliminate NO 2 F2 , we recognize that
the first elementary reaction is very fast and will have the same rate forward as
reverse: R f = k1 NO 2 F2 = k2 NO 2 F2 = Rr . We solve for the concentration of
intermediate: NO 2 F2 = k1 NO 2 F2 /k2 . We now substitute this expression for NO 2 F2
into the rate equation: Rate = k1k3 /k2 NO 2 F2 . Thus the predicted rate law agrees with
the experimental rate law.
616
Chapter 14: Chemical Kinetics
INTEGRATIVE EXAMPLE
A.
(M)
(a) The time required for the fixed (c) process of souring is three times as long at 3 °C refrigerator
temperature (276 K) as at 20 °C room temperature (293 K).
E 1 1 E 1
64 h
1 Ea
c / t2
t
4
ln
ln 1 ln
1.10 a a
( 2.10 10 )
3 64 h
c / t1
t2
R T1 T2 R 293 K 276 K R
1.10 R
1.10 8.3145 J mol1 K 1
4.4 104 J/mol 44 kJ/mol
2.10 104 K 1
2.10 104 K 1
(b) Use the Ea determined in part (a) to calculate the souring time at 40 °C = 313 K.
E 1 1
t
4.4 104 J/mol 1
1
t1
ln 1 a
1.15 ln
1
1
t2 R T1 T2 8.3145 J mol K 313 K 293 K
64 h
Ea
t1
e 1.15 0.317 t1 0.317 64 h 20. h
64 h
B.
(M) The species A* is a reactive intermediate. Let’s deal with this species by using a steady state
approximation.
d[A*]/dt = 0 = k1[A]2 – k-1[A*][A] – k2[A*]. Solve for [A*]. k-1[A*][A] + k2[A*] = k1[A]2
k1 [A]2
k 2 k1 [A]2
[A*] =
The rate of reaction is: Rate = k2[A*] =
k -1 [A] k 2
k -1 [A] k 2
At low pressures ([A] ~ 0 and hence k2 >> k-1[A]), the denominator becomes ~ k2 and the rate law is
k k [A]2
Rate = 2 1
= k1 [A]2 Second-order with respect to [A]
k2
At high pressures ([A] is large and k-1[A] >> K2), the denominator becomes ~ k-1[A] and the rate law
is
k k [A]2
k k [A]
Rate = 2 1
= 2 1
First-order with respect to [A]
k -1 [A]
k -1
EXERCISES
Rates of Reactions
1.
(M) 2A + B C + 3D
(a)
(b)
(c)
[A]
= 6.2 104 M s1
t
1 [A]
= 1/2(6.2 104 M s1) = 3.1 104 Ms1
2 t
1 [A]
Rate of disappearance of B =
= 1/2(6.2 104 M s1) = 3.1 104 Ms1
2 t
3 [A]
Rate of appearance of D =
= 3(6.2 104 M s1) = 9.3 104 Ms1
2 t
Rate =
617
Chapter 14: Chemical Kinetics
2.
(M) In each case, we draw the tangent line to the plotted curve.
H 2 O 2 1.7 M 0.6 M
(a) The slope of the line is
=
= 9.2 104 M s 1
t
400 s 1600 s
H 2O2
reaction rate =
= 9.2 104 M s 1
t
(b)
Read the value where the horizontal line [H2O2] = 0.50 M intersects the curve,
2150 s or 36 minutes
3.
(E) Rate =
4.
(M)
(a)
(b)
5.
(M)
(a)
(b)
A
0.474 M 0.485 M = 1.0 103 M s 1
=
t
82.4 s 71.5 s
Rate =
A
0.1498 M 0.1565 M
=
= 0.0067 M min 1
t
1.00 min 0.00 min
Rate =
A
0.1433 M 0.1498 M
=
= 0.0065 M min 1
t
2.00 min 1.00 min
The rates are not equal because, in all except zero-order reactions, the rate depends on the
concentration of reactant. And, of course, as the reaction proceeds, reactant is consumed
and its concentration decreases, so the rate of the reaction decreases.
A = A i + A = 0.588 M 0.013 M = 0.575 M
A = 0.565 M 0.588 M = 0.023 M
t
0.023 M
t = A
=
= 1.0 min
A 2.2 102 M/min
time = t + t = 4.40 +1.0 min = 5.4 min
6.
(M) Initial concentrations are HgCl 2 = 0.105 M and C 2 O 4 2 = 0.300 M.
The initial rate of the reaction is 7.1 105 M min 1 . Recall that the reaction is:
2 HgCl2 (aq) + C2 O 42 (aq) 2 Cl (aq) + 2 CO 2 (g) + Hg 2 Cl2 (aq) .
The rate of reaction equals the rate of disappearance of C2 O 24 . Then, after 1 hour, assuming that
the rate is the same as the initial rate,
(a)
(b)
mol C 2 O 4 2 2 mol HgCl2
60 min
1 h
0.096 M
2
L s
1h
1 mol C2 O 4
FG
H
mol
60 min
1 h
= 0.296 M
L min
1h
HgCl2 = 0.105 M 7.1 105
C 2 O 24 = 0.300 M 7.1 105
IJ
K
618
Chapter 14: Chemical Kinetics
7.
(M)
(a)
Rate =
C
(b)
(c)
8.
(M)
t
=
C
2t
= 1.76 105 M s 1
= 2 1.76 105 M s 1 = 3.52 105 M/s
t
A
C
=
= 1.76 105 M s1 Assume this rate is constant.
t
2 t
60 s
A = 0.3580 M + 1.76 105 M s1 1.00 min
= 0.357 M
1min
A
= 1.76 105 M s 1
t
A
0.3500 M 0.3580 M
t =
=
= 4.5 102 s
5
5
1.76 10 M/s
1.76 10 M/s
(a)
n O 2
5.7 104 mol H 2 O 2
1 mol O 2
= 1.00 L soln
= 2.9 104 mol O 2 /s
t
1 L soln s
2 mol H 2 O 2
(b)
n O 2
mol O 2 60 s
= 2.9 104
= 1.7 102 mol O 2 / min
1 min
t
s
(c)
9.
A
V O 2
mol O 2 22,414 mL O 2 at STP 3.8 102 mL O 2 at STP
= 1.7 102
=
1 mol O 2
t
min
min
(M) Notice that, for every 1000 mmHg drop in the pressure of A(g), there will be a corresponding
2000 mmHg rise in the pressure of B(g) plus a 1000 mmHg rise in the pressure of C(g).
(a)
We set up the calculation with three lines of information below the balanced equation: (1)
the initial conditions, (2) the changes that occur, which are related to each other by reaction
stoichiometry, and (3) the final conditions, which simply are initial conditions + changes.
A(g)
2B(g)
+
C(g)
Initial
1000. mmHg
0. mmHg
0. mmHg
Changes
–1000. mmHg
+2000. mmHg
+1000. mmHg
Final
0. mmHg
2000. MmHg
1000. mmHg
Total final pressure = 0. mmHg + 2000. mmHg +1000. mmHg = 3000. mmHg
(b)
Initial
Changes
Final
A(g)
1000. mmHg
–200. mmHg
800 mmHg
2B(g)
+
0. mmHg
+400. mmHg
400. mmHg
C(g)
0. mmHg
+200. mmHg
200. mmHg
Total pressure = 800. mmHg + 400. mmHg + 200. mmHg = 1400. mmHg
619
Chapter 14: Chemical Kinetics
10.
(M)
(a) We will use the ideal gas law to determine N 2 O5 pressure
1 mol N 2 O5
1.00 g× 108.0 g
nRT
P{N 2 O5 } =
=
V
(b)
(c)
L atm
×0.08206 mol K × 273 + 65 K
760 mmHg
×
= 13 mmHg
15 L
1 atm
After 2.38 min, one half-life passes. The initial pressure of N 2 O5 decreases by half to 6.5
mmHg.
From the balanced chemical equation, the reaction of 2 mol N 2 O5 g produces 4 mol
bg
bg
bg
NO 2 g and 1 mol O 2 g . That is, the consumption of 2 mol of reactant gas produces 5
mol of product gas. When measured at the same temperature and confined to the same
volume, pressures will behave as amounts: the reaction of 2 mmHg of reactant produces 5
mmHg of product.
5 mmHg(product)
Ptotal = 13 mmHg N 2 O5 (initially) 6.5 mmHg N 2 O5 (reactant) + 6.5 mmHg(reactant)
2 mmHg(reactant)
(13 6.5 16) mmHg 23 mmHg
Method of Initial Rates
11.
(M)
(a) From Expt. 1 to Expt. 3, [A] is doubled, while [B] remains fixed. This causes the rate to
6.75 104 M s1
increases by a factor of
= 2.01 2 .
3.35 104 M s1
Thus, the reaction is first-order with respect to A.
From Expt. 1 to Expt. 2, [B] doubles, while [A] remains fixed. This causes the rate to
1.35 103 M s1
= 4.03 4 .
increases by a factor of
3.35 104 M s1
Thus, the reaction is second-order with respect to B.
(b) Overall reaction order order with respect to A order with respect to B = 1 2 = 3. The
reaction is third-order overall.
2
(c) Rate = 3.35 104 M s1 = k 0.185 M 0.133 M
b
k=
12.
3.35 10
4
Ms
1
b0.185 Mgb0.133 Mg
2
gb
g
= 0.102 M 2 s1
(M) From Expt. 1 and Expt. 2 we see that [B] remains fixed while [A] triples. As a result, the
initial rate increases from 4.2 103 M/min to 1.3 102 M/min, that is, the initial reaction rate
triples. Therefore, the reaction is first-order in [A]. Between Expt. 2 and Expt. 3, we see that [A]
doubles, which would double the rate, and [B] doubles. As a consequence, the initial rate goes
from 1.3 102 M/min to 5.2 10 2 M/min, that is, the rate quadruples. Since an additional
doubling of the rate is due to the change in [B], the reaction is first-order in [B]. Now we
determine the value of the rate constant.
620
Chapter 14: Chemical Kinetics
Rate = k A
1
B
1
5.2 102 M / min
Rate
= 5.8 103 L mol 1min 1
=
k=
3.00 M 3.00 M
A B
c
h
The rate law is Rate = 5.8 103 L mol 1min 1 A
13.
1
B .
(M) From Experiment 1 to 2, [NO] remains constant while [Cl2] is doubled. At the same time the
initial rate of reaction is found to double. Thus, the reaction is first-order with respect to [Cl2],
since dividing reaction 2 by reaction 1 gives 2 = 2x when x = 1. From Experiment 1 to 3, [Cl2]
remains constant, while [NO] is doubled, resulting in a quadrupling of the initial rate of reaction.
Thus, the reaction must be second-order in [NO], since dividing reaction 3 by reaction 1 gives
4 = 2x when x = 2. Overall the reaction is third-order: Rate = k [NO]2[Cl2]. The rate constant
may be calculated from any one of the experiments. Using data from Exp. 1,
k=
14.
1
Rate
2.27 105 M s 1
=
= 5.70 M2 s1
[NO]2 [Cl2 ] (0.0125 M) 2 (0.0255 M)
(M)
(a) From Expt. 1 to Expt. 2, [B] remains constant at 1.40 M and [C] remains constant at
1.00 M, but [A] is halved 0.50 . At the same time the rate is halved 0.50 . Thus, the
b
b
g
g
x
reaction is first-order with respect to A, since 0.50 = 0.50 when x = 1.
From Expt. 2 to Expt. 3, [A] remains constant at 0.70 M and [C] remains constant at
1.00 M, but [B] is halved 0.50 , from 1.40 M to 0.70 M. At the same time, the rate is
b
b
g
g
quartered 0.25 . Thus, the reaction is second-order with respect to B, since 0.50 y = 0.25
when y = 2 .
From Expt. 1 to Expt. 4, [A] remains constant at 1.40 M and [B] remains constant at
1.40 M, but [C] is halved 0.50 , from 1.00 M to 0.50 M. At the same time, the rate is
increased by a factor of 2.0.
1
1
Rate
=
16
Rate
=
16
Rate
=
4
Rate
=
4
Rate1 = 2 Rate1.
4
3
2
2
4
2
Thus, the order of the reaction with respect to C is 1 , since 0.5z = 2.0 when z = 1 .
(b)
rate5 = k 0.70 M 0.70 M 0.50 M
1
k
11
2
2
1.40 M
1 12
2
1.40 M
2 1 1
2
1
1
1.00 M
1
621
2
1+2 1
1
= rate1
2
This is based on rate1 = k 1.40 M 1.40 M 1.00 M
1
2
1.40 M 1.40 M 1.00 M
= k
2 2 2
1
2
1
1
= rate1 = 14 rate1
2
Chapter 14: Chemical Kinetics
First-Order Reactions
15.
16.
17.
(E)
(a)
TRUE
The rate of the reaction does decrease as more and more of B and C are formed,
but not because more and more of B and C are formed. Rather, the rate decreases
because the concentration of A must decrease to form more and more of B and
C.
(b)
FALSE
The time required for one half of substance A to react—the half-life—is
independent of the quantity of A present.
(E)
(a)
FALSE
(b)
TRUE
For first-order reactions, a plot of ln [A] or log [A] vs. time yields a straight line.
A graph of [A] vs. time yields a curved line.
The rate of formation of C is related to the rate of disappearance of A by the
stoichiometry of the reaction.
(M)
(a) Since the half-life is 180 s, after 900 s five half-lives have elapsed, and the original quantity
of A has been cut in half five times.
5
final quantity of A = 0.5 initial quantity of A = 0.03125 initial quantity of A
About 3.13% of the original quantity of A remains unreacted after 900 s.
or
More generally, we would calculate the value of the rate constant, k, using
ln 2 0.693
=
= 0.00385 s 1 Now ln(% unreacted) = -kt = -0.00385 s-1×(900s) = k=
180 s
t1/2
3.465
(% unreacted) = 0.0313 × 100% = 3.13% of the original quantity.
b g
(b)
18.
Rate = k A = 0.00385 s1 0.50 M = 0.00193 M / s
(M)
(a) The reaction is first-order, thus
A t
0.100 M
ln
= kt = ln
= 54 min(k )
0.800 M
A 0
k=
2.08
= 0.0385 min 1
54 min
We may now determine the time required to achieve a concentration of 0.025 M
A t
3.47
0.025 M
t=
ln
= kt = ln
= 0.0385 min 1 (t )
= 90. min
0.0385 min 1
0.800 M
A 0
(b)
Since we know the rate constant for this reaction (see above),
1
Rate = k A = 0.0385 min 1 0.025 M = 9.6 104 M/min
622
Chapter 14: Chemical Kinetics
19.
(M)
(a) The mass of A has decreased to one fourth of its original value, from 1.60 g to
1 1 1
0.40 g. Since
= , we see that two half-lives have elapsed.
4 2 2
Thus, 2 t1/2 = 38 min, or t1/2 = 19 min .
A t
0.693
(b) k = 0.693/t1/2 =
= 0.036 min 1 ln
= kt = 0.036 min 1 60 min = 2.2
19 min
A 0
A t
=
A 0
20.
(M)
A t
A 0
e 2.2 = 0.11 or A t = A 0 e kt = 1.60 g A 0.11 = 0.18 g A
= kt = ln
k=
0.256
= 0.0160 min 1
16.0 min
ln
(b)
t1/2 =
(c)
We need to solve the integrated rate equation to find the elapsed time.
ln
(d)
ln
0.693
0.693
=
= 43.3 min
k
0.0160 min 1
A t
A 0
= kt = ln
0.235 M
= 1.245 = 0.0160 min 1 t
0.816 M
t=
1.245
= 77.8 min
0.0160 min 1
A
A = e kt which in turn becomes
= kt becomes
A0
A 0
A = A
0
21.
0.632 M
= 0.256
0.816 M
(a)
e kt = 0.816 M exp 0.0160 min 1 2.5 h
60 min
1h
= 0.816 0.0907 = 0.074 M
(M) We determine the value of the first-order rate constant and from that we can calculate the
half-life. If the reactant is 99% decomposed in 137 min, then only 1% (0.010) of the initial
concentration remains.
A t
4.61
0.010
k=
ln
= kt = ln
= 4.61 = k 137min
= 0.0336 min 1
137 min
1.000
A 0
t1/2 =
0.0693
0.693
=
= 20.6 min
k
0.0336 min 1
623
Chapter 14: Chemical Kinetics
(E) If 99% of the radioactivity of
32
22.
P is lost, 1% (0.010) of that radioactivity remains. First we
0.693 0.693
compute the value of the rate constant from the half-life. k =
=
= 0.0485 d 1
t1/2
14.3 d
Then we use the integrated rate equation to determine the elapsed time.
At
At
1
1
0.010
ln
= kt t = ln
=
= 95 days
ln
1
0.0485 d
1.000
k
A0
A0
23.
(D)
(a)
(b)
35
100 [A]0
3
1
ln
= ln(0.35) = kt = (4.81 10 min )t t = 218 min.
[A]
0
Note: We did not need to know the initial concentration of acetoacetic acid to answer the
question.
Let’s assume that the reaction takes place in a 1.00L container.
10.0 g acetoacetic acid
1 mol acetoacetic acid
102.090 g acetoacetic acid
= 0.09795 mol acetoacetic acid.
After 575 min. (~ 4 half lives, hence, we expect ~ 6.25% remains as a rough
approximation), use integrated form of the rate law to find [A]t = 575 min.
[A]t
3
1
ln
= kt = (4.81 10 min )(575 min) = 2.766
[A]0
[A]t
[A]t
= e2.766 = 0.06293 (~ 6.3% remains)
= 0.063 [A]t = 6.2 103
[A]0
0.09795 moles
moles.
[A]reacted = [A]o [A]t = (0.098 6.2 103) moles = 0.092 moles acetoacetic acid. The
stoichiometry is such that for every mole of acetoacetic acid consumed, one mole of CO2
forms. Hence, we need to determine the volume of 0.0918 moles CO2 at 24.5 C (297.65
K) and 748 torr (0.984 atm) by using the Ideal Gas law.
L atm
0.0918 mol 0.08206
297.65 K
K mol
nRT
= 2.3 L CO2
=
V=
P
0.984 atm
24.
(M)
ln
(a)
A t
A 0
= kt = ln
2.5 g
= 3.47 = 6.2 104 s 1t
80.0 g
3.47
= 5.6 103 s 93 min
6.2 104 s 1
We substituted masses for concentrations, because the same substance (with the same
molar mass) is present initially at time t , and because it is a closed system.
t=
624
Chapter 14: Chemical Kinetics
(b)
25.
(D)
(a)
1 mol N 2 O5
1 mol O 2
= 0.359 mol O 2
108.0 g N 2 O5 2 mol N 2 O5
L atm
0.359 mol O 2 0.08206
(45 273) K
nRT
mol K
V
9.56 L O 2
1 atm
P
745 mmHg
760 mmHg
amount O 2 = 77.5 g N 2 O5
If the reaction is first-order, we will obtain the same value of the rate constant from several
sets of data.
A t
0.497 M
0.188
ln
= kt = ln
= k 100 s = 0.188, k =
= 1.88 103 s 1
A
0.600
M
100
s
0
A t
0.344 M
= kt = ln
= k 300 s = 0.556,
0.600 M
A 0
A t
0.285 M
ln
= kt = ln
= k 400 s = 0.744,
0.600 M
A 0
A t
0.198 M
ln
= kt = ln
= k 600 s = 1.109,
0.600 M
A 0
A t
0.094 M
ln
= kt = ln
= k 1000 s = 1.854,
0.600 M
A 0
ln
(b)
(c)
26.
(D)
(a)
k=
0.556
= 1.85 103 s 1
300 s
k=
0.744
= 1.86 103 s 1
400 s
k=
1.109
= 1.85 103 s 1
600 s
k=
1.854
= 1.85 103 s 1
1000 s
The virtual constancy of the rate constant throughout the time of the reaction confirms that
the reaction is first-order.
For this part, we assume that the rate constant equals the average of the values obtained in
part (a).
1.88 +1.85 +1.86 +1.85
k=
103 s1 = 1.86 103 s1
4
We use the integrated first-order rate equation:
A 750 = A 0 exp kt = 0.600 M exp 1.86 103 s 1 750 s
A 750 = 0.600 M e 1.40 = 0.148 M
If the reaction is first-order, we will obtain the same value of the rate constant from several
sets of data.
0.167
P
264 mmHg
ln t = kt = ln
= k 390 s = 0.167 ,
k=
= 4.28 104 s1
390 s
P0
312 mmHg
0.331
P
224 mmHg
k=
= 4.26 104 s1
ln t = kt = ln
= k 777 s = 0.331 ,
777 s
312 mmHg
P0
0.512
P
187 mmHg
k=
= 4.28 104 s1
ln t = kt = ln
= k 1195 s = 0.512 ,
1195 s
312 mmHg
P0
625
Chapter 14: Chemical Kinetics
Pt
78.5 mmHg
1.38
= kt = ln
= k 3155 s = 1.38 ,
k=
= 4.37 104 s 1
312 mmHg
3155 s
P0
The virtual constancy of the rate constant confirms that the reaction is first-order.
ln
(b)
For this part we assume the rate constant is the average of the values in part
(a): 4.3 104 s1 .
(c)
At 390 s, the pressure of dimethyl ether has dropped to 264 mmHg. Thus, an amount of
dimethyl ether equivalent to a pressure of 312 mmHg 264 mmHg = 48 mmHg has
decomposed. For each 1 mmHg pressure of dimethyl ether that decomposes, 3 mmHg of
pressure from the products is produced. Thus, the increase in the pressure of the products is
3 48 = 144 mmHg. The total pressure at this point is 264 mmHg +144 mmHg = 408
mmHg. Below, this calculation is done in a more systematic fashion:
b
bCH g Obgg
3 2
Initial
312 mmHg
Changes – 48 mmHg
Final
264 mmHg
g
bg
CH 4 g
bg
+
0 mmHg
+ 48 mmHg
48 mmHg
H2 g
bg
+
CO g
0 mmHg
+ 48 mmHg
48 mmHg
0 mmHg
+ 48 mmHg
48 mmHg
Ptotal = PDME + Pmethane + Phydrogen + PCO
264 mmHg 48 mmHg 48 mmHg 48 mmHg 408 mmHg
(d) This question is solved in the same manner as part (c). The results are summarized below.
bCH g Obgg
3 2
bg
CH 4 g
Initial
312 mmHg
0 mmHg
Changes 312 mmHg 312 mmHg
Final
0 mmHg
312 mmHg
Ptotal = PDME + Pmethane + Phydrogen + PCO
bg
H2 g
bg
+ CO g
0 mmHg
312 mmHg
312 mmHg
0 mmHg
+ 312 mmHg
312 mmHg
0 mmHg 312 mmHg 312 mmHg 312 mmHg 936 mmHg
P
(e) We first determine PDME at 1000 s. ln 1000 = kt = 4.3 104 s1 1000 s = 0.43
P0
P1000
= e 0.43 = 0.65
P1000 = 312 mmHg 0.65 = 203 mmHg
P0
Then we use the same approach as was used for parts (c) and (d)
CH3 2 O g
Initial
312 mmHg
Changes 109 mmHg
CH 4 g
+
H2 g
0 mmHg
109 mmHg
0 mmHg
109 mmHg
Final
203 mmHg
109 mmHg
Ptotal = PDME + Pmethane +Phydrogen +PCO
109 mmHg
+
CO g
0 mmHg
109 mmHg
109 mmHg
= 203 mmHg +109 mmHg +109 mmHg +109 mmHg = 530. mmHg
626
Chapter 14: Chemical Kinetics
Reactions of Various Orders
27.
(M)
(a) Set II is data from a zero-order reaction. We know this because the rate of set II is constant.
0.25 M/25 s = 0.010 M s1 . Zero-order reactions have constant rates of reaction.
(b)
A first-order reaction has a constant half-life. In set I, the first half-life is slightly less than
75 sec, since the concentration decreases by slightly more than half (from 1.00 M to 0.47 M)
in 75 s. Again, from 75 s to 150 s the concentration decreases from 0.47 M to 0.22 M, again
by slightly more than half, in a time of 75 s. Finally, two half-lives should see the
concentration decrease to one-fourth of its initial value. This, in fact, is what we see. From
100 s to 250 s, 150 s of elapsed time, the concentration decreases from 0.37 M to 0.08 M,
i.e., to slightly less than one-fourth of its initial value. Notice that we cannot make the same
statement of constancy of half-life for set III. The first half-life is 100 s, but it takes more
than 150 s (from 100 s to 250 s) for [A] to again decrease by half.
(c)
For a second-order reaction,1/ A t 1/ A 0 = kt . For the initial 100 s in set III, we have
1
1
= 1.0 L mol1 = k 100 s,
k = 0.010 L mol1 s 1
0.50 M 1.00 M
For the initial 200 s, we have
1
1
= 2.0 L mol1 = k 200 s,
k = 0.010 L mol1 s 1
0.33 M 1.00 M
Since we obtain the same value of the rate constant using the equation for second-order
kinetics, set III must be second-order.
28.
(E) For a zero-order reaction (set II), the slope equals the rate constant:
k = A /t = 1.00 M/100 s = 0.0100 M/s
29.
(M) Set I is the data for a first-order reaction; we can analyze those items of data to determine the
half-life. In the first 75 s, the concentration decreases by a bit more than half. This implies a halflife slightly less than 75 s, perhaps 70 s. This is consistent with the other time periods noted in the
answer to Review Question 18 (b) and also to the fact that in the 150-s interval from 50 s to 200 s,
the concentration decreases from 0.61 M to 0.14 M, which is a bit more than a factor-of-four
decrease. The factor-of-four decrease, to one-fourth of the initial value, is what we would expect
for two successive half-lives. We can determine the half-life more accurately, by obtaining a
value of k from the relation ln A t / A 0 = kt followed by t1/2 = 0.693 / k For instance,
ln(0.78/1.00) = -k (25 s);
k = 9.94 ×10-3 s-1. Thus, t1/2 = 0.693/9.94 ×10-3 s-1 = 70 s.
30.
(E) We can determine an approximate initial rate by using data from the first 25 s.
A
0.80 M 1.00 M
Rate =
=
= 0.0080 M s 1
t
25 s 0 s
627
Chapter 14: Chemical Kinetics
31.
(M) The approximate rate at 75 s can be taken as the rate over the time period from 50 s to 100 s.
A
0.00 M 0.50 M
(a) Rate II =
=
= 0.010 M s1
100 s 50 s
t
(b)
Rate I =
A
0.37 M 0.61 M
=
= 0.0048 M s1
100
50
t
s s
A
0.50 M 0.67 M
=
= 0.0034 M s1
100 s 50 s
t
Alternatively we can use [A] at 75 s (the values given in the table) in the
m
relationship Rate = k A , where m = 0 , 1, or 2.
(c)
Rate III =
(a)
Rate II = 0.010 M s 1 0.25 mol/L = 0.010 M s 1
(b)
Since t1/2 70s, k 0.693 / 70s 0.0099s 1
0
Rate I 0.0099 s 1 (0.47 mol/L)1 0.0047 M s 1
(c)
32.
Rate III = 0.010 L mol1 s 1 0.57 mol/L = 0.0032 M s 1
2
(M) We can combine the approximate rates from Exercise 31, with the fact that 10 s have
elapsed, and the concentration at 100 s.
(a) A II = 0.00 M There is no reactant left after 100 s.
(b)
AI= A
(c)
A
III
100
= A
b
g
c
h
10 s rate = 0.37 M 10 s 0.0047 M s1 = 0.32 M
100
b
g
c
h
10 s rate = 0.50 M 10 s 0.0032 M s1 = 0.47 M
33.
(E) Substitute the given values into the rate equation to obtain the rate of reaction.
2
0
2
0
Rate = k A B = 0.0103 M 1min 1 0.116 M 3.83 M = 1.39 10 4 M / min
34.
(M)
(a) A first-order reaction has a constant half-life. Thus, half of the initial concentration remains
after 30.0 minutes, and at the end of another half-life—60.0 minutes total—half of the
concentration present at 30.0 minutes will have reacted: the concentration has decreased to
one-quarter of its initial value. Or, we could say that the reaction is 75% complete after two
half-lives—60.0 minutes.
c
(b)
gb
hb
g
A zero-order reaction proceeds at a constant rate. Thus, if the reaction is 50% complete in
30.0 minutes, in twice the time—60.0 minutes—the reaction will be 100% complete. (And
in one-fifth the time—6.0 minutes—the reaction will be 10% complete. Alternatively, we
can say that the rate of reaction is 10%/6.0 min.) Therefore, the time required for the
60.0 min
reaction to be 75% complete = 75%
= 45 min.
100%
628
Chapter 14: Chemical Kinetics
(M) For reaction: HI(g) 1/2 H2(g) 1/2 I2(g) (700 K)
Time
(s)
0
100
200
300
400
[HI] (M)
(D)
(a)
1.60
y = 0.00118x + 0.997
1.00
0.90
0.81
0.74
0.68
0
0.105
0.211
0.301
0.386
1.00
1.11
1.235
1.35
1.47
1.40
1.20
From data above, a plot of 1/[HI] vs. t yields
a straight line. The reaction is second-order
in HI at 700 K. Rate = k[HI]2. The slope of
the line = k = 0.00118 M1s1
1.00
0
250
Time(s)
500
We can graph 1/[ArSO2H] vs. time and obtain a straight line. We can also graph [ArSO2H]
vs. time and ln([ArSO2H]) vs. time to demonstrate that they do not yield a straight line.
Only the plot of 1/[ArSO2H] versus time is shown.
of 1/[ArSO
H] versus Time
Plot ofPlot
1/[ArSO
2H]2versus Time
50
-1
1/[ArSO2H] (M )
1/[ArSO2H] (M-1)
36.
Plot of 1/[HI] vs time
1/[HI](M1)
ln[HI]
1/[HI] (M-1)
35.
30
y = 0.137x + 9.464
10
0
50
100
150
Time (min)
200
250
300
The linearity of the line indicates that the reaction is second-order.
(b)
We solve the rearranged integrated second-order rate law for the rate constant, using the
longest time interval.
k=
(c)
1
1
= kt
At A0
F
GH
I
JK
1 1
1
=k
t At A0
1
1
1
1
1
= 0.137 L mol min
300 min 0.0196 M 0.100 M
We use the same equation as in part (b), but solved for t , rather than k .
1 1
1
1
1
1
t=
=
= 73.0 min
1
1
k A t A 0 0.137 L mol min 0.0500 M 0.100 M
629
Chapter 14: Chemical Kinetics
We use the same equation as in part (b), but solve for t , rather than k .
1 1
1
1
1
1
t=
=
= 219 min
1
1
k A t A 0 0.137 L mol min 0.0250 M 0.100 M
(e)
We use the same equation as in part (b), but solve for t , rather than k .
1 1
1
1
1
1
t=
=
= 136 min
1
1
k A t A 0 0.137 L mol min 0.0350 M 0.100 M
(M)
(a) Plot [A] vs t, ln[A] vs t, and 1/[A] vs t and see which yields a straight line.
0
0.8
y = -0.0050x + 0.7150
0.7
-0.5 0
R 2 = 1.0000
0.6
150
-1
0.5
ln[A]
[A] (mol/L)
0.4
0.3
-1.5
-2
-2.5
0.2
y = -0.0191x
- 0.1092
R2 = 0.9192
-3
0.1
-3.5
0
Time(s)
0
Time(s)
150
20
y = 0.1228x
- 0.9650
R2 = 0.7905
15
1/[A]
37.
(d)
10
5
0
0
Time(s)
150
Clearly we can see that the reaction is zero-order in reactant A with a rate constant of 5.0 ×10-3.
(b)
The half-life of this reaction is the time needed for one half of the initial [A] to react.
0.358 M
= 72 s.
Thus, A = 0.715 M 2 = 0.358 M and t1/2 =
5.0 103 M / s
630
Chapter 14: Chemical Kinetics
38.
(D)
(a)
We can either graph 1/ C 4 H 6 vs. time and obtain a straight line, or we can determine the
second-order rate constant from several data points. Then, if k indeed is a constant, the
reaction is demonstrated to be second-order. We shall use the second technique in this case.
First we do a bit of algebra.
1
1
= kt
At A0
F
GH
I
JK
1 1
1
=k
t At A0
IJ
FG
K
H
1
FG 1 1 IJ = 0.875 L mol min
k=
24.55 min H 0.0124 M 0.0169 M K
1
FG 1 1 IJ = 0.892 L mol min
k=
42.50 min H 0.0103 M 0.0169 M K
1
FG 1 1 IJ = 0.870 L mol min
k=
68.05 min H 0.00845 M 0.0169 M K
k=
1
1
1
= 0.843 L mol 1min 1
12.18 min 0.0144 M 0.0169 M
1
1
1
1
1
(b)
39.
1
The fact that each calculation generates similar values for the rate constant indicates that
the reaction is second-order.
The rate constant is the average of the values obtained in part (a).
0.843 + 0.875 + 0.892 + 0.870
k=
L mol1min 1 = 0.87 L mol1min 1
4
(c)
We use the same equation as in part (a), but solve for t , rather than k .
1 1
1
1
1
1
2
t=
=
= 2.0 10 min
1
1
k A t A 0 0.870 L mol min 0.00423 M 0.0169 M
(d)
We use the same equation as in part (a), but solve for t , rather than k .
1 1
1
1
1
1
2
t=
=
= 1.6 10 min
1
1
k A t A 0 0.870 L mol min 0.0050 M 0.0169 M
(E)
(a)
(b)
A
1.490 M 1.512 M
=
= +0.022 M / min
t
1.0 min 0.0 min
A
2.935 M 3.024 M
initial rate =
=
= +0.089 M / min
t
1.0 min 0.0 min
initial rate =
b g
When the initial concentration is doubled 2.0 , from 1.512 M to 3.024 M, the initial rate
b g
quadruples 4.0 . Thus, the reaction is second-order in A (since 2.0 x = 4.0 when x = 2 ).
631
Chapter 14: Chemical Kinetics
40.
(M)
(a) Let us assess the possibilities. If the reaction is zero-order, its rate will be constant. During
the first 8 min, the rate is 0.60 M 0.80 M /8 min = 0.03 M/min . Then, during the first
24 min, the rate is 0.35 M 0.80 M /24 min = 0.019 M/min . Thus, the reaction is not
zero-order. If the reaction is first-order, it will have a constant half-life that is consistent
with its rate constant. The half-life can be assessed from the fact that 40 min elapse while
the concentration drops from 0.80 M to 0.20 M, that is, to one-fourth of its initial value.
Thus, 40 min equals two half-lives and t1/2 = 20 min .
This gives k = 0.693 / t1/2 = 0.693 / 20 min = 0.035 min 1 . Also
At
0.35 M
0.827
kt = ln
= ln
= 0.827 = k 24 min
k=
= 0.034 min 1
0.80 M
24 min
A0
The constancy of the value of k indicates that the reaction is first-order.
(b)
The value of the rate constant is k = 0.034 min 1 .
(c)
Reaction rate =
ln
A
A 0
1
2
1
(rate of formation of B) = k A First we need [A] at t = 30 . min
= kt = 0.034 min 1 30. min = 1.02
A
A 0
= e 1.02 = 0.36
A = 0.36 0.80 M = 0.29 M
rate of formation of B = 2 0.034 min 1 0.29 M = 2.0 102 M min 1
41.
(M) The half-life of the reaction depends on the concentration of A and, thus, this reaction cannot
be first-order. For a second-order reaction, the half-life varies inversely with the reaction rate:
t1/2 = 1/ k A 0 or k = 1/ t1/2 A 0 . Let us attempt to verify the second-order nature of this
c
h
c
h
reaction by seeing if the rate constant is fixed.
1
k=
= 0.020 L mol1min 1
1.00 M 50 min
1
k=
= 0.020 L mol1min 1
2.00 M 25 min
1
k=
= 0.020 L mol 1 min 1
0.50 M 100 min
The constancy of the rate constant demonstrates that this reaction indeed is second-order. The rate
2
equation is Rate = k A and k = 0.020 L mol 1min 1 .
632
Chapter 14: Chemical Kinetics
42.
(M)
(a) The half-life depends on the initial NH 3 and, thus, the reaction cannot be first-order. Let
us attempt to verify second-order kinetics.
1
1
k=
for a second-order reaction k =
= 42 M 1min 1
NH
t
0.0031
M
7.6
min
3 0 1/2
1
1
= 180 M 1min 1
k=
= 865 M 1min 1
0.0015 M 3.7 min
0.00068 M 1.7min
The reaction is not second-order. But, if the reaction is zero-order, its rate will be constant.
A 0 / 2 0.0031 M 2
Rate =
=
= 2.0 104 M/min
t1/2
7.6 min
0.0015 M 2
Rate =
= 2.0 104 M/min
3.7 min
0.00068 M 2
Rate =
2.0 104 M / min
Zero-order reaction
1.7 min
k=
(b)
43.
44.
The constancy of the rate indicates that the decomposition of ammonia under these
conditions is zero-order, and the rate constant is k = 2.0 104 M/min.
[A]0
1
Second-order: t1/2 =
k[A]0
2k
A zero-order reaction has a half life that varies proportionally to [A]0, therefore, increasing [A]0
increases the half-life for the reaction. A second-order reaction's half-life varies inversely
proportional to [A]0, that is, as [A]0 increases, the half-life decreases. The reason for the
difference is that a zero-order reaction has a constant rate of reaction (independent of [A]0). The
larger the value of [A]0, the longer it will take to react. In a second-order reaction, the rate of
reaction increases as the square of the [A]0, hence, for high [A]0, the rate of reaction is large and
for very low [A]0, the rate of reaction is very slow. If we consider a bimolecular elementary
reaction, we can easily see that a reaction will not take place unless two molecules of reactants
collide. This is more likely when the [A]0 is large than when it is small.
(M) Zero-order: t1/2 =
(M)
(a)
[A]0 0.693
=
k
2k
Hence,
[A]0
= 0.693 or [A]0 = 1.39 M
2
(b)
[A]0
1
,
2k
k[A]0
Hence,
[A]0 2
= 1 or [A]02 = 2.00 M
2
(c)
0.693
1
1
, Hence, 0.693 =
or [A]0 = 1.44 M
k
k[A]o
[A]0
633
[A]0 = 1.414 M
Chapter 14: Chemical Kinetics
Collision Theory; Activation Energy
45.
46.
(M)
(a) The rate of a reaction depends on at least two factors other than the frequency of collisions.
The first of these is whether each collision possesses sufficient energy to get over the
energy barrier to products. This depends on the activation energy of the reaction; the higher
it is, the smaller will be the fraction of successful collisions. The second factor is whether
the molecules in a given collision are properly oriented for a successful reaction. The more
complex the molecules are, or the more freedom of motion the molecules have, the smaller
will be the fraction of collisions that are correctly oriented.
(b)
Although the collision frequency increases relatively slowly with temperature, the fraction
of those collisions that have sufficient energy to overcome the activation energy increases
much more rapidly. Therefore, the rate of reaction will increase dramatically with
temperature.
(c)
The addition of a catalyst has the net effect of decreasing the activation energy of the
overall reaction, by enabling an alternative mechanism. The lower activation energy of the
alternative mechanism, (compared to the uncatalyzed mechanism), means that a larger
fraction of molecules have sufficient energy to react. Thus the rate increases, even though
the temperature does not.
(M)
(a) The activation energy for the reaction of hydrogen with oxygen is quite high, too high, in
fact, to be supplied by the energy ordinarily available in a mixture of the two gases at
ambient temperatures. However, the spark supplies a suitably concentrated form of energy
to initiate the reaction of at least a few molecules. Since the reaction is highly exothermic,
the reaction of these first few molecules supplies sufficient energy for yet other molecules
to react and the reaction proceeds to completion or to the elimination of the limiting
reactant.
(b)
47.
A larger spark simply means that a larger number of molecules react initially. But the
eventual course of the reaction remains the same, with the initial reaction producing enough
energy to initiate still more molecules, and so on.
(M)
(a) The products are 21 kJ/mol closer in energy to the energy activated complex than are the
reactants. Thus, the activation energy for the reverse reaction is
84 kJ / mol 21 kJ / mol = 63 kJ / mol.
634
Chapter 14: Chemical Kinetics
Potential Energy (kJ)
(b) The reaction profile for the reaction in Figure 14-10 is sketched below.
A ....B
Transiton State
Ea(reverse) = 63 kJ
Ea(forward) = 84 kJ
Products
C+D
H = +21 kJ
Reactants
A+B
Progress of Reaction
48.
(M) In an endothermic reaction (right), Ea must be larger than the H for the reaction. For an
exothermic reaction (left), the magnitude of Ea may be either larger or smaller than that of H .
In other words, a small activation energy can be associated with a large decrease in the enthalpy,
or a large Ea can be connected to a small decrease in enthalpy.
products
reactants
ΔH<0
49.
products
reactants
ΔH>0
(E)
(a)
There are two intermediates (B and C).
(b)
There are three transition states (peaks/maxima) in the energy diagram.
(c)
The fastest step has the smallest Ea, hence, step 3 is the fastest step in the reaction with
step 2 a close second.
(d)
Reactant A (step 1) has the highest Ea, and therefore the slowest smallest constant
(e)
Endothermic; energy is needed to go from A to B.
(f)
Exothermic; energy is released moving from A to D.
635
Chapter 14: Chemical Kinetics
50.
(E)
(a)
There are two intermediates (B and C).
(b)
There are three transition states (peaks/maxima) in the energy diagram.
(c)
The fastest step has the smallest Ea, hence, step 2 is the fastest step in the reaction.
(d)
Reactant A (step 1) has the highest Ea, and therefore the slowest smallest constant
(e)
Endothermic; energy is needed to go from A to B.
(f)
Endothermic, energy is needed to go from A to D.
Effect of Temperature on Rates of Reaction
51.
(M)
k
E 1 1
5.4 104 L mol 1 s1 E a
1
1
ln 1 = a
= ln
=
2
1 1
k2
R T2 T1
2.8 10 L mol s
R 683 K 599 K
FG
H
IJ
K
FG
H
IJ
K
3.95R = Ea 2.05 104
3.95 R
Ea =
= 1.93 104 K 1 8.3145 J mol 1 K 1 = 1.60 105 J / mol = 160 kJ / mol
4
2.05 10
52.
(M)
E 1 1
k
5.0 103 L mol1 s 1
1.60 105 J/mol 1
1
ln 1 = a = ln
=
2
1 1
1
1
k2
R T2 T1
2.8 10 L mol s
8.3145 J mol K 683 K T
1
1
1.72
1
1
1.72 = 1.92 104
=
= 8.96 105
4
683
K
T
683
K
T
1.92
10
1
= 8.96 105 +1.46 103 = 1.55 103
T = 645 K
T
53.
(D)
(a)
First we need to compute values of ln k and 1/ T . Then we plot the graph of ln k versus
1/T.
T , C
0 C
10 C
20 C
30 C
T,K
273 K
283 K
293 K
303 K
1
1/ T , K
0.00366
0.00353
0.00341
0.00330
1
6
5
4
k, s
5.6 10
3.2 10
1.6 10
7.6 104
ln k
12.09
10.35
8.74
7.18
636
Chapter 14: Chemical Kinetics
Plot of ln k versus 1/T
0.00325
1/T (K-1)
0.00345
0.00365
-7.1
-8.1
ln k
-9.1
-10.1
-11.1
y = -13520x + 37.4
-12.1
(b)
The slope = Ea / R .
J
1 kJ
kJ
1.352 104 K
= 112
mol K
1000 J
mol
6 1
We apply the Arrhenius equation, with k = 5.6 10 s at 0 C (273 K), k = ? at 40 C
(313 K), and Ea = 113 103 J/mol.
Ea = R slope = 8.3145
(c)
E 1 1
k
= a
6 1
R T1 T2
5.6 10 s
k
e6.306 = 548 =
5.6 106 s 1
ln
t1/2 =
(D)
(a)
k = 548 5.6 106 s 1 = 3.07 103 s 1
0.693
0.693
=
= 2.3 10 2 s
3
1
k
3.07 10 s
Here we plot ln k vs. 1/ T . The slope of the straight line equals Ea / R . First we tabulate
the data to plot. (the plot is shown below).
T , C
T ,K
1/ T , K1
k , M1s1
ln k
15.83
288.98
0.0034604
5.03 105
9.898
32.02
305.17
0.0032769
3.68 104
7.907
59.75
332.90
0.0030039
6.71 103
5.004
90.61
363.76
0.0027491
0.119
2.129
Plot of ln k versus 1/T
0.0027
0.0029
-1
1/T (K )
0
-2
-4
ln k
54.
112 103 J/mol 1
1
=
= 6.306
1
1
8.3145 J mol K 273 K 313 K
-6
-8
-10
y = -10890x + 27.77
-12
637
0.0031
0.0033
0.0035
Chapter 14: Chemical Kinetics
The slope of this graph = 1.09 104 K = Ea / R
J
J
1 kJ
kJ
Ea = 1.089 104 K 8.3145
= 9.054 104
= 90.5
mol K
mol
1000 J
mol
(b)
We calculate the activation energy with the Arrhenius equation using the two extreme data
points.
E 1 1 E
k
0.119
1
1
ln 2 = ln
= +7.77 = a = a
5
k1
5.03 10
R T1 T2 R 288.98 K 363.76 K
Ea
7.769 8.3145 J mol1 K 1
Ea =
= 9.08 104 J/mol
4
1
R
7.1138 10 K
Ea = 91 kJ/mol. The two Ea values are in quite good agreement, within experimental limits.
= 7.1138 104 K 1
(c)
We apply the Arrhenius equation, with Ea = 9.080 104 J/mol,
k = 5.03 10 5 M 1 s1 at 15.83 C (288.98 K), and k = ? at 100.0 C (373.2 K).
Ea 1 1 90.80 103 J/mol
k
1
1
ln
=
=
5
1 1
1
1
5.03 10 M s
R T1 T 2 8.3145 J mol K 288.98 K 373.2 K
ln
k
5
1
1
e8.528 = 5.05 103 =
= 8.528
5.03 10 M s
k = 5.05 103 5.03 105 M 1 s 1 = 0.254 M 1 s 1
55.
5.03 10
M 1 s 1
(M) The half-life of a first-order reaction is inversely proportional to its rate constant:
k = 0.693 / t1/2 . Thus we can apply a modified version of the Arrhenius equation to find Ea.
(a)
(b)
t1/2 1 Ea 1 1
k2
46.2 min Ea 1
1
= ln
=
=
= ln
k1
2.6 min
R 298 K 102 + 273 K
t1/2 2 R T1 T2
E
2.88 8.3145 J mol1 K 1 1 kJ
2.88 = a 6.89 104 Ea =
= 34.8 kJ/mol
R
6.89 104 K 1
1000 J
10.0 min
34.8 103 J / mol 1
1
1
1
ln
=
= 1.53 = 4.19 103
1
1
46.2 min 8.3145 J mol K T 298
T 298
ln
FG
H
1
1.53
1
= 3.65 104
=
3
T
298
4.19
10
56.
k
5
IJ
K
1
= 2.99 103
T
FG
H
IJ
K
T = 334 K = 61 o C
(M) The half-life of a first-order reaction is inversely proportional to its rate
constant: k = 0.693 / t1/2 . Thus, we can apply a modified version of the Arrhenius equation.
(a)
t1/2 1 Ea 1 1
k2
22.5 h Ea 1
1
= ln
=
=
= ln
k1
1.5 h
R 293 K 40 + 273 K
t1/2 2 R T1 T2
Ea
2.71 8.3145 J mol1 K 1 1 kJ
4
2.71 =
2.18 10 , Ea =
= 103 kJ/mol
R
2.18 104 K 1
1000 J
ln
638
Chapter 14: Chemical Kinetics
(b)
b
The relationship is k = A exp Ea / RT
g
103 10 J mol1
13 1
40.9
k = 2.05 1013 s 1exp
3.5 105 s 1
= 2.05 10 s e
1
1
8.3145
J
mol
K
273
+
30
K
3
57.
(M)
(a) It is the change in the value of the rate constant that causes the reaction to go faster. Let k1
be the rate constant at room temperature, 20 C (293 K). Then, ten degrees higher (30° C
or 303 K), the rate constant k2 = 2 k1 .
E 1 1 E 1
k2
2 k1
1
4
1 Ea
= ln
= 0.693 = a = a
= 1.13 10 K
k1
k1
R T1 T2 R 293 303 K
R
0.693 8.3145 J mol 1 K 1
Ea =
= 5.1 104 J / mol = 51 kJ / mol
4
1
1.13 10 K
Since the activation energy for the depicted reaction (i.e., N2O + NO N2 + NO2) is 209
kJ/mol, we would not expect this reaction to follow the rule of thumb.
ln
(b)
58.
(M) Under a pressure of 2.00 atm, the boiling point of water is approximately 121 C or 394 K.
Under a pressure of 1 atm, the boiling point of water is 100 C or 373 K. We assume an
activation energy of 5.1 104 J / mol and compute the ratio of the two rates.
Rate 2 Ea 1 1
5.1 104 J / mol
1
1
ln
=
=
= 0.88
1
1
Rate1
R T1 T2
8.3145 J mol K 373 394 K
FG
H
IJ
K
FG
H
IJ
K
Rate 2 = e0.88 Rate1 = 2.4 Rate1 . Cooking will occur 2.4 times faster in the pressure cooker.
Catalysis
59.
(E)
(a)
(b)
60.
Although a catalyst is recovered unchanged from the reaction mixture, it does “take part in
the reaction.” Some catalysts actually slow down the rate of a reaction. Usually, however,
these negative catalysts are called inhibitors.
The function of a catalyst is to change the mechanism of a reaction. The new mechanism is
one that has a different (lower) activation energy (and frequently a different A value), than
the original reaction.
(M) If the reaction is first-order, its half-life is 100 min, for in this time period [S] decreases from
1.00 M to 0.50 M, that is, by one half. This gives a rate constant of
k = 0.693 / t1/2 = 0.693 / 100 min = 0.00693 min 1 .
The rate constant also can be determined from any two of the other sets of data.
A 0
1.00 M
0.357
kt = ln
k=
= ln
= 0.357 = k 60 min
= 0.00595 min 1
0.70 M
60 min
A t
This is not a very good agreement between the two k values, so the reaction is probably not firstorder in [A]. Let's try zero-order, where the rate should be constant.
639
Chapter 14: Chemical Kinetics
0.90 M 1.00 M
0.50 M 1.00 M
= 0.0050 M/min
Rate =
= 0.0050 M/min
20 min
100 min
0.20 M 0.90 M
0.50 M 0.90 M
Rate =
= 0.0050 M/min
Rate =
= 0.0050 M/min
160 min 20 min
100 min 20 min
Thus, this reaction is zero-order with respect to [S].
Rate =
61.
(E) Both platinum and an enzyme have a metal center that acts as the active site. Generally
speaking, platinum is not dissolved in the reaction solution (heterogeneous), whereas enzymes are
generally soluble in the reaction media (homogeneous). The most important difference, however,
is one of specificity. Platinum is rather nonspecific, catalyzing many different reactions. An
enzyme, however, is quite specific, usually catalyzing only one reaction rather than all reactions
of a given class.
62.
(E) In both the enzyme and the metal surface cases, the reaction occurs in a specialized location:
either within the enzyme pocket or on the surface of the catalyst. At high concentrations of
reactant, the limiting factor in determining the rate is not the concentration of reactant present but
how rapidly active sites become available for reaction to occur. Thus, the rate of the reaction
depends on either the quantity of enzyme present or the surface area of the catalyst, rather than on
how much reactant is present (i.e., the reaction is zero-order). At low concentrations or gas
pressures the reaction rate depends on how rapidly molecules can reach the available active sites.
Thus, the rate depends on concentration or pressure of reactant and is first-order.
63.
(E) For the straight-line graph of Rate versus [Enzyme], an excess of substrate must be present.
64.
(E) For human enzymes, we would expect the maximum in the curve to appear around 37C, i.e.,
normal body temperature (or possibly at slightly elevated temperatures to aid in the control of
diseases (37 - 41 C). At lower temperatures, the reaction rate of enzyme-activated reactions
decreases with decreasing temperature, following the Arrhenius equation. However, at higher
temperatures, these temperature sensitive biochemical processes become inhibited, probably by
temperature-induced structural modifications to the enzyme or the substrate, which prevent
formation of the enzyme-substrate complex.
Reaction Mechanisms
65.
(E) The molecularity of an elementary process is the number of reactant molecules in the process.
This molecularity is equal to the order of the overall reaction only if the elementary process in
question is the slowest and, thus, the rate-determining step of the overall reaction. In addition, the
elementary process in question should be the only elementary step that influences the rate of the
reaction.
66.
(E) If the type of molecule that is expressed in the rate law as being first-order collides with other
molecules that are present in much larger concentrations, the reaction will seem to depend only
on the amount of those types of molecules present in smaller concentration, since the larger
concentration will be essentially unchanged during the course of the reaction. Such a situation is
quite common, and has been given the name pseudo first-order. It is also possible to have
molecules which, do not participate directly in the reaction— including product molecules—
640
Chapter 14: Chemical Kinetics
strike the reactant molecules and impart to them sufficient energy to react. Finally, collisions of
the reactant molecules with the container walls may also impart adequate energy for reaction to
occur.
67.
(M) The three elementary steps must sum to give the overall reaction. That is, the overall reaction
is the sum of step 1 step 2 step 3. Hence, step 2 = overall reaction step 1 step 3. Note that
all species in the equations below are gases.
2 NO + 2 H 2 N 2 + 2 H 2O
Overall: 2 NO + 2H 2 N 2 + 2 H 2 O
First:
N 2 O 2
2 NO
2NO
N 2 O 2
Third
N 2 O + H 2 N 2 + H 2 O or N 2 + H 2 O N 2 O + H 2
H 2 + N 2O2 H 2O + N 2O
The result is the second step, which is slow:
The rate of this rate-determining step is: Rate = k2 H2 N2O2
Since N 2 O 2 does not appear in the overall reaction, we need to replace its concentration with the
concentrations of species that do appear in the overall reaction. To do this, recall that the first step
is rapid, with the forward reaction occurring at the same rate as the reverse reaction.
2
k1 NO = forward rate = reverse rate = k 1 N 2 O 2 . This expression is solved for N 2 O 2 ,
which then is substituted into the rate equation for the overall reaction.
2
k1 NO
k k
2
N 2O2 =
Rate = 2 1 H 2 NO
k 1
k 1
The reaction is first-order in H 2 and second-order in [NO]. This result conforms to the
experimentally determined reaction order.
68.
1
I 2 (g)
2 I(g)
k
k
(M) Proposed mechanism:
Observed rate law:
-1
k2
2 I(g) + H 2 (g)
2 HI(g)
Rate = k[I2][H2]
I2(g) + H2(g) 2 HI(g)
The first step is a fast equilibrium reaction and step 2 is slow. Thus, the predicted rate law is
Rate = k2[I]2[H2]. In the first step, set the rate in the forward direction for the equilibrium equal
to the rate in the reverse direction. Then solve for [I]2.
Rateforward = Ratereverse
Use: Rateforward = k1[I2] and Ratereverse = k-1[I]2
From this we see: k1[I2] = k-1[I]2. Rearranging (solving for [I]2)
k [I ]
k [I ]
[I]2 = 1 2
Substitute into Rate = k2[I]2[H2] = k2 1 2 [H2] = kobs[I2][H2]
k-1
k-1
Since the predicted rate law is the same as the experimental rate law, this mechanism is
plausible.
641
Chapter 14: Chemical Kinetics
69.
1
Cl2 (g)
2 Cl(g)
k
k
(M) Proposed mechanism:
Observed rate law:
-1
k2
2 Cl(g) + 2 NO(g)
2 NOCl(g)
Rate = k[Cl2][NO]2
Cl2(g) + 2NO(g) 2 NOCl(g)
The first step is a fast equilibrium reaction and step 2 is slow. Thus, the predicted rate law is
Rate = k2[Cl]2[NO]2 In the first step, set the rate in the forward direction for the equilibrium equal
to the rate in the reverse direction. Then express [Cl]2 in terms of k1, k-1 and [Cl2]. This
mechanism is almost certainly not correct because it involves a tetra molecular second step.
Rateforward = Ratereverse
Use: Rateforward = k1[Cl2] and Ratereverse = k-1[Cl]2
From this we see: k1[Cl2] = k-1[Cl]2. Rearranging (solving for [Cl]2)
k [Cl ]
k [Cl ]
[Cl]2 = 1 2
Substitute into Rate = k2[Cl]2[NO]2 = k2 1 2 [NO]2 = kobs[Cl2][NO2]2
k-1
k-1
k1
There is another plausible mechanism. Cl2 (g) + NO(g)
NOCl(g) + Cl(g)
k-1
k1
Cl(g) + NO(g)
NOCl(g)
k-1
Cl2(g) + 2NO(g) 2 NOCl(g)
Rateforward = Ratereverse
Use: Rateforward = k1[Cl2][NO] and Ratereverse = k-1[Cl][NOCl]
From this we see: k1[Cl2][NO] = k-1[Cl][NOCl]. Rearranging (solving for [Cl])
k [Cl ][NO]
k k [Cl ][NO]2
[Cl] = 1 2
Substitute into Rate = k2[Cl][NO] = 2 1 2
k-1 [NOCl]
k -1 [NOCl]
If [NOCl], the product is assumed to be constant (~ 0 M using method of initial rates), then
k2 k1
constant = k obs Hence, the predicted rate law is kobs [Cl2 ][NO]2 which agrees with
k-1 [NOCl]
the experimental rate law. Since the predicted rate law agrees with the experimental rate law,
both this and the previous mechanism are plausible, however, the first is dismissed as it has a
tetramolecular elementary reaction (extremely unlikely to have four molecules simultaneously
collide).
70.
(M) A possible mechanism is:
Step 1:
Step 2:
O 2 + O fast
O3
k
3 2 O slow
O + O3
2
The overall rate is that of the slow step: Rate = k 3 O O 3 . But O is a reaction intermediate, whose
concentration is difficult to determine. An expression for [O] can be found by assuming that the
forward and reverse “fast” steps proceed with equal speed.
2
k1 O3
k1 O3
k3 k1 O3
Rate = k3 O3 =
Rate1 = Rate 2 k1 O3 = k2 O 2 O O =
k2 O 2
k2 O 2
k2 O 2
Then substitute this expression into the rate law for the reaction. This rate equation has the same
form as the experimentally determined rate law and thus the proposed mechanism is plausible.
642
Chapter 14: Chemical Kinetics
71.
(M)
k1
S1 S2
S1 : S2
k
1
S1 : S2
k2
S1 : S2
d S1 : S2
k1 S1 S2 k 1 S1 : S2 k 2 S1 : S2
dt
k1 S1 S2 k 1 k 2 S1 : S2
S1 : S2
k1
S1 S2
k 1 k 2
k k S S
d S1 : S2
k 2 S1 : S2 2 1 1 2
dt
k 1 k 2
72.
(M)
k1
CH3 2 CO (aq) + OH
CH 3C O CH 2 (aq) + H 2 O (l)
k
1
k2
CH 3C O CH (aq) + CH 3 2 CO (aq)
Pr od
2
We note that CH3C(O)CH2– is an intermediate species. Using the steady state approximation,
while its concentration is not known during the reaction, the rate of change of its concentration is
zero, except for the very beginning and towards the end of the reaction. Therefore,
d CH 3C O CH 2
k1 CH 3 2 CO OH k 1 CH 3C O CH 2 H 2O
dt
k 2 CH 3C O CH 2 CH 3 2 CO 0
Rearranging the above expression to solve for CH3C(O)CH2– gives the following expression
k1 CH 3 2 CO OH
CH3C O CH 2
k 1 H 2 O k 2 CH 3 2 CO
The rate of formation of product, therefore, is:
d Pr od
k 2 CH 3C O CH 2 CH 3 2 CO
dt
k 2 CH 3 2 CO
k1 CH 3 2 CO OH
k 1 H 2 O k 2 CH 3 2 CO
k 2 k1 CH 3 2 CO OH
k 1 H 2 O k 2 CH 3 2 CO
2
643
Chapter 14: Chemical Kinetics
INTEGRATIVE AND ADVANCED EXERCISES
73.
(M) The data for the reaction starting with 1.00 M being first-order or second-order as well as that
for the first-order reaction using 2.00 M is shown below
Time
(min)
0
5
10
15
25
[A]o = 1.00 M
(second order)
1.00
0.63
0.46
[A]o = 1.00 M
(first order)
1.00
0.55
0.30
[A]o = 2.00 M
(second order)
2.00
0.91
0.59
[A]o = 2.00 M
(first order)
2.00
1.10
0.60
0.36
0.25
0.165
0.05
0.435
0.286
0.33
0.10
Clearly we can see that when [A]o = 1.00 M, the first-order reaction concentrations will always be
lower than that for the second-order case (assumes magnitude of the rate constant is the same). If,
on the other hand, the concentration is above 1.00 M, the second-order reaction decreases faster
than the first-order reaction (remember that the half-life shortens for a second-order reaction as
the concentration increases, whereas for a first-order reaction, the half-life is constant).
From the data, it appears that the crossover occurs in the case where [A]0 = 2.00 M at just over 10
minutes.
Second-order at 11 minutes:
1
1 0.12
(11 min)
[A] 2 M min
[A] = 0.549 M
0.12
ln[A] ln(2)
[A] = 0.534 M
(11 min)
min
A quick check at 10.5 minutes reveals,
1
1 0.12(10.5 min)
Second-order at 10.5 minutes:
[A] = 0.568 M
[A] 2
M min
0.12(10.5 min)
First-order at 10.5 minutes: ln[A] ln(2)
[A] = 0.567 M
M min
Hence, at approximately 10.5 minutes, these two plots will share a common point (point at which
the concentration versus time curves overlap).
First-order at 11 minutes:
74. (M)
(a) The concentration vs. time graph is not linear. Thus, the reaction is obviously not zero-order
(the rate is not constant with time). A quick look at various half lives for this reaction shows
the ~2.37 min (1.000 M to 0.5 M), ~2.32 min (0.800 M to 0.400 M), and ~2.38 min(0.400 M
to 0.200 M). Since the half-life is constant, the reaction is probably first-order.
(2.37 2.32 2.38)
0.693
0.693
2.36 min k
0.294 min 1
3
t1/2
2.36 min
–1
or perhaps better expressed as k = 0.29 min due to imprecision.
(b) average t1/ 2
644
Chapter 14: Chemical Kinetics
(c) When t 3.5 min, [A] 0.352 M. Then, rate = k[A] = 0.294 min –1 0.352 M 0.103 M/min.
(d) Slope
[A]
t
Rate
0.1480 M 0.339 M
6.00 min 3.00 min
0.0637 M / min
Rate 0.064 M/min.
(e) Rate = k[A] = 0.294 min–1 × 1.000 M = 0.294 M/min.
75. (M) The reaction being investigated is:
2 MnO 4- (aq) + 5 H 2O 2 (aq) + 6 H + (aq)
2 Mn 2+ (aq) +8 H 2O(l) + 5O 2 (g)
We use the stoichiometric coefficients in this balanced reaction to determine [H2O2].
0.1000 mmol MnO-4 5 mmol H 2 O 2
37.1 mL titrant
1 mL titrant
2 mmol MnO-4
[H 2 O 2 ]
1.86 M
5.00 mL
76.
(D) We assume in each case that 5.00 mL of reacting solution is titrated.
2.32 mmol H 2 O 2 2 mmol MnO-4
1 mL titrant
volume MnO-4 5.00 mL
1 mL
5 mmol H 2 O 2 0.1000 mmol MnO-4
At 200 s
At 400 s
At 600 s
At 1200 s
At 1800 s
At 3000 s
20.0 2.32 M H 2 O 2 46.4 mL 0.1000 M MnO-4 titrant at 0 s
Vtitrant = 20.0 × 2.01 M H2O2 = 40.2
38.5 mL
Vtitrant = 20.0 × 1.72 M H2O2 = 34.433 mL
27.5
Vtitrant = 20.0 × 1.49 M H2O2 = 29.822 mL
Vtitrant = 20.0 × 0.98 M H2O2 = 19.16.5
6 mL
11
Tangent line
Vtitrant = 20.0 × 0.62 M H2O2 = 12.5.5
4 mL
Vtitrant = 20.0 × 0.25 M H2O2 = 5.000mL
0
700
1400
2100
2800
The graph of volume of titrant vs. elapsed time is given above. This graph is of approximately the
same shape as Figure 14-2, in which [H2O2] is plotted against time. In order to determine the rate,
the tangent line at 1400 s has been drawn on the graph. The intercepts of the tangent line are at 34
mL of titrant and 2800 s. From this information we determine the rate of the reaction.
34 mL
1L
0.1000 mol MnO-4 5 mol H 2 O 2
2800 s 1000 mL
1 L titrant
2 mol MnO-4
6.1 104 M/s
Rate
0.00500 L sample
This is the same as the value of 6.1 × 10–4 obtained in Figure 14-2 for 1400 s. The discrepancy is
due, no doubt, to the coarse nature of our plot.
645
Chapter 14: Chemical Kinetics
77.
(M) First we compute the change in [H2O2]. This is then used to determine the amount, and
ultimately the volume, of oxygen evolved from the given quantity of solution. Assume the O2(g)
is dry.
[H 2 O 2 ]
60 s
[H 2 O 2 ]
t 1.7 103 M/s 1.00 min
0.102 M
t
1 min
0.102 mol H 2 O 2
1 mol O 2
amount O 2 0.175 L soln
0.00892 mol O 2
1L
2 mol H 2 O 2
L atm
0.00892 mol O 2 0.08206
(273 24) K
nRT
mol
K
Volume O 2
0.22 L O 2
1 atm
P
757 mmHg
760 mmHg
78. (M) We know that rate has the units of M/s, and also that concentration has the units of M. The
generalized rate equation is Rate k A 0 . In terms of units, this becomes
M/s = {units of k} M0. Therefore {Units of k}
M/s
M10 s 1
M0
79. (M)
(a) Comparing the third and the first lines of data, [I-] and [OH-]stay fixed, while [OCl–]
doubles. Also the rate for the third kinetics run is one half of the rate found for the first run.
Thus, the reaction is first-order in [OCl-]. Comparing the fourth and fifth lines, [OCl–] and
[I-] stay fixed, while [OH-] is halved. Also, the fifth run has a reaction rate that is twice that
of the fourth run. Thus, the reaction is minus first-order in [OH-]. Comparing the third and
second lines of data, [OCl-] and [OH-] stay fixed, while the [I-] doubles. Also, the second
run has a reaction rate that is double that found for the third run. Thus, the reaction is firstorder in [I-].
(b) The reaction is first-order in [OCl-] and [I-] and minus first-order in [OH-]. Thus, the overall
order = 1 + 1 – 1 = 1. The reaction is first-order overall.
Rate k
(c)
[OCl ][I ]
[OH ]
using data from first run:
Rate [OH ] 4.8 104 M/s 1.00 M
60. s 1
[OCl ][I ]
0.0040 M 0.0020 M
80. (M) We first determine the number of moles of N2O produced. The partial pressure of N2O(g) in
the “wet” N2O is 756 mmHg – 12.8 mmHg = 743 mmHg.
1 atm
743 mmHg
0.0500 L
760 mmHg
PV
amount N 2 O
0.00207 mol N 2 O
RT 0.08206 L atm mol 1 K 1 (273 15) K
Now we determine the change in [NH2NO2].
646
Chapter 14: Chemical Kinetics
1 mol NH 2 NO 2
1 mol N 2 O
[NH 2 NO 2 ]
0.0125 M
0.165 L soln
0.693
[NH 2 NO 2 ]final 0.105 M–0.0125 M 0.093 M
0.00563 min 1
k
123 min
[A]t
1
1
0.093 M
ln
22 min elapsed time
t ln
1
0.00563 min
0.105 M
k [A]0
0.00207 mol N 2 O
81. (D) We need to determine the partial pressure of ethylene oxide at each time in order to determine
the order of the reaction. First, we need the initial pressure of ethylene oxide. The pressure at
infinite time is the pressure that results when all of the ethylene oxide has decomposed. Because
two moles of product gas are produced for every mole of reactant gas, this infinite pressure is twice
the initial pressure of ethylene oxide. Pinitial = 249.88 mmHg 2 = 124.94 mmHg. Now, at each
(CH 2 ) 2 O(g)
CH 4 (g) CO(g)
time we have the following.
Initial: 124.94 mmHg Changes: –x mmHg +x mmHg +x mmHg Final: 124.94 + x mmHg
Thus, x = Ptot – 124.94 and PEtO = 124.94 – x = 124.94 – (Ptot – 124.94) = 249.88 – Ptot
Hence, we have, the following values for the partial pressure of ethylene oxide.
t, min
0
10
20
40
60
100
200
PEtO, mmHg 124.94
110.74
98.21 77.23 60.73 37.54 11.22
For the reaction to be zero-order, its rate will be constant.
(110.74 124.94) mmHg
The rate in the first 10 min is: Rate
1.42 mmHg/min
10 min
(77.23 124.94) mmHg
The rate in the first 40 min is: Rate
1.19 mmHg/min
40 min
We conclude from the non-constant rate that the reaction is not zero-order. For the reaction to be
first-order, its half-life must be constant. From 40 min to 100 min—a period of 60 min—the partial
pressure of ethylene oxide is approximately halved, giving an approximate half-life of 60 min.
And, in the first 60 min, the partial pressure of ethylene oxide is approximately halved. Thus, the
reaction appears to be first-order. To verify this tentative conclusion, we use the integrated firstorder rate equation to calculate some values of the rate constant.
1 P
1
110.74 mmHg
k ln
ln
0.0121 min 1
t P0
10 min 124.94 mmHg
k
1
100 min
ln
37.54 mmHg
124.94 mmHg
0.0120 min 1
k
1
60 min
ln
60.73 mmHg
124.94 mmHg
0.0120 min 1
The constancy of the first-order rate constant suggests that the reaction is first-order.
Pt
1 P
kt Elapsed time is computed as: t ln t
P0
k P0
We first determine the pressure of DTBP when the total pressure equals 2100 mmHg.
82. (M) For this first-order reaction
ln
647
Chapter 14: Chemical Kinetics
Reaction:
C8 H18O 2 (g)
2C3 H 6 O(g) C2 H 6 (g) [Equation 15.16]
Initial:
800.0 mmHg
Changes:
–x mmHg
2x mmHg
x mmHg
x mmHg
Final:
(800.0 x) mmHg
2x mmHg
Total pressure (800.0–x) 2x x 800.0 2 x 2100.
x 650. mmHg P{C8 H18O 2 (g)} 800. mmHg 650. mmHg 150. mmHg
1 P
1
150. mmHg
192 min 1.9 102 min
t ln t
ln
3
1
k P0
8.7 10 min
800. mmHg
83. (D) If we compare Experiment 1 with Experiment 2, we notice that [B] has been halved, and also
that the rate, expressed as [A]/ t , has been halved. This is most evident for the times 5 min,
10 min, and 20 min. In Experiment 1, [A] decreases from 1.000 × 10–3 M to 0.779 × 10–3 M in 5
min, while in Experiment 2 this same decrease in [A] requires 10 min. Likewise in Experiment 1,
[A] decreases from 1.000 × 10–3 M to 0.607 M× 10–3 in 10 min, while in Experiment 2 the same
decrease in [A] requires 20 min. This dependence of rate on the first power of concentration is
characteristic of a first-order reaction. This reaction is first-order in [B]. We now turn to the order
of the reaction with respect to [A]. A zero-order reaction will have a constant rate. Determine the
rate
(0.951 1.000) 10 3 M
Rate
4.9 10 5 M/min
After over the first minute:
1 min
(0.779 1.000) 10 3 M
Rate
4.4 10 5 M/min
After over the first five minutes:
5 min
(0.368 1.000) 10 3 M
3.2 10 5 M/min
After over the first twenty minutes: Rate
20 min
This is not a very constant rate; we conclude that the reaction is not zero-order. There are no clear
half-lives in the data with which we could judge the reaction to be first-order. But we can
determine the value of the first-order rate constant for a few data.
1
[A]
1
0.951 mM
0.0502 min 1
k ln
ln
t [A]0
1 min 1.000 mM
1
0.607 mM
1
0.368 mM
0.0499 min 1
0.0500 min 1
ln
k
ln
10 min 1.000 mM
20 min 1.000 mM
The constancy of the first-order rate constant indicates that the reaction indeed is first-order in [A].
(As a point of interest, notice that the concentrations chosen in this experiment are such that the
reaction is pseudo-zero-order in [B]. Here, then it is not necessary to consider the variation of [B]
with time as the reaction proceeds when determining the kinetic dependence on [A].)
k
648
Chapter 14: Chemical Kinetics
84. (M) In Exercise 79 we established the rate law for the iodine-hypochlorite ion reaction:
Rate k[OCl ][I ][OH ]1 . In the mechanism, the slow step gives the rate law;
Rate k3 [I ][HOCl] . We use the initial fast equilibrium step to substitute for [HOCl] in this rate
equation. We assume in this fast step that the forward rate equals the reverse rate.
k1[OCl ][H 2 O] k1[HOCl][OH ]
Rate k2 [I ]
k1[OCl ][H 2 O]
[HOCl]
k1[OH ]
k1[OCl ][H 2 O] k2 k1[H 2 O] [OCl ][I ]
[OCl ][I ]
k
k1[OH ]
k1
[OH ]
[OH ]
This is the same rate law that we established in Exercise 79. We have incorporated [H2O] in the
rate constant for the reaction because, in an aqueous solution, [H2O] remains effectively constant
during the course of the reaction. (The final fast step simply involves the neutralization of the acid
HOI by the base hydroxide ion, OH–.)
85. (M) It is more likely that the cis-isomer, compound (I), would be formed than the trans-isomer,
compound (II). The reason for this is that the reaction will involve the adsorption of both
CH3CCCH3 and H2 onto the surface of the catalyst. These two molecules will eventually be
adjacent to each other. At some point, one of the bonds in the CC bond will break, the H–H
bond will break, and two C–H bonds will form. Since these two C–H bonds form on the same side
of the carbon chain, compound (I) will be produced. In the sketches below, dotted lines (…)
indicate bonds forming or breaking.
86. (D) Hg2
2+
C C CH3
CH3
+ Tl 2 Hg + Tl Experimental rate law = k
3+
Possible mechanism:
2+
+
CH3
C C CH3
[Hg 2 2+ ][Tl3+ ]
[Hg 2+ ]
2+
3+
1
Hg22+ + Tl3+
Hg + HgTl (fast)
k1
k2
(slow)
HgTl3+ Hg2+ + Tl3+
2+
3+
2+
+
Rate = k2[HgTl3+]
Hg2 + Tl 2 Hg + Tl
k
k1[Hg22+][Tl3+] = k-1 [Hg2+][HgTl3+]
Rate = k2[HgTl3+] =
H H
···
CH3
H ··· H
C ··· C CH3
···
H H
rearrange [HgTl3+ ] =
k2 k1 [Hg 2 2+ ][Tl3+ ]
[Hg 2 2+ ][Tl3+ ]
=
k
obs
k-1
[Hg 2+ ]
[Hg 2+ ]
649
k1 [Hg 2 2+ ][Tl3+ ]
k-1
[Hg 2+ ]
Chapter 14: Chemical Kinetics
87.
(M)
ΔCCl 3
rateformation ratedisappearance 0
Δt
so rateformation ratedecomposition
k 2 [Cl(g)][CHCl 3 ] k 3[CCl 3 ][[Cl(g)] and, simplifying, [CCl 3 ]
k2
[CHCl 3 ]
k3
k
since rate k 3[CCl 3 ][Cl(g)] k3 2 [CHCl 3 ] [Cl(g)] k 2 [CHCl 3 ][Cl(g)]
k3
1/ 2
k
We know: [Cl(g)] 1 [Cl 2 (g)]
k -1
1/2
k
then rateoverall = k 2 [CHCl 3 ] 1 [Cl 2 (g)]
k -1
1/2
k
and the rate constant k will be: k k2 1
k -1
88.
1/2
4.8 103
(1.3 10 )
3
3.6 10
2
0.015
(D)
d[A]
d[A]
Rearrange:
-kt =
dt
[A]3
Integrate using the limits time (0 to t) and concentration ([A]0 to [A]t )
Rate = k[A]3 -
1 1
1 1
2
2 [A]t 2 [A]o 2
0
1
1
1
1
Simplify : kt
Rearrange :
kt
2
2
2
2[A]t 2[A]o
2[A]t
2[A]0 2
1
1
Multiply through by 2 to give the integrated rate law:
2kt
2
[A]t
[A]o 2
t
-k t =
[A]t
d[A]
[A]3
[A]o
kt (k (0))
To derive the half life(t1 2 ) substitute t =t1 2 and [A]t =
1
[A]0
2
2kt1 2
2
2kt1 2
1
1
4
2
2
[A]o
[A]o 2
[A]o
4
1
3
4
2
2
[A]o [A]o
[A]o 2
[A]o
2
Collect terms
Solve for t1 2
t1 2
3
2k[A]o 2
89. (D) Consider the reaction: A + B products (first-order in A, first-order in B). The initial
concentration of each reactant can be defined as [A]0 and [B]0 Since the stoichiometry is 1:1, we
can define x as the concentration of reactant A and reactant B that is removed (a variable that
changes with time).
The [A]t = ([A]o – x) and [B]t = ([B]o – x).
Algebra note: [A]o – x = – (x – [A]o) and [B]o – x = – (x – [B]o).
650
Chapter 14: Chemical Kinetics
As well, the calculus requires that we use the absolute value of |x – [A]o| and |x – [B]o| when taking
the integral of the reciprocal of |x – [A]o| and |x – [B]o|
Rate = -
d[A]
dt
=-
d[B] dx
=
k [A]t [B]t =k ([A]o -x)([B]o -x)
dt dt
or
dx
([A]o -x)([B]o -x)
= kdt
In order to solve this, partial fraction decomposition is required to further ease integration:
1
1
kdt
([B]o -[A]o ) ([A]o -x) ([B]o -x)
Note: this is a constant
dx
1
From the point of view of integration, a further rearrangement is desirable (See algebra note above).
-1
-1
kdt
([B]o -[A]o ) x-[A]o
x-[B]o
t
dx
1
-ln x-[A]o - (-ln x-[B]o ) = kt + C
([B]o -[A]o )
Substitute and simplify
x-[A]o = [A]o -x and x-[B]o = [B]o -x
Integrate both sides
1
([B]o -x)
1
ln
= kt + C Determine C by setting x = 0 at t = 0
([B]o -[A]o ) [A]o -x
[B]o
([B]o -x)
[B]o
1
1
Hence:
ln
ln
= kt +
ln
([B]o -[A]o ) [A]o
([B]o -[A]o ) ([A]o -x)
([B]o -[A]o ) [A]o
([B]o -x)
[B]o
Multiply both sides by ([B]o -[A]o ), hence, ln
= ([B]o -[A]o )×kt + ln
[A]o
([A]o -x)
([B]o -x)
([A] -x)
([B]o -x) [B]o
= ln [A]o ([B]o -x)
o
ln
-ln
=
t
=
ln
k
([B]
-[A]
)×
o
o
[B]o
[B]o ([A]o -x)
([A]o -x) [A]o
[A]o
C=
1
[A]o ×[B]t
= ([B]o -[A]o )×kt
[B]o ×[A]t
Set ([A]o -x) =[A]t and ([B]o -x) =[B]t to give ln
651
Chapter 14: Chemical Kinetics
(D) Let 250-2x equal the partial pressure of CO(g) and x be the partial pressure of CO2(g).
2CO CO 2 C(s)
Ptot =PCO + PCO2 = 250 – 2x +x = 250 – x
x
250-2x
Ptot
[torr]
250
238
224
210
Time
[sec]
0
398
1002
1801
PCO2
PCO [torr]
0
12
26
40
250
226
198
170
The plots that follow show that the reaction appears to obey a second-order rate law.
Rate = k[CO]2
250
0.006
5.52
240
230
220
5.32
210
0.005
-1
200
1/PCO (mmHg )
ln PCO
PCO (mmHg)
90.
190
180
2
R2 = 0.9965
5.12
R = 0.9867
170
0
0
1000
time (S)
PCO
250
226
198
170
1000
1500
2000
ZERO ORDER PLOT
T
0
398
1002
1801
500
time (S)
1st ORDER PLOT
T
0
398
1002
1801
lnPCO
5.521461
5.420535
5.288267
5.135798
652
2000
R2 = 1
0.004
0
1000
time (S)
2nd ORDER PLOT
T
0
398
1002
1801
1/CO
0.004
0.004425
0.005051
0.005882
2000
(Best
correlation
coefficient)
Chapter 14: Chemical Kinetics
91. (D) Let 100-4x equal the partial pressure of PH3(g), x be the partial pressure of P4(g)and 6x be the
partial pressure of H2 (g)
4 PH 3 (g) P 4 ( g )
6 H 2 (g)
100 4x
x
6x
Ptot =PPH3 + PP4 + PH2 = 100 4x + x + 6x = 100+ 3x
Ptot
[torr]
100
150
167
172
Time [sec]
PP4 [torr]
PPH3 [torr]
0
40
80
120
0
50/3
67/3
72/3
100
100-(4)(50/3)
100-(4)(67/3)
100-(4)(72/3)
The plots to follow show that the reaction appears to obey a first-order rate law.
Rate = k[PH3]
100
R2 = 0.8652
4.35
90
2
R = 0.8369
80
0.210
2
R = 0.9991
3.85
0.160
50
1/PPH3 (mmHg -1 )
3.35
60
2.85
ln PPH3
PPH3 (mmHg)
70
40
0.110
2.35
30
0.060
20
1.85
10
0.010
1.35
0
-50
50
time (S)
150
ZERO ORDER PLOT
T
0
40
80
120
PPH3
100
33.3
10.7
4
0
0
100
time (S)
1st ORDER PLOT
T
0
40
80
120
lnPPH3
4.61
3.51
2.37
1.39
653
50
100
200
time (S)
2nd ORDER PLOT
T
0
40
80
120
1/PPH3
0.010
0.030
0.093
0.250
150
Chapter 14: Chemical Kinetics
92.
(D) Consider the following equilibria.
k1
KI
k2
E + S
E + I
E+P
ES
EI
k -1
d[P]
= k 2 [ES]
dt
Product production
Use the steady state approximation for [ES]
d[ES]
= k1 [E][S] k -1 [ES] k 2 [ES] = k1 [E][S] [ES] k -1 k 2 = 0
dt
k k2
k [E][S] [E][S]
[ES] = 1
=
solve for [ES] Keep in mind K M = -1
k1
k -1 k 2
KM
[E][I]
[E][I]
Formation of EI: K I =
[EI] =
[EI]
KI
[E o ] = [E] + [ES] + [EI] = [E] +
[S]
[I]
[E o ] = [E] 1 +
+
KM
KI
From above: [ES] =
[ES] =
Solve for [E]
[I]K M
K M + [S] +
KI
d[P]
= k 2 [ES] =
Remember
dt
[E] =
[S]
[E o ]
KM
[ES] =
[S]
[I]
+
1 +
KM
KI
[E][S]
KM
[E o ][S]
[E][S] [E][I]
+
KM
KI
=
[E o ]
[S]
[I]
+
1 +
KM
KI
multiplication by
[E o ][S]
[I]
K M 1 +
+ [S]
KI
k 2 [E o ][S]
[I]
K M 1 +
+ [S]
KI
Vmax [S]
d[P]
If we substitute k 2 [E o ] = Vmax then
=
dt
[I]
K M 1 +
+ [S]
K
I
Vmax [S]
Thus, as [I] increases, the ratio
decreases;
[I]
K M 1 +
+ [S]
KI
i.e., the rate of product formation decreases as [I] increases.
654
KM
affords
KM
Chapter 14: Chemical Kinetics
93.
(D) In order to determine a value for KM, we need to rearrange the equation so that we may obtain
a linear plot and extract parameters from the slope and intercepts.
k [E ][S]
K M +[S]
KM
1
[S]
V= 2 o
=
=
+
K M +[S]
V
k 2 [E o ][S]
k 2 [E o ][S]
k 2 [E o ][S]
KM
1
1
=
+
V
k 2 [E o ][S]
k 2 [E o ]
Plot
We need to have this in the form of y = mx+b
1
1
on the y-axis and
on the x-axis. See result below:
V
[S]
1
[V]
x-intercept
= -1
KM
pe
o
l
s
K M o]
[E
= k2
y-intercept
= 1
k2[Eo]
1
[S]
The plot of 1/V vs 1/[S] should yield a slope of
y-intercept of
0=
KM
and a
k 2 [E o ]
1
. The x-intercept = -1/KM
k 2 [E o ]
KM 1
1
k 2 [E o ] [S] k 2 [E o ]
1
[S]
=
-k 2 [E o ]
k 2 [E o ] K M
-1
KM
To find the value of KM take the negative inverse of x-intercept.
To find k2, invert the y-intercept and divide by the [Eo].
94. (M)
k1
HOOBr is rate-determining if the reaction obeys
a) The first elementary step HBr O 2
reaction rate = k [HBr][O2] since the rate of this step is identical to that of the experimental rate
law.
b) No, mechanisms cannot be shown to be absolutely correct, only consistent with experimental
observations.
c) Yes; the sum of the elementary steps (3 HBr + O2 HOBr + Br2 + H2O) is not consistent with
the overall stoichiometry (since HOBr is not detected as a product) of the reaction and
therefore cannot be considered a valid mechanism.
95.
(M)
(a) Both reactions are first-order, because they involve the decomposition of one molecule.
(b) k2 is the slow reaction.
(c) To determine the concentration of the product, N2, we must first determine how much reactant
remains at the end of the given time period, from which we can calculate the amount of
reactant consumed and therefore the amount of product produced. Since this is a first-order
reaction, the concentration of the reactant, N2O after time t is determined as follows:
A t A 0 e kt
N 2O0.1 2.0 M exp 25.7 s1 0.1 s 0.153 M N 2O remaining
655
Chapter 14: Chemical Kinetics
The amount of N2O consumed = 2.0 M – 0.153 M =1.847 M
1 M N2
0.9235 M N 2
[N 2 ] 1.847 M NO
2 M NO
(d) The process is identical to step (c).
N 2O0.1 4.0 M exp 18.2 s1 0.025 s 2.538 M N 2O remaining
The amount of N2O consumed = 4.0 M – 2.538 M =1.462 M
1 M N2O
0.731 M N 2 O
[N 2 O] 1.462 M NO
2 M NO
FEATURE PROBLEMS
96.
(D)
(a)
To determine the order of the reaction, we need C 6 H 5 N 2 Cl at each time. To determine
this value, note that 58.3 mL N 2 g evolved corresponds to total depletion of C6 H 5 N 2 Cl ,
to C 6 H 5 N 2 Cl = 0.000 M .
F
GH
bg
Thus, at any point in time, C 6 H 5 N 2 Cl = 0.071 M volume N 2 g
I
b g JK
0.071 M C 6 H 5 N 2 Cl
58.3 mL N 2 g
0.071 M C6 H 5 N 2 Cl
Consider 21 min: C6 H 5 N 2 Cl = 0.071 M 44.3 mL N 2
= 0.017 M
58.3 mL N 2 g
The numbers in the following table are determined with this method.
Time, min
VN2 , mL
0
0
C H
71
6
(b)
5
N 2 Cl , mM
3
6
9
12
15
18
21
24
27
30
10.8 19.3 26.3 32.4 37.3 41.3 44.3 46.5 48.4 50.4 58.3
58
47
39
32
26
21
17
14
12
10
[The concentration is given in thousandths of a mole per liter (mM).]
time(min) 0 3 6 9 12 15 18 21 24
27
0
30
T(min)
3 3 3
3
3
3
3
3
3
3
[C6H5N2Cl](mM) 71 58 47 39 32 26
21 17
14
12 10
[C6H5N2Cl](mM)
-13 -11 -8
-1
Reaction Rate (mM min )
-7
4.3 3.7 2.7 2.3
656
-6
-5
2.0
1.7
-4
1.3
-3
-2
1.0
0.7
-2
0.7
Chapter 14: Chemical Kinetics
The two graphs are drawn on the same axes.
(m
L)
N
2
Cl]
(m
M)
or
V
2
N5
H6
[C
Plot of [C 6H 5N 2Cl] and VN versus time
[C6H5N2Cl (mM) or VN2 (mL)
(c)
2
60
mL of N2
40
20
[C6H5N2Cl]
0
0
5
10
15
20
25
30
time (min)
(d)
The rate of the reaction at t = 21 min is the slope of the tangent line to the
C 6 H 5 N 2 Cl curve. The tangent line intercepts the vertical axis at about
C 6 H 5 N 2 Cl = 39 mM and the horizontal axis at about 37 min
39 103 M
= 1.05 103 M min 1 = 1.1 103 M min 1
37 min
The agreement with the reported value is very good.
Reaction rate =
(e)
(f)
The initial rate is the slope of the tangent line to the C 6 H 5 N 2 Cl curve at t = 0 . The
intercept with the vertical axis is 71 mM, of course. That with the horizontal axis is about
13 min.
71103 M
Rate =
= 5.5 103 M min 1
13 min
The first-order rate law is Rate = k C6 H 5 N 2 Cl , which we solve for k:
k=
Rate
C6 H5 N 2Cl
k0 =
5.5 103 M min 1
= 0.077 min 1
3
71 10 M
1.1103 M min 1
k21 =
= 0.065 min 1
3
17 10 M
An average value would be a reasonable estimate: k avg = 0.071 min 1
(g)
The estimated rate constant gives one value of the half-life:
0.693
0.693
t1/2 =
=
= 9.8 min
0.071 min 1
k
The first half-life occurs when C6 H 5 N 2 Cl drops from 0.071 M to 0.0355 M. This occurs
at about 10.5 min.
(h)
The reaction should be three-fourths complete in two half-lives, or about 20 minutes.
657
Chapter 14: Chemical Kinetics
(i)
The graph plots ln C6 H 5 N 2 Cl (in millimoles/L) vs. time in minutes.
0
Time (min)
Plot of ln[C6H5N2Cl] versus time
10
20
30
ln[C6H5N2Cl]
-2.75
-3.25
-3.75
y = -0.0661x - 2.66
-4.25
-4.75
The linearity of the graph demonstrates that the reaction is first-order.
(j)
k = slope = 6.61102 min 1 = 0.0661 min 1
t1/2 =
97.
(D)
(a)
0.693
= 10.5 min , in good agreement with our previously determined values.
0.0661 min 1
In Experiments 1 & 2, [KI] is the same (0.20 M), while
bNH g S O
4 2
2
8
is halved, from 0.20
M to 0.10 M. As a consequence, the time to produce a color change doubles (i.e., the rate is
2
halved). This indicates that reaction (a) is first-order in S2 O8 . Experiments 2 and 3
produce a similar conclusion. In Experiments 4 and 5,
bNH g S O
4 2
2
8
is the same (0.20 M)
while [KI] is halved, from 0.10 to 0.050 M. As a consequence, the time to produce a color
change nearly doubles, that is, the rate is halved. This indicates that reaction (a) is also firstorder in I . Reaction (a) is (1 + 1) second-order overall.
(b)
The blue color appears when all the S2 O 23 has been consumed, for only then does reaction
(b) cease. The same amount of S2 O 2
3 is placed in each reaction mixture.
0.010 mol Na 2S2 O3 1 mol S2 O32
1L
amount S2 O32 = 10.0 mL
= 1.0 104 mol
1000 mL
1L
1 mol Na 2 S2 O3
Through stoichiometry, we determine the amount of each reactant that reacts before this
2
amount of S2 O3 will be consumed.
658
Chapter 14: Chemical Kinetics
amount S2 O8
2
1.0 10
4
mol S2 O3
2
= 5.0 105 mol S2 O8
2
amount I = 5.0 105 mol S2 O8
1 mol I3
1 mol S2 O8
2
2 mol S2 O3
1 mol I3
2
2
2 mol I
= 1.0 104 mol I
2
1 mol S2 O8
Note that we do not use “3 mol I ” from equation (a) since one mole has not been
oxidized; it simply complexes with the product I 2 . The total volume of each solution
is 25.0 mL + 25.0 mL +10.0 mL + 5.0 mL = 65.0 mL , or 0.0650 L.
The amount of S2 O8
2
that reacts in each case is 5.0 105 mol and thus
5.0 105 mol
= 7.7 104 M
0.0650 L
S2 O82
+7.7 104 M
Thus, Rate1 =
=
= 3.7 105 M s1
t
21 s
S2 O8 2 =
(c)
S2 O8
2
+7.7 104 M
= 1.8 105 M s1
t
42 s
To determine the value of k , we need initial concentrations, as altered by dilution.
25.0 mL
25.0 mL
S2 O82 = 0.20 M
I = 0.20 M
= 0.077 M
= 0.077 M
1
1
65.0 mL total
65.0 mL
For Experiment 2, Rate 2 =
1
Rate1 = 3.7 105 M s1 = k S2 O82
=
1
b
I = k 0.077 M
3.7 105 M s 1
k=
= 6.2 103 M 1 s 1
0.077 M 0.077 M
25.0 mL
S2 O82 = 0.10 M
= 0.038 M
2
65.0 mL total
g b0.077 Mg
1
1
25.0 mL
I = 0.20 M
= 0.077 M
2
65.0 mL
Rate 2 = 1.8 105 M s 1 = k S2 O82 I = k 0.038 M 0.077 M
1.8 10 5 M s1
k=
= 6.2 103 M 1 s1
0.038 M 0.077 M
1
(d)
1
1
1
First we determine concentrations for Experiment 4.
25.0 mL
25.0 mL
S2 O8 2 = 0.20 M
I = 0.10
=
0.077
M
= 0.038 M
4
4
65.0 mL total
65.0 mL
We have two expressions for Rate; let us equate them and solve for the rate constant.
2
S2 O8 +7.7 104 M
1
2 1
Rate 4 =
=
= k S2O8 I = k 0.077 M 0.038 M
4
4
t
t
k=
7.7 104 M
0.26 M 1
=
t 0.077 M 0.038 M
t
659
k3 =
0.26 M 1
= 0.0014 M 1 s 1
189 s
Chapter 14: Chemical Kinetics
(e)
k13 =
0.26 M 1
= 0.0030 M 1 s 1
88 s
k33 =
0.26 M 1
= 0.012 M 1 s 1
21 s
k24 =
0.26 M 1
= 0.0062 M 1 s 1
42 s
We plot ln k vs. 1/ T The slope of the line = Ea / R .
Plot of ln k versus 1/T
0.00325
-1
1/T (K )
0.0034
0.00355
-3.5
-4
ln k
-4.5
-5
y = -6135x + 16
-5.5
-6
Ea = +6135 K 8.3145 J mol1 K 1 = 51.0 103 J/mol = 51.0 kJ/mol
The scatter of the data permits only a two significant figure result: 51 kJ/mol
(f)
For the mechanism to agree with the reaction stoichiometry, the steps of the mechanism
must sum to the overall reaction, in the manner of Hess's law.
(slow) I + S2 O82 IS2 O83
(fast) IS2 O83 2 SO 24 + I +
(fast) I + + I I 2
(fast) I 2 + I I 3
(net) 3 I + S2 O82 2 SO 24 + I 3
Each of the intermediates cancels: IS2 O83 is produced in the first step and consumed in the
second, I + is produced in the second step and consumed in the third, I 2 is produced in the
third step and consumed in the fourth. The mechanism is consistent with the stoichiometry.
The rate of the slow step of the mechanism is
Rate1 = k1 S2 O82
1
I
1
This is exactly the same as the experimental rate law. It is reasonable that the first step be
slow since it involves two negatively charged species coming together. We know that like
charges repel, and thus this should not be an easy or rapid process.
660
Chapter 14: Chemical Kinetics
SELF-ASSESSMENT EXERCISES
98.
(E)
(a) [A]0: Initial concentration of reactant A
(b) k: Reaction rate constant, which is the proportionality constant between reaction rate and
reactant concentration
(c) t1/2: Half-life of the reaction, the amount of time that the concentration of a certain reactant is
reduced by half
(d) Zero-order reaction: A reaction in which the rate is not dependent on the concentration of the
reactant
(e) Catalyst: A substance which speeds up the reaction by lowering the activation energy, but it
does not itself get consumed
99.
(E)
(a) Method of initial rates: A study of the kinetics of the reaction by measuring the initial reaction
rates, used to determine the reaction order
(b) Activated complex: Species that exist in a transitory state between the reactants and the
products
(c) Reaction mechanism: Sequential elementary steps that show the conversion of reactant(s) to
final product(s)
(d) Heterogeneous Catalyst: A catalyst which is in a different physical phase than the reaction
medium
(e) Rate-determining step: A reaction which occurs more slowly than other reactions in a
mechanism and therefore usually controls the overall rate of the reaction
100. (E)
(a) First-order and second-order reactions: In a first-order reaction, the rate of the reaction
depends on the concentration of only one substrate and in a 1-to-1 manner (doubling the
concentration of the reactant doubles the rate of the reaction). In a second-order reaction, the
rate depends on two molecules reacting with each other at the elementary level.
(b) Rate law and integrated rate law: Rate law describes how the rate relates to the concentration
of the reactants and the overall rate of a reaction, whereas the integrated rate law expresses the
concentration of a reactant as a function of time
(c) Activation energy and enthalpy of reaction: Activation energy is the minimum energy
required for a particular reaction to take place, whereas enthalpy of reaction is the amount of
heat generated (or consumed) by a reaction when it happens
(d) Elementary process and overall reaction: Individual steps of a reaction mechanism, which
describes any molecular event that significantly alters a molecule’s energy or geometry or
produces a new molecule
(e) Enzyme and substrate: An enzyme is a protein that acts as a catalyst for a biological reaction.
A substrate is the reactant that is transformed in the reaction (in this context, by the enzyme).
101. (E) The answer is (c). The rate constant k is only dependent on temperature, not on the
concentration of the reactants
661
Chapter 14: Chemical Kinetics
102. (E) The answers are (b) and (e). Because half-life is 75 seconds, the quantity of reactant left at
two half-lives (75 + 75 = 150) equals one-half of the level at 75 seconds. Also, if the initial
concentration is doubled, after one half-life the remaining concentration would have to be twice
as much as the original concentration.
103. (E) The answer is (a). Half-life t½ = 13.9 min, k = ln 2/t½ = 0.050 min-1. Rate of a first-order
reaction is as follows:
d A
k A 0.050 min 1 0.40 M 0.020 M min 1
dt
104. (E) The answer is (d). A second-order reaction is expressed as follows:
d A
2
k A
dt
If the rate of the reaction when [A] =0.50 is k(0.50)2 = k(0.25). If [A] = 0.25 M, then the rate is
k(0.0625), which is ¼ of the rate at [A] =0.50.
105. (M) The answer is (b). Going to slightly higher temperatures broadens the molecular speed
distribution, which in turn increases the fraction of molecules at the high kinetic energy range
(which are those sufficiently energetic to make a reaction happen).
106. (E) The answer is (c). Since the reaction at hand is described as an elementary one, the rate of the
reaction is k[A][B].
107. (E) We note that from the given data, the half-life of the reaction is 100 seconds (at t = 0, [A] =
0.88 M/s, whereas at t = 100, [A] = 0.44 M/s). Therefore, the rate constant k is:
k = ln 2/100 s = 0.00693 s-1. We can now calculate instantaneous rate of the reaction:
d[A]/dt = (0.00693 s-1)(0.44 M) = 3.0×10-3 M·s-1
108. (M)
(a) For a first-order reaction,
ln 2 0.693
0.0231 min 1
k
t1/2
30
ln[A]t ln[A]0 kt
ln 0.25 0
ln[A]t ln[A]0
60.0 min
k
0.0231 min 1
(b) For a zero-order reaction,
0.5 0.5
0.0167 min 1
k
t1/2 30
t
[A]t [A]0 kt
t
[A]t [A]0
0.25 1.00
45.0 min
k
0.0167 min 1
662
Chapter 14: Chemical Kinetics
109. (M) The reaction is second-order, because the half-life doubles with each successive half-life
period.
110. (M)
(a) The initial rate = ΔM/Δt = (1.204 M – 1.180 M)/(1.0 min) = 0.024 M/min
(b) In experiment 2, the initial concentration is twice that of experiment A. For a second-order
reaction:
Rate = k [Aexp 2]2 = k [2×Aexp 1]2 = 4 k [Aexp 1]2
This means that if the reaction is second order, its initial rate of experiment 2 will be 4 times that
of experiment 1 (that is, 4 times as many moles of A will be consumed in a given amount of
time). The initial rate is 4×0.024 M/min = 0.096 M/s. Therefore, at 1 minute, [A] = 2.408 –
0.0960 = 2.312 M.
(c) The half-life of the reaction, obtained from experiment 1, is 35 minutes. If the reaction is
first-order, then k = ln 2/35 min = 0.0198 min-1.
For a first-order reaction,
A t A 0 e kt
A 35min 2.408 exp 0.0198 min -1 30 min 1.33 M
111. (D) The overall stoichiometry of the reaction is determined by adding the two reactions with each
other: A + 2B C + D
(a) Since I is made slowly but is used very quickly, its rate of formation is essentially zero. The
amount of I at any given time during the reaction can be expressed as follows:
d I
dt
I
0 k1 A B k 2 B I
k1
A
k2
Using the above expression for [I], we can now determine the overall reaction rate law:
d C
k
k 2 I B k 2 1 A B k1 A B
dt
k2
(b) Adding the two reactions given, we still get the same overall stoichiometry as part (a).
However, with the given proposed reaction mechanisms, the rate law for the product(s) is given
as follows:
663
Chapter 14: Chemical Kinetics
d B2
dt
B2
k1 B k 1 B2 k 2 A B 0
2
k1 B
2
k 1 k 2 A
Therefore,
d C
dt
k 2 A B2
k 2 k1 A B
2
k 1 k 2 A
which does not agree with the observed reaction rate law.
112. (M) The answer is (b), first-order, because only in a first-order reaction is the half-life
independent of the concentration of the reacting species.
113. (E) The answer is (a), zero-order, because in a zero-order reaction the relationship between
concentration and time is: [A]t = kt + [A]0
114. (M) The answer is (d). The relationship between rate constant (and thus rate) between two
reactions can be expressed as follows:
Ea 1 1
k2
exp
k1
R T1 T2
If T2 is twice T1, the above expression gets modified as follows:
k Ea 1
1
ln 2
R T1 2T1
k1
R k 2 T1 1
,
ln
E a k1 2T1
R
k Ea T 1
ln 2 1
2T1
k1
For reasonably high temperatures,
R
k Ea
T
1
ln 2 1
2T1 2
k1
Therefore,
R
k 2 Ea
1/2
e 1.64
k
1
664
Chapter 14: Chemical Kinetics
115. (E) The answer is (c), remain the same. This is because for a zero-order reaction,
d[A]/dt = k[A]0 = k. Therefore, the reaction rate is independent of the concentration of the
reactant.
116. (M) The overarching concept for this concept map is kinetics as a result of successful collision.
The subtopics are collision theory, molecular transition theory. Molecular speed and orientation
derive from collision theory. Transition complexes and partial bonds fall under the molecular
transition theory heading. Deriving from the collision theory is another major topic, the
Arrhenius relationship. The Arrhenius relationship encompasses the ideas of activation energy,
Arrhenius collision factor, and exponential relationship between temperature and rate constant.
665
CHAPTER 15
PRINCIPLES OF CHEMICAL EQUILIBRIUM
PRACTICE EXAMPLES
1A
(E) The reaction is as follows:
2Cu 2 (aq) Sn 2 (aq) 2Cu (aq) Sn 4 (aq)
Therefore, the equilibrium expression is as follows:
2
Cu Sn 4
K
2
Cu 2 Sn 2
Rearranging and solving for Cu2+, the following expression is obtained:
1/ 2
Cu + 2 Sn 4+
2+
Cu =
K Sn 2+
1B
1/2
x2 x
1.48 x
x
1.22
(E) The reaction is as follows:
2Fe3 (aq) Hg 22 (aq) 2Fe2 (aq) 2Hg 2 (aq)
Therefore, the equilibrium expression is as follows:
2
2
Fe2 Hg 2
0.0025 0.0018
9.14 106
K
2
3 2
2
0.015 x
Fe Hg 2
2
2
Rearranging and solving for Hg22+, the following expression is obtained:
2
2
Fe2 Hg 2
0.0025 0.0018
2
Hg 2
0.009847 0.0098 M
2
2
0.015 9.14 106
Fe3 K
2
2A
2
(E) The example gives Kc = 5.8×105 for the reaction N 2 g + 3 H 2 g
2 NH 3 g .
The reaction we are considering is one-third of this reaction. If we divide the reaction by
3, we should take the cube root of the equilibrium constant to obtain the value of the
equilibrium constant for the “divided” reaction: K c3 3 K c 3 5.8 105 8.3 102
2B
bg
(E) First we reverse the given reaction to put NO 2 g on the reactant side. The new
equilibrium constant is the inverse of the given one.
NO 2 g
NO g + 12 O 2 g
K c ' = 1/ (1.2 102 ) = 0.0083
bg
Then we double the reaction to obtain 2 moles of NO 2 g as reactant. The equilibrium
constant is then raised to the second power.
2 NO 2 g
2 NO g + O 2 g
K c = 0.00833 = 6.9 105
2
666
Chapter15: Principles of Chemical Equilibrium
3A
(E) We use the expression K p = K c RT
ngas
. In this case, ngas = 3 +1 2 = 2 and thus we
have
K p = K c RT = 2.8 109 0.08314 298 = 1.7 106
2
3B
2
(M) We begin by writing the Kp expression. We then substitute
P n / V RT = [ concentration ]RT for each pressure. We collect terms to obtain an expression
relating Kc and Kp , into which we substitute to find the value of Kc .
Kp
{ P( H 2 )}2 { P(S2 )} ([ H 2 ]RT ) 2 ([S2 ]RT ) [H 2 ]2 [S2 ]
RT Kc RT
{ P( H 2S)}2
([ H 2S]RT ) 2
[H 2S]2
b g
The same result can be obtained by using Kp = Kc RT
Kc =
Kp
RT
=
ngas
, since ngas = 2 +1 2 = +1.
1.2 102
= 1.1 104
0.08314 1065 + 273
But the reaction has been reversed and halved. Thus Kfinal=
4A
(E) We remember that neither solids, such as Ca 5 ( PO 4 ) 3 OH(s) , nor liquids, such as H 2 O(l) ,
appear in the equilibrium constant expression. Concentrations of products appear in the
numerator, those of reactants in the denominator. Kc =
4B
1
1
9091 95
1.1104
Kc
Ca 2+
5
HPO 4
H+
2 3
4
(E) First we write the balanced chemical equation for the reaction. Then we write the
equilibrium constant expressions, remembering that gases and solutes in aqueous solution
appear in the Kc expression, but pure liquids and pure solids do not.
3 Fe s + 4 H 2 O g
Fe3O 4 s + 4 H 2 g
b g
b g
{ P H 2 }4
Kp =
{ P H 2 O }4
5A
Kc =
H2
H 2O
4
4
Because ngas = 4 4 = 0, Kp = Kc
bg
(M) We compute the value of Qc. Each concentration equals the mass m of the substance
divided by its molar mass (this quotient is the amount of the substance in moles) and further
divided by the volume of the container.
667
Chapter15: Principles of Chemical Equilibrium
m
1 mol CO 2
m
1 mol H 2
1
44.0 g CO 2
2.0 g H 2
28.0 18.0
CO 2 H 2
V
V
Qc =
44.0 2.0
= 5.7 1.00 = K c
1
44.0 2.0
CO H 2 O m 1 mol CO m 1 mol H 2O
18.0 g H 2 O 28.0 18.0
28.0 g CO
V
V
(In evaluating the expression above, we cancelled the equal values of V , and we also
cancelled the equal values of m .) Because the value of Qc is larger than the value of Kc ,
the reaction will proceed to the left to reach a state of equilibrium. Thus, at equilibrium
there will be greater quantities of reactants, and smaller quantities of products than there
were initially.
5B
(M) We compare the value of the reaction quotient, Qp , to that of K p .
Qp =
{P (PCl3 )}{P(Cl2 )} 2.19 0.88
=
= 0.098
{P(PCl5 }
19.7
K p = K c RT
2 1
= K c RT = 0.0454 0.08206 (261 273) = 1.99
1
1
Because Qc Kc , the net reaction will proceed to the right, forming products and consuming
reactants.
6A
bg
(E) O 2 g is a reactant. The equilibrium system will shift right, forming product in an
attempt to consume some of the added O2(g) reactant. Looked at in another way, O 2 is
increased above its equilibrium value by the addition of oxygen. This makes Qc smaller
than Kc . (The O 2 is in the denominator of the expression.) And the system shifts right to
drive Qc back up to Kc, at which point equilibrium will have been achieved.
6B
(M)
(a) The position of an equilibrium mixture is affected only by changing the concentration
of substances that appear in the equilibrium constant expression, Kc = CO 2 . Since
CaO(s) is a pure solid, its concentration does not appear in the equilibrium constant
expression and thus adding extra CaO(s) will have no direct effect on the position of
equilibrium.
bg
(b)
The addition of CO 2 g will increase CO 2 above its equilibrium value. The reaction
will shift left to alleviate this increase, causing some CaCO 3 s to form.
(c)
Since CaCO3(s) is a pure solid like CaO(s), its concentration does not appear in the
equilibrium constant expression and thus the addition of any solid CaCO3 to an
equilibrium mixture will not have an effect upon the position of equilibrium.
bg
668
Chapter15: Principles of Chemical Equilibrium
7A
(E) We know that a decrease in volume or an increase in pressure of an equilibrium mixture
of gases causes a net reaction in the direction producing the smaller number of moles of gas.
In the reaction in question, that direction is to the left: one mole of N 2 O 4 g is formed when
two moles of NO 2 g combine. Thus, decreasing the cylinder volume would have the initial
bg
bg
effect of doubling both N 2 O 4 and NO 2 . In order to reestablish equilibrium, some NO2
will then be converted into N2O4. Note, however, that the NO2 concentration will still
ultimately end up being higher than it was prior to pressurization.
b g b g
7B
(E) In the balanced chemical equation for the chemical reaction, ngas = 1+1 1+1 = 0 .
As a consequence, a change in overall volume or total gas pressure will have no effect on the
position of equilibrium. In the equilibrium constant expression, the two partial pressures in
the numerator will be affected to exactly the same degree, as will the two partial pressures in
the denominator, and, as a result, Qp will continue to equal Kp .
8A
(E) The cited reaction is endothermic. Raising the temperature on an equilibrium mixture
favors the endothermic reaction. Thus, N 2 O 4 g should decompose more completely at
higher temperatures and the amount of NO 2 g formed from a given amount of N 2 O 4 g
will be greater at high temperatures than at low ones.
8B
bg
bg
(E) The NH3(g) formation reaction is
1
2
bg
N 2 g + 32 H 2 g NH 3 g , H o = 46.11 kJ/mol.
This reaction is an exothermic reaction. Lowering temperature causes a shift in the direction
of this exothermic reaction to the right toward products. Thus, the equilibrium NH 3 g will
bg
be greater at 100 C .
9A
(E) We write the expression for Kc and then substitute expressions for molar concentrations.
2
0.22 0.11
[H 2 ]2 [S2 ] 3.00 3.00
Kc
2.3 10 4
2
2
[H 2 S]
2.78
3.00
9B
(M) We write the equilibrium constant expression and solve for N 2 O 4 .
NO2
=
NO2
2
2
0.0236
2
=
= 0.121 M
N 2 O4 =
N 2 O 4
4.61 103 4.61 103
Then we determine the mass of N 2 O 4 present in 2.26 L.
0.121 mol N 2 O 4 92.01 g N 2 O 4
N 2 O 4 mass = 2.26 L
= 25.2 g N 2 O 4
1L
1 mol N 2 O 4
K c = 4.61 10
3
669
Chapter15: Principles of Chemical Equilibrium
10A (M) We use the initial-change-equilibrium setup to establish the amount of each substance
at equilibrium. We then label each entry in the table in the order of its determination (1st,
2nd, 3rd, 4th, 5th), to better illustrate the technique. We know the initial amounts of all
substances (1st). There are no products at the start.
Because ‘‘initial’’+ ‘‘change’’ = ‘‘equilibrium’’ , the equilibrium amount (2nd) of Br2 g
bg
enables us to determine “change” (3rd) for Br2 g . We then use stoichiometry to write other
entries (4th) on the “change” line. And finally, we determine the remaining equilibrium
amounts (5th).
2 NOBr g
2 NO g
st
Initial:
1.86 mol (1 )
0.00 mol (1st)
Change:
0.164 mol (4th)
+0.164 mol (4th)
th
Equil.:
1.70 mol (5 )
0.164 mol (5th)
2
0.164 0.082 mol
5.00
[NO]2 [Br2 ] 5.00
1.5 104
Kc
2
2
[NOBr]
1.70
5.00
Reaction:
Here, ngas = 2 +1 2 = +1.
+
Br2 g
0.00 mol (1st)
+0.082 mol (3rd)
0.082 mol (2nd)
K p = K c RT = 1.5 104 0.08314 298 = 3.7 103
+1
10B (M) Use the amounts stated in the problem to determine the equilibrium concentration for
each substance.
2 SO3 g
2 SO 2 g
+
O2 g
Reaction:
Initial:
0 mol
0.100 mol
0.100 mol
0.0916 mol
0.0916 / 2 mol
Changes:
+0.0916 mol
Equil.:
0.0916 mol
0.0084 mol
0.0542 mol
0.0084 mol
0.0542 mol
0.0916 mol
Concentrations:
1.52 L
1.52 L
1.52 L
Concentrations: 0.0603 M
0.0055 M
0.0357 M
We use these values to compute Kc for the reaction and then the relationship
b g
Kp = Kc RT
ngas
(with ngas = 2 +1 2 = +1) to determine the value of Kp .
2
SO 2 O 2 0.0055 0.0357
Kc =
=
= 3.0 104
2
2
SO3
0.0603
2
K p = 3.0 104 (0.08314 900) 0.022
670
Chapter15: Principles of Chemical Equilibrium
11A (M) The equilibrium constant expression is Kp = P{H 2 O} P{CO 2 } = 0.231 at 100 C . From
bg
the balanced chemical equation, we see that one mole of H 2 O g is formed for each mole of
bg
b
g
2
CO 2 g produced. Consequently, P{H 2 O} = P{CO 2 } and Kp = P{CO 2 } . We solve this
expression for P{CO 2 } : P{CO 2 } ( P{CO 2 }) 2 K p 0.231 0.481 atm.
11B (M) The equation for the reaction is NH 4 HS s
NH3 g + H 2S g , K p = 0.108 at 25 C.
The two partial pressures do not have to be equal at equilibrium. The only instance in which
they must be equal is when the two gases come solely from the decomposition of NH 4 HS s .
In this case, some of the NH 3 g has come from another source. We can obtain the pressure
of H 2S g by substitution into the equilibrium constant expression, since we are given the
equilibrium pressure of NH 3 g .
0.108
K p = P{H 2S} P{NH 3 } = 0.108 = P{H 2S} 0.500 atm NH 3 P{H 2S} =
= 0.216 atm
0.500
So, Ptotal = PH S PNH3 = 0.216 atm + 0.500 atm = 0.716 atm
bg
bg
bg
bg
2
12A (M) We set up this problem in the same manner that we have previously employed, namely
designating the equilibrium amount of HI as 2x . (Note that we have used the same
multipliers for x as the stoichiometric coefficients.)
Equation:
H2 g
Initial:
Changes:
Equil:
0.150 mol
x mol
0.150 x mol
b
+
g
I2 g
2 HI g
0.200 mol
x mol
0.200 x mol
0 mol
+2x mol
2 x mol
b
g
2
2x
2
2x
15.0
Kc
=
= 50.2
0.150 x 0.200 x 0.150 x 0.200 x
15.0
15.0
We substitute these terms into the equilibrium constant expression and solve for x.
b
gb
g
c
h
4 x 2 = 0.150 x 0.200 x 50.2 = 50.2 0.0300 0.350 x + x 2 = 1.51 17.6 x + 50.2 x 2
0 = 46.2 x 2 17.6 x +1.51
x=
Now we use the quadratic equation to determine the value of x.
b b 2 4ac 17.6 (17.6) 2 4 46.2 1.51 17.6 5.54
=
=
= 0.250 or 0.131
2a
2 46.2
92.4
671
Chapter15: Principles of Chemical Equilibrium
The first root cannot be used because it would afford a negative amount of H 2 (namely,
0.150-0.250 = -0.100). Thus, we have 2 0.131 = 0.262 mol HI at equilibrium. We check
by substituting the amounts into the Kc expression. (Notice that the volumes cancel.) The
slight disagreement in the two values (52 compared to 50.2) is the result of rounding error.
0.0686
b0.262g
K =
=
b0.150 0.131gb0.200 0.131g 0.019 0.069 = 52
2
c
12B (D)
(a)
3
The equation for the reaction is N 2 O 4 g
2 NO 2 g and K c = 4.6110 at 25 C .
In the example, this reaction is conducted in a 0.372 L flask. The effect of moving the
mixture to the larger, 10.0 L container is that the reaction will be shifted to produce a
greater number of moles of gas. Thus, NO 2 g will be produced and N 2 O 4 g will
dissociate. Consequently, the amount of N 2 O 4 will decrease.
bg
bg
(b) The equilibrium constant expression, substituting 10.0 L for 0.372 L, follows.
2
2x
[ NO 2 ]2
4x2
10.0
4.61 103
Kc
x
.
0
0240
[N 2 O 4 ]
10.0(0.0240 x )
10.0
This can be solved with the quadratic equation, and the sensible result is x = 0.0118
moles. We can attempt the method of successive approximations. First, assume that
x 0.0240 . We obtain:
FG IJ
H K
x
4.61 103 10.0 (0.0240 0)
4.61 103 2.50 (0.0240 0) 0.0166
4
Clearly x is not much smaller than 0.0240. So, second, assume x 0.0166 . We obtain:
x 4.61 103 2.50 (0.0240 0.0166) 0.00925
This assumption is not valid either. So, third, assume x 0.00925 . We obtain:
x 4.61 103 2.50 (0.0240 0.00925) 0.0130
Notice that after each cycle the value we obtain for x gets closer to the value obtained
from the roots of the equation. The values from the next several cycles follow.
Cycle
x value
5th
6th
7th
8th
9th
10th
11th
4th
0.0112 0.0121 0.0117 0.0119 0.01181 0.01186 0.01183 0.01184
The amount of N 2 O 4 at equilibrium is 0.0118 mol, less than the 0.0210 mol N 2 O 4 at
equilibrium in the 0.372 L flask, as predicted.
672
Chapter15: Principles of Chemical Equilibrium
13A (M) Again we base our solution on the balanced chemical equation.
Fe3+ aq + Ag s
K c = 2.98
Equation: Ag + aq + Fe 2+ aq
Initial:
0M
Changes: +x M
x M
Equil:
Kc =
Fe 3+
Ag
+
Fe
2+
0M
+x M
x M
= 2.98 =
1.20 x
x2
1.20 M
x M
1.20 x M
b
g
2.98 x 2 = 1.20 x
0 = 2.98 x 2 + x 1.20
We use the quadratic formula to obtain a solution.
2
b b 2 4ac 1.00 (1.00) + 4 2.98 1.20 1.00 3.91
x=
=
=
= 0.488 M or 0.824 M
2a
2 2.98
5.96
A negative root makes no physical sense. We obtain the equilibrium concentrations from x.
Ag + = Fe 2+ = 0.488 M
Fe3+ = 1.20 0.488 = 0.71 M
13B (M) We first calculate the value of Qc to determine the direction of the reaction.
Qc =
V 2+ Cr 3+
V
3+
Cr
2+
=
0.150 0.150
= 225 7.2 102 = Kc
0.0100 0.0100
Because the reaction quotient has a smaller value than the equilibrium constant, a net
reaction to the right will occur. We now set up this solution as we have others, heretofore,
based on the balanced chemical equation.
V 3+ aq
2+
3+
+ Cr 2+ aq
V aq + Cr aq
initial
0.0100 M
0.0100 M
0.150 M
0.150 M
changes
–x M
–x M
+x M
+x M
equil
(0.0100 – x)M (0.0100 – x)M (0.150 + x)M (0.150 + x)M
Kc =
V 2+ Cr 3+
V 3+ Cr 2+
b0.150 + xg b0.150 + xg = 7.2 10 = FG 0.150 + x IJ
=
H 0.0100 x K
b0.0100 xg b0.0100 xg
2
2
If we take the square root of both sides of this expression, we obtain
0.150 x
27
7.2 102
0.0100 x
0.150 + x = 0.27 – 27x which becomes 28 x = 0.12 and yields 0.0043 M. Then the
equilibrium concentrations are: V 3+ = Cr 2+ = 0.0100 M 0.0043 M = 0.0057 M
V 2+ = Cr 3+ = 0.150 M + 0.0043 M = 0.154 M
673
Chapter15: Principles of Chemical Equilibrium
INTEGRATIVE EXAMPLE
A.
(E) We will determine the concentration of F6P and the final enthalpy by adding the two
reactions:
C6 H12 O6 ATP G6P ADP
G6P F6P
C6 H12 O6 ATP ADP F6P
ΔHTOT = –19.74 kJ·mol-1 + 2.84 kJ·mol-1 = –16.9 kJ·mol-1
Since the overall reaction is obtained by adding the two individual reactions, then the overall
reaction equilibrium constant is the product of the two individual K values. That is,
K K1 K 2 1278
The equilibrium concentrations of the reactants and products is determined as follows:
Initial
Change
Equil
K
C6H12O6
1.20×10-6
-x
1.20×10-6-x
+
ATP
1×10-4
-x
1×10-4-x
ADP
1×10-2
+x
1×10-2+x
+
F6P
0
+x
x
ADP F6P
C6 H12O6 ATP
1278
110
1.20 10
6
2
x x
x 1 104 x
1.0 102 x x 2
1.2 1010 1.012 104 x x 2
Expanding and rearranging the above equation yields the following second-order polynomial:
1277 x2 – 0.1393 x + 1.534×10-7 = 0
Using the quadratic equation to solve for x, we obtain two roots: x = 1.113×10-6 and
1.080×10-4. Only the first one makes physical sense, because it is less than the initial value of
C6H12O6. Therefore, [F6P]eq = 1.113×10-6.
During a fever, the body generates heat. Since the net reaction above is exothermic,
Le Châtelier's principle would force the equilibrium to the left, reducing the amount of F6P
generated.
674
Chapter15: Principles of Chemical Equilibrium
B.
(E)
(a) The ideal gas law can be used for this reaction, since we are relating vapor pressure and
concentration. Since K = 3.3×10-29 for decomposition of Br2 to Br (very small), then it can be
ignored.
1
nRT 0.100 mol 0.08206 L atm K 298.15 K
V
8.47 L
P
0.289 atm
(b) At 1000 K, there is much more Br being generated from the decomposition of Br2.
However, K is still rather small, and this decomposition does not notably affect the volume
needed.
EXERCISES
Writing Equilibrium Constants Expressions
1.
(E)
CO2 CF4
2
COF2
(a)
2 COF2 g
CO 2 g + CF4 g
Kc =
(b)
Cu s + 2 Ag
Cu 2+
Kc =
2
Ag +
+
aq
Cu aq + 2 Ag s
2+
2
(c)
2.
2
2
3+
aq + 2 Fe2+ aq
2 SO4 aq + 2 Fe aq
2
(E)
(a)
(b)
(c)
3.
S2 O8
SO 4 2 Fe3+
Kc =
2
2+ 2
S2 O8 Fe
P N 2 P H 2 O
2
4 NH 3 g + 3 O 2 g
2 N 2 g + 6 H 2O g
KP =
P NH 3 P O 2
4
KP =
N 2 g + Na 2 CO3 s + 4 C s
2 NaCN s + 3 CO g
P NH 3 P H 2 O
P H 2 P NO 2
7
CO
=
3
Kc
N 2
(E)
2
(a)
(b) Kc =
675
Zn 2+
Ag +
2
3
2
7 H 2 g + 2 NO 2 g
2 NH 3 g + 4 H 2 O g
NO2
Kc =
2
NO O2
6
(c)
Kc =
OH
2
CO 32
2
4
Chapter15: Principles of Chemical Equilibrium
4.
(E)
(a)
5.
6.
7.
Kp =
P{CH 4 }P{H 2S}2
P{CS2 }P{H 2 }4
l q
Kp = P O 2
(b)
1/2
(c)
Kp = P{CO 2 } P{H 2 O}
(E) In each case we write the equation for the formation reaction and then the equilibrium
constant expression, Kc, for that reaction.
HF
1
1
(a)
Kc =
HF g
2 H 2 g + 2 F2 g
1/2
1/2
H 2 F2
(b)
N2 g + 3 H2 g
2 NH 3 g
(c)
2 N 2 g + O2 g
2 N2O g
(d)
1
2
Cl2 (g) + 23 F2 (g)
ClF3 (l)
NH3
Kc =
3
N 2 H 2
2
[N 2 O]2
[N 2 ]2 [O 2 ]
1
Kc
1/ 2
[Cl2 ] [F2 ]3 / 2
Kc =
(E) In each case we write the equation for the formation reaction and then the equilibrium
constant expression, Kp, for that reaction.
PNOCl
1
1
1
(a)
KP =
NOCl g
2 N 2 g + 2 O 2 g + 2 Cl 2 g
1/2
1/2
1/2
PN2 PO2 PCl2
(b)
N 2 g + 2 O 2 g + Cl2 g
2 ClNO 2 g
KP =
(c)
N2 g + 2 H2 g
N2 H4 g
KP =
(d)
1
2
N2 g + 2 H2 g +
(E) Since K p = K c RT
(a)
Kc =
ng
SO2 [Cl2 ] = K
SO 2 Cl
2
1
2
Cl2 g
NH 4 Cl s
KP =
, it is also true that K c = K p RT
p
RT
( 1)
ng
PClNO2
2
2
PN 2 PO2 PCl2
PN2 H4
2
PN2 PH2
1
PN2
1/2
2
.
= 2.9 10-2 (0.08206 303) -1 = 0.0012
NO2 = K RT ( 1) = 1.48 104
Kc =
p
2
NO [O2 ]
3
H 2S
0
Kc =
= K p RT = K p = 0.429
3
2
(b)
(c)
H 2
676
1/2
PH 2 PCl2
(0.08206 303) = 5.55 105
Chapter15: Principles of Chemical Equilibrium
8.
9.
(E) K p = K c RT
, with R = 0.08206 L atm mol1 K 1
(a)
Kp =
P{NO 2 }2
K c (RT) +1 = 4.61 × 10-3 (0.08206×298)1 0.113
P{N 2 O 4 }
(b)
Kp
P{C2 H 2 }P{H 2 }3
K c (RT)( 2) (0.154) (0.08206 2000) 2 4.15 103
2
P{CH 4 }
(c)
Kp
P{H 2 }4 P{CS2 }
K c (RT)( 2) (5.27 108 ) (0.08206 973) 2 3.36 104
2
P{H 2 S} P{CH 4 }
(E) The equilibrium reaction is H 2 O l
H 2 O g with ngas = +1. K p = K c RT
K c = K p RT
ng
K c = K p RT
10.
ng
1
Kp = P{H 2 O} = 23.8 mmHg
.
=
Kp
RT
=
ng
gives
1 atm
= 0.0313
760 mmHg
0.0313
= 1.28 10-3
0.08206 298
C6 H 6 g with ngas = +1. Using K p = K c RT
(E) The equilibrium rxn is C6 H 6 l
ng
K p =K c RT = 5.12 103 0.08206 298 = 0.125 = P{C6 H 6 }
P{C6 H 6 } = 0.125 atm
11.
760 mmHg
= 95.0 mmHg
1 atm
(E) Add one-half of the reversed 1st reaction with the 2nd reaction to obtain the desired
reaction.
1
1
1
Kc =
NO g
2 N2 g + 2 O2 g
2.11030
NO g + 12 Br2 g
K c = 1.4
NOBr g
net : 12 N 2 g + 12 O 2 g + 12 Br2 g
NOBr g
12.
Kc =
1.4
2.1 10
30
= 9.7 1016
(M) We combine the several given reactions to obtain the net reaction.
1
2 N 2 g O 2 (g)
Kc =
2 N 2 O g
2
2.7 1018
1
2 N 2 O 4 g
4 NO 2 g
Kc =
4NO 2 (g)
2 N 2 (g) + 4 O 2 (g)
K c (4.1 109 ) 4
4.6 10
2 N 2 O 4 g K c ( Net )
net : 2 N 2 O g + 3 O 2 g
677
3 2
4.1 10
=
2.7 10 4.6 10
9 4
18 2
3 2
= 1.8 106
,
Chapter15: Principles of Chemical Equilibrium
13.
(M) We combine the Kc values to obtain the value of Kc for the overall reaction, and then
convert this to a value for Kp.
2
2 CO 2 g + 2H 2 g
K c = 1.4
2 CO g + 2 H 2O g
2 C graphite + O2 g
2 CO g
K c = 1 108
4 CO g
2 C graphite + 2 CO 2 g
Kc =
net:
2H 2 (g) + O 2 (g)
2H 2 O(g)
0.64
K c(Net)
n
2
(1.4) 2 (1108 ) 2
5 106
2
(0.64)
=
2 NO 2 Cl g
2 NOCl g + O 2 g
1
Kp =
2
1.1 10
2 NO 2 g + Cl2 g
2 NO 2 Cl g
N 2 g + 2 O2 g
2 NO 2 g
K p = 0.3
K p = K c RT
n
Kc =
Kp
RT
n
=
2
2
K p = (1.0 109 ) 2
net: N 2 (g) + O 2 (g) + Cl2 (g)
2 NOCl(g)
15.
1
Kc
5 1016
=
= 5 1014
RT 0.08206 1200
(M) We combine the Kp values to obtain the value of Kp for the overall reaction, and then
convert this to a value for Kc.
K p = K c RT
14.
2
K p(Net)
7.4 1024
0.08206 298
2 3
(0.3) 2 (1.0 109 )2
7.4 1024
(1.1 102 ) 2
= 2 1022
(E) CO 2 (g) H 2 O(l) H 2CO3 (aq)
In terms of concentration, K = a(H2CO3)/a(CO2)
H CO
In terms of concentration and partial pressure, K 2 3
c
PCO2 P
16.
(E) 2 Fe(s) 3 O2 (g) Fe 2 O3 (s)
a Fe2O3
. Since activity of solids and liquids is defined as 1, then the expression
K
a Fe a O2
simplifies to K
1
a O2
Similarly, in terms of pressure and concentration, K 1 PO2 / P
678
Chapter15: Principles of Chemical Equilibrium
Experimental Determination of Equilibrium Constants
17.
(M) First, we determine the concentration of PCl5 and of Cl 2 present initially and at
equilibrium, respectively. Then we use the balanced equation to help us determine the
concentration of each species present at equilibrium.
1.00 103 mol PCl5
9.65 104 mol Cl2
= 0.00400 M
= 0.00386 M
PCl5 initial =
Cl2 equil =
0.250 L
0.250 L
Equation: PCl5 g
PCl3 (g)
+
Cl2 g
Initial:
0.00400M
Changes: –xM
Equil: 0.00400M-xM
0M
+xM
xM
0M
+xM
xM 0.00386 M (from above)
At equilibrium, [Cl2] = [PCl3] = 0.00386 M and [PCl5] = 0.00400M – xM = 0.00014 M
Kc
18.
PCl3 Cl2 (0.00386 M) (0.00386 M)
0.00014 M
PCl5
= 0.106
(M) First we determine the partial pressure of each gas.
Pinitial {H 2 g }=
nRT
=
V
Pinitial {H 2S g }=
1.00 g H 2
nRT
=
V
1mol H 2 0.08206 L atm
1670 K
2.016 g H 2
mol K
136 atm
0.500 L
1.06 g H 2S
1mol H 2S 0.08206 L atm
1670 K
34.08g H 2S
mol K
8.52 atm
0.500 L
0.08206 L atm
1670 K
mol K
2.19 103 atm
0.500 L
2 H 2S g
8.00 106 mol S2
Pequil {S2 g }=
nRT
=
V
Equation:
2 H2 g
Initial :
Changes :
Equil :
136 atm
0 atm
8.52 atm
+0.00438atm 0.00219 atm 0.00438 atm
136 atm
0.00219 atm 8.52 atm
P{H 2S g }2
+
S2 g
8.52
Kp =
=
=1.79
2
P{H 2 g } P{S2 (g)} 136 2 0.00219
2
679
Chapter15: Principles of Chemical Equilibrium
19.
20.
(M)
(a)
.
0105
g PCl 5 1 mol PCl5
PCl5
2.50 L
208.2 g
Kc = L
26.3
=
O
L
O
NM PCl 3 PQ NMCl 2 QP 0.220 g PCl 3 1 mol PCl 3 2.12 g Cl 2 1 mol Cl 2
2.50 L
137.3 g
2.50 L
70.9 g
(b)
Kp = Kc RT
b g
FG
H
n
b
g
= 26.3 0.08206 523
IJ FG
K H
1
IJ
K
= 0.613
(M)
1 mol ICl
162.36 g ICl
= 6.72×10-3 M
0.625 L
0.682 g ICl×
[ICl]initial =
Reaction:
Initial:
Change
Equilibrium
2 ICl(g)
6.72 × 10-3 M
-2x
6.72 × 10-3 M -2x
I2(g)
0M
+x
x
+
Cl2(g)
0M
+x
x
1 mol I 2
253.808 g I 2
= 2.41×10-4 M = x
0.625 L
0.0383 g I 2 ×
[I 2 ]equil =
xx
(2.41104 ) 2
Kc
9.31106
3
3
4
(6.72 10 2 x) (6.72 10 2(2.4110 ))
21.
(E)
Fe3
Fe3
3
K
9.110
3
3
H
1.0 107
Fe3 9.11018 M
22.
(E)
K
NH3 (aq)
PNH3 (g)
57.5
5 109
PNH3 (g)
PNH3 (g) 5 109 57.5 8.7 1011
680
Chapter15: Principles of Chemical Equilibrium
Equilibrium Relationships
SO3
2
SO3
=
2
(M) K c =281=
24.
0.37 mol I
I
V
1 0.14
(M) K c 0.011
1.00
mol
I
V
I2
2
V
2
SO 2 O 2
SO 2
2
SO2 =
0.185 L
0.00247 mol
23.
SO
0.185
= 0.516
0.00247 281
3
2
2
25.
V
0.14
13L
0.011
(M)
(a) A possible equation for the oxidation of NH 3 g to NO 2 g follows.
NO 2 g + 32 H 2 O g
NH 3 g + 74 O 2 g
b g
(b)
b g
We obtain K p for the reaction in part (a) by appropriately combining the values of
K p given in the problem.
NO g + 32 H 2 O g
NH 3 g + 54 O 2 g
K p = 2.111019
NO 2 g
NO g + 12 O 2 g
Kp =
NO 2 g + 23 H 2 O g
net: NH3 g + 74 O 2 g
26.
1
0.524
2.111019
Kp =
= 4.03 1019
0.524
(D)
(a)
We first determine [H2] and [CH4] and then [C2H2]. [CH4] = [H2] = 0.10 mol = 0.10 M
C2 H 2 H 2
3
Kc =
CH
4
2
C2 H 2 =
K c CH 4
1.0 L
2
H
2
3
0.154 0.10
= 1.54 M
0.103
2
=
In a 1.00 L container, each concentration numerically equals the molar quantities of
the substance.
1.54 mol C2 H 2
{C2 H 2 } =
= 0.89
1.54 mol C2 H 2 + 0.10 mol CH 4 + 0.10 mol H 2
(b)
The conversion of CH4(g) to C2H2(g) is favored at low pressures, since the conversion
reaction has a larger sum of the stoichiometric coefficients of gaseous products (4) than
of reactants (2).
(c)
Initially, all concentrations are halved when the mixture is transferred to a flask that is
twice as large. To re-establish equilibrium, the system reacts to the right, forming more
moles of gas (to compensate for the drop in pressure). We base our solution on the
balanced chemical equation, in the manner we have used before.
681
Chapter15: Principles of Chemical Equilibrium
Equation:
2 CH 4 (g)
0.10 mol
2.00 L
= 0.050 M
2 x M
Initial:
Changes:
C 2 H 2 (g)
1.5 mol
2.00 L
= 0.75 M
+x M
+
Kc
0.10 mol
2.00 L
= 0.050 M
+3x M
0.050 2 x M
0.0750 + x M
3
3
C H H 0.050 + 3x 0.750 + x = 0.154
= 2 2 22 =
2
CH 4
0.050 2 x
Equil:
3 H2
0.050 + 3x
M
We can solve this 4th-order equation by successive approximations.
First guess: x = 0.010 M .
Qc
0.050 + 3 0.010 0.750 + 0.010 0.080 0.760
=
=
= 0.433 0.154
0.030
0.050 2 0.010
Qc
0.050 + 3 0.020 0.750 + 0.020 0.110 0.770
=
=
= 10.2 0.154
0.010
0.050 2 0.020
Qc
0.050 + 3 0.005 0.750 + 0.005 0.065 0.755
=
=
= 0.129 0.154
0.040
0.050 2 0.005
Qc
0.050 + 3 0.006 0.750 + 0.006 0.068 0.756
=
=
= 0.165 0.154
0.038
0.050 2 0.006
3
x = 0.010
3
2
2
3
x = 0.020
3
2
2
3
x = 0.005
3
2
2
3
x = 0.006
3
2
2
This is the maximum number of significant figures our system permits. We have
x = 0.006 M . CH 4 = 0.038 M ; C2 H 2 = 0.756 M; H 2 = 0.068 M
Because the container volume is 2.00 L, the molar amounts are double the values of
molarities.
2.00 L
0.756 mol C 2 H 2
2.00 L
0.068 mol H 2
= 0.14 mol H 2
1L
1L
= 1.51 mol C 2 H 2
2.00 L
0.038 mol CH 4
1L
= 0.076 mol CH 4
Thus, the increase in volume results in the production of some additional C2H2.
682
Chapter15: Principles of Chemical Equilibrium
27.
(M)
(a)
(b)
n CO n H 2 O
CO H 2 O
V
V
Kc
CO2 H 2 n{CO2 } n H 2
V
V
Since V is present in both the denominator and the numerator, it can be stricken from
the expression. This happens here because ng = 0. Therefore, Kc is independent of V.
Note that Kp = Kc for this reaction, since ngas = 0 .
Kc = Kp =
28.
0.224 mol CO 0.224 mol H 2 O
= 0.659
0.276 mol CO 2 0.276 mol H 2
CO2(g) + H2(g) the value of Kp = 23.2
(M) For the reaction CO(g) + H2O(g)
[CO 2 ][H 2 ]
The expression for Qp is
. Consider each of the provided situations
[CO][H 2 O]
(a) PCO = PH2O = PH2 = PCO2 ;
Qp = 1 Not an equilibrium position
(b)
(c)
(d)
PH2
PH2O
P
CO
PH2
PCO
PCO2
=
PCO
Qp = x2 If x =
=x;
PH2O = PCO2 PH2 ; Qp = 1
=
PCO2
PH2O
Not an equilibrium position
Qp = x2 If x =
=x;
23.2 , this is an equilibrium position.
23.2 , this is an equilibrium position.
Direction and Extent of Chemical Change
29.
(M) We compute the value of Qc for the given amounts of product and reactants.
1.8 mol SO3
7.2 L
SO3
Qc =
2
SO2 O2 3.6 mol
2
2
0.82 K c 100
2
SO 2 2.2 mol O 2
7.2 L
7.2 L
The mixture described cannot be maintained indefinitely. In fact, because Qc K c , the
reaction will proceed to the right, that is, toward products, until equilibrium is established.
We do not know how long it will take to reach equilibrium.
683
Chapter15: Principles of Chemical Equilibrium
30.
(M) We compute the value of Qc for the given amounts of product and reactants.
2
0.0205 mol NO 2
2
NO 2
5.25 L
1.07 104 K 4.61 103
Qc
c
0.750
mol
N
O
N
O
2 4
2 4
5.25 L
The mixture described cannot be maintained indefinitely. In fact, because Qc < Kc, the
reaction will proceed to the right, that is, toward products, until equilibrium is established.
If Ea is large, however, it may take some time to reach equilibrium.
31.
(M)
(a) We determine the concentration of each species in the gaseous mixture, use these
concentrations to determine the value of the reaction quotient, and compare this value
of Qc with the value of Kc .
SO =
0.455 mol SO 2
SO =
0.568 mol SO 3
2
1.90 L
3
1.90 L
O =
= 0.239 M
0.183 mol O 2
2
1.90 L
SO
2
Qc =
= 0.299 M
3
SO 2
2
O 2
= 0.0963 M
0.299
=
= 16.3
2
0.239 0.0963
2
Since Qc = 16.3 2.8 102 = Kc , this mixture is not at equilibrium.
(b)
32.
Since the value of Qc is smaller than that of Kc , the reaction will proceed to the right,
forming product and consuming reactants to reach equilibrium.
bg
(M) We compute the value of Qc . Each concentration equals the mass m of the substance
divided by its molar mass and further divided by the volume of the container.
m
Qc
CO H
=
CO H O
2
2
2
m
1 mol CO 2
44.0 g CO 2
V
1 mol CO
28.0 g CO
V
m
1 mol H 2
2.0 g H 2
1
28.0 18.0
V
44.0 2.0
= 5.7 31.4(value of K c )
1 mol H 2 O
1
44.0 2.0
m
28.0 18.0
18.0 g H 2 O
V
(In evaluating the expression above, we cancelled the equal values of V , along with, the
equal values of m .) Because the value of Qc is smaller than the value of Kc , (a) the
reaction is not at equilibrium and (b) the reaction will proceed to the right (formation of
products) to reach a state of equilibrium.
684
Chapter15: Principles of Chemical Equilibrium
33.
(M) The information for the calculation is organized around the chemical equation. Let
x = mol H 2 (or I 2 ) that reacts. Then use stoichiometry to determine the amount of HI
formed, in terms of x , and finally solve for x .
Equation:
Initial:
Changes:
Equil:
bg
H2 g
0.150 mol
x mol
0.150 x
bg
bg
I2 g
0.150 mol
x mol
0.150 x
2 HI g
0.000 mol
+2 x mol
2x
Kc 50.2
Then take the square root of both sides:
2
2x
2
HI
3.25 L
Kc
H 2 I2 0.150 x 0.150 x
3.25 L
3.25 L
2x
7.09
x
0150
.
1.06
= 0.117 mol, amount HI = 2 x = 2 0.117 mol = 0.234 mol HI
9.09
amount H 2 = amount I 2 = 0.150 x mol = 0.150 0.117 mol = 0.033 mol H 2 (or I 2 )
2 x = 1.06 7.09 x
34.
x=
(M) We use the balanced chemical equation as a basis to organize the information
Equation:
SbCl5 g
SbCl3 g + Cl2 g
0.00 mol
2.50 L
0.000 M
+x M
xM
Initial:
Initial:
Changes:
Equil:
K c = 0.025=
b
0.280 mol
2.50 L
0.112 M
xM
0.112 x M
g
0.160 mol
2.50 L
0.0640 M
xM
0.0640 x M
b
g
SbCl3 Cl2 = 0.112 x 0.0640 x = 0.00717 0.176x + x 2
SbCl5
0.025 x = 0.00717 0.176 x + x 2
x
x
x 2 0.201x + 0.00717 = 0
b b 2 4ac 0.201 0.0404 0.0287
x=
=
= 0.0464 or 0.155
2a
2
The second of the two values for x gives a negative value of Cl2 = 0.091 M , and thus is
physically meaningless in our universe. Thus, concentrations and amounts follow.
SbCl5 = x = 0.0464 M
amount SbCl5 = 2.50 L 0.0464 M = 0.116 mol SbCl5
SbCl3 = 0.112 x = 0.066 M
amount SbCl3 = 2.50 L 0.066 M = 0.17 mol SbCl3
Cl2 = 0.0640 x = 0.0176 M
amount Cl2 = 2.50 L 0.0176 M = 0.0440 mol Cl2
685
Chapter15: Principles of Chemical Equilibrium
35.
(M) We use the chemical equation as a basis to organize the information provided about the
reaction, and then determine the final number of moles of Cl2 g present.
Equation: CO g
+
Cl2 g
COCl2 g
Initial: 0.3500 mol
0.0000 mol
Changes: + x mol
+ x mol
Equil.: 0.3500 + x mol
x mol
0.05500 mol
x mol
0.05500 x mol
(0.0550 x) mol
COCl2
3.050 L
K c 1.2 103
CO Cl2 (0.3500 + x)mol
3.050 L
1.2 103
0.05500 x
3.050
(0.3500 x) x
1.2 103 0.05500
=
3.050
0.3500 x
x mol
3.050 L
Assume x 0.0550 This produces the following expression.
x=
3.050 0.05500
= 4.0 104 mol Cl2
3
0.3500 1.2 10
We use the first value we obtained, 4.0 104 (= 0.00040), to arrive at a second value.
x=
3.050 0.0550 0.00040
= 4.0 104 mol Cl2
3
0.3500 + 0.00040 1.2 10
Because the value did not change on the second iteration, we have arrived at a solution.
36.
(M) Compute the initial concentration of each species present. Then determine the equilibrium
concentrations of all species. Finally, compute the mass of CO 2 present at equilibrium.
COint =
1.00 g 1 mol CO
= 0.0253 M
1.41 L 28.01 g CO
H 2 int =
1.00 g 1 mol H 2
= 0.352 M
1.41 L 2.016 g H 2
Equation : CO g
+
Initial :
0.0253 M
Changes : x M
Equil :
Kc =
0.0253 x M
H 2Oint =
H2 O g
0.0394 M
x M
0.0394 x
M
1.00 g 1 mol H 2 O
= 0.0394 M
1.41 L 18.02 g H 2 O
CO 2 g
+
H2 g
0.0000 M
+x M
0.352 M
+x M
xM
0.352 + x
x 0.352 + x
CO2 H 2 = 23.2 =
0.352 x + x 2
=
0.0253 x 0.0394 x 0.000997 0.0647 x + x 2
CO H 2O
0.0231 1.50 x + 23.2 x 2 = 0.352 x + x 2
22.2 x 2 1.852 x + 0.0231 = 0
686
M
Chapter15: Principles of Chemical Equilibrium
x=
b b 2 4ac 1.852 3.430 2.051
=
= 0.0682 M, 0.0153 M
2a
44.4
The first value of x gives negative concentrations for reactants ([CO] = -0.0429 M and [H2O]
= -0.0288 M). Thus, x = 0.0153 M = CO 2 . Now we can find the mass of CO 2 .
1.41 L
37.
0.0153 mol CO 2 44.01 g CO 2
= 0.949 g CO 2
1 L mixture
1 mol CO 2
(D) We base each of our solutions on the balanced chemical equation.
(a)
Equation :
Initial :
Changes :
Equil :
PCl5 g
PCl3 g
0.550 mol
2.50 L
x mol
2.50 L
0.550 x mol
Cl2 g
+
0.550 mol
2.50 L
+ x mol
2.50 L
0.550 + x mol
0 mol
2.50 L
+ x mol
2.50 L
x mol
2.50 L
2.50 L
2.50 L
0.550 x mol x mol
[PCl3 ][Cl2 ]
2.50L
2.50L
Kc
3.8 102
(0.550 x)mol
[PCl5 ]
2.50L
x 2 + 0.550 x = 0.052 0.095x
x(0.550 x)
3.8 102
2.50(0.550 x)
x 2 + 0.645x 0.052 = 0
b b 2 4ac 0.645 0.416 + 0.208
=
= 0.0725 mol, 0.717 mol
2a
2
The second answer gives a negative quantity. of Cl 2 , which makes no physical sense.
x=
n PCl5 = 0.550 0.0725 = 0.478 mol PCl5
n Cl2 = x = 0.0725 mol Cl2
(b)
Equation :
Initial :
Changes :
Equil :
PCl5 g
PCl3 g
0.610 mol
2.50 L
x mol
2.50 L
0.610 x mol
2.50 L
687
n PCl3 = 0.550 + 0.0725 = 0.623 mol PCl3
+
Cl2 g
0M
0M
+ x mol
2.50 L
x mol
+ x mol
2.50 L
x mol
2.50 L
2.50 L
Chapter15: Principles of Chemical Equilibrium
( x mol) ( x mol)
PCl3 Cl2
2.50 L 2.50 L
2
= 3.8 10
Kc =
0.610 x mol
PCl5
2.50 L
x2
0.095
2.50 3.8 10
0.610 x
2
x=
0.058 0.095 x = x 2
x 2 + 0.095 x 0.058 = 0
b b 2 4ac 0.095 0.0090 + 0.23
=
= 0.20 mol, 0.29 mol
2a
2
amount PCl 3 = 0.20 mol = amount Cl 2 ; amount PCl5 = 0.610 0.20 = 0.41 mol
38.
(D)
(a)
We use the balanced chemical equation as a basis to organize the information we have
about the reactants and products.
Equation:
Initial:
Initial:
2 COF2 g
CO 2 g
0.145 mol
5.00 L
0.0290 M
+
0.262 mol
5.00 L
0.0524 M
CF4 g
0.074 mol
5.00 L
0.0148 M
And we now compute a value of Qc and compare it to the given value of K c .
Qc =
CO2 CF4 = 0.0524 0.0148 = 0.922 2.00 = K
c
2
2
COF2
0.0290
Because Qc is not equal to K c , the mixture is not at equilibrium.
(b)
Because Qc is smaller than K c , the reaction will shift right, that is, products will be
formed at the expense of COF2, to reach a state of equilibrium.
(c)
We continue the organization of information about reactants and products.
Equation:
2 COF2 g
Initial:
Changes:
0.0290 M
2x M
Equil:
0.0290 2x M
CO 2 g
+
0.0524 M
+x M
0.0524 + x M
688
CF4 g
0.0148 M
+x M
0.0148 + x M
Chapter15: Principles of Chemical Equilibrium
Kc =
CO2 CF4 = 0.0524 + x 0.0148 + x = 2.00 = 0.000776 + 0.0672 x + x 2
2
2
0.000841 0.1160 x + 4 x 2
0.0290 2 x
COF2
0.00168 0.232 x + 8 x 2 = 0.000776 + 0.0672 x + x 2 7 x 2 0.299 x + 0.000904 = 0
x=
b b 2 4ac 0.299 0.0894 0.0253
=
= 0.0033 M ,0.0394 M
2a
14
The second of these values for x (0.0394) gives a negative COF2 = 0.0498 M ,
clearly a nonsensical result. We now compute the concentration of each species at
equilibrium, and check to ensure that the equilibrium constant is satisfied.
COF2 = 0.0290 2 x = 0.0290 2 0.0033 = 0.0224 M
CO2 = 0.0524 + x = 0.0524 + 0.0033 = 0.0557 M
CF4 = 0.0148 + x = 0.0148 + 0.0033 = 0.0181 M
Kc =
CO2 CF4 = 0.0557 M 0.0181 M = 2.01
2
2
COF2
0.0224 M
The agreement of this value of Kc with the cited value (2.00) indicates that this solution
is correct. Now we determine the number of moles of each species at equilibrium.
mol COF2 = 5.00 L 0.0224 M = 0.112 mol COF2
mol CO 2 = 5.00 L 0.0557 M = 0.279 mol CO 2
mol CF4 = 5.00 L 0.0181 M = 0.0905 mol CF4
But suppose we had incorrectly concluded, in part (b), that reactants would be formed
in reaching equilibrium. What result would we obtain? The set-up follows.
Equation :
2 COF2 g
Initial :
Changes :
0.0290 M
+2 y M
0.0524 M
y M
0.0148 M
yM
Equil :
0.0290 + 2 y M
0.0524 y M
0.0148 y
Kc =
CO2 g
+
CF4 g
M
CO2 CF4 = 0.0524 y 0.0148 y = 2.00 = 0.000776 0.0672 y + y 2
2
2
0.000841 + 0.1160 y + 4 y 2
COF2
0.0290 + 2 y
0.00168 + 0.232 y + 8 y 2 = 0.000776 0.0672 y + y 2 7 y 2 + 0.299 y + 0.000904 = 0
y=
b b 2 4ac 0.299 0.0894 0.0253
=
= 0.0033 M , 0.0394 M
2a
14
689
Chapter15: Principles of Chemical Equilibrium
b
g
The second of these values for x 0.0394 gives a negative COF2 = 0.0498 M ,
clearly a nonsensical result. We now compute the concentration of each species at
equilibrium, and check to ensure that the equilibrium constant is satisfied.
COF2 = 0.0290 + 2 y = 0.0290 + 2 0.0033 = 0.0224 M
CO2 = 0.0524 y = 0.0524 + 0.0033 = 0.0557 M
CF4 = 0.0148 y = 0.0148 + 0.0033 = 0.0181 M
These are the same equilibrium concentrations that we obtained by making the correct
decision regarding the direction that the reaction would take. Thus, you can be assured
that, if you perform the algebra correctly, it will guide you even if you make the
incorrect decision about the direction of the reaction.
39.
(D)
(a)
We calculate the initial amount of each substance.
1 mol C2 H 5OH
= 0.373 mol C 2 H 5OH
46.07 g C 2 H 5OH
n {C 2 H 5OH
= 17.2 g C2 H5OH
n{CH 3CO 2 H
= 23.8 g CH3CO2 H
1 mol CH 3CO 2 H
= 0.396 mol CH 3CO 2 H
60.05 g CH 3CO 2 H
n{CH 3 CO 2 C2 H 5 } = 48.6 g CH3 CO 2 C2 H 5
1 mol CH 3 CO 2 C2 H 5
88.11 g CH 3 CO 2 C 2 H 5
n{CH 3 CO 2 C2 H 5 } = 0.552 mol CH 3 CO 2 C2 H 5
n H 2O
= 71.2 g H 2O
1 mol H 2 O
= 3.95 mol H 2 O
18.02 g H 2 O
Since we would divide each amount by the total volume, and since there are the same
numbers of product and reactant stoichiometric coefficients, we can use moles rather
than concentrations in the Qc expression.
Qc =
n{CH 3CO 2 C2 H 5 n H 2 O
0.552 mol 3.95 mol
=
= 14.8 K c = 4.0
n{C2 H 5OH n CH 3CO 2 H 0.373 mol 0.396 mol
Since Qc K c the reaction will shift to the left, forming reactants, as it attains
equilibrium.
(b)
Equation:
Initial
Changes
Equil
C2 H5 OH +
0.373 mol
+x mol
(0.373+x) mol
CH 3 CO 2 C2 H 5 +
CH3 CO2 H
0.396 mol
0.552 mol
+x mol
–x mol
(0.396+x) mol
(0.552–x) mol
690
H2O
3.95 mol
–x mol
(3.95–x) mol
Chapter15: Principles of Chemical Equilibrium
Kc
(0.552 x)(3.95 x)
2.18 4.50 x x 2
4.0
(0.373 x)(0.396 x) 0.148 0.769 x x 2
x 2 4.50 x + 2.18 = 4 x 2 + 3.08 x + 0.59
3x 2 + 7.58 x 1.59 = 0
b b 2 4ac 7.58 57 +19
x=
=
= 0.19 moles, 2.72 moles
2a
6
Negative amounts do not make physical sense. We compute the equilibrium amount of each
substance with x = 0.19 moles.
n{C 2 H 5OH
= 0.373 mol + 0.19 mol = 0.56 mol C2 H5OH
mass C 2 H 5OH = 0.56 mol C 2 H 5OH
n{CH 3CO 2 H
46.07 g C 2 H 5OH
26 g C 2 H 5OH
1 mol C2 H 5OH
= 0.396 mol + 0.19 mol = 0.59 mol CH3CO2 H
mass CH 3CO 2 H = 0.59 mol CH3CO2 H
n{CH 3CO 2C 2 H 5
60.05g CH3CO 2 H
35g CH3CO2 H
1 mol CH 3CO 2 H
= 0.552 mol 0.19 mol = 0.36 CH3CO2C2 H5
mass CH 3CO 2 C 2 H 5 = 0.36 mol CH 3CO 2 C2 H 5
n{H 2 O
= 3.95 mol 0.19 mol = 3.76 mol H 2O
18.02 g H 2 O
68g H 2 O
1 mol H 2 O
mass H 2O = 3.76 mol H 2O
To check K c =
40.
88.10 g CH3CO 2 C2 H 5
32 g CH 3CO 2 C 2 H 5
1 mol CH 3CO 2 C2 H 5
n CH 3 CO 2 C 2 H 5
n
0.36 mol 3.76 mol
=
= 4.1
n C 2 H 5 OH n CH 3 CO 2 H 0.56 mol 0.59 mol
H2O
(M) The final volume of the mixture is 0.750 L + 2.25 L = 3.00 L . Then use the balanced
chemical equation to organize the data we have concerning the reaction. The reaction
should shift to the right, that is, form products in reaching a new equilibrium, since the
volume is greater.
Equation:
Initial:
Initial:
Changes:
Equil :
N 2 O4 g
0.971 mol
3.00 L
0.324 M
x M
(0.324 x)M
2 NO 2 g
0.0580 mol
3.00 L
0.0193M
2x M
(0.0193 2x)M
691
Chapter15: Principles of Chemical Equilibrium
NO2
=
2
Kc
N 2 O 4
=
0.0193 + 2 x
0.324 x
2
0.000372 + 0.0772 x + 4 x 2
=
= 4.61 103
0.324 x
0.000372 + 0.0772 x + 4 x 2 = 0.00149 0.00461x
4 x 2 + 0.0818 x 0.00112 = 0
b b 2 4ac 0.0818 0.00669 + 0.0179
x=
=
= 0.00938 M, 0.0298 M
2a
8
NO 2 = 0.0193 + 2 0.00938 = 0.0381 M
amount NO 2 = 0.0381 M 3.00 L = 0.114 mol NO 2
N 2O4 = 0.324 0.00938 = 0.3146 M
amount N 2 O 4 = 0.3146 M 3.00 L = 0.944 mol N 2 O 4
41.
(M) HCONH 2 init =
0.186 mol
= 0.0861 M
2.16 L
Equation:
NH 3 g +
HCONH 2 g
CO g
Initial :
0.0861 M
0M
0M
Changes:
x M
+x M
+x M
Equil :
(0.0861-x) M
xM
xM
Kc =
x=
NH3 CO =
HCONH
2
xx
= 4.84
0.0861 x
x 2 = 0.417 4.84 x
0 = x 2 + 4.84 x 0.417
b b 2 4ac 4.84 23.4 +1.67
=
= 0.084 M, 4.92 M
2a
2
The negative concentration obviously is physically meaningless. We determine the total
concentration of all species, and then the total pressure with x = 0.084.
total = NH 3 + CO + HCONH 2 = x + x + 0.0861 x = 0.0861+ 0.084 = 0.170 M
Ptot = 0.170 mol L1 0.08206 L atm mol1 K 1 400. K = 5.58 atm
42.
(E) Compare Qp to K p . We assume that the added solids are of negligible volume so that the
initial partial pressures of CO2(g) and H2O(g) do not significantly change.
1 atm
P{H 2 O = 715 mmHg
= 0.941 atm H 2 O
760 mmHg
Qp = P CO 2
P H 2 O = 2.10 atm
CO 2 0.941 atm H 2 O = 1.98 0.23 = K p
Because Qp is larger than K p , the reaction will proceed left toward reactants to reach
equilibrium. Thus, the partial pressures of the two gases will decrease.
692
Chapter15: Principles of Chemical Equilibrium
43.
(M)
We organize the solution around the balanced chemical equation.
Equation:
2 Cr 3+ aq + Cd(s)
2 Cr 2+ aq Cd 2 aq
Initial :
1.00 M
0M
0M
Changes:
2 x M
2 x
x
Equil:
(1.00 2 x) M
2x
x
2
Cr 2+ Cd 2+
(2 x) 2 ( x)
Kc =
=
= 0.288
2
2
1.00 2 x
Cr 3
Via successive approximations , one obtains
x = 0.257 M
Therefore, at equilibrium, [Cd2+] = 0.257 M, [Cr2+] = 0.514 M and [Cr3+] = 0.486 M
Minimum mass of Cd(s) = 0.350L 0.257 M 112.41 g/mol = 10.1 g of Cd metal
44.
(M) Again we base the set-up of the problem around the balanced chemical equation.
Equation:
Pb s + 2 Cr 3+ aq
Initial:
Changes:
Equil :
Kc =
10
Kc =3.210
0.100 M
2x M
(0.100 2x)M
x (2 x) 2
= 3.2 1010
2
(0.100)
Pb 2+ aq + 2 Cr 2+ aq
0M
xM
xM
0M
2x M
2x M
4 x 3 (0.100) 2 3.2 1010 3.2 1012
3.2 1012
x
9.3 105 M
Assumption that 2 x 0.100 , is valid and thus
4
Pb 2+ = x = 9.3 10 5 M , [Cr2+] = 1.9 10-4 M and [Cr3+] = 0.100 M
3
45.
SO2Cl2(g) has
(M) We are told in this question that the reaction SO2(g) + Cl2(g)
Kc = 4.0 at a certain temperature T. This means that at the temperature T, [SO2Cl2] = 4.0
[Cl2] [SO2]. Careful scrutiny of the three diagrams reveals that sketch (b) is the best
representation because it contains numbers of SO2Cl2, SO2, and Cl2 molecules that are
consistent with the Kc for the reaction. In other words, sketch (b) is the best choice because it
contains 12 SO2Cl2 molecules (per unit volume), 1 Cl2 molecule (per unit volume) and 3 SO2
molecules (per unit volume), which is the requisite number of each type of molecule needed
to generate the expected Kc value for the reaction at temperature T.
46.
2 NOBr(g) has
(M) In this question we are told that the reaction 2 NO(g) + Br2(g)
Kc = 3.0 at a certain temperature T. This means that at the temperature T, [NOBr]2 = 3.0
[Br2][NO]2. Sketch (c) is the most accurate representation because it contains 18 NOBr
molecules (per unit volume), 6 NO molecules (per unit volume), and 3 Br2 molecules (per
unit volume), which is the requisite number of each type of molecule needed to generate the
expected Kc value for the reaction at temperature T.
693
Chapter15: Principles of Chemical Equilibrium
47.
(E)
K
aconitate
citrate
Q
4.0 105
0.031
0.00128
Since Q = K, the reaction is at equilibrium,
48.
(E)
CO2 NADred oxoglut.
citrate NADox
0.00868 0.00132 0.00868 0.00895
Q
0.00128 0.00868
K
Since Q < K, the reaction needs to proceed to the right (products).
Partial Pressure Equilibrium Constant, Kp
49.
bg
(M) The I 2 s maintains the presence of I 2 in the flask until it has all vaporized. Thus, if
enough HI(g) is produced to completely consume the I 2 s , equilibrium will not be achieved.
1 atm
P{H 2S} = 747.6 mmHg
= 0.9837 atm
760 mmHg
2 HI g + S s
H 2S g + I 2 s
Equation:
Initial:
Changes:
Equil:
bg
0.9837 atm
x atm
0.9837 x atm
b
g
0 atm
+2x atm
2x atm
2x
P{HI}2
4x2
=
= 1.34 105 =
P{H 2 S} 0.9837 x
0.9837
2
Kp =
1.34 105 0.9837
x=
= 1.82 103 atm
4
The assumption that 0.9837 x is valid. Now we verify that sufficient I2(s) is present by
computing the mass of I 2 needed to produce the predicted pressure of HI(g). Initially, 1.85 g
I2 is present (given).
mass I 2 =
1.82 103 atm 0.725 L
1 mol I 2 253.8 g I 2
= 0.00613 g I 2
1
1
0.08206 L atm mol K 333 K 2 mol HI 1 mol I 2
Ptot = P{H 2S} + P{HI} = 0.9837 x + 2 x = 0.9837 + x = 0.9837 + 0.00182 = 0.9855 atm
Ptot = 749.0 mmHg
694
Chapter15: Principles of Chemical Equilibrium
50.
(M) We first determine the initial pressure of NH 3 .
P{NH 3 g }=
Equation:
Initial:
Changes:
Equil:
nRT 0.100 mol NH 3 0.08206 L atm mol1 K 1 298 K
=
=0.948 atm
V
2.58 L
+
H2S(g)
NH4HS(s)
NH3(g)
0.948 atm
0 atm
+x atm
+x atm
x atm
0.948 + x atm
b
g
Kp = P{NH 3} P{H 2S} = 0.108 = 0.948 + x x = 0.948 x + x 2 0 = x 2 + 0.948 x 0.108
x=
b b 2 4ac 0.948 0.899 + 0.432
=
= 0.103 atm, 1.05 atm
2a
2
The negative root makes no physical sense. The total gas pressure is obtained as follows.
Ptot = P{NH 3} + P{H 2 S} = 0.948 + x + x = 0.948 + 2 x = 0.948 + 2 0.103 = 1.154 atm
b
51.
g
(M) We substitute the given equilibrium pressure into the equilibrium constant expression
P{O 2 }3
P{O 2 }3
=
28.5
=
and solve for the other equilibrium pressure. K p =
2
P{CO 2 }2
0.0721 atm CO2
P{O 2 } 3 P{O 2 }3 3 28.5(0.0712 atm) 2 0.529 atm O 2
Ptotal = P{CO 2 } + P{O 2 } = 0.0721 atm CO 2 + 0.529 atm O 2 = 0.601 atm total
52.
(M) The composition of dry air is given in volume percent. Division of these percentages by
100 gives the volume fraction, which equals the mole fraction and also the partial pressure in
atmospheres, if the total pressure is 1.00 atm. Thus, we have P{O 2 } = 0.20946 atm and
P{CO 2 } = 0.00036 atm. We substitute these two values into the expression for Qp .
0.20946 atm O2 = 6.4 104 28.5 = K
P{O 2 }3
Qp =
=
p
2
2
P{CO 2 }
0.00036 atm CO2
3
The value of Qp is much larger than the value of K p . Thus this reaction should be
spontaneous in the reverse direction until equilibrium is achieved. It will only be
spontaneous in the forward direction when the pressure of O2 drops or that of CO 2 rises
(as would be the case in self-contained breathing devices).
695
Chapter15: Principles of Chemical Equilibrium
53.
(M)
(a) We first determine the initial pressure of each gas.
nRT 1.00 mol 0.08206 L atm mol 1 K 1 668 K
=
= 31.3 atm
V
1.75 L
Then we calculate equilibrium partial pressures, organizing our calculation around
the balanced chemical equation. We see that the equilibrium constant is not very large,
meaning that we must solve the polynomial exactly (or by successive approximations).
COCl2 g
+ Cl2 g
Kp = 22.5
Equation
CO(g)
P{CO} = P{Cl 2 } =
Initial:
Changes:
Equil:
Kp
31.3 atm
x atm
31.3 x atm
31.3 atm
x atm
31.3 x atm
0 atm
+x atm
x atm
P{COCl2 }
x
x
22.5
2
P{CO} P{Cl2 }
(31.3 x)
(979.7 62.6 x x 2 )
22.5(979.7 62.6 x x 2 ) x 22043 1408.5x 22.5 x 2 x
22043 1409.5x 22.5 x 2 0 (Solve by using the quadratic equation)
b b 2 4ac (1409.5) (1409.5) 2 4(22.5)(22043)
=
x=
2a
2(22.5)
x=
1409.5 2818
30.14, 32.5(too large)
45
P{CO} = P{Cl2 } = 31.3 atm 30.14 atm = 1.16 atm
P{COCl2 } = 30.14 atm
(b) Ptotal = P{CO}+ P{Cl2 }+ P{COCl2 } = 1.16 atm +1.16 atm + 30.14 atm = 32.46 atm
54.
(M) We first find the value of Kp for the reaction.
6
2 NO 2 g
2 NO g + O 2 g , K c = 1.8 10 at 184 C = 457 K .
For this reaction ngas = 2 +1 2 = +1.
K p K c (RT)
n g
= 1.8 10-6 (0.08206 457) 1 = 6.8 10-5
NO 2 g from the initial reaction,
To obtain the required reaction NO g + 12 O 2 g
that initial reaction must be reversed and then divided by two. Thus, in order to determine
the value of the equilibrium constant for the final reaction, the value of Kp for the initial
reaction must be inverted, and the square root taken of the result.
K p, final
1
6.8 105
1.2 102
696
Chapter15: Principles of Chemical Equilibrium
Le Châtelier's Principle
55.
(E) Continuous removal of the product, of course, has the effect of decreasing the
concentration of the products below their equilibrium values. Thus, the equilibrium system
is disturbed by removing the products and the system will attempt (in vain, as it turns out)
to re-establish the equilibrium by shifting toward the right, that is, to generate more
products.
56.
(E) We notice that the density of the solid ice is smaller than is that of liquid water. This
means that the same mass of liquid water is present in a smaller volume than an equal mass
of ice. Thus, if pressure is placed on ice, attempting to force it into a smaller volume, the
ice will be transformed into the less-space-occupying water at 0 C . Thus, at 0 C under
pressure, H2O(s) will melt to form H2O(l). This behavior is not expected in most cases
because generally a solid is more dense than its liquid phase.
57.
(M)
(a) This reaction is exothermic with H o = 150 . kJ. Thus, high temperatures favor the
reverse reaction (endothermic reaction). The amount of H 2 g present at high
temperatures will be less than that present at low temperatures.
(b)
bg
H 2 O g is one of the reactants involved. Introducing more will cause the equilibrium
position to shift to the right, favoring products. The amount of H 2 g will increase.
58.
(c)
Doubling the volume of the container will favor the side of the reaction with the
largest sum of gaseous stoichiometric coefficients. The sum of the stoichiometric
coefficients of gaseous species is the same (4) on both sides of this reaction.
Therefore, increasing the volume of the container will have no effect on the amount
of H 2 g present at equilibrium.
(d)
A catalyst merely speeds up the rate at which a reaction reaches the equilibrium position.
The addition of a catalyst has no effect on the amount of H 2 g present at equilibrium.
(M)
(a) This reaction is endothermic, with H o = +92.5 kJ. Thus, a higher temperature will
favor the forward reaction and increase the amount of HI(g) present at equilibrium.
(b)
The introduction of more product will favor the reverse reaction and decrease the
amount of HI(g) present at equilibrium.
(c)
The sum of the stoichiometric coefficients of gaseous products is larger than that for
gaseous reactants. Increasing the volume of the container will favor the forward
reaction and increase the amount of HI(g) present at equilibrium.
(d)
A catalyst merely speeds up the rate at which a reaction reaches the equilibrium position.
The addition of a catalyst has no effect on the amount of HI(g) present at equilibrium.
697
Chapter15: Principles of Chemical Equilibrium
(e)
59.
The addition of an inert gas to the constant-volume reaction mixture will not change any
partial pressures. It will have no effect on the amount of HI(g) present at equilibrium.
(M)
(a) The formation of NO(g) from its elements is an endothermic reaction ( H o = +181
kJ/mol). Since the equilibrium position of endothermic reactions is shifted toward
products at higher temperatures, we expect more NO(g) to be formed from the
elements at higher temperatures.
(b) Reaction rates always are enhanced by higher temperatures, since a larger fraction of the
collisions will have an energy that surmounts the activation energy. This enhancement of
rates affects both the forward and the reverse reactions. Thus, the position of equilibrium is
reached more rapidly at higher temperatures than at lower temperatures.
60.
(M) If the reaction is endothermic (H > 0), the forward reaction is favored at high
temperatures. If the reaction is exothermic (H < 0), the forward reaction is favored at
low temperatures.
(a)
H o = H fo PCl5 g H fo PCl3 g H fo Cl2 g
H o = 374.9 kJ 287.0 kJ 0.00 kJ = 87.9 kJ/mol
(b)
(favored at low temperatures)
H o = 2 Hfo [ H 2 O(g)] 3Hfo [S(rhombic)] Hfo [SO 2 (g)] 2 Hfo [ H 2S(g)]
H o = 2 241.8 kJ + 3 0.00 kJ 296.8 kJ 2 20.63 kJ
H o = 145.5 kJ/mol (favored at low temperatures)
(c)
H o = 4H fo NOCl g + 2H fo H 2 O g
2H fo N 2 g 3H fo O 2 g 4H fo HCl g
H o = 4 51.71 kJ + 2 241.8 kJ 2 0.00 kJ 3 0.00 kJ 4 92.31 kJ
H o = +92.5 kJ/mol (favored at higher temperatures)
61.
(E) If the total pressure of a mixture of gases at equilibrium is doubled by compression, the
equilibrium will shift to the side with fewer moles of gas to counteract the increase in
pressure. Thus, if the pressure of an equilibrium mixture of N2(g), H2(g), and NH3(g) is
2 NH3(g), will
doubled, the reaction involving these three gases, i.e., N2(g) + 3 H2(g)
proceed in the forward direction to produce a new equilibrium mixture that contains
additional ammonia and less molecular nitrogen and molecular hydrogen. In other words,
P{N2(g)} will have decreased when equilibrium is re-established. It is important to note,
however, that the final equilibrium partial pressure for the N2 will, nevertheless, be higher
than its original partial pressure prior to the doubling of the total pressure.
698
Chapter15: Principles of Chemical Equilibrium
62.
(M)
(a) Because H o = 0 , the position of the equilibrium for this reaction will not be
affected by temperature. Since the equilibrium position is expressed by the value of
the equilibrium constant, we expect Kp to be unaffected by, or to remain constant
with, temperature.
(b)
From part (a), we know that the value of Kp will not change when the temperature is
changed. The pressures of the gases, however, will change with temperature. (Recall
the ideal gas law: P = nRT / V .) In fact, all pressures will increase. The
stoichiometric coefficients in the reaction are such that at higher pressures the
formation of more reactant will be favored (the reactant side has fewer moles of gas).
Thus, the amount of D(g) will be smaller when equilibrium is reestablished at the
higher temperature for the cited reaction.
B s + 2 C g + 1 D g
A s
2
63.
(M) Increasing the volume of an equilibrium mixture causes that mixture to shift toward
the side (reactants or products) where the sum of the stoichiometric coefficients of the
gaseous species is the larger. That is: shifts to the right if ngas 0 , shifts to the left
if ngas 0 , and does not shift if ngas = 0 .
(a)
C s + H 2O g
CO g + H 2 g , ngas 0, shift right, toward products
(b)
Ca OH 2 s + CO 2 g
CaCO3 s + H 2 O g , ngas = 0, no shift, no change in
equilibrium position.
(c)
64.
65.
4 NH 3 g + 5 O 2 g
4 NO g + 6 H 2 O g , ngas 0, shifts right, towards products
(M) The equilibrium position for a reaction that is exothermic shifts to the left (reactants
are favored) when the temperature is raised. For one that is endothermic, it shifts right
(products are favored) when the temperature is raised.
(a)
NO g
H o = 90.2 kJ shifts left, % dissociation
12 N 2 g + 12 O 2 g
(b)
SO3 g
SO 2 g + 12 O 2 g
H o = +98.9 kJ shifts right, % dissociation
(c)
N2 H4 g
N2 g + 2 H2 g
H o = 95.4 kJ shifts left, % dissociation
(d)
COCl2 g
CO g + Cl2 g
H o = +108.3 kJ shifts right, % dissociation
(E)
(a) Hb:O2 is reduced, because the reaction is exothermic and heat is like a product.
(b) No effect, because the equilibrium involves O2 (aq). Eventually it will reduce the Hb:O2
level because removing O2(g) from the atmosphere also reduces O2 (aq) in the blood.
(c) Hb:O2 level increases to use up the extra Hb.
699
Chapter15: Principles of Chemical Equilibrium
66.
(E)
(a) CO2 (g) increases as the equilibrium is pushed toward the reactant side
(b) Increase CO2 (aq) levels, which then pushes the equilibrium to the product side
(c) It has no effect, but it helps establish the final equilibrium more quickly
(d) CO2 increases, as the equilibrium shifts to the reactants
67.
(E) The pressure on N2O4 will initially increase as the crystal melts and then vaporizes, but
over time the new concentration decreases as the equilibrium is shifted toward NO2.
68.
(E) If the equilibrium is shifted to the product side by increasing temperature, that means that
heat is a “reactant” (or being consumed). Therefore, HI decomposition is endothermic.
69.
(E) Since ΔH is >0, the reaction is endothermic. If we increase the temperature of the
reaction, we are adding heat to the reaction, which shifts the reaction toward the
decomposition of calcium carbonate. While the amount of calcium carbonate will decrease,
its concentration will remain the same because it is a solid.
70.
(E) The amount of N2 increases in the body. As the pressure on the body increases, the
equilibrium shifts from N2 gas to N2 (aq).
Integrative and Advanced Exercises
71. (E) In a reaction of the type I2(g) 2 I(g) the bond between two iodine atoms, the I—I
bond, must be broken. Since I2(g) is a stable molecule, this bond breaking process must be
endothermic. Hence, the reaction cited is an endothermic reaction. The equilibrium position
of endothermic reactions will be shifted toward products when the temperature is raised.
72. (M)
(a) In order to determine a value of Kc, we first must find the CO2 concentration in the gas phase.
Note, the total volume for the gas is 1.00 L (moles and molarity are numerically equal)
[CO2 ]
n
P
V RT
1.00 atm
0.0409 M
L atm
298 K
0.08206
mol K
Kc
[CO2 (aq)] 3.29 102 M
0.804
[CO2 (g)]
0.0409 M
(b) It does not matter to which phase the radioactive 14CO2 is added. This is an example of
a Le Châtelier’s principle problem in which the stress is a change in concentration of the
reactant CO2(g). To find the new equilibrium concentrations, we must solve and I.C.E.
table. Since Qc < Kc, the reaction shifts to the product, CO2 (aq) side.
Reaction:
CO 2 (g)
CO 2 (aq)
0.0409 mol
3.29 10-3 mol
Stress
Changes:
+0.01000 mol
x mol
Equilibrium:
(0.05090 x ) mol 3.29 10-3 x mol
Initial:
x mol
700
Chapter15: Principles of Chemical Equilibrium
3.29 10-3 x mol
[CO 2 (aq)]
3.29 10-2 10 x
0.1000 L
KC
0.804
[CO2 (g)] (0.05090 x) mol
0.05090 x
1.000 L
x 7.43 104 mol CO 2
Total moles of CO2 in the aqueous phase (0.1000 L)(3.29×10-2 +7.43×10-3) = 4.03 ×10-3 moles
Total moles of CO2 in the gaseous phase (1.000 L)(5.090×10-2 7.43×10-4) = 5.02 ×10-2 moles
Total moles of CO2 = 5.02 ×10-2 moles + 4.03 ×10-3 moles = 5.42×10-2 moles
There is continuous mixing of the 12C and 14C such that the isotopic ratios in the two phases is
the same. This ratio is given by the mole fraction of the two isotopes.
For
14
CO 2 in either phase its mole fraction is 0.01000-2mol 100 18.45 %
5.419 10 mol
CO 2 in the gaseous phase = 5.02×10-2 moles×0.1845 = 0.00926 moles
14
CO 2 in the aqueous phase = 4.03 ×10-3 moles×0.1845 = 0.000744 moles
14
Moles of
Moles of
73.
(M) Dilution makes Qc larger than Kc. Thus, the reaction mixture will shift left in order to
regain equilibrium. We organize our calculation around the balanced chemical equation.
Ag (aq)
Equation:
Equil:
0.31 M
Fe 2 (aq)
K c 2.98
0.19 M
0.21 M
Dilution:
0.12 M
0.084 M
Changes:
xM
xM
New equil: (0.12 x) M (0.084 x) M
Kc
Fe3 (aq) Ag(s)
3
0.076 M
x M
(0.076 x) M
[Fe ]
0.076 x
2.98
2
[Ag ][Fe ]
(0.12 x) (0.084 x)
0.076 x 0.030 0.61x 2.98 x 2
2.98 (0.12 x) (0.084 x) 0.076 x
2.98 x 2 1.61
x 0.046 0
1.61 2.59 0.55
0.027, 0.57
Note that the negative root makes no physical
5.96
sense; it gives [Fe 2 ] 0.084 0.57 0.49 M .
Thus, the new equilibrium concentrations are
x
[Fe 2 ] 0.084 0.027 0.111 M
[Ag ] 0.12 0.027 0.15 M
[Fe3 ] 0.076- 0.027 0.049 M We can check our answer by substitution.
Kc =
74.
0.049 M
= 2.94 2.98 (within precision limits)
0.111 M 0.15 M
(M) The percent dissociation should increase as the pressure is lowered, according to Le
Châtelier’s principle. Thus the total pressure in this instance should be more than in
Example 15-12, where the percent dissociation is 12.5%. The total pressure in Example
15-12 was computed starting from the total number of moles at equilibrium.
The total amount = (0.0240 – 0.00300) moles N2O4 + 2 × 0.00300 mol NO2 = 0.027 mol gas.
701
Chapter15: Principles of Chemical Equilibrium
Ptotal
nRT 0.0270mol × 0.08206 L atm K -1 mol-1 × 298 K
=
=
= 1.77 atm (Example 15-12)
V
0.372 L
We base our solution on the balanced chemical equation. We designate the initial pressure
of N2O4 as P. The change in P{N2O4}is given as –0.10 P atm. to represent the 10.0 %
dissociation.
Equation:
N 2 O 4 (g)
2 NO 2 (g)
Initial:
P atm
0 atm
0.10 P atm
Changes:
Equil:
+2(0.10 P atm)
0.90 P atm
0.20 P atm
P{NO 2 }2 (0.20 P) 2 0.040 P
0.113 × 0.90
=
=
= 0.113
P=
= 2.54 atm.
P{N 2 O 4 } 0.90 P
0.90
0.040
Thus, the total pressure at equilibrium is 0.90 P + 0.20 P and 1.10 P (where P = 2.54 atm)
Therefore, total pressure at equilibrium = 2.79 atm.
Kp =
75.
(M) Equation:
Initial:
2 SO3 (g)
1.00 atm
Changes:
2 x atm
Equil:
(1.00 2 x)atm
2 SO 2 (g)
0 atm
O 2 (g)
0 atm
2x atm
x atm
2x atm
x atm
Because of the small value of the equilibrium constant, the reaction does not proceed very
far toward products in reaching equilibrium. Hence, we assume that x << 1.00 atm and
calculate an approximate value of x (small K problem).
KP
P{SO 2 }2 P{O 2 }
(2 x) 2 x
4 x3
5
1.6
10
P{SO3 }2
(1.00 2 x) 2
(1.00) 2
x 0.016 atm
A second cycle may get closer to the true value of x.
4 x3
5
1.6 10
x 0.016 atm
(1.00 0.032) 2
Our initial value was sufficiently close. We now compute the total pressure at equilibrium.
Ptotal P{SO 3 } P{SO 2 } P{O 2 } (1.00 2 x) 2 x x 1.00 x 1.00 0.016 1.02 atm
76.
(M) Let us start with one mole of air, and let 2x be the amount in moles of NO formed.
N 2 (g)
Initial:
0.79 mol
0.21 mol
x mol
(0.79 x)mol
x mol
(0.21 x)mol
Changes:
Equil:
O 2 (g)
Equation:
2 NO(g)
0 mol
702
2x mol
2x mol
Chapter15: Principles of Chemical Equilibrium
NO
2x
2x
n{NO}
0.018
n{N 2 } n{O 2 } n{NO} (0.79 x) (0.21 x) 2 x
1.00
x 0.0090 mol
0.79 x 0.78 mol N 2
0.21 x 0.20 mol O 2
2x 0.018 mol NO
2
n{NO} RT
2
Vtotal
(0.018) 2
P{NO}
n{NO}2
2.1 103
Kp
{N
)
{O
}
n
RT
n
RT
P (N 2 } P{O 2 }
n{N 2 } n{O 2 } 0.78 0.20
2
2
Vtotal
Vtotal
77. (D) We organize the data around the balanced chemical equation. Note that the reaction is
stoichimoetrically balanced.
(a)
Equation:
Equil :
0.32 mol
Add SO 3
0.32 mol
Initial :
0.32 mol
10.0 L
Initial :
To right :
0.032 M
0.000 M
Changes :
Equil :
O 2 (g)
2 SO 2 (g)
0.16 mol
0.16 mol
0.16 mol
10.0 L
0.016 M
0.000 M
2x M
2x M
xM
xM
2 SO3 (g)
0.68 mol
1.68 mol
1.68 mol
10.0 L
0.168 M
0.200 M
2x M
(0.200 2 x)M
In setting up this problem, we note that solving this question exactly involves finding
the roots for a cubic equation. Thus, we assumed that all of the reactants have been
converted to products. This gives the results in the line labeled “To right.” We then
reach equilibrium from this position by converting some of the product back into
reactants. Now, we substitute these expressions into the equilibrium constant
expression, and we solve this expression approximately by assuming that 2 x 0.200 .
Kc
[SO 3 ] 2
(0.200 2 x) 2
(0.200) 2
2
2
.
8
10
or x 0.033
[SO 2 ] 2 [O 2 ]
( 2 x) 2 x
4x3
We then substitute this approximate value into the expression for Kc.
Kc
(0.200 0.066) 2
2.8 10 2 or x 0.025
3
4x
(0.200 0.050) 2
Let us try one more cycle. K c
2.8 10 2 or x 0.027
3
4x
703
Chapter15: Principles of Chemical Equilibrium
This gives the following concentrations and amounts of each species.
[SO 3 ] 0.200 (2 0.027) 0.146 M
amount SO 3 10.0 L 0.146 M 1.46 mol SO 3
[SO 2 ] 2 0.027 0.054 M
amount SO 2 10.0 L 0.054 M 0.54 mol SO 2
[O 2 ] 0.027 M
amount O 2 10.0 L 0.027 M 0.27 mol O 2
(b)
Equation:
Equil :
2 SO 2 (g)
0.32 mol
2 SO3 (g)
O 2 (g)
0.16 mol
0.68 mol
Equil :
0.10 V :
0.32 mol
10.0 L
0.032 M
0.32 M
0.16 mol
10.0 L
0.016 M
0.16 M
0.68 mol
10.0 L
0.068 M
0.68 M
To right :
Changes :
Equil :
0.00 M
2x M
2x M
0.00 M
x M
xM
1.00 M
2 x M
(1.00 2 x)M
Equil :
Again, notice that an exact solution involves finding the roots of a cubic. So we have
taken the reaction 100% in the forward direction and then sent it back in the reverse
direction to a small extent to reach equilibrium. We now solve the Kc expression for x,
obtaining first an approximate value by assuming 2x << 1.00.
Kc
[SO 3 ] 2
[SO 2 ] 2 [O 2 ]
(1.00 2 x) 2
(1.00) 2
2
2
.
8
10
or x 0.096
(2 x) 2 x
4x 3
We then use this approximate value of x to find a second approximation for x.
(1.00 0.19) 2
Kc
2.8 10 2
3
4x
x 0.084
or
(1.00 0.17) 2
2.8 10 2 or x 0.085
3
4x
Then we compute the equilibrium concentrations and amounts.
Another cycle gives K c
[SO3 ] 1.00 (2 0.085) 0.83 M
amount SO3 1.00 L 0.83 M 0.83 mol SO3
[SO 2 ] 2 0.085 0.17 M
amount SO 2 1.00 L 0.17 M 0.17 mol SO 2
[O 2 ] 0.085 M
amount O 2 1.00 L 0.085 M 0.085 mol O 2
78. (M)
Equation:
n co 2
HOC 6 H 4 COOH(g)
C 6 H 5 OH(g) CO 2 (g)
730mmHg
1L
48.2 48.5
PV
760mmHg/atm
2
1000 mL 1.93 10-3 mol CO
2
RT 0.0821L atm/mol K
293K
704
Chapter15: Principles of Chemical Equilibrium
Note that moles of CO 2 moles phenol
n
salicylic acid
0.300 g
2.17 10-3 mol salicylic acid
138 g / mol
2
1.93 mmol
[C 6 H 5 OH ] [CO 2 (g)]
50.0 mL
0.310
Kc
(2.17 1.93) mmol
[HOC 6 H 4 COOH]
50.0 mL
L atm
K p K c (RT)(2-1) (0.310) 0.08206
(473 K) 12.0
mol K
79. (D)
(a) This reaction is exothermic and thus, conversion of synthesis gas to methane is favored at
lower temperatures. Since n gas (1 1) (1 3) 2 , high pressure favors the products.
(b) The value of Kc is a large number, meaning that almost all of the reactants are converted
to products (note that the reaction is stoichiometrically balanced). Thus, after we set up
the initial conditions we force the reaction to products and then allow the system to
reach equilibrium.
Equation: 3 H 2 (g)
Initial:
Initial:
3.00 mol
15.0 L
0.200 M
To right: 0.000 M
Changes: 3x M
Equil:
3x M
CH 4 (g)
CO(g)
1.00 mol
15.0 L
0.0667 M
0.000 M
xM
xM
[CH 4 ][H 2 O] (0.0667 x) 2
190.
Kc
[H 2 ]3 [CO]
(3 x)3 x
190 27 x 2 0.0667 x 71.6 x 2
H 2 O(g)
0M
0M
0M
0M
0.0667 M
0.0667 M
x M
xM
(0.0667 x)M (0.0667 x)M
190.
0.0667 x
27 x 2
71.6 x 2 x 0.0667 0
b b 2 4ac 1.00 1.00 19.1
0.0244 M
2a
143
[CH 4 ] [H 2 O] 0.0667 0.0244 0.0423 M
x
[H 2 ] 3 0.0244 0.0732 M
[CO] 0.0244 M
We check our calculation by computing the value of the equilibrium constant.
(0.0423) 2
Kc
187
[H 2 ] 3 [CO]
(0.0732) 3 0.0244
[CH 4 ] [H 2 O]
705
Chapter15: Principles of Chemical Equilibrium
Now we compute the amount in moles of each component present at equilibrium, and
finally the mole fraction of CH4.
amount CH 4 amount H 2O 0.0423 M 15.0 L 0.635 mol
amount H 2 0.0732 M 15.0 L 1.10 mol
amount CO 0.0244 M 15.0 L 0.366 mol
CH
4
0.635 mol
0.232
0.635 mol 0.635 mol 1.10 mol 0.366 mol
80. (M) We base our calculation on 1.00 mole of PCl5 being present initially.
PCl3 (g)
Equation: PCl5 (g)
Initial:
1.00 mol
0M
Changes:
mol
mol
Equil:
(1.00 ) mol
mol
ntotal 1.00 1.00
Equation: PCl5 (g)
Mol fract.
1.00
1.00
PCl3 (g)
Cl2 (g)
0M
mol
mol
Cl2 (g)
1.00
1.00
P
P{Cl2 } P{PCl3 } [ {Cl2 }Ptotal ] [{PCl3 }Ptotal ] 1.00 total
2 Ptotal
2 Ptotal
Kp
1.00
P{PCl5 }
[ {PCl5 }Ptotal ]
(1.00 )(1.00 ) 1 2
Ptotal
1.00
2
81.
(M) We assume that the entire 5.00 g is N2O4 and reach equilibrium from this starting point.
1 mol N 2 O 4
5.00 g
[ N 2 O 4 ]i
0.109 M
0.500 L 92.01 g N 2 O 4
2 NO2 (g)
Equation: N2O4 (g)
Initial:
0.109
0M
+ 2x M
Changes: – x M
2x M
Equil:
(0.0109 – x) M
KC
[NO 2 ]2
(2 x) 2
4.61103
[N 2 O 4 ]
0.109 x
4x 2 5.02 104 4.61103 x
4 x 2 0.00461 x 0.000502 0
x
b b 2 4ac 0.00461 2.13 105 8.03 103
0.0106 M, 0.0118 M
2a
8
706
Chapter15: Principles of Chemical Equilibrium
(The method of successive approximations yields 0.0106 after two iterations)
amount N 2 O 4 0.500 L (0.109 0.0106) M 0.0492 mol N 2O 4
amount NO 2 0.500 L 2 0.0106 M 0.0106 mol NO 2
mol fraction NO 2
82.
0.0106 mol NO 2
0.177
0.0106 mol NO 2 0.0492 mol N 2 O 4
(M) We let P be the initial pressure in atmospheres of COCl2(g).
Equation: COCl 2 (g)
CO(g)
Cl2 (g)
Initial:
Changes:
P
x
Equil:
Px
0M
x
0M
x
x
x
Total pressure 3.00 atm P x x x P x
P{COCl 2 } P x 3.00 x x 3.00 2x
Kp
P 3.00 x
P{CO}P{Cl2 }
x x
0.0444
P(COCl2 }
3.00 2x
x 2 0.133 0.0888 x x 2 0.0888 x 0.133 0
x
b b 2 4ac 0.0888 0.00789 0.532
0.323, 0.421
2a
2
Since a negative pressure is physically meaningless, x = 0.323 atm.
(The method of successive approximations yields x = 0.323 after four iterations.)
P{CO} P{Cl2 } 0.323 atm
P{COCl2 } 3.00 2 0.323 2.35 atm
The mole fraction of each gas is its partial pressure divided by the total pressure. And the
contribution of each gas to the apparent molar mass of the mixture is the mole fraction of that
gas multiplied by the molar mass of that gas.
M avg
P{CO}
P{Cl2 }
P{COCl2 }
M {CO}
M {Cl 2 }
M {COCl2 }
Ptot
Ptot
Ptot
2.32 atm
0.323 atm
0.323 atm
28.01 g/mol
70.91 g/mol
98.92 g/mol
3.00 atm
3.00 atm
3.00 atm
87.1 g/mol
707
Chapter15: Principles of Chemical Equilibrium
83. (M) Each mole fraction equals the partial pressure of the substance divided by the total
pressure. Thus {NH 3 } P{NH 3 } / Ptot or P{NH 3 } {NH 3 } Ptot
Kp
P{NH 3 } 2
P{N 2 } P{H 2 }3
( {NH 3 }Ptot ) 2
( {N 2 }Ptot ) ( {H 2 }Ptot ) 3
{NH 3 }2
( Ptot ) 2
{N 2 } {H 2 }3 ( Ptot ) 4
{NH 3 } 2
1
3
{N 2 } {H 2 } ( Ptot ) 2
This is the expression we were asked to derive.
84. (D) Since the mole ratio of N2 to H2 is 1:3, {H 2 } 3 {N 2 } . Since Ptot = 1.00 atm, it follows.
{NH 3 } 2
1
Kp
9.06 10 2 0.0906
3
2
{N 2 } (3 {N 2 }) (1.00)
3 3 0.0906
{NH 3 } 2
{NH 3 } 2
{N 2 } {N 2 }3
{N 2 } 4
{NH 3 }
3 3 0.0906 1.56
2
{N 2 }
We realize that {NH 3 } {N 2 } {H 2 } 1.00 {NH 3 } {N 2 } 3 {N 2 }
And we have
This gives {NH 3 } 1.00 4 {N 2 }
1.00 4 {N 2 }
For ease of solving, we let x {N 2 }
{N 2 }2
1.00 4 x
1.56
1.56 x 2 1.00 4 x
1.56 x 2 4 x 1.00 0
2
x
1.56
b b 2 4 ac
4.00 16.00 6.24
0.229, 2.794
x
2a
3.12
Thus {N 2 } 0.229
Mole % NH3 = (1.000 mol –(4 × 0.229 mol)) × 100% = 8.4%
85. (M) Since the initial mole ratio is 2 H2S(g) to 1 CH4(g), the reactants remain in their
stoichiometric ratio when equilibrium is reached. Also, the products are formed in their
stoichiometric ratio.
1 mol CH 4
4.77 103 mol CH 4
amount CH 4 9.54 103 mol H 2 S
2 mol H 2S
1 mol CS2
1 mol S
amount CS2 1.42 103 mol BaSO4
7.10 104 mol CS2
1 mol BaSO 4
2 mol S
4 mol H 2
2.84 103 mol H 2
amount H 2 7.10 104 mol CS2
1 mol CS2
total amount 9.54 103 mol H 2 S 4.77 103 mol CH 4 7.10 104 mol CS2 2.84 103 mol H 2
17.86 103 mol
The partial pressure of each gas equals its mole fraction times the total pressure.
708
Chapter15: Principles of Chemical Equilibrium
9.54 103 mol H 2S
P{H 2S} 1.00 atm
0.534 atm
17.86 103 mol total
4.77 103 mol CH 4
P{CH 4 } 1.00 atm
0.267 atm
17.86 103 mol total
7.10 104 mol CS2
P{CS2 } 1.00 atm
0.0398 atm
17.86 103 mol total
2.84 103 mol H 2
P{H 2 } 1.00 atm
0.159 atm
17.86 103 mol total
P{H 2 }4 P{CS2 } 0.1594 0.0398
Kp
3.34 104
2
2
P{H 2S} P{CH 4 } 0.534 0.267
86. (D)
81. We base our calculation on an I.C.E.
the direction of the reaction by computing :
table,
after
we
first
determine
[Fe 2 ] 2 [Hg 2 ] 2
(0.03000) 2 (0.03000) 2
Qc
6.48 10 6
2
2
3 2
(0.5000) (0.5000)
[Fe ] [Hg 2 ]
Because this value is smaller than Kc, the reaction will shift to the right to reach equilibrium.
Since the value of the equilibrium constant for the forward reaction is quite small, let us
assume that the reaction initially shifts all the way to the left (line labeled “to left:”), and then
reacts back in the forward direction to reach a position of equilibrium.
Equation: 2 Fe3 (aq)
Initial:
To left:
Hg 2 2 (aq)
0.5000 M
0.5000 M
0.03000 M
0.0150 M
0.5300 M
0.5150 M
2 Fe 2 (aq)
0.03000 M
Changes: 2 x M
x M
Equil:
(0.5300 2 x)M (0.5150 x)M
Kc
0.03000 M
0 M
2x M
2x M
2 Hg 2 (aq)
0.03000 M
0.03000 M
0 M
2x M
2x M
[Fe 2 ]2 [Hg 2 ]2
(2 x) 2 (2 x) 2
4 x 2 4x 2
6
9.14
10
[Fe3 ]2 [Hg 2 2 ]
(0.5300 2 x) 2 (0.5150 x) (0.5300) 2 0.5150
Note that we have assumed that 2 x 0.5300 and x 0.5150
9.14 10 6 (0.5300) 2 (0.5150)
8.26 10 8
x 0.0170
4 4
Our assumption, that 2 x ( 0.0340) 0.5300 , is reasonably good.
x4
[Fe3 ] 0.5300 2 0.0170 0.4960 M
[Fe 2 ] [Hg 2 ] 2 0.0170 0.0340 M
709
[Hg 2 2 ] 0.5150 0.0170 0.4980
Chapter15: Principles of Chemical Equilibrium
We check by substituting into the K c expression.
9.14 106 K c
[Fe 2 ]2 [Hg 2 ]2 (0.0340) 2 (0.0340) 2
11106 Not a substantial difference.
3 2
2
2
[Fe ] [Hg 2 ]
(0.4960) 0.4980
Mathematica (version 4.0, Wolfram Research, Champaign, IL) gives a root of 0.0163.
87.
(D) Again we base our calculation on an I.C.E. table. In the course of solving the
Integrative Example, we found that we could obtain the desired equation by reversing
equation (2) and adding the result to equation (1)
(2)
H 2 O(g)
CH 4 (g)
(1)
CO(g)
H 2 O(g)
Equation:CH 4 (g)
(2)
H 2 O(g)
(1)
CO(g)
CO(g)
3 H 2 (g)
CO 2 (g) H 2 (g)
K 1/190
K 1.4
CO 2 (g) 4 H 2 (g) K 1.4 /190 0.0074
2 H 2 O(g)
CH 4 (g)
CO(g)
3 H 2 (g) K 1/190
CO 2 (g)
H 2 (g)
Equation: CH 4 (g) 2 H 2 O(g)
Initial:
0.100 mol
0.100 mol
CO 2 (g)
4 H 2 (g) K 1.4 /190 0.0074
To left:
Concns:
H 2 O(g)
0.100 mol
K 1.4
0.100 mol
0.025 mol
0.050 mol
0.025 mol
0.100 mol
0.125 mol
0.0250 M
0.150 mol
0.0300 M
0.075 mol
0.015 M
0.000 mol
0.000 M
Changes:
x M
2 x mol
Equil:
(0.0250 x) M (0.0300 2 x)
x mol
(0.015 x) M
4 x mol
4x mol
Notice that we have a fifth order polynomial to solve. Hence, we need to try to approximate
its final solution as closely as possible. The reaction favors the reactants because of the small
size of the equilibrium constant. Thus, we approach equilibrium from as far to the left as
possible.
K c 0.0074
x4
[CO 2 ][H 2 ] 4
[CH 4 ][H 2 O] 2
(0.0150 x)(4 x) 4
0.0150 (4 x) 4
(0.0250 x) (0.0300 2 x) 2 0.0250 (0.0300) 2
0.0250 (0.0300) 2 0.0074
0.014 M
0.0150 256
Our assumption is terrible. We substitute to continue successive approximations.
(0.0150 0.014) (4x) 4
(0.029)(4 x) 4
(0.0250 0.014) (0.0300 2 0.014) 2 (0.011)(0.002) 2
Next, try x2 = 0.0026
0.0074
0.074
(0.0150 0.0026)(4x)4
(0.0250 0.0026) (0.0300 2 0.0026)2
710
Chapter15: Principles of Chemical Equilibrium
then, try x3 = 0.0123.
After 18 iterations, the x value converges to 0.0080.
Considering that the equilibrium constant is known to only two significant figures, this is a
pretty good result. Recall that the total volume is 5.00 L. We calculate amounts in moles.
88.
CH 4 (g)
(0.0250 0.0080) 5.00 L 0.017 M 5.00 L 0.085 moles CH 4 (g)
H 2 O(g)
(0.0300 2 0.0080) M 5.00 L 0.014 M 5.00 L 0.070 moles H 2 O(g)
CO 2 (g)
(0.015 0.0080) M 5.00 L 0.023 M 5.00 L 0.12 mol CO 2
H 2 (g)
(4 0.0080) M 5.00 L 0.032 M 5.00 L 0.16 mol H 2
(M) The initial mole fraction of C2H2 is i 0.88 . We use molar amounts at equilibrium to
compute the equilibrium mole fraction of C2H2, eq . Because we have a 2.00-L container,
molar amounts are double the molar concentrations.
(2 0.756) mol C2 H 2
0.877
(2 0.756) mol C2 H 2 (2 0.038) mol CH 4 (2 0.068) mol H 2
Thus, there has been only a slight decrease in mole fraction.
eq
89.
(M)
(a) Keq = 4.6×104
P{NO} =
P{NOCl}2
(4.125) 2
=
P{NO}2 P{Cl2 } P{NO}2 (0.1125)
(4.125) 2
= 0.0573 atm
4.6 104 (0.1125)
(b) Ptotal = PNO+ PCl2 + PNOCl = 0.0573 atm + 0.1125 atm + 4.125 atm = 4.295 atm
90.
(M) We base our calculation on an I.C.E. table.
Reaction:
N 2 (g)
Initial:
0.424 mol
10.0 L
Change
- x mol
10.0 L
Equilibrium
(0.424-x) mol
10.0 L
+
3H 2 (g)
1.272 mol
10.0 L
-3x mol
10.0 L
(1.272-3x) mol
10.0 L
711
2NH 3 (g)
0 mol
10.0 L
+2x mol
10.0 L
2x mol
10.0 L
Chapter15: Principles of Chemical Equilibrium
2
2x mol
2
10.0 L
[NH 3 ]
Kc =
=152 =
3
[N 2 ][H 2 ]3
(0.424-x) mol (1.272-3x) mol
10.0 L
10.0 L
Kc =
100 2x mol
2
(0.424-x) mol 3 0.424-x) mol
2
2
100 2x mol
2x mol
Kc = 3
= 152
4
4
3 0.424-x) mol
0.424-x) mol
3
2x mol
0.424-x) mol
2
= 6.41
= 41.04
Take root of both sides
6.41(0.424-x) 2 = 2x
3.20(0.180 0.848 x x 2 ) x 3.20 x 2 2.71x 0.576
3.20 x 2 3.71x 0.576 0
Now solve using the quadratic equation: x = 0.1846 mol or 0.9756 mol (too large)
amount of NH 3 2 x 2(0.1846 mol) = 0.369 mol in 10.0 L or 0.0369 M
([H 2 ] = 0.0718 M and [ N 2 ] 0.0239 M )
91.
(D)
Equation: 2 H 2 (g)
K p K c (RT)
Δn
CO(g)
CH3 OH(g) K c 14.5 at 483 K
L-atm
14.5 0.08206
483K
mol-K
2
9.23 103
We know that mole percents equal pressure percents for ideal gases.
P
CO
0.350 100 atm 35.0 atm
P H 2 0.650 100 atm 65.0 atm
Equation: 2 H 2 (g)
CO(g)
CH 3 OH(g)
Initial:
65 atm
35 atm
Changes: 2P atm
P atm
P atm
65-2P
35-P
P
Equil:
P CH 3 OH
P
9.23 103
Kp
2
2
(35.0 P)(65.0 2P)
P CO P H 2
By successive approximations, P 24.6 atm = PCH3OH at equilibrium.
Mathematica (version 4.0, Wolfram Research, Champaign, IL) gives a root of 24.5.
712
Chapter15: Principles of Chemical Equilibrium
FEATURE PROBLEMS
92.
(M) We first determine the amount in moles of acetic acid in the equilibrium mixture.
0.1000 mol Ba OH 2 2 mol CH 3CO 2 H
1L
amount CH 3CO 2 H = 28.85 mL
1000 mL
1L
1 mol Ba OH 2
complete equilibrium mixture
= 0.5770 mol CH 3CO 2 H
0.01 of equilibrium mixture
0.423mol 0.423mol
CH 3 CO 2 C2 H 5 H 2 O
0.423 0.423
V
V
4.0
Kc
=
C2 H5 OH CH3 CO2 H 0.077 mol 0.577 mol 0.077 0.577
V
V
93.
(D) In order to determine whether or not equilibrium has been established in each bulb, we
need to calculate the concentrations for all three species at the time of opening. The results
from these calculations are tabulated below and a typical calculation is given beneath this table.
Bulb
No.
1
2
3
4
5
Time
Initial
Amount of I2(g) Amount HI(g) [HI] [I2] &
and H2(g) at
at Time of
(mM) [H2]
Bulb
Amount
Time of Opening
(mM)
Opening
Opened
HI(g)
(in mmol)
(in mmol)
(hours) (in mmol)
2
2.345
0.1572
2.03
5.08 0.393
4
2.518
0.2093
2.10
5.25 0.523
12
2.463
0.2423
1.98
4.95 0.606
20
3.174
0.3113
2.55
6.38 0.778
40
2.189
0.2151
1.76
4.40 0.538
[H 2 ][I 2 ]
[HI]2
0.00599
0.00992
0.0150
0.0149
0.0150
Consider, for instance, bulb #4 (opened after 20 hours).
1 mole HI
= 0.003174 mol HI(g) or 3.174 mmol
Initial moles of HI(g) = 0.406 g HI(g)
127.9 g HI
moles of I2(g) present in bulb when opened.
1 mol I 2
0.0150 mol Na 2S2O3
= 0.04150 L Na2S2O3
= 3.113 10-4 mol I 2
2 mol Na 2 S 2 O 3
1 L Na 2S2 O3
millimoles of I2(g) present in bulb when opened = 3.113 10-4 mol I 2
moles of H2 present in bulb when opened = moles of I2(g) present in bulb when opened.
2 mole HI
= 6.226 10-4 mol HI (0.6226 mmol HI)
1 mol I 2
moles of HI(g) in bulb when opened= 3.174 mmol HI 0.6226 mmol HI = 2.55 mmol HI
HI reacted = 3.113 10-4 mol I2
Concentrations of HI, I2, and H2
713
Chapter15: Principles of Chemical Equilibrium
[HI] = 2.55 mmol HI 0.400 L = 6.38 mM
[I2] = [H2] = 0.3113 mmol 0.400 L = 0.778 mM
[H 2 ][I 2 ] (0.778 mM)(0.778 mM)
= 0.0149
Ratio:
[HI]2
(6.38 mM) 2
As the time increases, the ratio
[H 2 ][I 2 ]
initially climbs sharply, but then plateaus at
[HI]2
0.0150 somewhere between 4 and 12 hours. Consequently, it seems quite reasonable to
H2(g) + I2(g) has a Kc ~ 0.015 at 623 K.
conclude that the reaction 2HI(g)
94.
(D) We first need to determine the number of moles of ammonia that were present in the
sample of gas that left the reactor. This will be accomplished by using the data from the
titrations involving HCl(aq).
Original number of moles of HCl(aq) in the 20.00 mL sample
= 0.01872 L of KOH
0.0523 mol KOH 1 mol HCl
1 L KOH
1 mol KOH
= 9.7906 10-4 moles of HCl(initially)
Moles of unreacted HCl(aq)
0.0523 mol KOH 1 mol HCl
=
1 L KOH
1 mol KOH
8.0647 10-4 moles of HCl(unreacted)
= 0.01542 L of KOH
Moles of HCl that reacted and /or moles of NH3 present in the sample of reactor gas
= 9.7906 10-4 moles 8.0647 10-4 moles = 1.73 10-4 mole of NH3 (or HCl).
The remaining gas, which is a mixture of N2(g) and H2(g) gases, was found to occupy 1.82 L
at 273.2 K and 1.00 atm. Thus, the total number of moles of N2 and H2 can be found via the
ideal gas law: nH 2 N2 =
PV
(1.00 atm)(1.82 L)
=
= 0.08118 moles of (N2 + H2)
RT (0.08206 L atm )(273.2 K)
K mol
According to the stoichiometry for the reaction, 2 parts NH3 decompose to give 3 parts H2
and 1 part N2. Thus the non-reacting mixture must be 75% H2 and 25% N2.
So, the number of moles of N2 = 0.25 0.08118 moles = 0.0203 moles N2 and the
number of moles of H2 = 0.75 0.08118 moles = 0.0609 moles H2.
714
Chapter15: Principles of Chemical Equilibrium
Before we can calculate Kc, we need to determine the volume that the NH3, N2, and H2
molecules occupied in the reactor. Once again, the ideal gas law (PV = nRT) will be
employed. ngas = 0.08118 moles (N2 + H2 ) + 1.73 10-4 moles NH3 = 0.08135 moles
nRT
Vgases =
=
P
So, Kc =
L atm
)(1174.2 K)
K mol
= 0.2613 L
30.0 atm
(0.08135 mol)( 0.08206
1.73 10-4 moles
0.2613 L
2
3
1
0.0609 moles 0.0203 moles
0.2613 L 0.2613 L
= 4.46 10-4
To calculate Kp at 901 C, we need to employ the equation K p K c RT
n gas
; ngas 2
Kp = 4.46 10-4 [(0.08206 L atm K-1mol-1)] (1174.2 K)]-2 = 4.80 10-8 at 901C for the
2 NH3(g)
reaction N2(g) + 3 H2(g)
95.
(M) For step 1, rate of the forward reaction = rate of the reverse reaction, so,
k1 [I]2
=
= Kc (step 1)
k-1 [I 2 ]
Like the first step, the rates for the forward and reverse reactions are equal in the second step
and thus,
k
[HI]2
= Kc (step 2)
k2[I]2[H2] = k-2[HI]2 or 2 = 2
k-2 [I] [H 2 ]
k1[I2] = k-1[I]2 or
Now we combine the two elementary steps to obtain the overall equation and its associated
equilibrium constant.
k [I]2
2 I(g)
Kc = 1 =
(STEP 1)
I2(g)
k-1 [I 2 ]
and
k2
[HI]2
(STEP 2)
H2(g) + 2 I(g) 2 HI(g)
Kc =
=
k-2 [I]2 [H 2 ]
2 HI(g)
H2(g) + I2(g)
Kc(overall) = Kc(step 1) Kc(step 2)
Kc(overall) =
k1k2
[I]2 [HI]2
[HI]2
Kc(overall) =
= 2
=
[I] [I 2 ][H 2 ] [I 2 ][H 2 ]
k-1k2
715
k
k1
[HI]2
[I]2
2 =
2
[I 2 ] [I] [H 2 ]
k-1 k -2
Chapter15: Principles of Chemical Equilibrium
96.
(M) The equilibrium expressions for the two reactions are:
H HCO3
H CO32
K1
; K2
H 2 CO3
HCO3
First, start with [H+] = 0.1 and [HCO3–] = 1. This means that [H+]/[HCO3–] = 0.1, which
means that [CO32–] = 10K2. By adding a small amount of H2CO3 we shift [H+] by 0.1 and
[HCO3–] by 0.1. This leads to [H+]/[HCO3–] 0.2, which means that [CO32–] = 5K2. Note that
[CO32–] has decreased as a result of adding H2CO3 to the solution.
97.
(D) First, it is most important to get a general feel for the direction of the reaction by
determining the reaction quotient:
Q
C(aq) 0.1 10
A(aq) B(aq) 0.1 0.1
Since Q>>K, the reaction proceeds toward the reactants. Looking at the reaction in the
aqueous phase only, the equilibrium can be expressed as follows:
Initial
Change
Equil.
A(aq)
0.1
-x
0.1 - x
+ B(aq)
0.1
-x
0.1 - x
C(aq)
0.1
+x
0.1 + x
We will do part (b) first, which assumes the absence of an organic layer for extraction:
K
0.1 x
0.01
0.1 x 0.1 x
Expanding the above equation and using the quadratic formula, x = -0.0996. Therefore, the
concentration of C(aq) and equilibrium is 0.1 + (-0.0996) = 4×10-4 M.
If the organic layer is present for extraction, we can add the two equations together, as shown
below:
A(aq) + B(aq)
C(aq)
A(aq) + B(aq)
C(aq)
C(or)
C(or)
K = K1 × K2 = 0.1 × 15 = 0.15.
Since the organic layer is present with the aqueous layer, and K2 is large, we can expect that
the vast portion of C initially placed in the aqueous phase will go into the organic phase.
716
Chapter15: Principles of Chemical Equilibrium
Therefore, the initial [C] = 0.1 can be assumed to be for C(or). The equilibrium can be
expressed as follows
Initial
Change
Equil.
A(aq)
0.1
-x
0.1 - x
+ B(aq)
0.1
-x
0.1 - x
C(or)
0.1
+x
0.1 + x
We will do part (b) first, which assumes the absence of an organic layer for extraction:
K
0.1 x
0.15
0.1 x 0.1 x
Expanding the above equation and using the quadratic formula, x = -0.0943. Therefore, the
concentration of C(or) and equilibrium is 0.1 + (-0.0943) = 6×10-4 M. This makes sense
because the K for the overall reaction is < 1, which means that the reaction favors the
reactants.
SELF-ASSESSMENT EXERCISES
98.
(E)
(a) Kp: The equilibrium constant of a reaction where the pressures of gaseous reactants and
products are used instead of their concentrations
(b) Qc: The reaction quotient using the molarities of the reactants and products
(c) Δngas: The difference between the number of moles (as determined from a balanced
reaction) of product and reactant gases
99.
(E)
(a) Dynamic equilibrium: In a dynamic equilibrium (which is to say, real equilibrium), the
forward and reverse reactions happen, but at a constant rate
(b) Direction of net chemical change: In a reversible reaction, if the reaction quotient Qc > Kc,
then the net reaction will go toward the reactants, and vice versa
(c) Le Châtelier’s principle: When a system at equilibrium is subjected to external change
(change in partial pressure of reactants/products, temperature or concentration), the
equilibrium shifts to a side to diminish the effects of that external change
(d) Effect of catalyst on equilibrium: A catalyst does not affect the final concentrations of the
reactants and products. However, since it speeds up the reaction, it allows for the
equilibrium concentrations to be established more quickly
100. (E)
(a) Reaction that goes to completion and reversible reaction: In a reversible reaction, the
products can revert back to the reactants in a dynamic equilibrium. In a reaction that goes
to completion, the formation of products is so highly favored that there is practically no
reverse reaction (or the reverse is practically impossible, such as a combustion reaction).
717
Chapter15: Principles of Chemical Equilibrium
(b) Kp and Kc: Kp is the equilibrium constant using pressures of products and reactants, while
Kc is the constant for reaction using concentrations.
(c) Reaction quotient (Q) and equilibrium constant expression (K): The reaction quotient Q is
the ratio of the concentrations of the reactants and products expressed in the same format
as the equilibrium expression. The equilibrium constant expression is the ratio of
concentrations at equilibrium.
(d) Homogeneous and heterogeneous reaction: In a homogeneous reaction, the reaction
happens within a single phase (either aqueous or gas). In a heterogeneous reaction, there
is more than one phase present in the reaction.
101. (E) The answer is (c). Because the limiting reagent is I2 at one mole, the theoretical yield of
HI is 2 moles. However, because there is an established equilibrium, there is a small amount
of HI which will decompose to yield H2 and I2. Therefore the total moles of HI created is
close, but less than 2.
102. (E) The answer is (d). The equilibrium expression is:
K
P SO3
2
P SO 2 P O 2
2
100
If equilibrium is established, moles of SO3 and SO2 cancel out of the equilibrium expression.
Therefore, if K = 100, the moles of O2 have to be 0.01 to make K = 100.
103. (E) The answer is (a). As the volume of the vessel is expanded (i.e., pressure is reduced), the
equilibrium shifts toward the side with more moles of gas.
104. (E) The answer is (b). At half the stoichiometric values, the equilibrium constant is K1/2. If
the equation is reversed, it is K-1. Therefore, the K’ = K-1/2 = (1.8×10-6)-1/2 = 7.5×10-2.
105. (E) The answer is (a). We know that Kp = Kc (RT)Δn. Since Δn = (3–2) = 1, Kp = Kc (RT).
Therefore, Kp > Kc.
106. (E) The answer is (c). Since the number of moles of gas of products is more than the
reactants, increasing the vessel volume will drive the equilibrium more toward the product
side. The other options: (a) has no effect, and (b) drives the equilibrium to the reactant side.
107. (E) The equilibrium expression is:
2
C
0.43
K
1.9
2
2
B A 0.55 0.33
2
108. (E)
(a) As more O2 (a reactant) is added, more Cl2 is produced.
(b) As HCl (a reactant) is removed, equilibrium shifts to the left and less Cl2 is made.
(c) Since there are more moles of reactants, equilibrium shifts to the left and less Cl2 is made.
(d) No change. However, the equilibrium is reached faster.
(e) Since the reaction is exothermic, increasing the temperature causes less Cl2 to be made.
718
Chapter15: Principles of Chemical Equilibrium
109. (E) SO2 (g) will be less than SO2 (aq), because K > 1, so the equilibrium lies to the product
side, SO2 (aq).
110. (E) Since K >>1, there will be much more product than reactant
111. (M) The equilibrium expression for this reaction is:
SO3 35.5
K
2
SO2 O2
2
(a) If [SO3]eq = [SO2]eq, then [O2] = 1/35.5 = 0.0282 M.
moles of O2 = 0.0282 × 2.05 L = 0.0578 moles
(b) Plugging in the new concentration values into the equilibrium expression:
SO3 2 SO2 4 35.5
K
2
2
SO2 O2 SO2 O2 O2
2
2
[O2] = 0.113 M
moles of O2 = 0.113 × 2.05 L = 0.232 moles
112. (M) This concept map involves the various conditions that affect equilibrium of a reaction,
and those that don’t. Under the category of conditions that do cause a change, there is
changing the partial pressure of gaseous products and reactants, which includes pressure and
vessel volume. The changes that do not affect partial pressure are changing the concentration
of reactants or products in an aqueous solution, through dilution or concentration. Changing
the temperature can affect both aqueous and gaseous reactions. Under the category of major
changes that don’t affect anything is the addition of a non-reactive gas.
719
CHAPTER 16
ACIDS AND BASES
PRACTICE EXAMPLES
1A
(E)
(a)
In the forward direction, HF is the acid (proton donor; forms F), and H2O is the base
(proton acceptor; forms H3O). In the reverse direction, F is the base (forms HF),
accepting a proton from H3O, which is the acid (forms H2O).
(b) In the forward direction, HSO4 is the acid (proton donor; forms SO42), and NH3 is
the base (proton acceptor; forms NH4). In the reverse direction, SO42 is the base
(forms HSO4), accepting a proton from NH4, which is the acid (forms NH3).
(c)
1B
In the forward direction, HCl is the acid (proton donor; forms Cl), and C2H3O2 is
the base (proton acceptor; forms HC2H3O2). In the reverse direction, Cl is the base
(forms HCl), accepting a proton from HC2H3O2, which is the acid (forms C2H3O2).
(E) We know that the formulas of most acids begin with H. Thus, we identify HNO 2 and
HCO 3 as acids.
NO 2 aq + H 3O + aq ;
HNO 2 aq + H 2O(l)
CO32 aq + H 3O + aq
HCO3 aq + H 2O(l)
A negatively charged species will attract a positively charged proton and act as a base.
3
3
Thus PO 4 and HCO 3 can act as bases. We also know that PO 4 must be a base
because it cannot act as an acid—it has no protons to donate—and we know that all three
species have acid-base properties.
HPO 4 2 aq + OH aq ;
aq + H 2O(l)
H 2 CO3 (aq) CO 2 H 2 O aq + OH aq
HCO3 aq + H 2O(l)
PO 4
3
Notice that HCO 3 is the amphiprotic species, acting as both an acid and a base.
2A
(M) H 3O + is readily computed from pH: H 3O + = 10 pH
H 3O + = 102.85 = 1.4 103 M.
OH can be found in two ways: (1) from Kw = H 3O + OH , giving
Kw
1.0 1014
OH =
=
= 7.11012 M, or (2) from pH + pOH = 14.00 , giving
3
+
H 3O 1.4 10
pOH = 14.00 pH = 14.00 2.85 = 11.15 , and then OH = 10 pOH = 1011.15 = 7.1 1012 M.
720
Chapter 16: Acids and Bases
2B
(M) H 3O + is computed from pH in each case: H 3O + = 10 pH
H 3O +
conc .
= 102.50 = 3.2 103 M
H 3O +
dil .
= 103.10 = 7.9 10 4 M
All of the H 3O + in the dilute solution comes from the concentrated solution.
3.2 103 mol H 3O +
= 3.2 103 mol H 3O +
1 L conc. soln
Next we calculate the volume of the dilute solution.
1 L dilute soln
volume of dilute solution = 3.2 103 mol H 3O +
= 4.1 L dilute soln
7.9 104 mol H 3O +
Thus, the volume of water to be added is = 3.1 L.
Infinite dilution does not lead to infinitely small hydrogen ion concentrations. Since
dilution is done with water, the pH of an infinitely dilute solution will approach that of
pure water, namely pH = 7.
amount H 3O + = 1.00 L conc. soln
3A
(E) pH is computed directly from the equation below.
H 3O + , pH = log H 3O + = log 0.0025 = 2.60 .
We know that HI is a strong acid and, thus, is completely dissociated into H 3O + and I .
The consequence is that I = H 3O + = 0.0025 M. OH is most readily computed from
pH: pOH = 14.00 pH = 14.00 2.60 = 11.40;
3B
OH = 10 pOH = 1011.40 = 4.0 1012 M
(M) The number of moles of HCl(g) is calculated from the ideal gas law. Then H 3O + is
calculated, based on the fact that HCl(aq) is a strong acid (1 mol H 3O + is produced from
each mole of HCl).
1atm
747 mmHg
0.535 L
760 mmHg
moles HCl(g) =
0.08206 L atm
(26.5 273.2) K
mol K
moles HCl(g) 0.0214 mol HCl(g) = 0.0214 mol H 3O + when dissolved in water
[H 3O + ] 0.0214 mol
4A
pH = -log(0.0214) = 1.670
b g
(E) pH is most readily determined from pOH = log OH . Assume Mg OH
base.
OH =
2
is a strong
9.63 mg Mg OH 2 1000 mL
1 mol Mg OH 2
1g
2 mol OH
100.0 mL soln
1 L
1000 mg 58.32 g Mg OH 2 1 mol Mg OH 2
OH = 0.00330 M; pOH = log 0.00330 = 2.481
pH = 14.000 pOH = 14.000 2.481 = 11.519
721
Chapter 16: Acids and Bases
4B
(E) KOH is a strong base, which means that each mole of KOH that dissolves produces one
mole of dissolved OH aq . First we calculate OH and the pOH. We then use
b g
pH + pOH = 14.00 to determine pH.
OH =
3.00 g KOH
1 mol KOH 1 mol OH 1.0242 g soln 1000 mL
= 0.548 M
100.00 g soln 56.11 g KOH 1 mol KOH
1 mL soln
1L
pOH = log 0.548 = 0.261
5A
pH = 14.000 pOH = 14.000 0.261 = 13.739
(M) H 3O + = 10 pH = 104.18 = 6.6 105 M.
Organize the solution using the balanced chemical equation.
Equation:
Initial:
Changes:
Equil:
HOCl aq
+
H 2 O(l)
—
0.150 M
6.6 105 M
≈ 0.150 M
—
—
b g
b g
OCl aq
0M
+ 6.6 105 M
6.6 105 M
+
H 3O + aq
0 M
+ 6.6 105 M
6.6 105 M
H 3O + OCl 6.6 105 6.6 105
Ka
=
2.9 108
0.150
HOCl
5B
(M) First, we use pH to determine OH . pOH = 14.00 pH = 14.00 10.08 = 3.92 .
OH = 10 pOH = 10 3.92 = 1.2 104 M. We determine the initial concentration of cocaine
and then organize the solution around the balanced equation in the manner we have used
before.
0.17 g C17 H 21O 4 N 1000 mL
1mol C17 H 21O 4 N
[C17 H 21O 4 N] =
= 0.0056 M
100 mL soln
1L
303.36 g C17 H 21O 4 N
Equation:
Initial:
Changes:
Equil:
Kb
C17 H 21O 4 N(aq) + H 2 O(l)
0.0056 M
1.2 104 M
≈ 0.0055 M
C17 H 21O 4 NH + OH
C17 H 21O 4 N
—
—
—
b g
C17 H 21O 4 NH + aq
0M
+1.2 104 M
1.2 104 M
c1.2 10 hc1.2 10 h 2.6 10
4
4
0.0055
722
6
+
b g
OH aq
0 M
+1.2 104 M
1.2 104 M
Chapter 16: Acids and Bases
6A
(M) Again we organize our solution around the balanced chemical equation.
Equation:
Initial:
HC2 H 2 FO 2 (aq) + H 2 O(l)
—
0.100 M
—
x M
—
0.100 x M
Changes:
Equil:
b
g
H 3O + (aq) +
0M
+x M
x M
C 2 H 2 FO 2 (aq)
0M
+x M
x M
H 3O + C2 H 2 FO 2
xx
; therefore, 2.6 103
Ka =
HC2 H 2 FO 2
0.100 x
We can use the 5% rule to ignore x in the denominator. Therefore, x = [H3O+] = 0.016 M,
and pH = –log (0.016) = 1.8. Thus, the calculated pH is considerably lower than 2.89
(Example 16-6).
6B (M) We first determine the concentration of undissociated acid. We then use this value in
a set-up that is based on the balanced chemical equation.
0.500 g HC9 H 7 O 4
1 mol HC9 H 7 O 4
1
2 aspirin tablets×
×
×
= 0.0171M
tablet
180.155g HC9 H 7 O 4 0.325 L
Equation:
HC9 H 7 O 4 (aq) + H2O(l)
H 3O + (aq) + C9 H 7 O 4 (aq)
—
Initial:
0M
0.0171 M
0M
—
Changes:
x M
+x M
+x M
—
Equil:
x
M
x M
0.0171 x M
b
g
H 3 O + C9 H 7 O 4
xx
= 3.3 104 =
Ka =
HC9 H 7 O 4
0.0171 x
x 2 + 3.3 104 5.64 106 = 0 (find the physically reasonable roots of the quadratic equation)
3.3 104 1.1 107 2.3 105
x
0.0022 M;
2
7A
pH log (0.0022) 2.66
(M) Again we organize our solution around the balanced chemical equation.
Equation: HC 2 H 2 FO 2 (aq) +
Initial:
0.015 M
Changes
x M
Equil:
0.015 x M
b
g
H 2 O(l)
—
—
—
H3O+ (aq) +
0M
+x M
x M
C 2 H 2 FO 2 (aq)
0M
+x M
x M
2
H 3 O + C2 H 2 FO 2
= 2.6 103 = x x x
Ka =
0.015 x 0.015
HC2 H 2 FO2
x = x 2 0.015 2.6 103 0.0062 M = [H 3O + ]
723
Our assumption is invalid:
Chapter 16: Acids and Bases
0.0062 is not quite small enough compared to 0.015 for the 5% rule to hold. Thus we use
another cycle of successive approximations.
Ka =
x x
= 2.6 103 x = (0.015 0.0062) 2.6 103 0.0048 M = [H 3O + ]
0.015 0.0062
x x = 2.6 103 x = (0.015 0.0048) 2.6 103 0.0051 M = [H O+ ]
Ka =
3
0.015 0.0048
Ka =
x x
0.015 0.0051
= 2.6 103 x = (0.015 0.0051) 2.6 103 0.0051 M = [H 3O + ]
Two successive identical results indicate that we have the solution.
b
g
pH = log H 3O + = log 0.0051 = 2.29 . The quadratic equation gives the same result
(0.0051 M) as this method of successive approximations.
7B
(M) First we find [C5H11N]. We then use this value as the starting base concentration in a
set-up based on the balanced chemical equation.
114 mg C5 H 11N 1 mmol C5 H 11N
= 0.00425 M
C5 H 11N =
315 mL soln
85.15 mg C5 H 11N
Equation: C5 H11 N(aq) + H 2 O(l)
C5 H11 NH + (aq) + OH (aq)
—
0 M
0.00425 M
0M
Initial:
—
+x M
+x M
x M
Changes:
x M
—
x M
0.00425 x M
Equil:
C5 H11 NH + OH
xx
xx
= 1.6 103 =
Kb =
0.00425 x 0.00425
C5 H11 N
x 0.0016 0.00425 0.0026 M
We assumed that x 0.00425
The assumption is not valid. Let’s assume x 0.0026
x 0.0016 (0.00425 0.0026) 0.0016
Let’s try again, with x 0.0016
x 0.0016 (0.00425 0.0016) 0.0021
Yet another try, with x 0.0021
x 0.0016 (0.00425 0.0021) 0.0019
The last time, with x 0.0019
x 0.0016 (0.00425 0.0019) 0.0019 M = [OH ]
pOH= log[H3 O ] log(0.0019) 2.72 pH = 14.00 pOH = 14.00 2.72 = 11.28
We could have solved the problem with the quadratic formula roots equation rather than by
successive approximations. The same answer is obtained. In fact, if we substitute
2
x = 0.0019 into the Kb expression, we obtain 0.0019 / 0.00425 0.0019 = 1.5 103
b
3
g b
g
compared to Kb = 1.6 10 . The error is due to rounding, not to an incorrect value.
Using x = 0.0020 gives a value of 1.8 103 , while using x = 0.0018 gives 1.3 103 .
724
Chapter 16: Acids and Bases
8A
(M) We organize the solution around the balanced chemical equation; a M is [HF]initial.
Equation: HF(aq) + H 2 O(l)
H 3 O + (aq) + F (aq)
—
a M
0M
0 M
Initial:
—
+x M
+x M
x M
Changes:
—
xM
x M
a x M
Equil:
H 3O + F x x x 2
x = a 6.6 104
=
= 6.6 104
Ka =
ax
a
HF
x = 0.20 6.6 104 0.011 M
For 0.20 M HF, a = 0.20 M
% dissoc=
0.011 M
100%=5.5%
0.20 M
x = 0.020 6.6 104 0.0036 M
For 0.020 M HF, a = 0.020 M
We need another cycle of approximation: x = (0.020 0.0036) 6.6 104 0.0033 M
Yet another cycle with x 0.0033 M : x = (0.020 0.0033) 6.6 104 0.0033 M
0.0033M
% dissoc =
100% = 17%
0.020M
As expected, the weak acid is more dissociated.
8B
(E) Since both H 3O + and C 3 H 5O 3 come from the same source in equimolar amounts,
their concentrations are equal. H 3O + = C 3 H 5O 3 = 0.067 0.0284 M = 0.0019 M
Ka =
9A
H 3 O + C 3 H 5O 3
LM HC3 H 5O 3 OP
Q
N
=
b0.0019gb0.0019g = 1.4 10
4
0.0284 0.0019
(M) For an aqueous solution of a diprotic acid, the concentration of the divalent anion is
very close to the second ionization constant: OOCCH 2 COO K a 2 = 2.0 106 M . We
organize around the chemical equation.
Equation: CH 2 COOH (aq) + H 2 O(l)
2
Initial:
—
1.0 M
Change:
—
x M
Equil:
—
(1.0 –x) M
Ka
[H 3O ][HCH 2 (COO) 2 ]
[CH 2 (COOH) 2 ]
H 3 O + (aq)
0M
+x M
xM
+
HCH 2 COO 2 (aq)
0M
+x M
xM
( x)( x) x 2
1.4 103
1.0 x 1.0
x 1.4 103 3.7 102 M [H3 O+ ] [HOOCCH 2 COO ] x 1.0M is a valid assumption.
725
Chapter 16: Acids and Bases
9B
(M) We know K a 2 doubly charged anion for a polyprotic acid. Thus,
K a 2 = 5.3 105 C2 O 4 2 . From the pH, H 3O + = 10 pH = 100.67 = 0.21 M . We also
recognize that HC 2 O 4 = H 3O + , since the second ionization occurs to only a very small
extent. We note as well that HC 2 O 4 is produced by the ionization of H 2 C 2 O 4 . Each
mole of HC 2 O 4 present results from the ionization of 1 mole of H 2 C2 O 4 . Now we have
sufficient information to determine the K a1 .
H 3O + HC2O 4 0.21 0.21
=
K a1 =
= 5.3 102
1.05 0.21
H 2 C 2 O 4
10A (M) H 2SO 4 is a strong acid in its first ionization, and somewhat weak in its second, with
K a 2 = 1.1102 = 0.011 . Because of the strong first ionization step, this problem involves
determining concentrations in a solution that initially is 0.20 M H 3O + and 0.20 M HSO 4 .
We base the set-up on the balanced chemical equation.
Equation:
Initial:
Changes:
Equil:
b g
HSO 4 aq + H 2 O (l)
—
0.20 M
—
x M
0.20 x
M
—
b g
H 3O + aq
0.20 M
+x M
0.20 + x
+
M
2-
SO 4 (aq)
0M
+x M
x M
H 3O + SO 4 2 0.20 + x x
0.20 x
=
= 0.011
, assuming that x 0.20M.
Ka2 =
0.20 x
0.20
HSO 4
Try one cycle of approximation:
x = 0.011M
0.011
b0.20 + 0.011gx = 0.21x
b0.20 0.011g 0.19
x=
0.19 0.011
= 0.010 M
0.21
2
The next cycle of approximation produces the same answer 0.010M = SO 4 ,
HSO 4 = 0.20 0.010 M = 0.19 M
H 3O + = 0.010 + 0.20 M = 0.21 M,
10B (M) We know that H 2SO 4 is a strong acid in its first ionization, and a somewhat weak
acid in its second, with K a 2 = 1.1 102 = 0.011 . Because of the strong first ionization step,
the problem essentially reduces to determining concentrations in a solution that initially is
0.020 M H 3O + and 0.020 M HSO 4 . We base the set-up on the balanced chemical
equation. The result is solved using the quadratic equation.
726
Chapter 16: Acids and Bases
Equation:
b g
HSO 4 aq + H 2 O (l)
—
0.020 M
—
x M
0.020 x M —
Initial:
Changes:
Equil:
b g
H 3O + aq
0.020 M
+x M
0.020 + x M
H 3O + SO 4 2 0.020 + x x
=
= 0.011
Ka2 =
0.020 x
HSO 4
2
+
b g
SO 4 aq
0M
+x M
x M
0.020 x + x 2 = 2.2 104 0.011x
0.031 0.00096 + 0.00088
2
= 0.0060 M = SO 4
2
HSO 4 = 0.020 0.0060 = 0.014 M
H 3O + = 0.020 + 0.0060 = 0.026 M
x 2 + 0.031x 0.00022 = 0
x=
(The method of successive approximations converges to x = 0.006 M in 8 cycles.)
11A (E)
(a)
+
CH 3 NH 3 NO 3 is the salt of the cation of a weak base. The cation, CH 3 NH 3+ , will
+
CH 3 NH 2 + H 3O + , while
hydrolyze to form an acidic solution CH 3 NH 3 + H 2 O
–
NO3 , by virtue of being the conjugate base of a strong acid will not hydrolyze to a
detectable extent. The aqueous solutions of this compound will thus be acidic.
(b) NaI is the salt composed of the cation of a strong base and the anion of a strong
acid, neither of which hydrolyzes in water. Solutions of this compound will be pH
neutral.
(c)
NaNO 2 is the salt composed of the cation of a strong base that will not hydrolyze in
water and the anion of a weak acid that will hydrolyze to form an alkaline solution
HNO 2 + OH . Thus aqueous solutions of this compound will be
NO 2 + H 2 O
basic (alkaline).
11B (E) Even without referring to the K values for acids and bases, we can predict that the
reaction that produces H 3O + occurs to the greater extent. This, of course, is because the pH
is less than 7, thus acid hydrolysis must predominate.
We write the two reactions of H 2 PO 4 with water, along with the values of their equilibrium
constants.
H 3O + aq + HPO 4 2 aq
H 2 PO 4 aq + H 2 O(l)
K a 2 = 6.3 108
OH aq + H 3 PO4 aq
H 2 PO 4 aq + H 2 O(l)
Kb =
As predicted, the acid ionization occurs to the greater extent.
727
K w 1.0 1014
=
= 1.4 1012
3
K a1 7.1 10
Chapter 16: Acids and Bases
12A (M) From the value of pKb we determine the value of Kb and then Ka for the cation.
pK
= 10
8.41
= 3.9 10
9
cocaine:
Kb = 10
codeine:
Kb = 10 pK = 107.95 = 1.1 108
Kw 1.0 1014
= 2.6 106
Ka =
=
9
Kb 3.9 10
Ka =
Kw 1.0 1014
= 9.1 107
=
8
Kb 1.1 10
(This method may be a bit easier:
pKa = 14.00 pKb = 14.00 8.41 = 5.59, Ka = 105.59 = 2.6 106 ) The acid with the larger
Ka will produce the higher H + , and that solution will have the lower pH. Thus, the
solution of codeine hydrochloride will have the higher pH (i.e., codeine hydrochloride is
the weaker acid).
b g
12B (E) Both of the ions of NH 4 CN aq react with water in hydrolysis reactions.
+
+
NH 4 aq + H 2 O(l)
NH3 aq + H 3O aq
CN aq + H 2 O(l)
HCN aq + OH aq
Ka =
K w 1.0 1014
=
= 5.6 1010
K b 1.8 105
K w 1.0 1014
=
= 1.6 105
Kb =
10
K a 6.2 10
Since the value of the equilibrium constant for the hydrolysis reaction of cyanide ion is
larger than that for the hydrolysis of ammonium ion, the cyanide ion hydrolysis reaction
will proceed to a greater extent and thus the solution of NH 4 CN aq will be basic
(alkaline).
b g
13A (M) NaF dissociates completely into sodium ions and fluoride ions. The released fluoride
ion hydrolyzes in aqueous solution to form hydroxide ion. The determination of the
equilibrium pH is organized around the balanced equation.
Equation: F- (aq) + H 2 O(l)
HF(aq) + OH (aq)
—
Initial:
0M
0M
0.10 M
—
Changes:
+
xM
+
x
M
x M
Equil:
—
xM
xM
0.10 x M
K w 1.0 1014
[HF][OH ]
( x)( x)
x2
11
Kb
1.5 10
K a 6.6 104
[F- ]
(0.10 x) 0.10
x = 0.10 1.5 1011 1.2 106 M=[OH ]; pOH= log(1.2 106 ) 5.92
pH = 14.00 pOH = 14.00 5.92 = 8.08 (As expected, pH > 7)
728
Chapter 16: Acids and Bases
13B (M) The cyanide ion hydrolyzes in solution, as indicated in Practice Example 16-12B. As a
consequence of the hydrolysis, OH = HCN . OH can be found from the pH of the
solution, and then values are substituted into the Kb expression for CN , which is then
solved for CN .
pOH = 14.00 pH = 14.00 10.38 = 3.62
OH = 10-pOH = 103.62 = 2.4 104 M = HCN
Kb =
CN
c2.4 10 h
=
4 2
HCN OH
= 1.6 10
5
CN
CN
c2.4 10 h
=
4 2
1.6 10
5
= 3.6 103 M
14A (M) First we draw the Lewis structures of the four acids. Lone pairs have been omitted
since we are interested only in the arrangements of atoms.
O
O
H O N O
H O
H O
H O Cl O
F
C
H
O
C O H
Br
C C O H
H
HClO 4 should be stronger than HNO 3 . Although Cl and N have similar electronegativities,
there are more terminal oxygen atoms attached to the chlorine in perchloric acid than to the
nitrogen in nitric acid. By virtue of having more terminal oxygens, perchloric acid, when
ionized, affords a more stable conjugate base. The more stable the anion, the more easily it is
formed and hence the stronger is the conjugate acid from which it is derived. CH 2 FCOOH
will be a stronger acid than CH 2 BrCOOH because F is a more electronegative atom than Br.
The F atom withdraws additional electron density from the O — H bond, making the bond
easier to break, which leads to increased acidity.
14B (M) First we draw the Lewis structures of the first two acids. Lone pairs are not depicted
since we are interested in the arrangements of atoms.
O
H O
P
O
O H
H O
S O H
O H
H 3 PO 4 and H 2SO 3 both have one terminal oxygen atom, but S is more electronegative
than P. This suggests that H 2SO3 K a1 = 1.3 102 should be a stronger acid than
H 3PO 4 K a1 = 7.1103 , and it is. The only difference between CCl 3CH 2 COOH and
CCl 2 FCH 2 COOH is the replacement of Cl by F. Since F is more electronegative than Cl,
CCl 3CH 2 COOH should be a weaker acid than CCl 2 FCH 2 COOH .
729
Chapter 16: Acids and Bases
(M) We draw Lewis structures to help us decide.
(a) Clearly, BF3 is an electron pair acceptor, a Lewis acid, and NH 3 is an electron pair
donor, a Lewis base.
15A
F
F H
H
F
B N H
F
H
F
B N H
O H
F H
H
H 2 O certainly has electron pairs (lone pairs) to donate and thus it can be a Lewis
base. It is unlikely that the cation Cr 3+ has any accessible valence electrons that can
be donated to another atom, thus, it is the Lewis acid. The product of the reaction,
[Cr(H2O)6]3+, is described as a water adduct of Cr3+.
(b)
OH2
Cr3+(aq) +
6 O H
H2 O
H
H2 O
Cr
3+
OH2
OH2
OH2
15B (M) The Lewis structures of the six species follow.
O
H
O
H
+
Al
-
H
Cl
O
O
H
H
O
O
Al
O
H
Cl
Sn
Cl
O
H
2 Cl
Cl Cl
Cl
Cl
Sn
Cl
Cl
2-
H
Both the hydroxide ion and the chloride ion have lone pairs of electrons that can be
donated to electron-poor centers. These two are the electron pair donors, or the Lewis
bases. Al OH 3 and SnCl 4 have additional spaces in their structures to accept pairs of
b g
-
electrons, which is what occurs when they form the complex anions [Al(OH)4] and
2[SnCl6] . Thus, Al OH 3 and SnCl 4 are the Lewis acids in these reactions.
b g
730
Cl
2-
Chapter 16: Acids and Bases
INTEGRATIVE EXAMPLE
A.
(D) To confirm the pH of rainwater, we have to calculate the concentration of CO2(aq) in
water and then use simple acid-base equilibrium to calculate pH.
Concentration of CO2 in water at 1 atm pressure and 298 K is 1.45 g/L, or
1.45 g CO 2 1 mol CO 2
0.0329 M CO 2
L
44.0 g CO 2
Furthermore, if the atmosphere is 0.037% by volume CO2, then the mole fraction of CO2 is
0.00037, and the partial pressure of CO2 also becomes 0.00037 atm (because
PCO2 CO2 Patm )
From Henry’s law, we know that concentration of a gas in a liquid is proportional to its
partial pressure.
C(mol / L) H PCO2 , which can rearrange to solve for H:
H
0.0329 M
0.0329 mol L1 atm 1
1 atm
Therefore, concentration of CO2(aq) in water under atmospheric pressures at 298 K is:
CCO2 (mol / L) 0.0329 mol L1 atm 1 0.00037 atm = 1.217 105 M
Using the equation for reaction of CO2 with water:
CO2 (aq)
+
-5
1.217×10
-x
1.217×10-5–x
2 H2O (l)
H3O+ (aq) + HCO3 (aq)
0
0
+x
+x
x
X
x2
K a1
4.4 107
5
1.217 10 x
Using the quadratic formula, x = 2.104 106 M
pH log H 3O log 2.104 106 5.68 5.7
We can, of course, continue to refine the value of [H3O+] further by considering the
dissociation of HCO3–, but the change is too small to matter.
731
Chapter 16: Acids and Bases
B. (D)
(a) For the acids given, we determine values of m and n in the formula EOm(OH)n. HOCl or
Cl(OH) has m = 0 and n = 1. We expect Ka 10-7 or pKa 7, which is in good
agreement with the accepted pKa = 7.52. HOClO or ClO(OH) has m = 1 and n = 1. We
expect Ka 10-2 or pKa 2, in good agreement with the pKa = 1.92. HOClO2 or
ClO2(OH) has m = 2 and n = 1. We expect Ka to be large and in good agreement with
the accepted value of pKa = -3, Ka = 10-pKa = 103. HOClO3 or ClO3(OH) has m = 3 and
n = 1. We expect Ka to be very large and in good agreement with the accepted Ka = -8,
pKa = -8, Ka = 10-pKa = 108 which turns out to be the case.
(b) The formula H3AsO4 can be rewritten as AsO(OH)3, which has m = 1 and n = 3. The
expected value is Ka = 10-2.
(c) The value of pKa = 1.1 corresponds to Ka = 10-pKa = 10-1.1 = 0.08, which indicates that
m = 1. The following Lewis structure is consistent with this value of m.
OH
O
P
H
OH
EXERCISES
Brønsted-Lowry Theory of Acids and Bases
1.
2.
(E)
(a)
HNO 2 is an acid, a proton donor. Its conjugate base is NO 2 .
(b)
OCl is a base, a proton acceptor. Its conjugate acid is HOCl.
(c)
NH 2 is a base, a proton acceptor. Its conjugate acid is NH 3 .
(d)
NH 4 is an acid, a proton donor. It’s conjugate base is NH 3 .
(e)
CH 3 NH 3 is an acid, a proton donor. It’s conjugate base is CH 3 NH 2 .
+
+
(E) We write the conjugate base as the first product of the equilibrium for which the acid is
the first reactant.
(a)
+
HIO3 aq + H 2 O(l)
IO3 aq + H 3O aq
(b)
+
C6 H 5COOH aq + H 2 O(l)
C6 H 5COO aq + H 3O aq
(c)
HPO 4
(d)
+
C2 H 5 NH 3 aq + H 2 O(l)
C 2 H 5 NH 2 aq + H 3O aq
2
3
+
aq + H 2O(l)
PO 4 aq + H3O aq
732
Chapter 16: Acids and Bases
3.
(E) The acids (proton donors) and bases (proton acceptors) are labeled below their
formulas. Remember that a proton, in Brønsted-Lowry acid-base theory, is H + .
(a)
(b)
2
+
HSO 4 (aq) + H 2 O(l)
H 3O (aq) + SO 4 (aq)
acid
base
acid
base
(c)
HS (aq) + H 2 O(l)
H 2S(aq)
base
acid
acid
(d)
4.
+
HOBr(aq) + H 2O(l)
H 3O (aq) + OBr (aq)
acid
base
acid
base
base
C6 H 5 NH 3 (aq) + OH (aq)
C6 H 5 NH 2 (aq) + H 2 O(l)
acid
base
base
acid
(E) For each amphiprotic substance, we write both its acid and base hydrolysis reaction.
Even for the substances that are not usually considered amphiprotic, both reactions are
written, but one of them is labeled as unlikely. In some instances we have written an
oxygen as to keep track of it through the reaction.
2
+
H + H 2 O
H + H 2 O
+ H 3O unlikely
H 2 + OH
+
NH 4 + H 2 O
NH 3 + H 3O
NH 4 + has no e pairs that can be donated
+
H 2 + H 2O
H + H 3O
+
H 2 + H 2 O
H 3 + OH
2
+
HS + H 2 O
S + H 3O
HS + H 2 O
H 2S + OH
NO 2 cannot act as an acid, (no protons)
NO 2 + H 2 O
HNO 2 + OH
5.
+ OH (aq)
2
+
HCO3 + H 2 O
CO3 + H 3O
HCO3 + H 2 O
H 2 CO3 + OH
+
HBr + H 2 O
H 3O + Br
+
HBr + H 2 O
H 2 Br + OH unlikely
(E) Answer (b), NH 3 , is correct. HC 2 H 3O 2 will react most completely with the strongest
base. NO3 and Cl are very weak bases. H 2 O is a weak base, but it is amphiprotic,
acting as an acid (donating protons), as in the presence of NH 3 . Thus, NH 3 must be the
strongest base and the most effective in deprotonating HC 2 H 3O 2 .
6.
(M) Lewis structures are given below each equation.
NH 4 + + NH 2
(a) 2NH 3 l
H
2H N H
H N H + H N H
H
H
733
Chapter 16: Acids and Bases
(b)
+
2HF l
H2 F + F
H
H
(c)
+
F
H
F
2H C
H H
O H
H
H C O
H
H
+
2HC2 H 3O 2 l
H 2 C 2 H 3O 2 + C 2 H 3 O 2
H
O
2H C
C
O
H
H C C O
H
2 H 2 SO 4 (l )
2 H3SO 4 HSO 4
H
O
O
O
S O H + H O
S O H
O
O
H O
H C C O H
H O
H O H
H
7.
H
H C O H
H
(e)
F
+
2CH 3OH l
CH 3OH 2 + CH 3O
H
(d)
+
F
O
+
O
S O H + H O
S O H
O
O
H
(E) The principle we will follow here is that, in terms of their concentrations, the weaker
acid and the weaker base will predominate at equilibrium. The reason for this is that a
strong acid will do a good job of donating its protons and, having done so, its conjugate
base will be left behind. The preferred direction is:
strong acid + strong base weak (conjugate) base + weak (conjugate) acid
(a)
The reaction will favor the forward direction because OH (a strong base) NH 3 (a
weak base) and NH 4 (relatively strong weak acid) H 2 O (very weak acid) .
(b)
The reaction will favor the reverse direction because HNO 3 HSO 4 (a weak acid
in the second ionization) (acting as acids), and SO 4
(c)
2
NO 3 (acting as bases).
The reaction will favor the reverse direction because HC2 H 3O 2 CH 3OH (not
usually thought of as an acid) (acting as acids), and CH 3O C 2 H 3O 2 (acting as
bases).
734
Chapter 16: Acids and Bases
8.
(M) The principle we follow here is that, in terms of their concentrations, the weaker acid
and the weaker base predominate at equilibrium. This is because a strong acid will do a
good job of donating its protons and, having done so, its conjugate base will remain in
solution. The preferred direction is:
strong acid + strong base weak (conjugate) base + weak (conjugate) acid
(a)
The reaction will favor the forward direction because HC 2 H 3O 2 (a moderate acid)
HCO 3 (a rather weak acid) (acting as acids) and CO 32 C 2 H 3O 2 (acting as
bases).
(b)
The reaction will favor the reverse direction because HClO 4 (a strong acid)
HNO 2 (acting as acids), and NO 2 ClO 4 (acting as bases).
(c)
The reaction will favor the forward direction, because H 2 CO3 HCO3 (acting as
acids) (because K1 K2 ) and CO 3
2
HCO 3 (acting as bases).
Strong Acids, Strong Bases, and pH
9.
(M) All of the solutes are strong acids or strong bases.
(a)
1 mol H 3O +
H 3O + = 0.00165 M HNO3
0.00165 M
1 mol HNO3
Kw
1.0 1014
OH =
=
= 6.11012 M
+
H 3O 0.00165M
(b)
1 mol OH
OH = 0.0087 M KOH
0.0087 M
1 mol KOH
1.0 1014
H 3 O =
=
= 1.1 1012 M
OH 0.0087M
+
(c)
Kw
2 mol OH
OH = 0.00213 M Sr OH 2
0.00426 M
1mol Sr OH
2
14
Kw
1.0 10
H 3 O + =
=
= 2.3 1012 M
OH 0.00426 M
(d)
1mol H 3O +
H 3O + = 5.8 104 M HI
5.8 104 M
1mol HI
K
1.0 1014
w
OH =
=
1.7 1011 M
4
+
H 3 O 5.8 10 M
735
Chapter 16: Acids and Bases
10.
(M) Again, all of the solutes are strong acids or strong bases.
+
(a) H O + = 0.0045 M HCl 1 mol H 3O = 0.0045 M
3
1 mol HCl
pH = log 0.0045 = 2.35
(b)
1 mol H 3O +
H 3O + = 6.14 104 M HNO3
= 6.14 104 M
1 mol HNO
pH = log 6.14 10
(c)
4
3
= 3.21
1 mol OH
OH = 0.00683 M NaOH
= 0.00683M
1 mol NaOH
pOH = log 0.00683 = 2.166 and pH = 14.000 pOH = 14.000 2.166 = 11.83
(d)
2 mol OH
OH = 4.8 103 M Ba OH 2
= 9.6 103 M
1 mol Ba OH
pOH = log 9.6 10
11.
(E)
OH =
3
= 2.02
2
pH = 14.00 2.02 = 11.98
3.9 g Ba OH 2 8H 2 O 1000 mL 1 mol Ba OH 2 8H 2 O
2 mol OH
100 mL soln
1L
315.5 g Ba OH 2 8H 2 O 1 mol Ba OH 2 8H 2 O
= 0.25 M
K
1.0 1014
w
H 3O + =
=
= 4.0 1014 M
OH 0.25 M OH
12.
pH = log 4.0 1014 = 13.40
b g
(M) The dissolved Ca OH 2 is completely dissociated into ions.
pOH = 14.00 pH = 14.00 12.35 = 1.65
OH = 10 pOH = 101.65 = 2.2 102 M OH = 0.022 M OH
0.022 mol OH 1 mol Ca OH 2 74.09 g Ca OH 2 1000 mg Ca OH 2
solubility =
1 L soln
2 mol OH
1 mol Ca OH 2
1 g Ca OH 2
=
8.1 102 mg Ca OH 2
1 L soln
8.1102 mg Ca OH 2
1L
In 100 mL the solubility is 100 mL
= 81 mg Ca OH 2
1000 mL
1 L soln
736
Chapter 16: Acids and Bases
13.
(M) First we determine the moles of HCl, and then its concentration.
1 atm
751 mmHg
0.205 L
760 mmHg
PV
moles HCl =
8.34 103 mol HCl
1
1
RT
0.08206 L atm mol K 296 K
H 3O + =
14.
8.34 103 mol HCl 1 mol H 3O +
= 1.96 103 M
4.25 L soln
1 mol HCl
(E) First determine the concentration of NaOH, then of OH .
0.125 L
OH =
0.606 mol NaOH 1 mol OH
1L
1 mol NaOH
= 0.00505 M
15.0 L final solution
pOH = log 0.00505 M = 2.297
15.
(E) First determine the amount of HCl, and then the volume of the concentrated solution
required.
102.10 mol H 3O + 1 mol HCl
amount HCl = 12.5 L
= 0.099 mol HCl
1 L soln
1 mol H 3O +
Vsolution = 0.099 mol HCl
16.
pH = 14.00 2.297 = 11.703
36.46 g HCl 100.0 g soln 1 mL soln
= 8.5 mL soln
1 mol HCl
36.0 g HCl 1.18 g soln
(E) First we determine the amount of KOH, and then the volume of the concentrated solution
required.
pOH = 14.00 pH = 14.00 11.55 = 2.45
amount KOH = 25.0 L
0.0035 M mol OH 1 mol KOH
= 0.088 mol KOH
1 L soln
1 mol OH
Vsolution = 0.088 mol KOH
17.
OH = 10 pOH = 102.45 = 0.0035 M
56.11 g KOH 100.0 g soln 1 mL soln
= 29 mL soln
1 mol KOH 15.0 g KOH 1.14 g soln
b g
(E) The volume of HCl(aq) needed is determined by first finding the amount of NH 3 aq
present, and then realizing that acid and base react in a 1:1 molar ratio.
VHCl = 1.25 L base
0.265 mol NH 3 1 mol H 3 O + 1 mol HCl
1 L acid
+
1 L base
1 mol NH 3 1 mol H 3 O 6.15 mol HCl
= 0.0539 L acid or 53.9 mL acid.
737
Chapter 16: Acids and Bases
18.
(M) NH 3 and HCl react in a 1:1 molar ratio. According to Avogadro’s hypothesis, equal
volumes of gases at the same temperature and pressure contain equal numbers of moles.
Thus, the volume of H2(g) at 762 mmHg and 21.0 C that is equivalent to 28.2 L of H2(g)
at 742 mmHg and 25.0 C will be equal to the volume of NH3(g) needed for stoichiometric
neutralization.
VNH3 g = 28.2 L HCl g
742 mmHg 273.2 + 21.0 K 1 L NH 3 g
= 27.1 L NH 3 g
762 mmHg 273.2 + 25.0 K 1 L HCl g
Alternatively, we can solve the ideal gas equation for the amount of each gas, and then, by
equating these two expressions, we can find the volume of NH 3 needed.
n{HCl} =
742 mmHg 28.2 L
R 298.2 K
n{NH 3}
762 mmHg V {NH 3}
R 294.2 K
742 mmHg 28.2 L 762 mmHg V {NH 3}
This yields V {NH 3} 27.1 L NH 3 (g)
R 298.2 K
R 294.2 K
19.
(M) Here we determine the amounts of H 3O + and OH and then the amount of the one
that is in excess. We express molar concentration in millimoles/milliliter, equivalent to
mol/L.
0.0155 mmol HI 1 mmol H 3O +
50.00 mL
= 0.775 mmol H 3O +
1 mL soln
1 mmol HI
75.00 mL
0.0106 mmol KOH 1 mmol OH
= 0.795 mmol OH
1 mL soln
1 mmol KOH
The net reaction is H 3O + aq + OH aq 2H 2 O(l) .
There is an excess of OH of ( 0.795 0.775 = ) 0.020 mmol OH .
Thus, this is a basic solution. The total solution volume is ( 50.00 + 75.00 = ) 125.00 mL.
0.020 mmol OH
OH =
= 1.6 104 M, pOH = log 1.6 104 = 3.80,
125.00 mL
20.
pH = 10.20
(M) In each case, we need to determine the H 3O + or OH of each solution being
mixed, and then the amount of H 3O + or OH , so that we can determine the amount in
excess. We express molar concentration in millimoles/milliliter, equivalent to mol/L.
H 3O + = 102.12 = 7.6 103 M
pOH = 14.00 12.65 = 1.35
moles H 3O + = 25.00 mL 7.6 103 M = 0.19 mmol H 3O +
OH = 101.35 = 4.5 10 2 M
amount OH = 25.00 mL 4.5 10 2 M = 1.13 mmol OH
738
Chapter 16: Acids and Bases
There is excess OH in the amount of 0.94 mmol (=1.13 mmol OH– – 0.19 mmol H3O+).
0.94 mmol OH
OH =
= 1.88 102 M
25.00 mL + 25.00 mL
pOH = log 1.88 102 = 1.73
pH=12.27
Weak Acids, Weak Bases, and pH
21.
(M) We organize the solution around the balanced chemical equation.
+
Equation: HNO2 (aq)
+
H 2 O(l)
+
NO 2 (aq)
H 3 O (aq)
—
Initial:
0.143 M
0 M
0M
—
Changes:
x M
+x M
+x M
Equil:
—
x
M
xM
0.143
x
M
H 3 O + NO 2
x2
x2
4
=
7.2×10
(if x 0.143)
Ka
HNO 2
0.143 x
0.143
x = 0.143 7.2 104 0.010 M
We have assumed that x 0.143 M , an almost acceptable assumption. Another cycle of
approximations yields:
x = (0.143 0.010) 7.2 104 0.0098 M [H 3 O + ]
pH log(0.0098) 2.01
This is the same result as is determined with the quadratic formula roots equation.
22.
(M) We organize the solution around the balanced chemical equation, and solve first for
OH
C2 H 5 NH 3+ (aq)
Initial :
0.085 M
0M
0M
xM
+xM
+xM
xM
xM
Equil :
0.085 x M
H 2 O(l)
OH (aq) +
C 2 H 5 NH 2 (aq)
Changes :
+
Equation :
OH C2 H 5 NH 3+
x2
x2
=
4
Kb =
=
4.3
10
assuming x 0.085
C H NH 2
0.085 x
0.085
2 5
x = 0.085 4.3 104 0.0060 M
We have assumed that x 0.085 M, an almost acceptable assumption. Another cycle of
approximations yields:
x = (0.085 0.0060) 4.3 104 0.0058 M = [OH ]
739
Yet another cycle produces:
Chapter 16: Acids and Bases
x (0.085 0.0058) 4.3 10 4 0.0058 M [OH ]
pH = 14.00 2.24 = 11.76
pOH log(0.0058) 2.24
H 3O + = 10 pH = 1011.76 = 1.7 1012 M
This is the same result as determined with the quadratic formula roots equation.
23.
(M)
(a) The set-up is based on the balanced chemical equation.
Equation: HC8 H 7 O 2 (aq) + H 2 O(l)
H 3 O + (aq) + C8 H 7 O 2 (aq)
—
Initial:
0.186 M
0 M
0M
—
Changes:
x M
+x M
+x M
Equil:
—
xM
xM
0.186 x M
K a 4.9 10
5
[H 3 O ][C8 H 7 O 2 ]
xx
x2
[HC8 H 7 O 2 ]
0.186 x 0.186
x = 0186
.
4.9 105 0.0030 M = [H 3O + ] [C8 H 7 O 2 ]
0.0030 M is less than 5 % OF 0.186 M, thus, the approximation is valid.
(b)
The set-up is based on the balanced chemical equation.
Equation:
HC8 H 7 O 2 (aq)
Initial:
Changes:
Equil:
0.121 M
x M
0.121 x M
+
H 2 O(l)
—
—
—
H 3 O + (aq)
0 M
+x M
xM
H 3O + C8 H 7 O 2
x2
x2
=
K a = 4.9 10 =
0.121 x 0.121
HC8 H 7 O 2
5
b
g
(E)
0.275 mol 1000 mL
= 0.440 M
625 mL 1 L
HC3H5O2 initial =
HC3H5O2 equilibrium = 0.440 M 0.00239 M = 0.438 M
Ka =
H 3O + C 3 H 5O 2
LM HC3 H 5O 2 OP
N
Q
b0.00239g
=
0.438
2
= 1.30 105
740
C8 H 7 O 2 (aq)
0M
+x M
xM
x = 0.0024 M = H 3O +
Assumption x 0.121, is correct. pH = log 0.0024 = 2.62
24.
Chapter 16: Acids and Bases
25.
(M) We base our solution on the balanced chemical equation.
H 3O + = 101.56 = 2.8 102 M
CH 2 FCOO aq + H 3 O + aq
CH 2 FCOOH aq + H 2 O(l)
Equation :
Initial :
Changes :
Equil :
Ka =
26.
0.318 M
0.028 M
0.290 M
H 3O + CH 2 FCOO
CH 2 FCOOH
(M) HC 6 H 11O 2 =
=
0M
+ 0.028 M
0.028 M
0M
+ 0.028 M
0.028 M
b0.028gb0.028g = 2.7 10
3
0.290
11 g 1 mol HC 6 H 11O 2
= 9.5 102 M = 0.095 M
1 L 116.2 g HC 6 H 11O 2
H 3 O +
= 102.94 = 1.1 103 M = 0.0011 M
equil
The stoichiometry of the reaction, written below, indicates that H 3O + = C 6 H 11O 2
C6 H11O 2 (aq)
HC 6 H11O 2 (aq) + H 2 O(l)
Equation :
Initial :
Changes :
27.
H 3O + (aq)
0.095 M
0.0011 M
0M
+0.0011 M
0 M
+ 0.0011 M
0.094 M
0.0011 M
0.0011 M
Equil :
Ka =
+
C 6 H 11O 2
H 3O +
LM HC6 H11O 2 OP
Q
N
b0.0011g
=
2
0.094
= 1.3 10 5
(M) Here we need to find the molarity S of the acid needed that yields H 3 O + = 102.85 = 1.4 103 M
Equation:
HC7 H 5O 2 (aq) + H 2 O(l)
Initial:
Changes:
Equil:
S
0.0014 M
S 0.0014M
Ka =
H 3O + C 7 H 5O 2
LM HC7 H 5O 2 OP
Q
N
—
—
—
b0.0014g
=
H 3O + (aq)
0M
+0.0014M
0.0014 M
2
S 0.0014
= 6.3 10
5
+
C7 H 5O 2 (aq)
0M
+0.0014M
0.0014 M
b0.0014g
S 0.0014 =
2
6.3 105
= 0.031
S = 0.031+ 0.0014 = 0.032 M = HC7 H 5O 2
350.0 mL
0.032 mol HC7 H 5O 2 122.1 g HC7 H 5O 2
1L
= 1.4 g HC7 H 5O 2
1000 mL
1 L soln
1 mol HC7 H 5O 2
741
Chapter 16: Acids and Bases
28.
(M) First we find OH and then use the Kb expression to find (CH 3 )3 N
OH = 10 pOH = 102.88 = 0.0013 M
pOH = 14.00 pH = 14.00 11.12 = 2.88
K b = 6.3 10
5
2
(CH 3 )3 NH + OH
0.0013
=
=
(CH 3 )3 N equil
(CH 3 )3 N equil
(CH3 )3 N equil = 0.027 M (CH 3 )3 N
CH 3 3 N
= 0.027 M(equil. concentration) + 0.0013 M(concentration ionized)
initial
CH 3 3 N
= 0.028 M
initial
29.
(M) We use the balanced chemical equation, then solve using the quadratic formula.
+ ClO 2 (aq)
Initial :
0.55 M
0M
0M
x M
+x M
+x M
xM
xM
Equil :
0.55 x M
H 2 O(l)
H3O+ (aq)
HClO 2 (aq)
Changes :
+
Equation :
2
H 3O + ClO 2
= x
Ka =
= 1.1102 = 0.011
x
HClO
0.55
2
x 2 = 0.0061 0.011x
x 2 + 0.011x 0.0061 = 0
b b 2 4ac 0.011 0.000121+ 0.0244
0.073 M H 3O +
2a
2
The method of successive approximations converges to the same answer in four cycles.
pH = log H 3O + = log 0.073 = 1.14
pOH = 14.00 pH = 14.00 1.14 = 12.86
x
b
30.
g
OH = 10 pOH = 1012.86 = 1.4 1013 M
(M) Organize the solution around the balanced chemical equation, and solve first for
OH .
Equation :
CH 3 NH 2 (aq)
Initial :
0.386 M
Changes :
x M
Equil :
0.386 x
M
+
OH (aq)
H 2O(l)
0M
+ CH 3 NH 3+ (aq)
0M
+x M
+x M
xM
xM
OH CH 3 NH 3+
x2
x2
=
4
Kb =
=
4.2
10
0.386 x
0.386
CH 3 NH 2
x = 0.386 4.2 104 0.013 M = [OH - ]
742
assuming x 0.386
pOH = log(0.013) = 1.89
Chapter 16: Acids and Bases
H 3O + = 10 pH = 1012.11 = 7.8 1013 M
pH = 14.00 1.89 = 12.11
This is the same result as is determined with the quadratic equation roots formula.
31.
(M)
C10 H 7 NH 2 =
1 g C10 H 7 NH 2 1.00 g H 2 O 1000 mL 1 mol C10 H 7 NH 2
590 g H 2O
1 mL
1L
143.2 g C10 H 7 NH 2
= 0.012 M C10 H 7 NH 2
K b = 10 pKb = 103.92 = 1.2 104
Equation:
C10 H 7 NH 2 (aq) + H 2 O(l)
Initial:
Changes:
Equil:
0.012 M
x M
0.012 x M
—
—
—
OH (aq) +
0 M
+x M
xM
OH C10 H 7 NH 3
x2
x2
=
4
Kb =
=
1.2
10
0.012 x
0.012
C10 H 7 NH 2
C10 H 7 NH 3 (aq)
0M
+x M
xM
assuming x 0.012
x = 0.012 12
. 104 0.0012 This is an almost acceptable assumption. Another
approximation cycle gives:
x = (0.012 0.0012) 12
. 104 0.0011
Yet another cycle seems necessary.
x = (0.012 0.0011) 1.2 104 0.0011M [OH ]
The quadratic equation roots formula provides the same answer.
b
g
pOH = log OH = log 0.0011 = 2.96
pH = 14.00 pOH = 11.04
H 3 O 10 pH 1011.04 9.1 1012 M
32.
(M) The stoichiometry of the ionization reaction indicates that
H 3O + = OC6 H 4 NO 2 , K a = 10 pKb = 107.23 = 5.9 108 and
H 3O + = 104.53 = 3.0 105 M . We let S = the molar solubility of o -nitrophenol.
Equation:
HOC6 H 4 NO 2 (aq) + H 2 O(l)
Initial:
Changes:
Equil:
S
3.0 105 M
( S 3.0 105 ) M
—
—
—
743
OC6 H 4 NO 2 (aq) +
0M
+3.0 105 M
3.0 105 M
H 3O + (aq)
0 M
+3.0 105 M
3.0 105 M
Chapter 16: Acids and Bases
Ka = 5.9 108 =
OC 6 H 4 NO 2 H 3O +
LM HOC 6 H 5 NO 2 OP
Q
N
c3.0 10 h
=
c3.0 10 h
=
5 2
S 3.0 105
5 2
S 3.0 10
5
5.9 10
8
= 1.5 102 M
S = 1.5 102 M + 3.0 105 M = 1.5 102 M
Hence,
solubility =
33.
1.5 102 mol HOC 6 H 4 NO 2 139.1 g HOC 6 H 4 NO 2
= 2.1 g HOC 6 H 4 NO 2 / L soln
1 L soln
1 mol HOC 6 H 4 NO 2
(M) Here we determine H 3O + which, because of the stoichiometry of the reaction,
equals C2 H 3O 2 . H 3O + = 10 pH = 104.52 = 3.0 105 M = C2 H 3O 2
We solve for S , the concentration of HC 2 H 3O 2 in the 0.750 L solution before it dissociates.
Equation:
HC2 H 3 O 2 (aq) +
Initial:
Changes:
Equil:
SM
3.0 105 M
S 3.0 105 M
H 2 O(l)
—
—
—
C2 H 3 O 2 (aq)
0M
+3.0 10 5 M
3.0 105 M
+
H 3 O + (aq)
0M
+3.0 105 M
3.0 105 M
[H 3O ][C2 H 3O 2 ]
(3.0 105 ) 2
5
1.8 10
Ka
[HC2 H 3O 2 ]
( S 3.0 105 )
c3.0 10 h
5 2
S=
c
h
= 1.8 105 S 3.0 105 = 9.0 1010 = 1.8 105 S 5.4 1010
9.0 1010 + 5.4 1010
= 8.0 105 M Now we determine the mass of vinegar needed.
1.8 105
mass vinegar = 0.750 L
8.0 105 mol HC2 H 3O 2 60.05 g HC 2 H 3O 2 100.0 g vinegar
1 L soln
1 mol HC2 H 3O 2 5.7 g HC2 H 3O 2
= 0.063 g vinegar
34.
(M) First we determine OH which, because of the stoichiometry of the reaction,
equals NH 4 .
pOH = 14.00 pH = 14.00 11.55 = 2.45
OH = 10 pOH = 102.45 = 3.5 103 M = NH 4
We solve for S , the concentration of NH 3 in the 0.625 L solution prior to dissociation.
744
Chapter 16: Acids and Bases
Equation:
NH 3 (aq ) + H 2 O(l)
Initial:
Changes:
Equil:
SM
0.0035 M
S 0.0035 M
Kb =
NH 4
OH
LM NH 3 OP
N
Q
0.0035
S=
2
NH 4 (aq)
0M
+0.0035 M
0.0035 M
—
—
—
5
= 1.8 10 =
b0.0035g
+
OH (aq)
0 M
+0.0035 M
0.0035 M
2
bS 0.0035g
= 1.8 105 S 0.0035 = 1.225 105 = 1.8 105 S 6.3 108
1.225 105 + 6.3 108
= 0.68 M
1.8 105
Now we determine the volume of household ammonia needed
Vammonia = 0.625 L soln
0.68 mol NH 3 17.03 g NH 3 100.0 g soln 1 mL soln
1 L soln
1 mol NH 3
6.8 g NH 3 0.97 g soln
= 1.1 10 2 mL household ammonia solution
35.
(D)
1 atm
316 Torr
0.275 L
PV
760 Torr
= 4.67 103 mol propylamine
=
(a) nproplyamine =
-1
-1
RT 0.08206 L atm K mol 298.15 K
[propylamine] =
n 4.67 10-3 moles
= 9.35 103 M
=
V
0.500 L
Kb = 10pKb = 103.43 = 3.7 104
CH3CH2CH2NH2(aq) + H2O(l)
—
3
Initial
9.35 10 M
—
Change
x M
—
Equil.
(9.35 103 x) M
Kb = 3.7 104 =
CH3CH2CH2NH3+(aq) + OH (aq)
0M
0M
+x M
+x M
xM
xM
x2
or 3.5 106 3. 7 104(x) = x2
9.35 10-3 x
x2 + 3.7 104x 3.5 106 = 0
745
Chapter 16: Acids and Bases
Find x with the roots formula:
x
3.7 10-4
(3.7 10-4 ) 2 4(1)( 3.5 10-6 )
2(1)
Therefore x = 1.695 103 M = [OH ] pOH = 2.77 and pH = 11.23
(b)
[OH] = 1.7 103 M = [NaOH]
(MMNaOH = 39.997 g mol1)
nNaOH = (C)(V) = 1.7 103 M 0.500 L = 8.5 104 moles NaOH
mass of NaOH = (n)(MMNaOH) = 8.5 104 mol NaOH 39.997 g NaOH/mol NaOH
mass of NaOH = 0.034 g NaOH(34 mg of NaOH)
36.
(M) Kb = 10pKb = 109.5 = 3.2 1010
Solubility(25 C) =
MMquinoline = 129.16 g mol1
0.6 g
100 mL
Molar solubility(25 C) =
0.6 g
1 mol
1000 mL
= 0.046 M (note: only 1 sig fig)
100 mL 129.16 g
1L
+ H2O(l) C9H7NH+(aq) + OH(aq)
—
0M
Initial
0.046 M
0M
—
Change
+
x
M
+x M
x M
—
xM
xM
Equil.
(0.046 x) M
C9H7N(aq)
3.2 1010 =
x2
x2
,
0.046 x
0.046
Therefore, [OH ] = 4 106 M
x = 3.8 106 M (x << 0.046, valid assumption)
pOH = 5.4 and pH = 8.6
37.
(M) If the molarity of acetic acid is doubled, we expect a lower initial pH (more H3O+(aq)
in solution) and a lower percent ionization as a result of the increase in concentration. The
ratio between [H3O+] of concentration “M” and concentration 2 M is 2 1.4 . Therefore,
(b), containing 14 H3O+ symbols best represents the conditions (~(2)1/2 times greater).
38.
(M) If NH3 is diluted to half its original molarity, we expect a lower pH (a lower [OH]) and a
higher percent ionization in the diluted sample. The ratio between [OH–] of concentration “M”
and concentration 0.5 M is 0.5 0.71 . Since the diagram represent M has 24 OH– symbols,
then diagram (c), containing 17 OH– symbols would be the correct choice.
746
Chapter 16: Acids and Bases
Percent Ionization
39.
(M) Let us first compute the H 3O + in this solution.
H O + (aq)
Equation: HC3 H 5 O 2 (aq) + H 2 O(l)
3
—
Initial:
0.45 M
0 M
—
Changes: x M
+x M
Equil:
—
xM
0.45 x M
+ C3 H 5 O 2 (aq)
0 M
+x M
xM
H 3 O + C3 H 5 O 2
x2
x2
4.89
5
Ka =
=
= 10
= 1.3 10
HC3 H 5 O 2
0.45 x
0.45
x = 2.4 103 M ; We have assumed that x 0.45 M , an assumption that clearly is correct.
40.
(a)
H 3O +
2.4 103 M
equil
=
=
= 0.0053 = degree of ionization
HC3 H 5O 2 initial
0.45 M
(b)
% ionization = 100% = 0.0053 100% 0.53%
(M) For C2H5NH2 (ethylamine), Kb = 4.3 104
C2H5NH2(aq) + H2O(l)
Initial
Change
Equil.
0.85 M
x M
(0.85 x) M
4.3 104 =
C2H5NH3+(aq) + OH(aq)
—
0M
+x M
x M
—
—
x2
x2
(0.85 x) M
0.85 M
Degree of ionization =
41.
0M
+x M
x M
x = 0.019 M (x << 0.85, thus a valid assumption)
0.019 M
= 0.022
0.85 M
Percent ionization = 100% = 2.2 %
(M) Let x be the initial concentration of NH3, hence, the amount dissociated is 0.042 x
NH3(aq)
Initial
xM
Change
0.042x M
Equil.
(1x0.042x) M
= 0.958 x
+ H2O(l)
Ka = 1.8 10
-5
—
NH4+(aq)
0M
+0.042x M
0.042x M
—
—
747
+
OH-(aq)
0M
+0.042x M
0.042x M
Chapter 16: Acids and Bases
NH 4 OH 0.042 x 2
K b = 1.8 105 =
=
= 0.00184x
[0.958 x]
NH 3 equil
NH3 initial = x
42.
1.8 105
0.00978 M 0.0098 M
0.00184
(M)
HC3H5O2(aq) + H2O(l)
Initial
0.100 M
Change
x M
Equil.
(0.100 x) M
Ka = 1.3 10
-5
—
+
C3H5O2(aq) + H3O (aq)
0M
+x M
xM
—
—
0M
+x M
xM
x2
x2
1.3 10 =
x = 1.1 103 (x << 0.100, thus a valid assumption)
0.100 M x
0.100
0.0011
100% = 1.14 %
Percent ionization =
0.100
Next we need to find molarity of acetic acid that is 1.1% ionized
5
HC2H3O2(aq)
Initial
Change
Equil.
XM
1.14
XM
100
1.14
X
X
100
+ H2O(l)
Ka =1.8 10
-5
—
—
—
C2H3O2(aq) +
H3O+(aq)
0M
0M
1.14
XM
+
100
1.14
XM
100
1.14
+
XM
100
1.14
XM
100
= 0.9886 X M
= 0.0114 M
= 0.0114 M
2
1.14
X
= 1.3 104(X);
5 100
1.8 10 =
X = 0.138 M
0.9886 X
Consequently, approximately 0.14 moles of acetic acid must be dissolved in one liter of
water in order to have the same percent ionization as 0.100 M propionic acid.
43.
(E) We would not expect these ionizations to be correct because the calculated degree of
ionization is based on the assumption that the [HC2H3O2]initial [HC2H3O2]initial
[HC2H3O2]equil., which is invalid at the 13 and 42 percent levels of ionization seen here.
44.
(M) HC2Cl3O2 (trichloroacetic acid) pKa = 0.52
Ka = 100.52 = 0.30
For a 0.035 M solution, the assumption will not work (the concentration is too small and
Ka is too large) Thus, we must solve the problem using the quadratic equation.
HC2Cl3O2(aq) + H2O(l)
—
Initial
0.035 M
—
Change
x M
—
Equil.
(0.035 x) M
+
C2Cl3O2(aq) + H3O (aq)
0M
0M
+x M
+x M
xM
xM
748
Chapter 16: Acids and Bases
0.30 =
x
x2
or 0.0105 0.30(x) = x2 or x2 + 0.30x 0.0105 = 0 ; solve quadratic:
0.035 x
-0.30
(0.30) 2 4(1)(0.0105)
2(1)
Degree of ionization =
Therefore x = 0.032 M = [H3O+]
0.032 M
= 0.91
0.035 M
Percent ionization = 100% = 91. %
Polyprotic Acids
45.
(E) Because H3PO4 is a weak acid, there is little HPO42- (produced in the 2nd ionization)
compared to the H3O+ (produced in the 1st ionization). In turn, there is little PO43- (produced
in the 3rd ionization) compared to the HPO42-, and very little compared to the H3O+.
46.
(E) The main estimate involves assuming that the mass percents can be expressed as 0.057
g of 75% H 3 PO 4 per 100. mL of solution and 0.084 g of 75% H 3 PO 4 per 100. mL of
solution. That is, that the density of the aqueous solution is essentially 1.00 g/mL. Based
on this assumption, the initial minimum and maximum concentrations of H 3 PO 4 is:
75g H 3 PO 4
1mol H 3 PO 4
100 g impure H3 PO4 98.00 g H3 PO 4
0.0044 M
1L
100 mL soln
1000 mL
0.057 g impure H 3 PO 4
[H 3 PO 4 ]
75g H 3 PO 4
1mol H 3 PO 4
100 g impure H 3 PO 4 98.00 g H 3 PO 4
0.0064 M
1L
100 mL soln
1000 mL
0.084 g impure H 3 PO 4
[H 3 PO 4 ]
Equation:
H 3 PO 4 (aq)
+
Initial:
Changes:
Equil:
0.0044 M
x M
( 0.0044 x ) M
H 2 O(l)
—
—
—
H 2 PO 4 H 3O +
x2
=
= 7.1103
K a1 =
x
H
PO
0.0044
4
3
H 2 PO 4 (aq)
0M
+x M
x M
+
H 3 O + (aq)
0M
+x M
x M
x 2 + 0.0071x 3.1 105 = 0
b b 2 4ac 0.0071 5.0 105 +1.2 104
x=
=
= 3.0 103 M = H 3O +
2a
2
749
Chapter 16: Acids and Bases
The set-up for the second concentration is the same as for the first, with the exception that
0.0044 M is replaced by 0.0064 M.
H 2 PO 4 H 3O +
x2
K a1 =
x 2 + 0.0071x 4.5 105 = 0
=
= 7.1103
x
H
PO
0.0064
4
3
x=
b b 2 4ac 0.0071 5.0 105 +1.8 104
=
= 4.0 103 M = H 3O +
2a
2
The two values of pH now are determined, representing the pH range in a cola drink.
pH = log 3.0 103 = 2.52
47.
(D)
(a)
Equation: H 2S(aq)
pH = log 4. 103 = 2.40
+
Initial:
0.075 M
Changes: x M
Equil:
( 0.075 x ) M
H 2O(l)
—
HS (aq)
0M
+x M
x M
—
—
HS H 3O +
x2
x2
K a1 =
= 1.0 107 =
0.075 x 0.075
H 2S
+
H3O + (aq)
0M
+x M
+x M
x = 8.7 105 M = H 3O +
5
19
2
HS = 8.7 10 M and S = K a 2 = 110 M
(b)
The set-up for this problem is the same as for part (a), with 0.0050 M replacing 0.075
M as the initial value of H 2S .
HS H 3O +
x2
x2
K a1 =
x = 2.2 105 M = H 3O +
= 1.0 107 =
x
H
S
0.0050
0.0050
2
5
2
19
HS = 2.2 10 M and S = K a 2 = 1 10 M
(c)
The set-up for this part is the same as for part (a), with 1.0 105 M replacing 0.075
M as the initial value of H 2S . The solution differs in that we cannot
assume x 1.0 105 . Solve the quadratic equation to find the desired equilibrium
concentrations.
HS H 3 O +
K a1 =
= 1.0 107 =
H S
2
x2
1.0 10 5 x
x 2 1.0 107 x 1.0 1012 0
750
Chapter 16: Acids and Bases
1.0 107
1.0 1014
x=
2 1
x = 9.5 107 M = H 3O +
48.
4 1.0 10
12
HS = 9.5 107 M
S2 = K a 2 = 1 1019 M
(M) For H 2 CO 3 , K1 = 4.4 107 and K2 = 4.7 1011
(a) The first acid ionization proceeds to a far greater extent than does the second and
determines the value of H 3O + .
Equation:
H 2 CO3 (aq) +
Initial:
Changes:
Equil:
0.045 M
x M
0.045 x M
K1 =
b
H 3O + HCO 3
LM H 2 CO 3 OP
Q
N
H 2 O(l)
—
—
g
—
H 3O + (aq)
0 M
+x M
x M
x2
x2
7
=
= 4.4 10
0.045 x
0.045
+
HCO3 (aq)
0M
+x M
x M
x = 1.4 104 M = H 3O +
(b)
Since the second ionization occurs to only a limited extent,
HCO3 = 1.4 104 M = 0.00014 M
(c)
We use the second ionization to determine CO 32 .
2
Equation: HCO3 (aq)
+
H 2 O(l)
H 3 O + (aq)
CO3 (aq)
—
0.00014 M
0.00014 M
0M
—
+x M
x M
+x M
—
0.00014 + x M x M
0.00014 x M
Initial:
Changes:
Equil:
H 3O + CO32 0.00014 + x x
0.00014 x
=
K2 =
= 4.7 1011
0.00014 x
0.00014
HCO3
x = 4.7 1011 M = CO 32
Again we assumed that x 0.00014 M, which clearly is the case. We also note that
our original assumption, that the second ionization is much less significant than the
first, also is a valid one. As an alternative to all of this, we could have recognized
that the concentration of the divalent anion equals the second ionization constant:
CO 32 = K2 .
751
Chapter 16: Acids and Bases
49.
(D) In all cases, of course, the first ionization of H 2SO 4 is complete, and establishes the
initial values of H 3O + and HSO 4 . Thus, we need only deal with the second ionization
in each case.
(a) Equation:
HSO 4 (aq) +
H 2 O(l)
SO 4 2 (aq)
—
Initial:
0M
0.75 M
—
Changes: x M
+x M
—
Equil:
x M
( 0.75 x ) M
SO 4 2 H 3O +
x 0.75 + x 0.75 x
Ka2 =
= 0.011 =
0.75 x
0.75
HSO 4
+
H 3O + (aq)
0.75 M
+x M
( 0.75 + x ) M
2
x = 0.011 M = SO 4
We have assumed that x 0.75 M, an assumption that clearly is correct.
HSO 4 = 0.75 0.011 = 0.74 M
H 3O + = 0.75 + 0.011 = 0.76 M
(b)
The set-up for this part is similar to part (a), with the exception that 0.75 M is
replaced by 0.075 M.
SO 4 2 H 3O +
x 0.075 + x
Ka2 =
=
0.011
=
0.075 x
HSO 4
0.011 0.075 x = 0.075 x + x 2
x 2 + 0.086 x 8.3 104 = 0
b b 2 4ac 0.086 0.0074 + 0.0033
=
= 0.0087 M
2a
2
2
x = 0.0087 M = SO 4
x=
HSO 4 = 0.075 0.0087 = 0.066M
(c)
H 3O + = 0.075 + 0.0088M = 0.084 M
Again, the set-up is the same as for part (a), with the exception that 0.75 M is
replaced by 0.00075 M
2
Ka2
x=
x(0.00075 x)
[SO 4 ][H 3O ]
0.011
0.00075 x
[HSO 4 ]
0.011(0.00075 x) 0.00075 x x 2
x 2 0.0118 x 8.3 106 0
b b 2 4ac 0.0118 1.39 104 + 3.3 105
= 6.6 104
=
2
2a
2
x = 6.6 104 M = SO 4
HSO 4 = 0.00075 0.00066 = 9 105 M
H 3O + = 0.00075 + 0.00066M = 1.41 103 M
H 3O + is almost twice the initial
value of H 2SO 4 . Thus, the second ionization of H 2SO 4 is nearly complete in this
dilute solution.
752
Chapter 16: Acids and Bases
50.
(D) First we determine H 3O + , with the calculation based on the balanced chemical
equation.
+
Equation: HOOC CH 2 4 COOH aq + H 2 O(l)
HOOC CH 2 4 COO aq + H 3 O aq
Initial:
Changes:
Equil:
Ka1 =
—
0.10 M
x M
0.10 x M
b
H 3O +
g
HOOCbCH g COO
—
2 4
HOOCFH CH 2 IK 4 COOH
= 3.9 105 =
0M
+x M
x M
0M
+x M
x M
—
xx
x2
0.10 x 0.10
x 010
. 3.9 105 2.0 103 M = [H 3O + ]
We see that our simplifying assumption, that x 0.10 M, is indeed valid. Now we
consider the second ionization. We shall see that very little H 3O + is produced in this
b g
ionization because of the small size of two numbers: Ka 2 and HOOC CH 2 4 COO .
Again we base our calculation on the balanced chemical equation.
+
Equation: HOOC CH 2 4 COO aq + H 2 O(l)
OOC CH 2 4 COO aq + H 3O aq
Initial:
Changes:
Equil:
Ka 2 =
b
H 3O +
2.0 103 M
y M
0.0020 y M
g
OOCbCH g COO
2 4
HOOCFH CH 2 IK 4 COO
0M
+y M
y M
—
—
—
= 3.9 106 =
b
b
2.0 10 3 M
+y M
0.0020 + y M
g
g
y 0.0020 + y
0.0020 y
0.0020 y
0.0020
y 3.9 106 M = [ OOC(CH 2 ) 4 COO ]
Again, we see that our assumption, that y 0.0020 M, is valid. In addition, we also note
that virtually no H 3O + is created in this second ionization. The concentrations of all
species have been calculated above, with the exception of OH .
Kw
1.0 1014
OH =
=
= 5.0 1012 M;
H 3O + 2.0 103
HOOC CH 2 4 COOH = 0.10 M
H 3O + = HOOC CH 2 4 COO = 2.0 103 M;
OOC CH 2 4 COO = 3.9 106 M
753
Chapter 16: Acids and Bases
51.
(M)
(a) Recall that a base is a proton acceptor, in this case, accepting H+ from H 2 O .
C 20 H 24 O 2 N 2 H + + OH
First ionization : C 20 H 24O 2 N 2 + H 2O
pK b1 = 6.0
C 20 H 24O 2 N 2 H 2 + OH pK b = 9.8
Second ionization : C 20 H 24O 2 N 2 H + + H 2O
2
2+
(b)
1mol quinine
324.4 g quinine
1.62 103 M
1L
1900 mL
1000 mL
1.00 g quinine
[C20 H 24 O 2 N 2 ]
K b1 106.0 1 106
Because the OH- produced in (a) suppresses the reaction in (b) and since Kb1 >> Kb2,
the solution’s pH is determined almost entirely by the first base hydrolysis of the
first reaction. Once again, we set-up the I.C.E. table and solve for the [OH ] in this
case:
C20 H 24 O 2 N 2 H + (aq) + OH (aq)
Equation: C20 H 24 O 2 N 2 (aq) + H 2 O(l)
Initial:
Changes:
Equil:
—
0.00162 M
x M
0.00162 x M
b
0M
+x M
x M
—
g
—
C20 H 24 O 2 N 2 H + OH
x2
x2
6
K b1 =
=
1
10
=
0.00162 x 0.00162
C 20 H 24 O 2 N 2
The assumption x 0.00162 , is valid. pOH = log 4 105 = 4.4
0 M
+x M
x M
x 4 105 M
pH = 9.6
52. (M) Hydrazine is made up of two NH2 units held together by a nitrogen–nitrogen single
bond:
H
H
N
N
H
H
(a)
N2H5 (aq) + OH-(aq)
N2H4(aq) + H2O(l)
(b)
N2H62+(aq) + OH-(aq)
N2H5+(aq) + H2O(l)
Kb2 = 10-15.05 = 8.9 ×10-16
Because the OH produced in (a) suppresses the reaction in (b) and since Kb1 >> Kb2,
the solution’s pH is determined almost entirely by the first base hydrolysis reaction (a).
Thus we need only solve the I.C.E. table for the hydrolysis of N2H4 in order to find the
[OH-]equil, which will ultimately provide us with the pH via the relationship
[OH-]×[H3O+] = 1.00 × 10-14.
+
754
Kb1 = 10-6.07 = 8.5 ×10-7
Chapter 16: Acids and Bases
Reaction:
N2H4(aq) +
0.245 M
x M
(0.245 x ) M
Initial:
Changes:
Equil:
H2O(l)
—
—
—
N 2 H 5 + OH
x2
x2
7
=
8.5
10
=
=
K b1 =
0.245 x 0.245
N 2 H 4
1.0 1014
Thus, [H 3O ] =
= 2.19 1011 M
4
4.56 10
+
53.
N2H5+(aq) +
OH-(aq)
0M
+x M
x M
0 M
+x M
x M
x = 4.56 104 M = [OH - ]equil
pH = -log(2.19 1011 ) = 10.66
(E) Protonated codeine hydrolyzes water according to the following reaction:
C18 H 21O3 NH H 2 O C18 H 21O3 N H 3O
pK a 6.05
pK b 14 pK a 7.95
54. (E) Protonated quinoline is, of course, an acid, which hydrolyzes water to give a
hydronium ion. The reaction is as follows:
C9 H 7 NH H 2 O C9 H 7 N H 3O
Kb
KW
1 1014
1.6 105
K a 6.3 1010
pK b log 1.6 105 4.8
55.
+
(E) The species that hydrolyze are the cations of weak bases, namely, NH 4 and
+
C 6 H 5 NH 3 , and the anions of weak acids, namely, NO 2 and C 7 H 5O 2 .
(a)
NH 3 aq + NO3 aq + H 3O + aq
NH 4 aq + NO3 aq + H 2O(l)
(b)
Na + aq + HNO 2 aq + OH aq
Na + aq + NO 2 aq + H 2O(l)
(c)
K + aq + HC7 H 5O 2 aq + OH aq
K + aq + C7 H 5O 2 aq + H 2 O(l)
(d)
K + aq + Cl aq + Na + aq + I aq + H 2 O(l) no reaction
(e)
+
C6 H 5 NH 2 aq + Cl aq + H 3O + aq
C6 H 5 NH 3 aq + Cl aq + H 2 O(l)
+
755
Chapter 16: Acids and Bases
56.
57.
(E) Recall that, for a conjugate weak acid–weak base pair, Ka Kb = Kw
K w 1.0 1014
=
= 6.7 106 for C5 H 5 NH +
9
K b 1.5 10
(a)
Ka =
(b)
K w 1.0 1014
Kb =
=
5.6 1011 for CHO 2
4
K a 1.8 10
(c)
Kb =
(E)
(a)
K w 1.0 1014
=
1.0 104 for C6 H 5O
10
K a 1.0 10
Because it is composed of the cation of a strong base, and the anion of a strong acid,
KCl forms a neutral solution (i.e., no hydrolysis occurs).
(b) KF forms an alkaline (basic) solution. The fluoride ion, being the conjugate base of
a relatively strong weak acid, undergoes hydrolysis, while potassium ion, the weakly
polarizing cation of a strong base, does not react with water.
HF(aq) + OH (aq)
F (aq) + H 2 O(l)
(c)
NaNO 3 forms a neutral solution, being composed of a weakly polarizing cation and
anionic conjugate base of a strong acid. No hydrolysis occurs.
(d)
Ca OCl
b g
2
forms an alkaline (basic) solution, being composed of a weakly polarizing
cation and the anion of a weak acid Thus, only the hypochlorite ion hydrolyzes.
HOCl(aq) + OH (aq)
OCl (aq) + H 2 O(l)
(e)
NH 4 NO 2 forms an acidic solution. The salt is composed of the cation of a weak base
and the anion of a weak acid. The ammonium ion hydrolyzes:
+
NH 4 (aq) + H 2O(l)
NH 3 aq + H 3O (aq)
+
K a = 5.6 1010
as does the nitrite ion:
NO 2 (aq) + H 2 O(l)
HNO 2 (aq) + OH (aq)
K b = 1.4 1011
+
Since NH 4 is a stronger acid than NO 2 is a base (as shown by the relative sizes of
the K values), the solution will be acidic. The ionization constants were computed
from data in Table 16-3 and with the relationship Kw = Ka Kb .
For NH 4 + , Ka =
1.0 1014
1.0 1014
10
= 1.4 1011
=
5.6
10
,
=
For
NO
K
b
2
5
4
7.2 10
1.8 10
Where 1.8 105 = K b of NH3 and 7.2 104 = K a of HNO2
756
Chapter 16: Acids and Bases
58.
(E) Our list in order of increasing pH is also in order of decreasing acidity. First we look
for the strong acids; there is just HNO 3 . Next we look for the weak acids; there is only
one, HC 2 H 3O 2 . Next in order of decreasing acidity come salts with cations from weak
bases and anions from strong acids; NH 4 ClO 4 is in this category. Then come salts in
which both ions hydrolyze to the same degree; NH 4 C 2 H 3O 2 is an example, forming a pHneutral solution. Next are salts that have the cation of a strong base and the anion of a
weak acid; NaNO 2 is in this category. Then come weak bases, of which NH 3 aq is an
example. And finally, we terminate the list with strong bases: NaOH is the only one.
Thus, in order of increasing pH of their 0.010 M aqueous solutions, the solutes are:
HNO 3 HC 2 H 3O 2 NH 4 ClO 4 NH 4 C 2 H 3O 2 NaNO 2 NH 3 NaOH
b g
59.
b g
(M) NaOCl dissociates completely in aqueous solution into Na + aq , which does not
hydrolyze, and OCl-(aq), which undergoes base hydrolysis. We determine [OH-] in a 0.089 M
solution of OCl , finding the value of the hydrolysis constant from the ionization constant of
HOCl, Ka = 2.9 108 .
Equation :
OCl (aq)
Initial :
0.089 M
0M
0M
Changes :
x M
+x M
+x M
xM
0.089 x M
Equil :
Kb =
+
H 2O(l)
HOCl(aq)
OH (aq)
+
xM
HOCl OH
Kw 1.0 1014
x2
x2
7
=
3.4
10
=
=
=
0.089 x 0.089
Ka 2.9 108
OCl
1.7 104 M 0.089, the assumption is valid.
x = 1.7 104 M = OH ; pOH = log 1.7 104 = 3.77 , pH = 14.00 3.77 = 10.23
60.
b g
(M) dissociates completely in aqueous solution into NH 4 + aq , which hydrolyzes,
b g
and Cl aq , which does not. We determine H 3O
+
+
in a 0.123 M solution of NH 4 , finding
the value of the hydrolysis constant from the ionization constant of NH 3 , K b = 1.8 105 .
Equation :
NH 4 + (aq)
Initial :
0.123M
Changes :
x M
Equil :
Kb =
0.123 x M
+
H 2 O(l)
NH 3 (aq)
H 3O + (aq)
0M
0M
+x M
+x M
xM
xM
Kw 1.0 1014
x2
x2
[ NH 3 ][ H 3O ]
10
=
5.6
10
=
=
0123
.
x 0123
.
Ka 1.8 105
[ NH 4 ]
757
Chapter 16: Acids and Bases
8.3 106 M << 0.123, the assumption is valid
x = 8.3 106 M = H 3O + ;
61.
pH = log 8.3 106 = 5.08
(M) KC 6 H 7 O 2 dissociates completely in aqueous solution into K + (aq), which does not
hydrolyze, and the ion C 6 H 7 O 2 (aq) , which undergoes base hydrolysis. We determine
OH in 0.37 M KC 6 H 7 O 2 solution using an I.C.E. table.
Note: K a = 10 pK = 104.77 = 1.7 105
C6 H 7 O 2 (aq)
Equation :
Initial :
Changes :
Kb
HC6 H 7 O 2 (aq)
0M
0M
xM
+ xM
+x M
xM
xM
[ HC 6 H 7 O 2 ][OH ]
Kw 10
x2
x2
. 1014
10
5
.
9
10
. 105
0.37 x 0.37
Ka 17
[C6 H 7 O 2 ]
pH = 14.00 4.82 = 9.18
(M)
Equation :
C5 H 5 NH + (aq) + H 2O(l)
C5 H 5 N(aq)
Initial :
0.0482 M
0M
0M
xM
+ xM
+ xM
xM
xM
Changes :
0.0482 x M
Equil :
Ka =
OH (aq)
x = 1.5 105 M = OH , pOH = log 1.5 105 = 4.82,
62.
+
0.37 M
0.37 x M
Equil :
+ H 2O(l)
+
H 3O + (aq)
C5 H5 N H3O+
K w 1.0 1014
x2
x2
6
=
=
6.7
10
=
=
K b 1.5 109
0.0482 x 0.0482
C5 H 5 NH +
5.7 104 M << 0.0482 M, the assumption is valid
x = 5.7 104 M = H 3O + ,
63.
pH = log 5.7 104 = 3.24
(M)
H 3O + + SO32
(a) HSO3 + H 2 O
K a = K a 2 ,Sulfurous = 6.2 108
OH + H 2SO3 K b =
HSO3 + H 2 O
Kw
K a1 ,Sulfurous
Since Ka Kb , solutions of HSO 3 are acidic.
758
=
1.0 1014
= 7.7 1013
2
1.3 10
Chapter 16: Acids and Bases
(b)
H 3O + + S2
HS + H 2 O
K a = K a 2 ,Hydrosulfuric = 11019
OH + H 2S
HS + H 2 O
Kb =
Kw
K a1 ,Hydrosulfuric
=
1.0 1014
= 1.0 107
1.0 107
Since Ka Kb , solutions of HS are alkaline, or basic.
(c)
2
H 3O + + PO 4
HPO 4 + H 2 O
3
2
OH + H 2 PO 4
HPO 4 + H 2 O
Since Ka Kb , solutions of HPO 4
64.
K a = K a 3 ,Phosphoric = 4.2 1013
2
Kb =
Kw
=
K a 2 ,Phosphoric
1.0 1014
= 1.6 107
8
6.3 10
are alkaline, or basic.
(M) pH = 8.65 (basic). We need a salt made up of a weakly polarizing cation and an
anion of a weak acid, which will hydrolyze to produce a basic solution. The salt (c) KNO2
satisfies these requirements. (a) NH 4 Cl is the salt of the cation of a weak base and the anion
of a strong acid, and should form an acidic solution. (b) KHSO 4 and (d) NaNO 3 have
weakly polarizing cations and anions of strong acids; they form pH-neutral solutions.
pOH = 14.00 8.65 = 5.35 OH = 105.35 = 4.5 106 M
NO 2 (aq)
Equation:
+
S
Initial:
Changes:
4.5 106 M
Equil:
S 4.5 10
6
M
H 2O(l)
HNO 2 (aq)
+
OH (aq)
0M
0 M
+4.5 106 M
+4.5 106 M
4.5 106 M
4.5 106 M
c
h
2
HNO 2 OH
4.5 106
K
1.0 1014
11
Kb = w =
=
1.4
10
=
=
Ka 7.2 104
S 4.5 106
NO 2
4.5 10
=
6 2
S 4.5 10 6
1.4 1011
= 1.4 M
S = 1.4 M + 4.5 10 6 = 1.4 M = KNO 2
Molecular Structure and Acid-Base Behavior
65.
(M)
(a) HClO 3 should be a stronger acid than is HClO 2 . In each acid there is an H – O – Cl
grouping. The remaining oxygen atoms are bonded directly to Cl as terminal O
atoms. Thus, there are two terminal O atoms in HClO 3 and only one in HClO 2 .
For oxoacids of the same element, the one with the higher number of terminal
oxygen atoms is the stronger. With more oxygen atoms, the negative charge on the
conjugate base is more effectively spread out, which affords greater stability.
HClO 3 : K a1 = 5 102 and HClO 2 : Ka = 1.1 102 .
759
Chapter 16: Acids and Bases
(b)
HNO 2 and H 2 CO 3 each have one terminal oxygen atom. The difference? N is more
electronegative than C, which makes HNO2 K a = 7.2 104 , a stronger acid than
H2CO3 K a1 = 4.4 107 .
(c)
H 3 PO 4 and H 2SiO 3 have the same number (one) of terminal oxygen atoms.
They differ in P being more electronegative than Si, which makes H 3 PO 4
K
a1
= 7.1 103 a stronger acid than H 2SiO3 K a1 = 1.7 1010 .
66.
(E) CCl 3COOH is a stronger acid than CH 3COOH because in CCl 3COOH there are
electronegative (electro-withdrawing) Cl atoms bonded to the carbon atom adjacent to the
COOH group. The electron-withdrawing Cl atoms further polarize the OH bond and
enhance the stability of the conjugate base, resulting in a stronger acid.
67.
(E)
(a)
HI is the stronger acid because the H — I bond length is longer than the H — Br
bond length and, as a result, H — I is easier to cleave.
(b)
HOClO is a stronger acid than HOBr because
(i) there is a terminal O in HOClO but not in HOBr
(ii) Cl is more electronegative than Br.
(c)
H 3CCH 2 CCl 2 COOH is a stronger acid than I 3CCH 2 CH 2 COOH both because Cl is
more electronegative than is I and because the Cl atoms are closer to the acidic
hydrogen in the COOH group and thus can exert a stronger e withdrawing effect on
the OH bond than can the more distant I atoms.
68.
(M) The weakest of the five acids is CH 3CH 2 COOH . The reasoning is as follows. HBr is a
strong acid, stronger than the carboxylic acids. A carboxylic acid—such as CH 2ClCOOH
and CH 2 FCH 2 COOH —with a strongly electronegative atom attached to the hydrocarbon
chain will be stronger than one in which no such group is present. But the I atom is so
weakly electronegative that it barely influences the acid strength. Acid strengths of some
of these acids (as values of pKa ) follow. (Larger values of pKa indicate weaker acids.)
Strength: CH 3CH 2 COOH 4.89 CI3COOH CH 2 FCH 2 COOH CH 2 ClCOOH HBr 8.72
69.
(E) The largest Kb (most basic) belongs to (c) CH3CH2CH2NH2 (hydrocarbon chains have
the lowest electronegativity). The smallest Kb (least basic) is that of (a) o-chloroaniline
(the nitrogen lone pair is delocalized (spread out over the ring), hence, less available to
accept a proton (i.e., it is a poorer Brønsted base)).
70.
(E) The most basic species is (c) CH3O (methoxide ion). Methanol is the weakest acid,
thus its anion is the strongest base. Most acidic: (b) ortho-chlorophenol. In fact, it is the
only acidic compound shown.
760
Chapter 16: Acids and Bases
Lewis Theory of Acids and Bases
71.
(E)
(a) acid is CO2 and base is H2O
O
O
C
O
O
+
O
H
+
C
H
O
O
H
H
H
H
(b) acid is BF3 and base is H2O
F
F
B
F
H
F
H
2-
H
2–
O
O
+
2
H
H
2–
(d) acid is SO3 and base is S
+
S
H
2-
O
S
S
O
72.
O
H
O
2S
F
O
F
(c) acid is H2O and base is O
O
B
O
+
O
O
O
S
(E)
(a) SOI2 is the acid, and BaSO3 the base.
(b) HgCl3 – is the acid, Cl– the base.
73. (E) A Lewis base is an electron pair donor, while a Lewis acid is an electron pair acceptor.
We draw Lewis structures to assist our interpretation.
(a)
The lone pairs on oxygen can readily be donated;
this is a Lewis base.
O H
(b)
H
H
H
C
C
C H2 H
H
H
H
C
C
B
H
H
H
The incomplete octet of B provides a site for
acceptance of an electron pair, this is a Lewis acid.
C H3
761
Chapter 16: Acids and Bases
H H
(c)
H N
The incomplete octet of B provides a site for
acceptance of an electron pair, this is a Lewis acid.
C H
H
74.
(M) A Lewis base is an electron pair donor, while a Lewis acid is an electron pair acceptor.
We draw Lewis structures to assist our interpretation.
According to the following Lewis structures, SO 3 appears to be the electron pair
acceptor (the Lewis acid), and H 2 O is the electron pair donor (the Lewis base).
(Note that an additional sulfur-to-oxygen bond is formed upon successful attachment
of water to SO3.)
(a)
O
H
O
H O
O
S
H
O
S
O
H
O
O
b g
b g
(b) The Zn metal center in Zn OH 2 accepts a pair of electrons. Thus, Zn OH
2
is a
Lewis acid. OH donates the pair of electrons that form the covalent bond.
Therefore, OH is a Lewis base. We have assumed here that Zn has sufficient
covalent character to form coordinate covalent bonds with hydroxide ions.
2-
H
O
2 O H
+
H O
Zn O H
H O Zn O H
O
Base: e pair donor
75.
Acid: e pair acceptor
H
(M) (a)
-
H
O
O
H
+
Base: e pair donor
H
O
B
O
H
H
O
B
O
O
H
H
Acid: e pair acceptor
762
O
H
Chapter 16: Acids and Bases
(b)
H
H N N H
H
H N N H
H H
H O H
H H
H
H O
The actual Lewis acid is H+ , which is supplied by H3O+
(c)
F
H H
H H
H C C O C C H
H H
F
H H
B
B
F
H H
F
H H
H C C O C C H
Base: electron pair donor
76.
F
F
Acid: electron pair acceptor
H H
H H
-
(E) The C in a CO 2 molecule can accept a pair of electrons. Thus, CO2 is the Lewis acid.
The hydroxide anion is the Lewis base as it has pairs of electrons it can donate.
O
C O
acid
+
O
H
base
77.
I3 (aq)
(E) I 2 (aq) I (aq)
78.
(E)
(a)
F
H
F
Lewis base
(e- pair donor)
(b)
F
O
O
H
C
O
I I
I
Lewis acid
Lewis base
(e- pair acceptor) (e- pair donor)
F
F
Sb F
F
Lewis acid
(e- pair acceptor)
H
F
F
F
Sb
F
F
F
H
F
H
B F
F
F
F
B F
F
Lewis acid
Lewis base
(e- pair donor) (e- pair acceptor)
763
I
I
I
Chapter 16: Acids and Bases
79.
(E)
O H
H O
H O H
O
S
O
H O
H O
S
O
O
Lewis base
Lewis acid
(e pair donor) (e pair acceptor)
80.
S
(E)
H
H
Ag+
2 H N
+
H N Ag N H
H
electron pair donor
H
H
electron pair acceptor
H
ammonia adduct
aduct
ammonia
INTEGRATIVE AND ADVANCED EXERCISES
81. (E) We use (ac) to represent the fact that a species is dissolved in acetic acid.
(a) C2 H 3 O 2 (ac) HC 2 H 3 O 2 (l)
HC 2 H 3 O 2 (l) C 2 H 3 O 2 (ac)
C2H3O2– is a base in acetic acid.
(b) H 2 O(ac) HC 2 H 3 O 2 (l )
H 3 O (ac) C 2 H 3 O 2 (ac)
Since H2O is a weaker acid than HC2H3O2 in aqueous solution, it will also be weaker in
acetic acid. Thus H2O will accept a proton from the solvent acetic acid, making H2O a
base in acetic acid.
(c) HC2 H 3 O 2 (l) HC2 H 3 O 2 (l )
H 2 C 2 H 3 O 2 (ac) C 2 H 3 O 2 (aq)
Acetic acid can act as an acid or a base in acetic acid.
(d) HClO 4 (ac) HC2 H 3 O 2 (l)
ClO 4 (ac) H 2 C2 H 3 O 2 (ac)
Since HClO4 is a stronger acid than HC2H3O2 in aqueous solution, it will also be
stronger in acetic acid. Thus, HClO4 will donate a proton to the solvent acetic acid,
making HClO4 an acid in acetic acid.
2+
82. (E) Sr(OH)2 (sat'd)
Sr (aq) + 2 OH (aq)
pOH 14 pH=14-13.12 0.88
[OH ] 10-0.88 0.132 M
764
Chapter 16: Acids and Bases
[OH ]dilute 0.132M (
[HCl]
83.
10.0 ml concentrate
250.0 mL total
(10.0 ml ) (5.28 10 3 M)
25.1 mL
) 5.28 10 3 M
2.10 103 M
(M)
(a) H2SO4 is a diprotic acid. A pH of 2 assumes no second ionization step and the second
ionization step alone would produce a pH of less than 2, so it is not matched.
(b) pH 4.6 matched; ammonium salts hydrolyze to form solutions with pH < 7.
(c) KI should be nearly neutral (pH 7.0) in solution since it is the salt of a strong acid (HI)
and a strong base (KOH). Thus it is not matched.
(d) pH of this solution is not matched as it's a weak base. A 0.002 M solution of KOH has a
pH of about 11.3. A 0.0020 M methylamine solution would be lower (~10.9).
(e) pH of this solution is matched, since a 1.0 M hypochlorite salt hydrolyzes to form a
solution of ~ pH 10.8.
(f) pH of this solution is not matched. Phenol is a very weak acid (pH >5).
(g) pH of this solution is 4.3. It is close, but not matched.
(h) pH of this solution is 2.1, matched as it's a strong organic acid.
(i) pH of this solution is 2.5, it is close, but not matched.
84. (M) We let [H3O+]a = 0.500 [H3O+]i
pH = –log[H3O+]i
pHa = –log[H3O+]a = –log(0.500 [H3O+]i) = –log [H3O+]i – log (0.500) = pHi + 0.301
The truth of the statement has been demonstrated. Remember, however, that the extent of
ionization of weak acids and weak bases in water increases as these solutions become more
and more dilute. Finally, there are some solutions whose pH doesn’t change with dilution,
such as NaCl(aq), which has pH = 7.0 at all concentrations.
85. (M) Since a strong acid is completely ionized, increasing its concentration has the effect of
increasing [H3O+] by an equal degree. In the case of a weak acid, however, the solution of
the Ka expression gives the following equation (if we can assume that [H3O+] is negligible
compared to the original concentration of the undissociated acid, [HA]). x2 = Ka[HA] (where
x = [H3O+]), or x K a [HA] . Thus, if [HA] is doubled, [H3O+] increases by 2 .
86. (E) H2O(l) + H2O(l) H3O+(aq) + OH-(aq) Ho = (-230.0 kJ mol) – (-285.8 kJ mol-1)
= 55.8 kJ
Since Ho is positive, the value of Keq is larger at higher temperatures. Therefore, the ionic
products increase with increasing temperature. Le Ch âtelier’s principle or the Van’t Hoff
equation would show this to be true.
765
Chapter 16: Acids and Bases
87. (M) First we assume that molarity and molality are numerically equal in this dilute aqueous
solution. Then compute the molality that creates the observed freezing point, with Kf = 1.86
°C/m for water.
m
Tf
0.096 C
0.052 m
K f 1.86 C/m
concentration 0.052 M
This is the total concentration of all species in solution, ions as well as molecules. We base
our remaining calculation on the balanced chemical equation.
C 4 H 5 O 2 (aq) H 3 O (aq)
Equation: HC4 H 5 O 2 (aq) H 2 O
Initial:
0.0500 M
0M
0M
Changes:
xM
Equil:
(0.0500 x) M
xM
xM
xM
xM
Total conc. 0.0516 M (0.0500 x) M x M x M 0.0500 M xM
Ka
88.
x 0.0016 M
[C4 H 5 O 2 ][H3 O ]
(0.0016 )
xx
5 105
[HC4 H5 O2 ]
0.0500 x 0.0500 0.0016
2
(D) [H3O+] = 10–pH = 10–5.50 = 3.2 × 10–6 M
(a) To produce this acidic solution we could not use 15 M NH3(aq), because aqueous
solutions of NH3 have pH values greater than 7 owing to the fact that NH3 is a weak
base.
(b)
VHCl 100.0 mL
3.2 10 6 mol H 3 O
1L
1 mol HCl
1 mol H 3 O
1 L soln
12 mol HCl
2.7 10 5 mL 12 M HCl soln
(very unlikely, since this small a volume is hard to measure)
(c)
Ka = 5.6 × 10–10 for NH4+(aq)
Equation:
NH 4 (aq)
Initial:
c M
6
Changes:
3.2 10
Equil:
(c 3.2 106 ) M
K a 5.6 1010
M
H 2 O(l)
H3 O (aq)
NH 3 (aq)
0M
0M
6
M 3.2 106 M
3.2 10
3.2 106 M
[NH 3 ][H 3 O ] (3.2 106 )(3.2 106 )
[NH 4 ]
c 3.2 106
766
3.2 106 M
Chapter 16: Acids and Bases
(3.2 10 6 ) 2
1.8 10 2 M
[NH 4 ] 0.018 M
c 3.2 10
10
5.6 10
0.018 mol NH 4 1 mol NH 4 Cl 53.49 g NH 4 Cl
NH 4 Cl mass 100.0 mL
0.096 g NH 4 Cl
1000 mL
1 mol NH 4 Cl
1 mol NH 4
This would be an easy mass to measure on a laboratory three decimal place balance.
6
(d) The set-up is similar to that for NH4Cl.
Equation:
HC2 H 3 O 2 (aq)
c M
Initial:
3.2 10
Changes:
Equil:
C2 H3 O2 (aq)
H 2 O (l)
0 M
6
M
(c 3.2 106 ) M
K a 1.8 105
3.2 10
6
H3 O (aq)
0 M
3.2 106 M
M
3.2 106 M
3.2 106 M
[C2 H3O 2 ][H 3O ] (3.2 106 )(3.2 106 )
[HC2 H 3O 2 ]
c 3.2 106
(3.2 106 ) 2
5.7 107 c [HC2 H 3O 2 ] 3.8 106
5
1.8 10
3.8 106 mol HC2 H 3O 2 60.05 g HC2 H 3O 2
HC2 H 3O 2 mass 100.0 mL
1 mol HC2 H 3O 2
1000 mL
c 3.2 106
2.3 105 g HC2 H3O2 (an almost impossibly small mass to measure
using conventional laboratory scales)
89.
(M)
(a) 1.0 × 10-5 M
HCN:
10
HCN(aq)
Initial:
Change:
+
K a 6.210
+
H2O(l)
H3O (aq)
1.0 × 10-5 M
–x
Equilibrium: (1.0 × 10-5-x) M
6.2 1010
x(1.0 107 x)
(1.0 105 x)
–
1.0 × 10-7 M
–
+x
–
+
(1.0 × 10-7+x) M
6.2 1015 6.2 1010 x 1.0 107 x x 2
x 2 1.006 107 6.2 1015 0
1.006 107 (1.006 107 ) 2 4(1)( 6.2 1015 )
1.44 107
2(1)
+
Final pH = -log[H3O ] = -log(1.0 × 10-7+ 1.44 × 10-7) = 6.61
x
767
CN-(aq)
0M
+x
x
Chapter 16: Acids and Bases
(b) 1.0 × 10-5 M C6H5NH2:
10
K a 7.410
C6H5NH2(aq) + H2O(l)
-5
Initial:
1.0 × 10 M
–
Change:
–x
–
-5
Equilibrium: (1.0 × 10 –x) M
–
7.4 1010
x(1.0 107 x)
(1.0 105 x)
OH-(aq) +
C6H5NH3+(aq)
1.0 × 10-7 M
0M
+x
+x
-7
(1.0 × 10 +x) M
x
7.4 1015 7.4 1010 x 1.0 107 x x 2
x 2 1.0074 107 7.4 1015 0
(1.0074 107 ) 2 4(1)( 7.4 1015 )
4.93 108 M
2(1)
Final pOH = -log[OH ] = -log(1.0 × 10-8+ 4.93 × 10-7) = 6.83 Final pH = 7.17
x
90.
1.0074 107
(D)
(a) For a weak acid, we begin with the ionization constant expression for the weak acid, HA.
[H O ][A ] [H ]2
[H ]2
[H ]2
Ka 3
[HA]
[HA] M [H ]
M
The first modification results from realizing that [H+] = [A–]. The second modification
is based on the fact that the equilibrium [HA] equals the initial [HA] minus the amount
that dissociates. And the third modification is an approximation, assuming that the
amount of dissociated weak acid is small compared to the original amount of HA in
solution. Now we take the logarithm of both sides of the equation.
[H ] 2
log K a log
log [H ] 2 log M 2 log [H ] log M
c
2 log [H ] log K a log M
2 pH pK a log M
pH 12 pK a 12 log M
For a weak base, we begin with the ionization constant expression for a weak base, B.
[BH ][OH ] [OH ] 2
[OH ] 2
[OH ] 2
Kb
[B]
[B]
M
M [OH ]
The first modification results from realizing that [OH–] = [BH+]. The second
modification is based on the fact that the equilibrium [B] equals the initial [B] minus
the amount that dissociates. And the third modification is an approximation, assuming
that the amount of dissociated weak base is small compared to the original amount of B
in solution. Now we take the logarithm of both sides of the equation.
[OH ]2
log K b log
log [OH ]2 log M 2 log [OH ] log M
M
2 pOH pK b log M
pOH 12 pK b 12 logM
2 log [OH ] log K b log M
pH 14.00 pOH 14.00 12 pK b 12 log M
The anion of a relatively strong weak acid is a weak base. Thus, the derivation is the
same as that for a weak base, immediately above, with the exception of the value for
Kb. We now obtain an expression for pKb.
768
Chapter 16: Acids and Bases
Kw
Ka
Kb
log K b log
Kw
log K w log K a
Ka
log K b log K w log K a pK b pK w pK a
Substitution of this expression for pKb into the expression above gives the following result.
pH 14.00 12 pK w 12 pK a 12 log M
(b) 0.10 M HC2H3O2
pH = 0.500 × 4.74 – 0.500 log 0.10 = 2.87
Equation: HC2 H 3 O 2 (aq)
Initial:
Changes:
Equil:
H 2 O(l)
0.10 M
xM
(0.10 x) M
C2 H 3 O 2 (aq)
0M
H 3 O (aq)
x M
0M
xM
xM
xM
[H 3 O ][C2 H 3 O 2 ]
x x
x2
5
Ka
1.8 10
[HC2 H 3 O 2 ]
0.10 x 0.10
x 0.10 1.8 105 1.3 103 M, pH log(1.3 103 ) 2.89
0.10 M NH 3 pH 14.00 0.500 4.74 0.500 log 0.10 11.13
Equation:
NH 3 (aq)
Initial:
Changes:
Equil:
0.10 M
xM
(0.10 x) M
Kb
H 2 O(l)
NH 4 (aq)
0M
x M
xM
0M
xM
x M
[NH 4 ][OH ]
x x
x2
1.8 105
x 0.10 1.8 105
[NH 3 ]
0.10 x 0.10
pOH log(1.3 103 ) 2.89
1.3 103 M
pH 14.00 2.89 11.11
pH 14.00 0.500 14.00 0.500 4.74 0.500 log 0.10 8.87
0.10 M NaC2 H 3O 2
Equation: C 2 H 3 O 2 (aq) H 2 O(l)
HC2 H 3 O 2 (aq)
Initial:
0.10 M
0M
Changes: x M
xM
xM
Equil:
(0.10 x) M
Ka
OH (aq)
OH (aq)
0M
xM
xM
[OH ][HC2 H 3 O 2 ] 1.0 1014
x x
x2
[C2 H 3 O 2 ]
1.8 105 0.10 x 0.10
0.10 1.0 1014
7.5 106 M
5
1.8 10
pH 14.00 5.12 8.88
x
769
pOH log(7.5 106 ) 5.12
Chapter 16: Acids and Bases
91.
(D)
[H 3 O ] eq [A ] eq
[A ]eq
,
and finally % ionized = × 100%.
[HA]i
[HA]eq
Let us see how far we can get with no assumptions. Recall first what we mean by [HA]eq.
[HA] eq [HA]i [A ] eq or [HA]i [HA]eq [A ] eq
(a) We know K a
We substitute this expression into the expression for .
[ A ]eq
[HA]eq [ A ]eq
We solve both the and the Keq expressions for [HA]eq, equate the results, and solve for .
[H 3O ][A _ ]
[A ] α[A ]
[HA]eq
[A ] [HA]eq [A ]
[HA]eq
Ka
[H 3O ][A _ ] [A ](1 )
R
Ka
R 1
R 1 (1 R)
1
1 R
Momentarily, for ease in writing, we have let R
[H 3 O] 10 pH
pK 10 ( pK pH)
Ka
10
Once we realize that 100 = % ionized, we note that we have proven the cited formula.
Notice that no approximations were made during the course of this derivation.
(b) Formic acid has pKa = 3.74. Thus 10(pK–pH) = 10(3.74–2.50) = 101.24 = 17.4
% ionization
100
5.43%
1 17.4
(c) [H3O+] = 10–2.85 = 1.4 × 10–3 M
% ionization
1.4 10 3 M
100% 0.93%
0.150 M
100
100
1 10 ( pK pH)
108
10 ( pK pH) 107
( pK pH)
0.93
1 10
2.03 pK pH
pK pH 2.03 2.85 2.03 4.88 K a 10 4.88 1.3 10 5
0.93
92.
(M) We note that, as the hydrocarbon chain between the two COOH groups increases in
length, the two values of pKa get closer together. The reason for this is that the ionized
COOH group, COO–, is negatively charged. When the two COOH groups are close together,
the negative charge on the molecule, arising from the first ionization, inhibits the second
ionization, producing a large difference between the two ionization constants. But as the
hydrocarbon chain lengthens, the effect of this negative COO- group on the second
ionization becomes less pronounced, due to the increasing separation between the two
carbonyl groups, and the values of the two ionization constants are more alike.
770
Chapter 16: Acids and Bases
93. (M) Consider only the 1st dissociation.
6.210
H 2 C4 H 4 O 4 (aq) H 2 O(l)
H 3 O (aq) HC4 H 4 O 4 (aq)
0.100 M
0M
0M
xM
xM
xM
(0.100 x) M
xM
xM
-5
Equation:
Initial:
Changes:
Equil:
x2
6.2 105 x 0.00246 M
Ka
0.100 x
Solve using the quadratic equation. Hence pH = log 0.00246 = 2.61
Consider only the 2nd dissociation.
Equation:
HC4 H 4 O 4 (aq)
Initial:
0.0975 M
Changes:
Equil:
x M
(0.0975 x) M
2.310
H 2 O(l)
-5
H3O (aq)
C4 H 4 O4 2 (aq)
x M
0.00246 + x M
x(0.00246 x)
x(0.00246)
2.3 106
0.0975 x
0.0975
x 0.000091 M or x 0.000088 M (solve quadratic)
0M
0.00246 M
x M
xM
Ka
Hence pH 2.59
Thus, the pH is virually identical, hence, we need only consider the 1st ionization.
94. (M) To have the same freezing point, the two solutions must have the same total
concentration of all particles of solute—ions and molecules. We first determine the
concentrations of all solute species in 0.150 M HC2H2ClO2, Ka = 1.4 × 10–3
Equation:
Initial:
Changes:
Equil:
HC2 H 2 ClO 2 (aq) H 2 O(l)
0.150 M
xM
(0.150 x) M
H 3 O (aq) C2 H 2 ClO 2 (aq)
0M
xM
xM
0M
xM
xM
[H 3 O ][C2 H 2 ClO 2 ]
x x
1.4 103
Ka
[HC2 H 2 ClO 2 ]
0.150 x
x 2 2.1 104 1.4 103 x
x
x 2 1.4 103 x 2.1 104 0
b b 2 4ac 1.4 103 2.0 106 8.4 104
0.014 M
2a
2
total concentration = (0.150 – x) + x + x = 0.150 + x = 0.150 + 0.014 = 0.164 M
Now we determine the [HC2H3O2] that has this total concentration.
771
Chapter 16: Acids and Bases
Equation:
Initial:
Changes:
Equil:
Ka
HC 2 H 3 O 2 (aq) H 2 O(l)
zM
yM
(z y ) M
H3 O (aq) C2 H3 O 2 (aq )
0M
yM
yM
0M
yM
yM
[H 3 O ][C2 H 3 O 2 ]
y y
1.8 105
[HC2 H 3 O 2 ]
z y
We also know that the total concentration of all solute species is 0.164 M.
y2
z y y y z y 0.164
z 0.164 y
1.8 10 5
0.164 2 y
y 2 0.164 1.8 10 5 3.0 10 6
We assume that 2 y 0.164
y 1.7 10 3
The assumption is valid: 2 y 0.0034 0.164
Thus, [HC2H3O2] = z = 0.164 – y = 0.164 – 0.0017 = 0.162 M
mass HC 2 H 3 O 2 1.000 L
0.162 mol HC 2 H 3 O 2 60.05 g HC 2 H 3 O 2
9.73 g HC 2 H 3 O 2
1 L soln
1 mol HC 2 H 3 O 2
95. (M) The first ionization of H2SO4 makes the major contribution to the acidity of the
solution.
Then the following two ionizations, both of which are repressed because of the presence of
H3O+ in the solution, must be solved simultaneously.
Equation: HSO 4 - (aq)
Initial:
Changes:
0.68 M
x M
Equil:
(0.68 x) M
H 2 O(l)
2
K2
[H 3 O ][SO 4 ]
0.011
[HSO 4 ]
SO 4 2 (aq) H 3O (aq)
0M
x M
xM
0.68 M
x M
(0.68 x)M
x(0.68 x)
0.68 x
Let us solve this expression for x. 0.011 (0.68 – x) = 0.68x + x2 = 0.0075 – 0.011x
x 2 0.69 x 0.0075 0
x
b b2 4ac 0.69 0.48 0.030
0.01 M
2a
2
772
Chapter 16: Acids and Bases
This gives [H3O+] = 0.68 + 0.01 = 0.69 M. Now we solve the second equilibrium.
Equation: HCHO 2 (aq) H 2 O(l )
Initial:
1.5 M
Changes:
Equil:
xM
(1.5 x) M
CHO 2 (aq) H 3 O (aq)
0M
xM
xM
0.69 M
xM
(0.69 x) M
[H 3 O ][CHO 2 ]
x(0.69 x) 0.69 x
1.8 104
Ka
[HCHO 2 ]
1.5 x
1.5
x 3.9 104 M
We see that the second acid does not significantly affect the [H3O+], for which a final value is
now obtained. [H3O+] = 0.69 M, pH = –log(0.69) = 0.16
96.
(D)
[ H ][ A1 ]
[ H ]2 [ H ]2
[ HA1 ] [ H ] [ HA1 ]
M
Let [ HA1 ] M
K HA1
Let [ HA2 ] M
K HA2 2 K HA1
K HA1
2 K HA1
[ H ]2
M
[ H ]2
M
[ H ]overall K HA1 M
1/2
[ H ]2 2 K HA1 M
[ H ] 2 K HA1 M 2 K HA1 M
1/2
2 K HA1 M
HA1
M
1/2
1/2
1/2
HA1
1/ 2
1/2
HA1
1/2
(take the negative log10 of both sides)
log 2 K M
1
1
M log K M log K M log 2 K
2
2
[ H ] K HA1 M
1
log K HA1 M log 2 K HA1 M
2
1
1
pH 3K HA1 M 3MK HA1
2
2
pH
K
[ H ]2 K HA1 M
log([ H ]overall ) log K HA1 M
pH log K HA1
[ H ][ A2 ]
[ H ]2 [ H ]2
[ HA2 ] [ H ] [ HA2 ]
M
HA1
773
1
log K HA1 M 2 K HA1 M
2
HA1
M
Chapter 16: Acids and Bases
97.
(M) HSbF6 H+ + SbF6H+
F
F
F
Sb
FFhas
2pone
orbital
uses
of its 2p orbitals
2 of its sp3d2 hybrid orbitals
uses
all3dsix
SbSbhas
6 sp
hybrid orbitals
F
F
The six bonds are all sigma bonds
(Sbsp3d2 - F2p)
F
HBF4 H+ + BF4H+
F
hasone
2p orbital
FFuses
of its 2p orbitals
3
BBuses
all
four
of its sp
hybrid orbitals
has 4 sp3 hybrid
orbitals
B
F
F
F
The four bonds are all sigma bonds
(Bsp3 - F2p)
H
98. (M) The structure of H3PO3 is shown on the right.
The two ionizable protons are bound to oxygen.
99.
(M)
(a) H2SO3 > HF >N2H5+ > CH3NH3+ > H2O
(b) OH– > CH3NH2 > N2H4 > F–> HSO3–
(c) (i) to the right and (ii) to the left.
774
H O P O H
O
Chapter 16: Acids and Bases
FEATURE PROBLEMS
100. (D)
(a)
From the combustion analysis we can determine the empirical formula. Note that the
mass of oxygen is determined by difference.
1 mol CO 2
1 mol C
amount C = 1.599 g CO 2
= 0.03633 mol C
44.01 g CO 2 1 mol CO 2
12.011 g C
= 0.4364 g C
1 mol C
1 mol H 2 O
2 mol H
amount H = 0.327 g H 2 O
= 0.03629 mol H
18.02 g H 2 O 1 mol H 2 O
mass of C = 0.03633 mol C
mass of H = 0.03629 mol H
1.008 g H
= 0.03658 g H
1 mol H
b
g
1 mol O
= 0.0364 mol O
16.00 g O
There are equal moles of the three elements. The empirical formula is CHO.
The freezing-point depression data are used to determine the molar mass.
Tf
0.82 C
Tf = Kf m m =
=
= 0.21 m
Kf 3.90 C / m
1 kg 0.21 mol solute
amount of solute = 25.10 g solvent
= 0.0053 mol
1000 g
1 kg solvent
0.615 g solute
M=
= 1.2 102 g/mol
0.0053 mol solute
The formula mass of the empirical formula is: 12.0g C +1.0g H +16.0g O = 29.0g / mol
Thus, there are four empirical units in a molecule, the molecular formula is C4 H 4 O 4 ,
and the molar mass is 116.1 g/mol.
amount O = 1.054 g sample 0.4364 g C 0.03568 g H
(b) Here we determine the mass of maleic acid that reacts with one mol OH ,
mass
0.4250 g maleic acid
58.03g/mol OH
1L
0.2152
mol
KOH
mol OH
34.03mL base
1000 mL
1L base
This means that one mole of maleic acid (116.1 g/mol) reacts with two moles of
hydroxide ion. Maleic acid is a diprotic acid: H 2 C 4 H 2 O 4 .
(c)
Maleic acid has two—COOH groups joined together by a bridging C2H2 group. A
plausible Lewis structure is shown below:
O H H O
H
O
C
C
C
C
O
775
H
Chapter 16: Acids and Bases
(d)
We first determine H 3O + = 10 pH = 101.80 = 0.016 M and then the initial
concentration of acid.
0.215g 1000 mL 1 mol
CHCOOH 2
=
= 0.0370 M
initial
50.00 mL
1L
116.1 g
We use the first ionization to determine the value of K a1
H CHCOO (aq) +
Equation: CHCOOH 2 (aq) + H 2 O(l)
2
Initial:
0.0370 M
Changes: 0.016 M
Equil:
0.021M
0M
+ 0.016 M
0.016 M
H 3 O + (aq)
0 M
+ 0.016 M
0.016 M
H CHCOO H 3O + 0.016 0.016
2
Ka =
=
= 1.2 102
0.021
CHCOOH 2
Ka2 could be determined if we had some way to measure the total concentration of all
2
ions in solution, or if we could determine CHCOO 2 K a 2
(e)
Practically all the H 3O + arises from the first ionization.
+
Equation: CHCOOH 2 (aq) + H 2O(l)
H CHCOO 2 (aq) + H 3O (aq)
Initial:
0.0500 M
Changes: x M
Equil:
0.0500 x M
0M
+x M
0 M
+x M
xM
xM
Ka
[H(CHCOO) 2 ][H 3O ]
x2
1.2 102
[(CHCOOH)2 ]
0.0500 x
x 2 0.00060 0.012 x
x 2 0.012 x 0.00060 0
b b 2 4ac 0.012 0.00014 + 0.0024
=
= 0.019 M = H 3O +
2a
2
pH = log 0.019 = 1.72
x=
b
g
776
Chapter 16: Acids and Bases
101. (D)
(a)
x2
x2
4.2 104
x1 0.00102
0.00250 x 0.00250
x2
x2
4.2 104
x2 0.000788
0.00250 0.00102 0.00148
x2
x2
4.2 104
x3 0.000848
0.00250 0.000788 0.00171
x2
x2
4.2 104
x4 0.000833
0.00250 0.000848 0.00165
x2
x2
4.2 104
x5 0.000837
0.00250 0.000833 0.00167
x2
x2
4.2 104
x6 0.000836
0.00250 0.000837 0.00166
x6 0.000836
or 8.4 104 ,
which is the same as the value obtained using the quadratic equation.
(b) We organize the solution around the balanced chemical equation, as we have done before.
+
Equation:
HClO 2 (aq) + H 2 O(l)
ClO 2 (aq) + H 3 O (aq)
Initial:
Changes:
Equil:
0.500 M
xM
xM
0M
+x M
xM
ClO 2 H 3O +
x2
x2
Ka =
=
= 1.1 102
HClO 2
0.500 x
0.500
x=
x 2 0.500 1.1 102 0.074 M
0 M
+x M
xM
Assuming x 0.500 M ,
Not significantly smaller than 0.500 M.
Assume x 0.074 x (0.500 0.0774) 1.1102 0.068 M Try once more.
Assume x 0.068 x (0.500 0.068) 1.110 2 0.069 M
One more time.
Assume x 0.069 x (0.500 0.069) 1.1102 0.069 M
Final result!
H 3O + = 0.069 M, pH = log H 3O + = log 0.069 = 1.16
777
Chapter 16: Acids and Bases
102. (D)
(a) Here, two equilibria must be satisfied simultaneously. The common variable,
H 3O + = z .
Equation: HC2 H3O2 (aq) +
H 2 O(l)
C2 H 3O 2 (aq)
+
H 3O + (aq)
Initial:
0.315 M
0M
0M
Changes:
+x M
+x M
+z M
xM
zM
Equil:
0.315 x M
H 3O + C2 H 3O 2
xz
Ka =
=
= 1.8 105
0.315 x
HC 2 H 3O 2
CHO 2 (aq) +
H 3 O + (aq)
HCHO 2 (aq)
Initial:
Changes:
0.250 M
y M
0M
+y M
0M
+z M
(0.250 y ) M
yM
zM
Equil:
+ H 2 O(l)
Equation:
H 3 O CHO 2
yz
Ka =
=
= 1.8 104
0.250 y
HCHO 2
+
In this system, there are three variables, x = C 2 H 3O 2 , y = CHO 2 , and z = H 3O + .
These three variables also represent the concentrations of the only charged species in
the reaction, and thus x + y = z . We solve the two Ka expressions for x and y. (Before
we do so, however, we take advantage of the fact that x and y are quite small:
x 0.315 and y 0.250 .) Then we substitute these results into the expression for z ,
and solve to obtain a value of that variable.
xz = 1.8 105 0.315 = 5.7 106
yz = 1.8 104 0.250 = 4.5 105
5.7 106
4.5 105
y=
z
z
6
5
5
5.7 10
4.5 10
5.07 10
z = x+ y =
+
=
z = 5.07 105 = 7.1103 M = [H3O + ]
z
z
z
pH = log 7.1 103 = 2.15 We see that our assumptions about the sizes of x and y
x=
c
h
must
be valid, since each of them is smaller than z . (Remember that x + y = z .)
(b)
We first determine the initial concentration of each solute.
12.5g NH 2 1 mol NH 3
= 1.96M
NH3 =
0.375L soln 17.03g NH 3
1.55 g CH 3 NH 2 1 mol CH 3 NH 2
= 0.133M
CH3 NH 2 =
0.375L soln
31.06 g CH 3 NH 2
K b = 1.8 105
K b = 4.2 104
Now we solve simultaneously two equilibria, which have a common variable, OH = z .
778
Chapter 16: Acids and Bases
Equation: NH 3 aq
+
Initial:
1.96 M
Changes: x M
Equil:
1.96 x M
NH 4 + aq
H2O
OH aq
+
0M
+x M
0M
+z M
xM
zM
NH 4 + OH
xz
xz
5
Kb =
=
1.8
10
=
1.96 x 1.96
NH 3
CH 3 NH 2 aq
Eqn:
+
H 2 O(l)
Initial:
0.133M
Changes: y M
Equil:
(0.133 y ) M
CH 3 NH3+ aq
+
OH aq
0M
z M
zM
0M
yM
yM
CH 3 NH 3+ OH
yz
yz
4
Kb =
=
4.2
10
=
=
0.133 y 0.133
CH 3 NH 2
In this system, there are three variables, x = NH 4 + , y = CH 3 NH 3 + , and z = OH .
These three variables also represent the concentrations of the only charged species in
solution in substantial concentration, and thus x + y = z . We solve the two Kb expressions
for x and y . Then we substitute these results into the expression for z , and solve to obtain
the value of that variable.
xz = 1.96 1.8 105 = 3.53 105
x=
yz = 0.133 4.2 105 = 5.59 105
3.53 105
z
y=
5.59 105
z
3.53 105 5.59 105 9.12 105
+
=
z 2 = 9.12 105
z
z
z
3
z = 9.5 10 M = OH pOH = log 9.5 103 = 2.02 pH = 14.00 2.02 = 11.98
z = x+ y =
We see that our assumptions about x and y (that x 1.96 M and y 0.133 M) must be
valid, since each of them is smaller than z . (Remember that x + y = z .)
(c)
b g
b g
b g
In 1.0 M NH 4 CN there are the following species: NH 4 + aq , NH 3 aq , CN aq , HCN(aq),
+
b g
b g
H 3O aq , and OH aq . These species are related by the following six equations.
1 K w = 1.0 1014 = H3O+ OH
b2g K = 6.2 10
a
10
=
H 3O + CN
1.0 1014
OH =
H 3O +
b3g
HCN
779
5
Kb = 1.8 10 =
NH 4
+
OH
LM NH 3 OP
Q
N
Chapter 16: Acids and Bases
4 NH3 + NH 4 + = 1.0 M
5 HCN + CN = 1.0 M
6 NH 4 + + H3O+ = CN + OH
NH3 = 1.0 NH 4 +
HCN = 1.0 CN
or NH 4 + CN
Equation (6) is the result of charge balance, that there must be the same quantity of
positive and negative charge in the solution. The approximation is the result of
remembering that not much H 3O + or OH will be formed as the result of hydrolysis of
ions. Substitute equation (4) into equation (3), and equation (5) into equation (2).
b3’gK
+
OH
H 3O + CN
b2’gK = 4.0 10 = 1.0 CN
Now substitute equation (6) into equation b2’g , and equation (1) into equation b3’g .
NH
NH
1.8 10
=
b2’’gK = 6.2 10 = H1.0O NH
b3’’g 1.0
10
H O e1.0 NH j
5
= 1.8 10 =
b
NH 4
1.0 NH 4
10
+
+
10
a
+
3
a
+
5
4
4
14
+
+
+
4
3
4
Now we solve both of these equations for NH 4 + .
b‘2g:6.2 10
b‘3g:1.8 10
10
9
6.2 10
10
NH 4
+
= H 3O
H 3O 1.8 10 H 3O
+
9
+
+
NH 4
NH 4
+
+
= NH 4
NH 4
+
NH 4
+
+
6.2 1010
=
6.2 1010 + H 3O +
=
1.8 109 H 3O +
1.00 +1.8 109 H 3O +
We equate the two results and solve for
1.8 109 H 3O +
6.2 1010
+
H 3O .
=
6.2 10 10 + H 3O +
1.00 +1.8 109 H 3O +
6.2 1010 +1.1 H 3O + = 1.1 H 3O + +1.8 109 H 3O +
2
6.2 1010 = 1.8 109 H 3O +
6.2 10 10
H 3O =
5.9 1010 M pH = log 5.9 1010 = 9.23
9
18
. 10
Ka K w
Note that H 3 O + =
or
pH=0.500 (pK a pK w pK b )
Kb
c
+
780
h
2
Chapter 16: Acids and Bases
SELF-ASSESSMENT EXERCISES
103. (E)
(a) KW: Dissociation constant of water, which is defined as [H3O+][OH¯]
(b) pH: A logarithmic scale of expressing acidity of a solution (H3O+); it is defined as
-log [H3O+]
(c) pKa: A logarithmic scale expressing the strength of an acid (which is how much an acid
dissociates in water at equilibrium); it is defined as –log Ka
(d) Hydrolysis: A reaction involving ionization of water
(e) Lewis acid: A compound or species that accepts electrons
104. (E)
(a) Conjugate base: A base that is derived by removing an abstractable proton from the acid
(b) Percent ionization of acid/base: The ratio between the conjugate of the acid or base
under study and the acid/base itself times 100
(c) Self-ionization: A reaction involving ionization of two molecules of water to give
[H3O+] and [OH¯]
(d) Amphiprotic behavior: A substance acting as either an acid or a base
105. (E)
(a) Brønstead–Lowry acid and base: A Brønsted–Lowry acid is a substance that when
placed in water will cause the formation of hydronium ion by donating a proton to
water. The base is one that abstracts a proton from water and results in a hydroxyl ion.
(b) [H3O+] and pH: [H3O+] is the molar concentration of hydronium ion in water, whereas
pH is the –log [H3O+]
(c) Ka for NH4+ and Kb for NH3: Kb of NH3 is the equilibrium constant for hydrolysis of
water with NH3 to generate NH4+ and OH¯. Ka of NH4+ is for deprotonation of NH4+
with water to give NH3 and OH¯.
(d) Leveling effect and electron-withdrawing effect: The leveling effect is the result of
reaction of water with any acid or base such that no acid or base can exist in water that
are more acidic than H3O+ or more basic than OH¯. The electron-withdrawing effect is
the tendency for a central atom with high electronegativity in an oxoacid to withdraw
electrons from the O–H bond.
106. (E) The answer is (a), HCO3–, because it can donate a proton to water to become CO32–, or
it can be protonated to give H2CO3 (i.e., CO2(aq)).
107. (E) The answer is (c). CH3CH2COOH is a weak acid, so the concentration of H3O+ ions it
will generate upon dissociation in water will be significantly less than 0.1 M. Therefore, its
pH will be higher the 0.10 M HBr, which dissociates completely in water to give 0.10 M
H3O+.
108. (E) The answer is (d). CH3NH2 is a weak base.
781
Chapter 16: Acids and Bases
109. (M) The answer is (e) because CO32- is the strongest base and it drives the dissociation of
acetic acid furthest.
110. (M) The answer is (c). H2SO4 dissociation is nearly complete for the first proton, giving a
[H3O+] = 0.10 M (pH = 1). The second dissociation is not complete. Therefore, (c) is the
only choice that makes sense.
111. (M) The answer is (b). You can write down the stepwise equations for dissociation of
H2SO3 and then the dissociation of HSO3¯ and calculate the equilibrium concentration of
each species. However, you can determine the answer without detailed calculations. The
two dissociation equations are given below:
Eq.1
H 2SO3 H 2 O HSO3 H 3O
Ka1 = 1.3×10-2
Eq.2
HSO3 H 2 O SO32 H 3O
Ka2 = 6.3×10-8
For the first equation, the equilibrium concentrations, of HSO3¯ and H3O+ are the same.
They become the initial concentration values for the second dissociation reaction. Since
Ka2 is 6.3×10-8 (which is small), the equilibrium concentrations of HSO3¯ and H3O+ won’t
change significantly. So, the equation simplifies as follows:
SO32- H 3O +
SO32- H 3O + +x
K a2 6.8 10
=
HSO3- -x
HSO3-
8
K a2 =6.8 108 SO32-
112. (E) Since both the acid and the base in this reaction are strong, they dissociate completely.
The pH is determined as follows:
mol H3O+: 0.248 M HNO3 × 0.02480 L = 0.00615 mol
mol OH¯: 0.394 M KOH × 0.01540 L = 0.00607 mol
Final: 0.00615 – 0.00607 = 8.00×10-5 mol H3O+
8.00 105 mol
H 3O
0.00200 M
0.02480 L + 0.01540 L
pH log 0.00200 2.70
113. (M) Since pH = 3.25, the [H3O+] = 10-3.25 = 5.62×10-4 M. Using the reaction below, we
can determine the equilibrium concentrations of other species
Initial
Change
Equil.
HC2H3O2
C
-x
C-x
+ H2O
782
C2H3O2¯
0
+x
x
+ H3O+
0
+x
x
Chapter 16: Acids and Bases
Since x = 5.62×10-4 M, the equilibrium expression becomes
H 3O C2 H 3O 2 5.62 104 M 5.62 104 M
Ka
1.8 105
4
HC2 H3O2
C 5.62 10 M
Solving C gives a concentration of 0.0181 M.
Since concentrated acetic acid is 35% by weight and the density is 1.044 g/mL, the
concentration of acetic acid is:
35 g HAc
1.044 g Conc. HAc 1 mol HAc
6.084 M HAc
100 g Conc. HAc sol'n
0.001 L sol'n
60.06 g HAc
Therefore, volume of concentrated solution required to make 12.5 L of 0.0181 M HAc
solution is:
M V dilute M V conc.
0.0181 M 12500 mL 37.2 mL
V
conc
6.084 M
114. (M) Table 16.3 has the Ka value for chloroacetic acid, HC2H2ClO2. Ka = 1.4×10-3. The
solution is made of the chloroacetate salt, which hydrolyzes water according to the
following reaction:
Initial
Change
Equil.
C2H2ClO2¯ + H2O
2.05
-x
2.05 – x
HC2H2ClO2 + OH¯
0
0
+x
+x
x
x
K W 1.0 1014
7.14 1012
Kb
3
K a 1.4 10
Kb
HC2 H 2ClO2 OH -
C2 H 2 ClO 2
xx
7.14 1012
2.05 x
Solving the above simplified expression for x, we get x = OH - =3.826×10-6
pH=14-pOH=14- -log 3.826×10-6 =8.58
783
Chapter 16: Acids and Bases
115. (M) 0.10 M HI < 0.05 M H2SO4 < 0.05 M HC2H2ClO2 < 0.50 M HC2H3O2 < 0.50 M
NH4Cl < 1.0 M NaBr < 0.05 M KC2H3O2 < 0.05 M NH3 < 0.06 M NaOH < 0.05 M
Ba(OH)2
116. (M) Since pH = 5 pOH, pH must be significantly larger than pOH, which means that the
solution is basic. The pH can be determined by solving two simultaneous equations:
pH 5 pOH 0
pH + pOH = 14
pOH = 2.333
pH = 14 – 2.33 = 11.67.
Therefore, [H3O+] = 2.14×10-12 M (and [OH¯] = 4.67×10-3 M)
The solute must be NH3, because it is the only basic species (the other two are acidic and
neutral, respectively). Since the Kb of NH3 is 1.8×10-5, the concentration of NH3 can be
determined as follows:
OH - NH 4+ 4.67 103 4.67 103
1.8 105
Kb =
x
NH3
x 1.2 M
117. (M) The answer is (a). If HC3H5O2 is 0.42% ionized for a 0.80 M solution, then the
concentration of the acid at equilibrium is (0.80×0.0042) = 0.00336. The equilibrium
expression for the dissociation of HC3H5O2 is:
H 3O C3 H 5O 2 0.00336 0.00336
Ka
1.42 105
HC
H
O
0.800
0.00336
3 5 2
118. (E) The answer is (b), because H2PO4¯ is a result of addition of one proton to the base
HPO42-.
119. (M) The answer is (d). This is because, in the second equation, HNO2 will give its proton
to ClO¯ (rather than HClO giving its proton to NO2¯), which means that HNO2 is a
stronger acid than HClO. Also, in the first equation, ClO¯ is not a strong enough base to
abstract a proton from water, which means that HClO is in turn a stronger acid than H2O.
120. (M) The dominant species in the solution is ClO2¯. Therefore, determine its concentration
and ionization constant first:
3.00 mol CaClO 2 2 mol ClO 2
ClO
= 2.40 M
2.50 L
1 mol CaClO 2
2
K W 1.0 1014
9.1 1013
Kb
2
Ka
1.1 10
784
Chapter 16: Acids and Bases
The reaction and the dissociation of ClO2¯ is as follows:
Initial
Change
Equil.
ClO2¯
2.40
-x
2.40 – x
+ H2O
HClO2
0
+x
x
+ OH¯
0
+x
x
HClO 2 OH -
Kb
ClO-2
9.1 1013
xx
2.40 x
Solving the above simplified expression for x, we get x = OH - =1.478×10-6 .
pH=14-pOH=14- -log 1.478×10-6 =8.2
121. (M) Section 16-8 discusses the effects of structure on acid/base behavior. The major
subtopics are effects of structure on binary acids, oxo acids, organic acids, and amine
bases. For binary acids, acidity can be discussed in terms of electronegativity of the main
atom, bond length, and heterolytic bond dissociation energy. For oxo acids, the main
things affecting acidity are the electronegativity of the central atom and number of oxygen
atoms surrounding the central atom. For organic acids, the main topic is the electronwithdrawing capability of constituent groups attached to the carboxyl unit. For amines, the
basicity is discussed in terms of electron withdrawing or donating groups attached to the
nitrogen in the amine, and the number of resonance structures possible for the base
molecule.
785
CHAPTER 17
ADDITIONAL ASPECTS OF ACID–BASE EQUILIBRIA
PRACTICE EXAMPLES
1A
(D) Organize the solution around the balanced chemical equation, as we have done before.
Equation: HF(aq) H 2 O(l)
Initial:
0.500 M
Changes:
xM
Equil:
(0.500 x) M
Ka
H 3 O + (aq) F (aq)
0M
+x M
xM
0M
+x M
xM
[ H 3O + ][ F ] ( x ) ( x )
x2
6.6 104
[ HF]
0.500 x
0.500
x 0.500 6.6 104 0.018 M
assuming x 0.500
One further cycle of approximations gives:
x (0.500 0.018) 6.6 104 0.018 M [H 3 O + ]
Thus, [HF] = 0.500 M 0.018 M 0.482 M
Recognize that 0.100 M HCl means H 3O +
0.100 M , since HCl is a strong acid.
initial
Equation: HF(aq) + H 2 O(l)
Initial:
0.500 M
Changes: x M
Equil:
(0.500 x ) M
H 3 O + (aq) + F (aq)
0.100 M
+x M
(0.100+x) M
0M
+x M
xM
0100
.
x
[ H 3O + ][ F ] ( x ) (0100
.
+ x)
Ka
assuming x 0.100
6.6 104
0.500
0.500 x
[HF]
6.6 104 0.500
= 3.3 103 M = F
x=
0.100
The assumption is valid.
HF = 0.500 M 0.003M = 0.497 M
[H3O+] = 0.100 M + x = 0.100 M + 0.003 M = 0.103 M
786
Chapter 17: Additional Aspects of Acid–Base Equilibria
1B
(M) From Example 17-6 in the text, we know that [ H 3O + ] [C 2 H 3O 2 ] 13
. 103 M in
0.100 M HC 2 H 3O 2 . We base our calculation, as usual, on the balanced chemical
equation. The concentration of H 3O + from the added HCl is represented by x.
+
Equation: HC 2 H 3 O 2 (aq) + H 2 O(l)
H 3 O (aq) + C 2 H 3 O 2 (aq)
Changes: 0.00010 M
From HCl:
0M
+0.00010 M
+x M
Equil:
(0.00010 x) M 0.00010 M
Initial:
Ka =
0.100 M
0.100 M
0M
+0.00010 M
[ H 3O + ][C 2 H 3O 2 ] (0.00010 + x ) 0.00010
. 105
18
[ HC 2 H 3O 2 ]
0100
.
0.00010 x
1.8 105 0.100
0.018 M
0.00010
V12 M HCl =1.00 L
0.018 mol H 3O +
1L
x 0.018 M 0.00010 M 0.018 M
1mol HCl
1mol H 3 O
+
1 L soln
12 mol HCl
1000 mL
1L
1 drop
0.050 mL
30.drops
Since 30. drops corresponds to 1.5 mL of 12 M solution, we see that the volume of solution
does indeed remain approximately 1.00 L after addition of the 12 M HCl.
2A
(M) We again organize the solution around the balanced chemical equation.
+
Equation: HCHO 2 aq + H 2 O(l)
CHO 2 aq + H 3 O aq
Initial:
0.100 M
0.150 M
0M
Changes:
xM
Equil:
(0.100 x) M
+x M
(0.150 + x) M
+x M
xM
[CHO 2 ][H 3O ] (0.150 + x)( x)
0.150 x
1.8 104
Ka
[HCHO 2 ]
0.100 x
0.100
assuming x 0.100
0.100 1.8 104
= 1.2 104 M = H 3 O + , x <<0.100, thus our assumption is valid
0.150
CHO 2 = 0.150 M + 0.00012 M = 0.150 M
x=
787
Chapter 17: Additional Aspects of Acid–Base Equilibria
2B
(M) This time, a solid sample of a weak base is being added to a solution of its
conjugate acid. We let x represent the concentration of acetate ion from the added
sodium acetate. Notice that sodium acetate is a strong electrolyte, thus, it completely
dissociates in aqueous solution. H 3O + = 10 pH = 10 5.00 = 1.0 10 5 M = 0.000010 M
C 2 H 3 O 2 aq + H 3 O + aq
Equation: HC 2 H 3 O 2 aq + H 2 O(l)
Initial:
0.100 M
0M
0M
Changes: 0.000010 M
+0.000010 M
+0.000010 M
From NaAc:
+x M
Equil:
0.100 M
(0.000010 + x) M 0.000010 M
H 3O + C2 H 3O 2 0.000010 0.000010 + x
=
Ka =
= 1.8 105
0.100
HC 2 H 3O 2
0.000010 + x
1.8 105 0.100
0.18 M
0.000010
x 0.18 M 0.000010 M = 0.18 M
mass of NaC2 H 3O 2 = 1.00 L
0.18 mol C2 H 3O 2 1mol NaC2 H3O 2 82.03g NaC2 H3O 2
1L
1mol NaC2 H 3O 2
1mol C2 H 3O 2
=15g NaC2 H 3O 2
3A
(M) A strong acid dissociates essentially completely, and effectively is a source of
H 3O + . NaC2 H 3O 2 also dissociates completely in solution. The hydronium ion and the acetate
ion react to form acetic acid: H 3O (aq) C2 H 3O 2 (aq)
HC 2 H 3O 2 (aq) H 2 O(l)
All that is necessary to form a buffer is to have approximately equal amounts of a weak
acid and its conjugate base together in solution. This will be achieved if we add an
amount of HCl equal to approximately half the original amount of acetate ion.
3B
(M) HCl dissociates essentially completely in water and serves as a source of hydronium
ion. This reacts with ammonia to form ammonium ion:
NH3 aq + H3O+ aq
NH 4+ aq + H 2 O(l) .
Because a buffer contains approximately equal amounts of a weak base (NH3) and its
conjugate acid (NH4+), to prepare a buffer we simply add an amount of HCl equal to
approximately half the amount of NH3(aq) initially present.
788
Chapter 17: Additional Aspects of Acid–Base Equilibria
4A
(M) We first find the formate ion concentration, remembering that NaCHO 2 is a strong
b g
electrolyte, existing in solution as Na + aq and CHO 2 aq .
[CHO 2 ]
23.1g NaCHO 2 1000 mL 1 mol NaCHO 2
1 mol CHO 2
0.679 M
500.0 mL soln
1L
68.01g NaCHO 2 1 mol NaCHO 2
As usual, the solution to the problem is organized around the balanced chemical equation.
+
Equation: HCHO 2 aq + H 2 O(l)
CHO 2 aq + H 3 O aq
0M
Initial:
0.432 M
0.679 M
Changes: x M
Equil:
(0.432 x) M
+x M
(0.679 + x) M
+x M
xM
H 3O + CHO 2 x 0.679 + x
0.679 x
=
Ka =
= 1.8 104
HCHO 2
0.432 x
0.432
x=
0.432 1.8 104
0.679
This gives H 3O + = 1.14 104 M . The assumption that x 0.432 is clearly correct.
pH log H 3O log 1.14 104 3.94 3.9
4B
(M) The concentrations of the components in the 100.0 mL of buffer solution are found
via the dilution factor. Remember that NaC2 H 3O 2 is a strong electrolyte, existing in
solution as
b g
b g
Na + aq and C 2 H 3O 2 aq .
[HC 2 H 3O 2 ] 0.200 M
63.0 mL
100.0 mL
0.126 M [C 2 H 3O 2 ] 0.200 M
37.0 mL
100.0 mL
0.0740 M
As usual, the solution to the problem is organized around the balanced chemical equation.
Equation:
Initial:
Changes:
Equil:
+
HC2 H 3O 2 aq + H 2O(l)
C2 H3O 2 aq + H 3O aq
0.0740 M
0 M
0.126 M
x M
x M
x M
(0126
.
x) M
0.0740 + x M
xM
b
g
[H O + ][C2 H 3O 2 ] x (0.0740 x)
0.0740 x
Ka 3
1.8 105
[HC2 H3O 2 ]
0.126 x
0.126
x
18
. 105 0126
.
31
. 105 4.51
. 105 M = [H 3O + ] ; pH = log [ H 3O + ] log 31
0.0740
Note that the assumption is valid: x 0.0740 0.126.
Thus, x is neglected when added or subtracted
789
Chapter 17: Additional Aspects of Acid–Base Equilibria
5A
(M) We know the initial concentration of NH 3 in the buffer solution and can use the pH
to find the equilibrium [OH-]. The rest of the solution is organized around the balanced
chemical equation. Our first goal is to determine the initial concentration of NH 4 .
pOH = 14.00 pH = 14.00 9.00 5.00
NH3 aq
Equation:
Initial:
Changes:
Equil:
+
H 2 O(l)
0.35 M
1.0 105 M
(0.35 1.0 105 ) M
[OH ] 10 pOH 105.00 10
. 105 M
NH 4 + aq
+
xM
1.0 105 M
( x 10
. 105 ) M
OH aq
0 M
1.0 105 M
10
. 105 M
Kb
[NH 4 ][OH ]
( x 1.0 105 )(1.0 105 ) 1.0 105 x
1.8 105
[NH 3 ]
0.35 1.0 105
0.35
Assume x 1.0 105 x =
0.35 1.8 105
= 0.63 M = initial NH 4 + concentration
5
1.0 10
mass NH 4 2 SO 4 = 0.500 L×
0.63 mol NH 4 + 1 mol NH 4 2 SO 4 132.1 g NH 4 2 SO 4
×
×
1 L soln
1 mol NH 4 2 SO 4
2 mol NH 4+
Mass of (NH4)2SO4 = 21 g
5B
(M) The solution is composed of 33.05 g NaC2H3O2 3 H2O dissolved in 300.0 mL of 0.250 M
HCl. NaC2H3O2 3 H2O , a strong electrolyte, exists in solution as Na+(aq) and C2H3O2-(aq)
ions. First we calculate the number of moles of NaC2H3O2 3 H2O, which, based on the 1:1
stoichiometry, is also equal to the number of moles of C2HO- that are released into solution.
From this we can calculate the initial [C2H3O2-] assuming the solution's volume remains at
300. mL.
moles of NaC2 H 3O 2 3 H2O (and moles of C2H3O2-)
33.05 g NaC2 H 3 O 2 3H 2 O
=
0.243moles NaC 2 H 3 O 2 3H 2 O moles C 2 H 3 O 2
1 mole NaC2 H3 O2 3H 2 O
136.08 g NaC2 H3 O2 3H 2 O
0.243 mol C2 H 3O 2
0.810 M
[C2 H 3O 2 ]
0.300 L soln
(Note: [HCl] is assumed to remain unchanged at 0.250 M)
We organize this information around the balanced chemical equation, as before. We
recognize that virtually all of the HCl has been hydrolyzed and that hydronium ion will
react to produce the much weaker acetic acid.
790
Chapter 17: Additional Aspects of Acid–Base Equilibria
+
HC2 H 3O 2 (aq) + H 2 O (l)
C2 H 3O 2 (aq) + H 3O (aq)
0.810 M
0.250 M
0M
+0.250 M
–0.250 M
–0.250 M
0.250 M
0.560 M
0 M
+x M
+x M
Changes:
–x M
Equil:
(0.250 – x) M
(0.560 + x) M
+x M
+
[H O ][C2 H 3O 2 ] x (0.560 x)
0.560 x
Ka 3
1.8 105
[HC2 H 3O 2 ]
0.250 x
0.250
Equation:
Initial:
Form HAc:
1.8 105 0.250
8.0 106 M = [H 3O + ]
0.560
(The approximation was valid since x << both 0.250 and 0.560 )
pH = log [H3 O + ] log 8.0 106 5.09 5.1
x
6A
(D)
(a)
(b)
For formic acid, pKa log (18
. 104 ) 3.74. The Henderson-Hasselbalch equation
provides the pH of the original buffer solution:
[CHO 2 ]
0.350
pH = pKa log
3.74 log
354
.
[ HCHO 2 ]
0.550
The added acid can be considered completely reacted with the formate ion to produce
formic acid. Each mole/L of added acid consumes 1 M of formate ion and forms 1 M of
+
HCHO2 (aq) + H 2 O(l). Kneut = Kb/Kw ≈
formic acid: CHO2 (aq) + H3 O (aq)
5600. Thus, CHO 2 0.350 M 0.0050 M = 0.345 M and
HCHO2 = 0.550 M + 0.0050 M = 0.555 M . By using the Henderson-Hasselbalch
equation
[CHO 2 ]
0.345
pH = pK a + log
= 3.74 + log
= 3.53
[HCHO2 ]
0.555
(c) Added base reacts completely with formic acid producing, an equivalent amount of
formate ion. This continues until all of the formic acid is consumed.
Each 1 mole of added base consumes 1 mol of formic acid and forms 1 mol of
formate ion: HCHO 2 + OH
CHO 2 + H 2O . Kneut = Ka/Kw ≈ 1.8 × 1010. Thus,
CHO 2 = 0.350 M + 0.0050 M = 0.355 M
HCHO2 = 0.550 0.0050 M = 0.545 M . With the Henderson-Hasselbalch equation
CHO 2
0.355
we find pH = pK a + log
= 3.74 + log
= 3.55
HCHO
2
791
0.545
Chapter 17: Additional Aspects of Acid–Base Equilibria
6B
(D) The buffer cited has the same concentration as weak acid and its anion, as does the
buffer of Example 17-6. Our goal is to reach pH = 5.03 or
c
h
H 3O + = 10 pH = 105.03 = 9.3 106 M . Adding strong acid H 3O + , of course, produces
HC2 H 3O 2 at the expense of C 2 H 3O 2 . Thus, adding H+ drives the reaction to the left.
Again, we use the data around the balanced chemical equation.
Equation:
HC2 H 3O 2 aq + H 2O(l)
+ H 3O + aq
C 2 H 3O 2 aq
b g
b g
Initial:
Add acid:
Form HAc:
0.250 M
0.560 M
+y M
y M
b0.250 + yg M
Changes:
Equil:
b
x M
0.250 + y x M
g
b0.560 yg M
b
x M
0.560 y + x M
g
8.0 106 M
+y M
y M
0 M
x M
9.3 106 M
Ka
[H 3 O + ][C 2 H 3 O 2 ] 9.3 106 (0.560 y x)
9.3 10 6 (0.560 y )
1.8 105
[HC 2 H 3 O 2 ]
0.250 y x
0.250 y
(Assume that x is negligible compared to y)
b
g
1.8 105 0.250 + y
0.560 0.484
= 0.484 +1.94 y = 0.560 y
y=
= 0.026 M
6
1.94 +1.00
9.3 10
Notice that our assumption is valid: x 0.250 + y = 0.276 0.560 y = 0.534 .
VHNO3 300.0 mL buffer
0.026 mmol H 3O + 1 mL HNO3 aq
= 1.3 mL of 6.0 M HNO3
1 mL buffer
6.0 mmol H 3O +
Instead of the algebraic solution, we could have used the Henderson-Hasselbalch
equation, since the final pH falls within one pH unit of the pKa of acetic acid. We let z
indicate the increase in HC 2 H 3O 2 , and also the decrease in C 2 H 3O 2
pH = pKa + log L
C 2 H 3O 2
0.560 z
0.560 z
= 105.03 4.74 = 1.95
0.250 + z
= 4.74 + log
= 5.03
0.250 + z
MN HC 2 H 3O 2 OPQ
0.560 0.488
= 0.024 M
1.95 +1.00
This is, and should be, almost exactly the same as the value of y we obtained by the
I.C.E. table method. The slight difference is due to imprecision arising from rounding
errors.
b
g
0.560 z = 1.95 0.250 + z = 0.488 +1.95 z
7A
(D)
(a)
z=
The initial pH is the pH of 0.150 M HCl, which we obtain from H 3O + of that
strong acid solution.
[H 3O + ]
.
mol HCl 1 mol H 3O +
0150
0150
.
M,
1 L soln
1 mol HCl
pH = log [H 3O + ] log (0150
. ) 0.824
792
Chapter 17: Additional Aspects of Acid–Base Equilibria
(b)
To determine H 3O + and then pH at the 50.0% point, we need the volume of the
solution and the amount of H 3O + left unreacted. First we calculate the amount of
hydronium ion present and then the volume of base solution needed for its
complete neutralization.
amount H 3O + = 25.00 mL
Vacid = 3.75 mmol H 3O +
0.150 mmol HCl 1 mmol H 3O +
= 3.75 mmol H 3O +
1 mL soln
1 mmol HCl
1 mmol OH 1 mmol NaOH
1 mL titrant
+
1 mmol H3O
1 mmol OH
0.250 mmol NaOH
15.0 mL titrant
At the 50.0% point, half of the H3O+ (1.88 mmol H 3O + ) will remain unreacted and
only half (7.50 mL titrant) of the titrant solution will be added. From this information,
and the original 25.00-mL volume of the solution, we calculate H 3O + and then pH.
1.88 mmol H 3O left
H 3O + =
= 0.0578 M
25.00 mL original + 7.50 mL titrant
pH = log 0.0578 = 1.238
(c)
Since this is the titration of a strong acid by a strong base, at the equivalence
point, the pH = 7.00 . This is because the only ions of appreciable concentration in
the equivalence point solution are Na+(aq) and Cl-(aq), and neither of these
species undergoes detectable hydrolysis reactions.
(d)
Beyond the equivalence point, the solution pH is determined almost entirely by
the concentration of excess OH-(aq) ions. The volume of the solution is
40.00 mL +1.00 mL = 41.00 mL. The amount of hydroxide ion in the excess
titrant is calculated and used to determine OH , from which pH is computed.
amount of OH = 1.00 mL
OH =
0.250 mmol NaOH
= 0.250 mmol OH
1 mL
0.250 mmol OH
= 0.006098 M
41.00 mL
pOH = log 0.006098 = 2.215; pH = 14.00 2.215 = 11.785
7B
(D)
(a)
b g
The initial pH is simply the pH of 0.00812 M Ba OH 2 , which we obtain from
OH
for the solution.
OH =
b g
0.00812 mol Ba OH
1 L soln
2
2 mol OH
1 mol Ba OH
b g
= 0.01624 M
2
pOH = log OH = log 0.0162 = 1.790; pH = 14.00 pOH = 14.00 1.790 = 12.21
793
Chapter 17: Additional Aspects of Acid–Base Equilibria
(b)
To determine OH and then pH at the 50.0% point, we need the volume of the
solution and the amount of OH unreacted. First we calculate the amount of
hydroxide ion present and then the volume of acid solution needed for its complete
neutralization.
amount OH = 50.00 mL
Vacid = 0.812 mmol OH
0.00812 mmol Ba OH 2
1mL soln
2 mmol OH
1mmol Ba OH 2
= 0.812 mmol OH
+
1mmol H 3 O
1mmol HCl
1mL titrant
= 32.48 mL titrant
+
1mmol OH
1mmol H 3 O 0.0250 mmol HCl
At the 50.0 % point, half (0.406 mmol OH ) will remain unreacted and only half
(16.24 mL titrant) of the titrant solution will be added. From this information, and
the original 50.00-mL volume of the solution, we calculate OH and then pH.
OH =
0.406 mmol OH left
= 0.00613 M
50.00 mL original +16.24 mL titrant
pOH = log 0.00613 = 2.213; pH = 14.00 pOH = 11.79
(c)
8A
(D)
(a)
Since this is the titration of a strong base by a strong acid, at the equivalence
point,pH = 7.00. The solution at this point is neutral because the dominant ionic
species in solution, namely Ba2+(aq) and Cl-(aq), do not react with water to a
detectable extent.
c
h
Initial pH is just that of 0.150 M HF ( pKa = log 6.6 104 = 3.18 ).
[Initial solution contains 20.00 mL
Equation : HF aq
+
Initial : 0.150 M
Changes : x M
Equil:
(0.150 x) M
0.150 mmol HF
=3.00 mmol HF]
1 mL
+
H 2O(l)
H 3O aq
0M
+x M
+
xM
F aq
0M
+x M
xM
H 3 O F
xx
x2
Ka =
=
= 6.6 104
0.150 x 0.150
HF
+
-
x 0.150 6.6 104 9.9 103 M
x 0.05 0.150 . The assumption is invalid. After a second cycle of
approximation, H 3O + = 9.6 103 M ; pH = log 9.6 103 = 2.02
794
Chapter 17: Additional Aspects of Acid–Base Equilibria
(b)
When the titration is 25.0% complete, there are (0.25×3.00=) 0.75 mmol F for every
3.00 mmol HF that were present initially. Therefore, (3.00-0.75=) 2.25 mmol HF
remain untitrated. We designate the solution volume (the volume holding these
3.00 mmol total) as V and use the Henderson-Hasselbalch equation to find the
pH.
F
0.75 mmol / V
= 2.70
pH = pKa + log
= 3.18 + log
HF
2.25 mmol / V
(c)
At the midpoint of the titration of a weak base, pH = pKa = 3.18.
(d)
At the endpoint of the titration, the pH of the solution is determined by the
conjugate base hydrolysis reaction. We calculate the amount of anion and the
volume of solution in order to calculate its initial concentration.
0.150 mmol HF 1 mmol F
amount F = 20.00 mL
= 3.00 mmol F
1 mL soln
1 mmol HF
1mmol OH
1 mL titrant
volume titrant = 3.00 mmol HF
= 12.0 mL titrant
1mmol HF 0.250 mmol OH
3.00 mmol F
F =
= 0.0938 M
20.00 mL original volume +12.0 mL titrant
We organize the solution of the hydrolysis problem around its balanced equation.
F aq
+ H 2O(l)
0.0938M
xM
Equation :
Initial:
Changes :
0.0938 x M
Equil :
Kb =
HF OH
F
HF aq
+ OH aq
0M
+x M
0M
+x M
xM
xM
K w 1.0 1014
x x
x2
11
=
=
= 1.5 10
=
Ka
6.6 104
0.0938 x 0.0938
x 0.0934 1.5 10 11 1.2 10 6 M = [OH ]
The assumption is valid (x 0.0934).
pOH = log 1.2 106 = 5.92; pH = 14.00 pOH = 14.00 5.92 = 8.08
8B
(D)
(a)
The initial pH is simply that of 0.106 M NH 3 .
+
Equation: NH 3 aq + H 2 O(l)
NH 4 aq + OH aq
Initial:
0.106 M
Changes: x M
Equil:
0.106 x M
b
0M
+x M
xM
g
795
0 M
+x M
x M
Chapter 17: Additional Aspects of Acid–Base Equilibria
NH 4 + OH
=
Kb =
NH
3
xx
0.106 x
x2
0.106
= 1.8 105
x 0.106 1.8 10 4 1.4 103 M = [OH ]
-
The assumption is valid (x << 0.106).
pOH = log 0.0014 = 2.85 pH = 14.00 – pOH = 14.00 – 2.85 = 11.15
(b)
+
When the titration is 25.0% complete, there are 25.0 mmol NH 4 for every 100.0
mmol of NH 3 that were present initially (i.e., there are 1.33 mmol of NH4+ in
solution), 3.98 mmol NH 3 remain untitrated. We designate the solution volume
(the volume holding these 5.30 mmol total) as V and use the basic version of the
Henderson-Hasselbalch equation to find the pH.
1.33 mmol
NH 4 +
= 4.74 + log
V
pOH = pK b + log
= 4.26
3.98
mmol
NH
3
V
pH = 14.00 4.26 = 9.74
(c)
At the midpoint of the titration of a weak base, pOH = pK b = 4.74 and pH = 9.26
(d)
At the endpoint of the titration, the pH is determined by the conjugate acid
hydrolysis reaction. We calculate the amount of that cation and the volume of the
solution in order to determine its initial concentration.
+
amount NH 4 + = 50.00 mL
0.106 mmol NH 3 1 mmol NH 4
1 mL soln
1 mmol NH 3
amount NH 4 + = 5.30 mmol NH 4
Vtitrant = 5.30 mmol NH 3
+
1 mmol H 3O +
1 mL titrant
= 23.6 mL titrant
1 mmol NH 3 0.225 mmol H 3O +
+
5.30 mmol NH 4
NH 4 + =
50.00 mL original volume + 23.6 mL titrant = 0.0720 M
We organize the solution of the hydrolysis problem around its balanced chemical
equation.
+
+
Equation: NH 4 aq + H 2 O(l)
NH3 aq + H 3O aq
0M
0 M
Initial:
0.0720 M
Changes:
x M
+x M
+x M
Equil:
0.0720 x M
xM
x M
b
g
796
Chapter 17: Additional Aspects of Acid–Base Equilibria
Kb =
NH 3 H 3O +
NH 4
+
Kw 1.0 1014
xx
x2
10
=
5.6
10
=
=
=
Kb 1.8 10 5
0.0720 x 0.0720
x 0.0720 5.6 1010 6.3 10 6 M=[H 3 O + ]
The assumption is valid (x << 0.0720).
c
h
pH = log 6.3 106 = 5.20
9A
(M) The acidity of the solution is principally the result of the hydrolysis of the carbonate
ion, which is considered first.
Equation:
Initial:
Changes:
Equil:
+ H 2 O(l)
HCO3 aq + OH aq
0M
0 M
1.0 M
x M
+x M
+x M
1.0 x M
x M
x M
CO3
b
2
aq
g
HCO3 OH
xx
x2
1.0 1014
Kw
4
Kb =
=
=
2.1
10
=
11
1.0 x 1.0
CO32
K a (HCO3 ) 4.7 10
x 1.0 2.1 104 1.5 102 M 0.015 M [OH ] The assumption is valid (x 1.0 M ).
Now we consider the hydrolysis of the bicarbonate ion.
Equation:
Initial:
Changes:
Equil:
HCO3 aq + H 2 O(l)
0.015 M
y M
0.015 y M
b
g
H 2CO3 aq + OH aq
0M
+y M
y M
0.015 M
+y M
0.015 + y M
b
g
b
g
H 2 CO 3 OH
y 0.015 + y
0.015 y
1.0 1014
Kw
8
=y
=
2.3
10
=
=
=
Kb = F
7
0.015 x
0.015
Ka H H 2 CO 3 IK 4.4 10
HCO 3
The assumption is valid (y << 0.015) and y = H 2 CO 3 = 2.3 108 M. Clearly, the
second hydrolysis makes a negligible contribution to the acidity of the solution. For the
entire solution, then
b
g
pOH = log OH = log 0.015 = 1.82
9B
pH = 14.00 1.82 = 12.18
(M) The acidity of the solution is principally the result of hydrolysis of the sulfite ion.
2
Equation:
SO3 aq + H 2 O(l)
HSO3 aq + OH aq
Initial:
Changes:
Equil:
0.500 M
x M
0.500 x M
b
g
0M
+x M
xM
797
0 M
+x M
x M
Chapter 17: Additional Aspects of Acid–Base Equilibria
HSO3 OH
Kw
x x
x2
1.0 1014
7
Kb =
=
=
1.6
10
6.2 108
[SO32 ]
0.500 x 0.500
K a HSO3
x 0.500 1.6 107 2.8 104 M = 0.00028 M = [OH ]
The assumption is valid (x << 0.500).
Next we consider the hydrolysis of the bisulfite ion.
H 2SO3 aq + OH aq
Equation: HSO3 aq + H 2O(l)
Initial:
0.00028 M
0M
0.00028 M
+y M
+y M
Changes: y M
Equil:
0.00028 y M
y M
0.00028 + y M
b
Kb =
b
g
g
Kw
1.0 1014
=
= 7.7 1013
2
K a H 2SO3 1.3 10
K b = 7.7 10
13
=
H 2SO3 OH
HSO3
=
y 0.00028 + y 0.00028 y
=y
0.00028 y
0.00028
The assumption is valid (y << 0.00028) and y = H 2 SO3 = 7.7 1013 M. Clearly, the
second hydrolysis makes a negligible contribution to the acidity of the solution. For the
entire solution, then
b
g
pOH = log OH = log 0.00028 = 3.55
pH = 14.00 3.55 = 10.45
INTEGRATIVE EXAMPLE
A.
(D)
From the given information, the following can be calculated:
pH of the solution = 2.716 therefore,
[H+] = 1.92×10-3
pH at the halfway point = pKa
pH = 4.602 = pKa
pKa = -log Ka therefore Ka = 2.50×10-5
FP = 0 C + Tf
Tf = -i K f m
Tf -1 1.86 C / m molality
798
Chapter 17: Additional Aspects of Acid–Base Equilibria
molality =
# moles solute
# moles solute
=
kg solvent
0.500 kg - 0.00750 kg
To determine the number of moles of solute, convert 7.50 g of unknown acid to moles by
using its molar mass. The molar mass can be calculated as follows:
[H+] = 1.92×10-3
pH = 2.716
HA
7.50 g
MM
Initial
Change
Equilibrium
0.500 L
–x
7.50 g
MM
1.92 103
0.500 L
2.50 10-5 =
H+
+
0
0
x
x
1.92×10-3
1.92×10-3
[H + ][A - ]
[HA]
2.50 10-5
(1.92 103 ) 2
7.50
MM 1.92 103
0.500
MM =100.4 g/mol
# moles of solute = 7.50 g
Molality =
1 mol
0.0747 mol
100.4 g
# moles solute
0.0747 mol
0.152 m
=
Kg solvent
0.500 Kg - 0.00750 Kg
Tf = -i K f m
Tf -1 1.86 C / m 0.152 m
Tf -0.283 C
FP = 0C + Tf = 0C 0.283 C = 0.283 C
799
A-
Chapter 17: Additional Aspects of Acid–Base Equilibria
B.
(M) By looking at the titration curve provided, one can deduce that the titrant was a strong acid.
The pH before titrant was added was basic, which means that the substance that was titrated was
a base. The pH at the end of the titration after excess titrant was added was acidic, which means
that the titrant was an acid.
Based on the titration curve provided, the equivalence point is at approximately 50 mL of
titrant added. At the halfway point, of approximately 25 mL, the pH = pKa. A pH ~8 is
obtained by extrapolation a the halfway point.
pH = 8 = pKa
Ka = 10-8 = 1 × 10-8
Kb = Kw/Ka = 1 × 10-6
~50 mL of 0.2 M strong acid (1×10-2 mol) was needed to reach the equivalence point. This
means that the unknown contained 1×10-2 mol of weak base. The molar mass of the unknown
can be determined as follows:
0.800 g
80 g/mol
1 102 mol
EXERCISES
The Common-Ion Effect
1.
(M)
(a) Note that HI is a strong acid and thus the initial H 3O + = HI = 0.0892 M
Equation :
Initial :
Changes :
Equil :
HC3 H 5 O 2 +
0.275 M
x M
0.275 x M
+
H2O
C3 H 5 O 2 + H 3 O
0M
0.0892M
+x M
+x M
xM
0.0892 + x M
C3 H 5 O 2 H 3O +
x 0.0892 + x 0.0892 x
Ka =
= 1.3 105 =
HC3 H 5O 2
0.275 x
0.275
The assumption that x 0.0892 M is correct. H 3O + = 0.0892 M
Kw
1.0 1014
=
= 1.1 1013 M
+
0.0892
H 3O
(b)
OH =
(c)
C3 H 5O 2 = x = 4.0 105 M
800
x 4.0 105 M
Chapter 17: Additional Aspects of Acid–Base Equilibria
(d)
2.
I = HI int = 0.0892 M
(M)
(a) The NH 4 Cl dissociates completely, and thus, NH +4 = Cl = 0.102 M
int
int
Equation: NH 3 (aq) + H 2 O(l)
NH 4 (aq) + OH (aq)
Initial:
0.164 M
Changes:
x M
Equil:
0.164 x M
b
b
g
0.102 M
+x M
0.102 + x M
g
0 M
+x M
x M
NH +4 OH 0.102 + x x
0.102 x
5
Kb =
=
=
1.8
10
; x = 2.9 105 M
NH 3
0.164 x
0.164
Assumed x 0.102 M , a clearly valid assumption. OH = x = 2.9 10 5 M
(b)
(c)
(d)
3.
NH 4 = 0.102 + x = 0.102 M
Cl = 0.102 M
H 3O
+
1.0 1014
=
= 3.4 1010 M
5
2.9 10
(M)
(a) We first determine the pH of 0.100 M HNO 2 .
Equation HNO 2 (aq) + H 2 O(l)
NO 2 (aq) +
Initial :
0.100 M
xM
Changes :
Equil :
0.100 x M
H 3 O + (aq)
0M
0M
+x M
+ xM
xM
xM
NO 2 H 3 O +
x2
4
Ka =
=
7.2
10
=
HNO 2
0.100 x
Via the quadratic equation roots formula or via successive approximations,
x = 8.1 103 M = H 3O + .
Thus pH = log 8.1 103 = 2.09
When 0.100 mol NaNO 2 is added to 1.00 L of a 0.100 M HNO 2 , a solution with
NO 2 = 0.100M = HNO 2 is produced. The answer obtained with the Henderson-
c
h
Hasselbalch equation, is pH = pKa = log 7.2 104 = 3.14 . Thus, the addition has
caused a pH change of 1.05 units.
801
Chapter 17: Additional Aspects of Acid–Base Equilibria
(b) NaNO 3 contributes nitrate ion, NO 3 , to the solution. Since, however, there is no
molecular HNO 3 aq in equilibrium with hydrogen and nitrate ions, there is no
equilibrium to be shifted by the addition of nitrate ions. The [H3O+] and the pH are thus
unaffected by the addition of NaNO 3 to a solution of nitric acid. The pH changes are not
the same because there is an equilibrium system to be shifted in the first solution, whereas
there is no equilibrium, just a change in total ionic strength, for the second solution.
b g
4.
(M) The explanation for the different result is that each of these solutions has acetate
ion present, C 2 H 3O 2 , which is produced in the ionization of acetic acid. The presence
of this ion suppresses the ionization of acetic acid, thus minimizing the increase in
H 3O + . All three solutions are buffer solutions and their pH can be found with the aid
of the Henderson-Hasselbalch equation.
(a)
pH = pKa + log
% ionization =
(b)
C 2 H 3O 2
HC 2 H 3O 2
H 3O
HC 2 H 3O 2
= 4.74 + log
100%
0.10
= 3.74
1.0
H 3O + = 103.74 = 1.8 104 M
1.8 104 M
100% 0.018%
1.0M
C 2 H 3O 2
= 4.74 + log 0.10 = 4.74
pH = pK a + log
0.10
HC 2 H 3 O 2
H 3O + = 104.74 = 1.8 105 M
H 3O
1.8 105 M
% ionization =
100%
100% 0.018%
HC 2 H 3O 2
0.10M
(c)
C 2 H 3O 2
= 4.74 + log 0.10 = 5.74
pH = pK a + log
0.010
HC 2 H 3O 2
H 3O + = 105.74 = 1.8 106 M
% ionization =
5.
H 3O
HC 2 H 3O 2
100%
1.8 106 M
100% 0.018%
0.010M
(M)
(a) The strong acid HCl suppresses the ionization of the weak acid HOCl to such an
extent that a negligible concentration of H 3O + is contributed to the solution by
HOCl. Thus, H 3O + = HCl = 0.035 M
802
Chapter 17: Additional Aspects of Acid–Base Equilibria
(b)
This is a buffer solution. Consequently, we can use the Henderson-Hasselbalch
equation to determine its pH.
pK a = log 7.2 104 = 3.14;
NO 2
= 3.14 + log 0.100 M = 3.40
pH = pK a + log
HNO 2
0.0550 M
H 3O + = 103.40 = 4.0 104 M
(c)
This also is a buffer solution, as we see by an analysis of the reaction between the
components.
H 3O + aq, from HCl + C 2 H 3O 2 aq, from NaC 2 H 3 O 2 HC2 H 3O 2 aq + H 2 O(l)
Equation:
In soln:
Produce HAc:
Initial:
0.0525 M
0.0525 M
0M
0.0768 M
0.0525 M
0.0243 M
0M
+0.0525 M
0.0525 M
Now the Henderson-Hasselbalch equation can be used to find the pH.
c
h
pKa = log 1.8 105 = 4.74
C2 H3 O2
= 4.74 + log 0.0243M = 4.41
pH = pK a + log
HC2 H 3 O 2
0.0525 M
H 3 O + = 104.41 = 3.9 105 M
6.
(M)
(a) Neither Ba 2+ aq nor Cl aq hydrolyzes to a measurable extent and hence they
b g
b g
have no effect on the solution pH. Consequently, OH is determined entirely
b g
by the Ba OH
OH =
(b)
2
solute.
0.0062 mol Ba OH 2
1 L soln
2 mol OH
= 0.012 M
1mol Ba OH 2
We use the Henderson-Hasselbalch equation to find the pH for this buffer
solution.
+
2 mol NH 4
NH 4 + = 0.315 M NH 4 SO 4
= 0.630 M
2
1 mol NH 4 2 SO 4
pH = pK a + log
NH3
NH 4 +
= 9.26 + log
OH = 104.85 = 1.4 105 M
803
+
pK a = 9.26 for NH 4 .
0.486 M
= 9.15 pOH = 14.00 9.15 = 4.85
0.630 M
Chapter 17: Additional Aspects of Acid–Base Equilibria
(c)
This solution also is a buffer, as analysis of the reaction between its components shows.
Equation: NH 4 aq, from NH 4 Cl + OH aq, from NaOH NH 3 aq + H 2 O(l)
In soln:
Form NH 3 :
Initial:
0.264 M
0.196 M
0.068 M
0.196 M
0.196 M
0M
0M
+0.196 M
0.196 M
NH3 = 9.26 + log 0.196 M = 9.72
pH = pK a + log
NH +4
pOH = 14.00 9.72 = 4.28
0.068 M
Buffer Solutions
7.
(M) H 3O + = 104.06 = 8.7 105 M . We let S = CHO 2
int
Equation: HCHO 2 aq + H 2 O(l)
Initial :
0.366 M
5
CHO 2 aq
0M
SM
5
+8.7 105 M
Changes : 8.7 10 M
+ 8.7 10 M
Equil :
S + 8.7 10 M
Ka =
0.366 M
H 3O + CHO 2
5
LM HCHO 2 OP
Q
N
4
8.7 105 M
cS + 8.7 10 h8.7 10
=
5
= 1.8 10
H 3 O + aq
+
5
0.366
8.7 105 S
; S = 0.76 M
0.366
To determine S , we assumed S 8.7 105 M , which is clearly a valid
assumption. Or, we could have used the Henderson-Hasselbalch equation (see
below). pKa = log 1.8 104 = 3.74
c
h
CHO 2
;
4.06 = 3.74 + log
HCHO 2
CHO 2
= 2.1;
HCHO 2
CHO 2 = 2.1 0.366 = 0.77 M
The difference in the two answers is due simply to rounding.
8.
(E) We use the Henderson-Hasselbalch equation to find the required [NH3].
c
h
pKb = log 1.8 105 = 4.74
pKa = 14.00 pKb = 14.00 4.74 = 9.26
NH3
NH 4
+
= 100.14 = 0.72
pH = 9.12 = 9.26 + log
NH 3
NH 4
+
NH3 = 0.72 NH 4+ = 0.72 0.732 M = 0.53M
804
Chapter 17: Additional Aspects of Acid–Base Equilibria
9.
(M)
(a) Equation: HC7 H 5O 2 aq + H 2 O(l)
C7 H 5O 2 (aq) +
0.033 M
Initial:
0.012 M
+x M
Changes:
x M
0.033 + x M
Equil:
0.012 x M
b
H 3O + (aq)
0 M
+x M
x M
g
H 3O + C7 H 5O 2
x 0.033 + x 0.033x
5
Ka =
=
6.3
10
=
HC7 H 5O 2
0.012 x
0.012
x = 2.3 105 M
To determine the value of x , we assumed x 0.012 M, which is an assumption
that clearly is correct.
H 3O + = 2.3 105 M
pH = log 2.3 105 = 4.64
c
(b)
Initial :
Changes :
NH 3 aq + H 2 O(l)
0.408 M
x M
Equil :
0.408 x M
Equation:
Kb =
NH 4
+
OH
LM NH 3 OP
N
Q
NH 4 (aq)
+ OH (aq)
0.153 M
+x M
0M
+xM
0.153 + x M
= 1.8 105 =
h
b
g
x 0.153 + x
0.153x
0.408 x
0.408
xM
x = 4.8 105 M
To determine the value of x , we assumed x 0.153 , which clearly is a valid
assumption.
OH = 4.8 105 M; pOH = log 4.8 105 = 4.32;
10.
pH = 14.00 4.32 = 9.68
(M) Since the mixture is a buffer, we can use the Henderson-Hasselbalch equation to
determine Ka of lactic acid.
1.00 g NaC3 H 5 O3
1000 mL
1 mol NaC3 H 5 O3
1 mol C3 H 5 O 3
C3 H 5 O3 =
= 0.0892 M
100.0 mL soln
1 L soln 112.1 g NaC3 H 5 O 3 1 mol NaC3 H 5 O 3
C3 H 5 O 3
= pK + log 0.0892 M = pK + 0.251
pH = 4.11 = pK a + log
a
a
0.0500 M
HC3 H 5 O3
pK a = 4.11 0.251 = 3.86; K a = 103.86 = 1.4 104
11.
(M)
(a) 0.100 M NaCl is not a buffer solution. Neither ion reacts with water to a
detectable extent.
(b)
0.100 M NaCl—0.100 M NH 4 Cl is not a buffer solution. Although a weak acid,
NH 4+ , is present, its conjugate base, NH3, is not.
805
Chapter 17: Additional Aspects of Acid–Base Equilibria
12.
(c)
0.100 M CH 3 NH 2 and 0.150 M CH 3 NH 3+ Cl is a buffer solution. Both the weak
base, CH 3 NH 2 , and its conjugate acid, CH 3NH 3+ , are present in approximately
equal concentrations.
(d)
0.100 M HCl—0.050 M NaNO 2 is not a buffer solution. All the NO 2 has
converted to HNO 2 and thus the solution is a mixture of a strong acid and a weak
acid.
(e)
0.100 M HCl—0.200 M NaC2 H 3O 2 is a buffer solution. All of the HCl reacts
with half of the C 2 H 3O 2 to form a solution with 0.100 M HC2 H 3O 2 , a weak acid,
and 0.100 M C 2 H 3O 2 , its conjugate base.
(f)
0.100 M HC 2 H 3O 2 and 0.125 M NaC 3 H 5O 2 is not a buffer in the strict sense
because it does not contain a weak acid and its conjugate base, but rather the
conjugate base of another weak acid. These two weak acids (acetic,
Ka = 1.8 10 5 , and propionic, Ka = 1.35 105 ) have approximately the same
strength, however, this solution would resist changes in its pH on the addition of
strong acid or strong base, consequently, it could be argued that this system
should also be called a buffer.
(M)
2
(a) Reaction with added acid: HPO 4 + H 3O + H 2 PO 4 + H 2O
Reaction with added base: H 2 PO 4 + OH
HPO 4
(b)
2
+ H2O
We assume initially that the buffer has equal concentrations of the two ions,
H 2 PO 4 = HPO 4 2
HPO 4 2
pH = pK a 2 + log
= 7.20 + 0.00 = 7.20 (pH at which the buffer is most effective).
H 2 PO4
(c)
13.
HPO 4 2
= 7.20 + log 0.150 M = 7.20 + 0.48 = 7.68
pH = 7.20 + log
0.050 M
H 2 PO 4
1 mol C6 H 5 NH 3+ Cl
1 mol C6 H 5 NH 3+
1g
(M) moles of solute = 1.15 mg
1000 mg
129.6 g
1mol C6 H 5 NH 3+ Cl
= 8.87 106 mol C6 H 5 NH 3+
C 6 H 5 NH 3
+
8.87 106 mol C 6 H 5 NH 3 +
=
= 2.79 10 6 M
3.18 L soln
Equation:
C6 H 5 NH 3+ aq +
C6 H 5 NH 2 aq + H 2 O(l)
Initial:
Changes:
Equil:
0.105 M
x M
0.105 x M
b
g
2.79 106 M
+x M
2.79 106 + x M
c
806
h
OH aq
0M
+x M
xM
Chapter 17: Additional Aspects of Acid–Base Equilibria
Kb =
C 6 H 5 NH 3
+
OH
LMC6 H 5 NH 2 OP
Q
N
= 7.4 10
10
c2.79 10
=
6
h
+x x
0.105 x
7.4 1010 0.105 x = 2.79 106 + x x; 7.8 1011 7.4 1010 x = 2.79 106 x + x 2
x 2 + 2.79 106 + 7.4 1010 x 7.8 1011 = 0;
x=
b b 2 4ac 2.79 106 7.78 1012 + 3.1 1010
=
= 7.5 106 M = OH
2a
2
c
h
pOH = log 7.5 106 = 5.12
14.
x 2 + 2.79 106 x 7.8 1011 0
pH = 14.00 5.12 = 8.88
(M) We determine the concentration of the cation of the weak base.
1 mmol C6 H 5 NH 3 Cl
1 mmol C6 H 5 NH 3
+
129.6 g
1 mmol C6 H 5 NH 3 Cl
+
[C6 H 5 NH 3 ]
0.0874 M
1L
750 mL
1000 mL
In order to be an effective buffer, each concentration must exceed the ionization
constant Kb = 7.4 1010 by a factor of at least 100, which clearly is true. Also, the
+
+
8.50 g
c
h
ratio of the two concentrations must fall between 0.1 and 10:
+
C 6 H 5 NH 3
0.0874 M
LMC6 H 5 NH 2 OP = 0.215M = 0.407 .
Q
N
Since both criteria are met, this solution will be an effective buffer.
15.
(M)
(a) First use the Henderson-Hasselbalch equation. pK b = log 1.8 105 = 4.74,
pK a = 14.00 4.74 = 9.26 to determine NH 4 + in the buffer solution.
pH = 9.45 = pK a + log
NH3
NH
+
4
NH
3
NH 4
= 100.19 = 1.55
= 9.26 + log
NH
3
; log
NH
3
NH 4
NH 4 +
NH3 = 0.258M = 0.17M
NH 4 + =
1.55
1.55
+
+
= 9.45 9.26 = +0.19
We now assume that the volume of the solution does not change significantly when
the solid is added.
+
1 L soln 0.17 mol NH 4 1 mol NH 4 2 SO 4
mass NH 4 2 SO 4 = 425 mL
+
1000 mL
1 L soln
2 mol NH 4
132.1 g NH 4 2 SO 4
1 mol NH 4 2 SO 4
807
= 4.8 g NH 4 2 SO 4
Chapter 17: Additional Aspects of Acid–Base Equilibria
(b)
We can use the Henderson-Hasselbalch equation to determine the ratio of
concentrations of cation and weak base in the altered solution.
pH = 9.30 = pK a + log
NH3
NH
+
4
= 100.04
NH
3
= 9.26 + log
NH 4
0.258
= 1.1 =
0.17 M + x M
+
NH
3
NH 4
+
log
NH
3
NH 4 +
= 9.30 9.26 = +0.04
0.19 1.1x 0.258
x 0.062 M
The reason we decided to add x to the denominator follows. (Notice we cannot
remove a component.) A pH of 9.30 is more acidic than a pH of 9.45 and
therefore the conjugate acid’s NH 4 + concentration must increase. Additionally,
d
i
mathematics tells us that for the concentration ratio to decrease from 1.55 to 1.1,
its denominator must increase. We solve this expression for x to find a value of
0.062 M. We need to add NH4+ to increase its concentration by 0.062 M in 100
mL of solution.
NH 4 2 SO 4
mass = 0.100 L
0.062 mol NH 4+
1L
= 0.41g NH 4 2 SO 4
16.
1mol NH 4 2 SO 4
2 mol NH +4
132.1g
1mol
NH 4 2 SO 4
NH 4 2 SO 4
Hence, we need to add 0.4 g
(M)
(a)
nHC7 H5O2 = 2.00 g HC7 H 5O 2
1mol HC7 H 5O 2
= 0.0164 mol HC7 H 5O 2
122.1g HC7 H 5O 2
nC H O = 2.00 g NaC7 H 5O 2
1mol NaC7 H 5O 2
1mol C7 H 5O 2
144.1g NaC7 H 5O 2 1mol NaC7 H 5O 2
7
5
2
= 0.0139 mol C7 H 5O 2
C 7 H 5 O 2
= log 6.3 10 5 + log 0.0139 mol C7 H 5 O 2 /0.7500 L
pH = pK a + log
HC H O
0.0164 mol HC 7 H 5 O 2 /0.7500 L
7
5 2
= 4.20 0.0718 = 4.13
(b)
To lower the pH of this buffer solution, that is, to make it more acidic, benzoic acid
must be added. The quantity is determined as follows. We use moles rather than
concentrations because all components are present in the same volume of solution.
4.00 = 4.20 + log
0.0139 mol C 7 H 5O 2
x mol HC 7 H 5O 2
0.0139 mol C7 H 5O 2
= 100.20 = 0.63
x mol HC7 H 5O 2
log
x=
808
0.0139 mol C 7 H 5O 2
= 0.20
x mol HC 7 H 5O 2
0.0139
= 0.022 mol HC7 H 5O 2 (required)
0.63
Chapter 17: Additional Aspects of Acid–Base Equilibria
HC7 H 5O 2 that must be added = amount required amount already in solution
HC7 H 5O 2 that must be added = 0.022 mol HC7 H 5O 2 0.0164 mol HC7 H 5O 2
HC7 H 5O 2 that must be added = 0.006 mol HC7 H 5O 2
122.1g HC 7 H 5O 2
added mass HC 7 H 5O 2 = 0.006 mol HC 7 H 5O 2
= 0.7g HC 7 H 5O 2
1 mol HC 7 H 5O 2
17.
(M) The added HCl will react with the ammonia, and the pH of the buffer solution will
decrease. The original buffer solution has NH 3 = 0.258 M and NH 4 = 0.17 M .
We first calculate the [HCl] in solution, reduced from 12 M because of dilution. [HCl]
0.55 mL
added = 12 M
= 0.066 M We then determine pKa for ammonium ion:
100.6 mL
c
h
pKb = log 1.8 105 = 4.74 pKa = 14.00 4.74 = 9.26
+
H 3O + aq
NH 4 aq + H 2 O l
Buffer:
0.258 M
≈0M
0.17 M
Added:
+0.066 M
Changes:
0.066 M
0.066 M
+0.066 M
Final:
0.192 M
0M
0.24 M
NH 3
0.192
pH = pKa + log L
= 9.26 + log
= 9.16
+O
0.24
MN NH 4 PQ
Equation:
18.
NH 3 aq +
(M) The added NH 3 will react with the benzoic acid, and the pH of the buffer solution
will increase. Original buffer solution has [C7H5O2] = 0.0139 mol C7H5O2/0.750 L =
0.0185 M and HC7 H 5O 2 = 0.0164 mol HC7 H 5O 2 /0.7500 L = 0.0219 M . We first
calculate the NH 3 in solution, reduced from 15 M because of dilution.
NH3 added = 15 M
0.35 mL
= 0.0070 M
750.35 mL
c
h
For benzoic acid, pKa = log 6.3 105 = 4.20
Equation: NH 3 aq
Buffer:
0M
Added: 0.0070 M
Changes: 0.0070M
Final:
0.000 M
+ HC7 H 5O 2 aq
NH 4+ aq + C7 H 5O 2 aq
0.0219 M
0M
0.0070 M
0.0149 M
+ 0.0070 M
0.0070 M
C7 H 5 O 2
= 4.20 + log 0.0255 = 4.43
pH = pK a + log
HC7 H 5O 2
0.0149
809
0.0185 M
+ 0.0070 M
0.0255 M
Chapter 17: Additional Aspects of Acid–Base Equilibria
19.
(M) The pKa ’s of the acids help us choose the one to be used in the buffer. It is the acid
with a pKa within 1.00 pH unit of 3.50 that will do the trick. pKa = 3.74 for
HCHO 2 , pKa = 4.74 for HC2 H 3O 2 , and pK a1 = 2.15 for H 3 PO 4 . Thus, we choose
HCHO 2 and NaCHO 2 to prepare a buffer with pH = 3.50 . The Henderson-Hasselbalch
equation is used to determine the relative amount of each component present in the buffer
solution.
CHO 2
pH = 3.50 = 3.74 + log
HCHO
2
CHO 2
= 3.50 3.74 = 0.24
log
HCHO
2
CHO 2
= 100.24 = 0.58
HCHO
2
This ratio of concentrations is also the ratio of the number of moles of each component
in the buffer solution, since both concentrations are a number of moles in a certain
volume, and the volumes are the same (the two solutes are in the same solution). This
ratio also is the ratio of the volumes of the two solutions, since both solutions being
mixed contain the same concentration of solute. If we assume 100. mL of acid solution,
Vacid = 100. mL . Then the volume of salt solution is
Vsalt = 0.58 100. mL = 58 mL 0.100 M NaCHO2
20.
(D) We can lower the pH of the 0.250 M HC 2 H 3O 2 — 0.560 M C 2 H 3O 2 buffer
solution by increasing HC 2 H 3O 2 or lowering C 2 H 3O 2 . Small volumes of NaCl
b g
solutions will have no effect, and the addition of NaOH(aq) or NaC 2 H 3O 2 aq will
raise the pH. The addition of 0.150 M HCl will raise HC 2 H 3O 2 and lower C 2 H 3O 2
through the reaction H 3O + (aq) + C 2 H 3O 2 (aq)
HC 2 H 3O 2 (aq) + H 2 O(l) and bring
about the desired lowering of the pH. We first use the Henderson-Hasselbalch equation
to determine the ratio of the concentrations of acetate ion and acetic acid.
C 2 H 3O 2
pH = 5.00 = 4.74 + log L
O
NM HC 2 H 3O 2 QP
C 2 H 3O 2
C 2 H 3O 2
= 100.26 = 1.8
log
= 5.00 4.74 = 0.26;
HC2 H 3O 2
HC2 H 3O 2
Now we compute the amount of each component in the original buffer solution.
amount of C2 H 3O 2 = 300. mL
amount of HC2 H 3O 2 = 300. mL
0.560 mmol C 2 H 3O 2
1 mL soln
= 168 mmol C 2 H 3O 2
0.250 mmol HC2 H 3O 2
= 75.0 mmol HC2 H 3O 2
1mL soln
810
Chapter 17: Additional Aspects of Acid–Base Equilibria
Now let x represent the amount of H 3O + added in mmol.
1.8 =
x=
168 x
; 168 x = 1.8 75 + x = 135 +1.8 x
75.0 + x
168 135 = 2.8 x
168 135
= 12 mmol H 3O +
2.7
Volume of 0.150 M HCl = 12 mmol H 3 O +
1mmol HCl
1mL soln
+
1mmol H 3 O 0.150 mmol HCl
= 80 mL 0.150 M HCl solution
21.
(M)
(a) The pH of the buffer is determined via the Henderson-Hasselbalch equation.
C 3 H 5O 2
0.100M
=
4.89
+
= 4.89
log
O
0.100M
NM HC 3 H 5O 2 QP
pH = pKa + log L
The effective pH range is the same for every propionate buffer: from pH = 3.89
to pH = 5.89 , one pH unit on either side of pKa for propionic acid, which is 4.89.
(b)
To each liter of 0.100 M HC3 H 5O 2 — 0.100M NaC 3 H 5O 2 we can add 0.100
mol OH before all of the HC3 H 5O 2 is consumed, and we can add 0.100 mol
H 3O + before all of the C 3 H 5O 2 is consumed. The buffer capacity thus is 100.
millimoles (0.100 mol) of acid or base per liter of buffer solution.
22. (M)
(a) The solution will be an effective buffer one pH uniton either side of the pKa of
methylammonium ion, CH 3 NH 3 , K b = 4.2 104 for methylamine,
c
h
pKb = log 4.2 104 = 3.38 . For methylammonium cation,
(b)
pKa = 14.00 3.38 = 10.62 . Thus, this buffer will be effective from a pH of 9.62 to
a pH of 11.62.
The capacity of the buffer is reached when all of the weak base or all of the
conjugate acid has been neutralized by the added strong acid or strong base.
Because their concentrations are the same, the number of moles of base is equal to
the number of moles of conjugate acid in the same volume of solution.
0.0500 mmol
amount of weak base = 125 mL
= 6.25 mmol CH 3 NH 2 or CH 3 NH 3
1 mL
Thus, the buffer capacity is 6.25 millimoles of acid or base per 125 mL buffer solution.
811
Chapter 17: Additional Aspects of Acid–Base Equilibria
23.
(M)
(a) The pH of this buffer solution is determined with the Henderson-Hasselbalch equation.
CHO 2
= log 1.8 104 + log 8.5 mmol/75.0 mL
pH = pK a + log
15.5 mmol/75.0 mL
HCHO 2
= 3.74 0.26 = 3.48
[Note: the solution is not a good buffer, as CHO 2 = 1.1 101 , which is only ~
600 times Ka ]
(b)
b g
Amount of added OH = 0.25 mmol Ba OH 2
2 mmol OH
1 mmol Ba OH
b g
= 0.50 mmol OH
2
The OH added reacts with the formic acid and produces formate ion.
Equation:
HCHO 2 (aq) +
OH (aq)
CHO 2 (aq) + H 2 O(l)
Buffer:
0M
15.5 mmol
Add base:
8.5 mmol
+0.50 mmol
React:
0.50 mmol
0.50 mmol
+0.50 mmol
Final:
15.0 mmol
0 mmol
9.0 mmol
CHO 2
= log 1.8 104 + log 9.0 mmol/75.0 mL
pH = pK a + log
15.0 mmol/75.0 mL
HCHO 2
= 3.74 0.22 = 3.52
(c)
12 mmol HCl 1 mmol H 3 O +
= 13 mmol H 3O +
Amount of added H 3O = 1.05 mL acid
1 mL acid
1 mmol HCl
+
The H 3O added reacts with the formate ion and produces formic acid.
Equation:
CHO 2 (aq)
Buffer :
8.5 mmol
H 3 O + (aq)
0 mmol
HCHO 2 (aq) + H 2 O(l)
15.5 mmol
+8.5 mmol
13mmol
Add acid :
React :
+
8.5 mmol
8.5 mmol
Final :
0 mmol
4.5 mmol
24.0 mmol
The buffer's capacity has been exceeded. The pH of the solution is determined by
the excess strong acid present.
4.5 mmol
H 3O + =
= 0.059 M; pH = log 0.059 = 1.23
75.0 mL +1.05 mL
812
Chapter 17: Additional Aspects of Acid–Base Equilibria
24.
c
h
(D) For NH 3 , pKb = log 1.8 105 For NH 4 , pKa = 14.00 pKb = 14.00 4.74 = 9.26
(a)
NH3 =
1.68g NH 3 1 mol NH 3
= 0.197 M
0.500 L 17.03 g NH 3
NH 4 =
4.05 g NH 4 2 SO 4
0.500 L
1 mol NH 4 2 SO 4
2 mol NH 4
= 0.123 M
132.1 g NH 4 2 SO 4 1mol NH 4 2 SO 4
NH 3
0.197 M
pH = pKa + log L
= 9.26 + log
= 9.46
O
0.123 M
MN NH 4 PQ
(b)
b g
b g
b g
The OH aq reacts with the NH 4 aq to produce an equivalent amount of NH 3 aq .
OH =
i
0.88 g NaOH
1mol NaOH
1mol OH
= 0.044 M
0.500 L
40.00 g NaOH 1mol NaOH
NH 4 aq + OH aq
Equation:
Initial :
Add NaOH :
React :
Final :
0M
0.044 M
0.044 M
0.0000 M
0.123M
0.044 M
0.079 M
NH 3 aq +
H 2 O(l)
0.197 M
+0.044 M
0.241M
NH 3
0.241 M
log
pH = pKa + log L
=
9.26
+
= 9.74
0.079 M
MN NH 4 OPQ
(c)
NH 4 aq
NH 3 aq + H 3O + aq
0.197 M
0.123M
0M
+x M
+ xM
x M
x M
Equation:
Initial :
Add HCl :
React :
Final :
0.197 x M
b
b
b
b
b
b
g
g
NH 3
gM
gM
g M = 10
gM
b
g
0.197 x = 0.55 0.123 + x = 0.068 + 0.55 x
x=
H 2 O(l)
0.123 + x M
1109 M 0
0.197 x
log
=
9.26
+
0.123 + x
MN NH 4 OPQ
0.197 x
0.197 x M
= 9.00 9.26 = 0.26
log
0.123 + x
0.123 + x M
pH = 9.00 = pKa + log L
+
0.26
= 0.55
1.55 x = 0.197 0.068 = 0.129
0.129
= 0.0832 M
1.55
volume HCl = 0.500 L
0.0832 mol H 3O + 1mol HCl 1000 mL HCl
= 3.5 mL
1L soln
1mol H 3O + 12 mol HCl
813
Chapter 17: Additional Aspects of Acid–Base Equilibria
25.
(D)
(a)
We use the Henderson-Hasselbalch equation to determine the pH of the solution.
The total solution volume is
36.00 mL + 64.00 mL = 100.00 mL. pK a = 14.00 pK b = 14.00 + log 1.8 105 = 9.26
NH 3 =
36.00 mL 0.200 M NH 3 7.20 mmol NH 3
= 0.0720 M
=
100.0 mL
100.00 mL
NH 4 =
64.00 mL 0.200 M NH 4 12.8 mmol NH 4
=
= 0.128 M
100.00 mL
100.0 mL
pH = pK a log
(b)
[NH 3 ]
0.0720 M
9.26 log
9.01 9.00
0.128 M
[NH 4 ]
The solution has OH = 104.99 = 1.0 105 M
The Henderson-Hasselbalch equation depends on the assumption that:
NH3 1.8 105 M NH 4
If the solution is diluted to 1.00 L, NH 3 = 7.20 103 M , and
NH 4 = 1.28 102 M . These concentrations are consistent with the assumption.
However, if the solution is diluted to 1000. L, NH 3 = 7.2 106 M , and
NH 3 = 1.28 105 M , and these two concentrations are not consistent with the
assumption. Thus, in 1000. L of solution, the given quantities of NH 3 and NH 4
will not produce a solution with pH = 9.00 . With sufficient dilution, the solution
will become indistinguishable from pure water (i.e.; its pH will equal 7.00).
(c)
The 0.20 mL of added 1.00 M HCl does not significantly affect the volume of the
solution, but it does add 0.20 mL 1.00 M HCl = 0.20 mmol H 3O + . This added
H 3O + reacts with NH 3 , decreasing its amount from 7.20 mmol NH 3 to 7.00
mmol NH 3 , and increasing the amount of NH 4 from 12.8 mmol NH 4 to 13.0
mmol NH 4 , as the reaction: NH 3 + H 3O + NH 4 + H 2 O
pH = 9.26 + log
(d)
7.00 mmol NH 3 / 100.20 mL
= 8.99
13.0 mmol NH 4 / 100.20 mL
We see in the calculation of part (c) that the total volume of the solution does not
affect the pOH of the solution, at least as long as the Henderson-Hasselbalch
equation is obeyed. We let x represent the number of millimoles of H 3O + added,
814
Chapter 17: Additional Aspects of Acid–Base Equilibria
through 1.00 M HCl. This increases the amount of NH 4 and decreases the
amount of NH 3 , through the reaction NH 3 + H 3O + NH 4 + H 2 O
7.20 x
7.20 x
pH = 8.90 = 9.26 + log
; log
= 8.90 9.26 = 0.36
12.8 + x
12.8 + x
Inverting, we have:
12.8 + x
= 100.36 = 2.29; 12.8 + x = 2.29 7.20 x = 16.5 2.29 x
7.20 x
x=
16.5 12.8
= 1.1 mmol H 3O +
1.00 + 2.29
vol 1.00 M HCl = 1.1mmol H 3O +
26.
1mmol HCl
1mmol H 3O
+
1mL soln
1.00 mmol HCl
= 1.1mL 1.00 M HCl
(D)
(a)
C2 H 3 O 2 =
12.0 g NaC 2 H 3 O 2 1mol NaC 2 H 3 O 2
1mol C 2 H 3 O 2
= 0.488 M C2 H 3 O 2
0.300 L soln
82.03g NaC 2 H 3 O 2 1mol NaC 2 H 3 O 2
C2 H 3 O 2 +
HC2 H 3 O 2 + H 2 O
0M
0M
Equation :
Initial :
H3O+
0.200 M
0M
0.488 M
Consume H 3 O +
+ 0.200 M
0.200 M
0.200 M
Buffer :
0.200 M
0.288 M
0M
Add C2 H 3 O 2
0M
Then use the Henderson-Hasselbalch equation to find the pH.
pH = pKa + log L
C 2 H 3O 2
0.288 M
= 4.74 + log
= 4.74 + 0.16 = 4.90
0.200 M
MN HC2 H 3O 2 OPQ
(b)
b g
We first calculate the initial OH due to the added Ba OH 2 .
OH =
b g
1.00 g Ba OH
0.300 L
2
b g
171.3 g BabOH g
1 mol Ba OH
2
2
2 mol OH
1 mol Ba OH
b g
= 0.0389 M
2
Then HC 2 H 3O 2 is consumed in the neutralization reaction shown directly below.
C2 H 3O 2 + H 2 O
HC2 H 3O 2 + OH
Initial:
0.200 M
0.0389 M
0.288 M
—
Consume OH : -0.0389 M -0.0389 M +0.0389 M
—
Buffer:
0.161 M
~ 0M
0.327 M
—
Equation:
815
Chapter 17: Additional Aspects of Acid–Base Equilibria
Then use the Henderson-Hasselbalch equation.
C 2 H 3O 2
0.327 M
=
4.74
+
log
= 4.74 + 0.31 = 5.05
O
0.161 M
NM HC 2 H 3O 2 QP
pH = pKa + log L
(c)
b g can be added up until all of the HC H O is consumed.
BabOH g + 2 HC H O
BabC H O g + 2 H O
Ba OH
2
2
2
2
3
2
2
moles of Ba OH 2 = 0.300 L
3
2 2
b g
0.36 g Ba OH
2
2
2
0.200 mol HC 2 H 3 O 2
1L soln
mass of Ba OH 2 = 0.0300 mol Ba OH 2
(d)
3
1mol Ba OH 2
2 mol HC 2 H 3 O 2
171.3g Ba OH 2
1mol Ba OH 2
= 0.0300 mol Ba OH 2
= 5.14 g Ba OH 2
is too much for the buffer to handle and it is the excess of OH-
originating from the Ba(OH)2 that determines the pOH of the solution.
OH =
b g
0.36 g Ba OH
0.300 L soln
c
2
b g
171.3 g BabOH g
1 mol Ba OH
h
pOH = log 1.4 102 = 1.85
2
2
2 mol OH
1 mol Ba OH
b g
= 1.4 102 M OH
2
pH = 14.00 1.85 = 12.15
Acid–Base Indicators
27.
(E)
(a)
The pH color change range is 1.00 pH unit on either side of pKHln . If the pH color
change range is below pH = 7.00 , the indicator changes color in acidic solution. If
it is above pH = 7.00 , the indicator changes color in alkaline solution. If pH = 7.00
falls within the pH color change range, the indicator changes color near the neutral
point.
Indicator
K HIn
Bromphenol blue 1.4 104
Bromcresol green 2.1 105
Bromthymol blue 7.9 108
2,4-Dinitrophenol 1.3 104
Chlorophenol red 1.0 106
Thymolphthalein 1.0 1010
(b)
pK HIn
3.85
4.68
7.10
3.89
6.00
10.00
pH Color Change Range
Changes Color in?
2.9 (yellow) to 4.9 (blue)
acidic solution
3.7 (yellow) to 5.7 (blue)
acidic solution
6.1 (yellow) to 8.1 (blue)
neutral solution
2.9 (colorless) to 4.9 (yellow) acidic solution
5.0 (yellow) to 7.0 (red)
acidic solution
9.0 (colorless) to 11.0 (blue) basic solution
If bromcresol green is green, the pH is between 3.7 and 5.7, probably about pH = 4.7 .
If chlorophenol red is orange, the pH is between 5.0 and 7.0, probably about pH = 6.0 .
816
Chapter 17: Additional Aspects of Acid–Base Equilibria
28.
(M) We first determine the pH of each solution, and then use the answer in Exercise
27 (a) to predict the color of the indicator. (The weakly acidic or basic character of the
indicator does not affect the pH of the solution, since very little indicator is added.)
(a)
(b)
(c)
1 mol H 3O +
H 3O + = 0.100 M HCl
= 0.100 M; pH = log 0.100 M = 1.000
1 mol HCl
2,4-dinitrophenol is colorless.
Solutions of NaCl(aq) are pH neutral, with pH = 7.000 . Chlorphenol red assumes
its neutral color in such a solution; the solution is red/orange.
+
Equation: NH 3 (aq) + H 2 O(l)
NH 4 (aq) + OH (aq)
Initial:
1.00 M
—
0M
0 M
—
+x M
+x M
Changes: x M
x M
x M
Equil: 1.00 x M —
NH 4
OH
+
x2
x2
LM NH 3 OP
1.00 x 1.00
N
Q
(x << 1.00 M, thus the approximation was valid).
Kb =
c
= 1.8 105 =
h
pOH = log 4.2 103 = 2.38
x = 4.2 10 3 M = OH
pH = 14.00 2.38 = 11.62
Thus, thymolphthalein assumes its blue color in solutions with pH 11.62 .
(d)
29.
(M)
(a)
(b)
From Figure 17-8, seawater has pH = 7.00 to 8.50. Bromcresol green solution is
blue.
In an Acid–Base titration, the pH of the solution changes sharply at a definite pH
that is known prior to titration. (This pH change occurs during the addition of a
very small volume of titrant.) Determining the pH of a solution, on the other hand,
is more difficult because the pH of the solution is not known precisely in
advance. Since each indicator only serves to fix the pH over a quite small region,
often less than 2.0 pH units, several indicators—carefully chosen to span the
entire range of 14 pH units—must be employed to narrow the pH to 1 pH unit
or possibly lower.
An indicator is, after all, a weak acid. Its addition to a solution will affect the
acidity of that solution. Thus, one adds only enough indicator to show a color
change and not enough to affect solution acidity.
817
Chapter 17: Additional Aspects of Acid–Base Equilibria
30.
(E)
(a)
We use an equation similar to the Henderson-Hasselbalch equation to determine the
relative concentrations of HIn, and its anion, In , in this solution.
In
In
In
pH = pK HIn + log
; 4.55 = 4.95 + log
; log
= 4.55 4.95 = 0.40
HIn
HIn
HIn
In
x
= 100.40 = 0.40 =
100 x
HIn
(b)
x = 40 0.40 x
x=
40
= 29% In and 71% HIn
1.40
When the indicator is in a solution whose pH equals its pKa (4.95), the ratio
In / HIn = 1.00 . And yet, at the midpoint of its color change range (about
pH = 5.3 ), the ratio In /[HIn] is greater than 1.00. Even though HIn In at
this midpoint, the contribution of HIn to establishing the color of the solution is
about the same as the contribution of In . This must mean that HIn (red) is more
strongly colored than In (yellow).
31.
32.
(E)
(a)
0.10 M KOH is an alkaline solution and phenol red will display its basic color in
such a solution; the solution will be red.
(b)
0.10 M HC 2 H 3O 2 is an acidic solution, although that of a weak acid, and phenol
red will display its acidic color in such a solution; the solution will be yellow.
(c)
0.10 M NH 4 NO 3 is an acidic solution due to the hydrolysis of the ammonium ion.
Phenol red will display its acidic color, that is, yellow, in this solution.
(d)
0.10 M HBr is an acidic solution, the aqueous solution of a strong acid. Phenol red
will display its acidic color in this solution; the solution will be yellow.
(e)
0.10 M NaCN is an alkaline solution because of the hydrolysis of the cyanide ion.
Phenol red will display its basic color, red, in this solution.
(f)
An equimolar acetic acid–potassium acetate buffer has pH = pKa = 4.74 for acetic
acid. In this solution phenol red will display its acid color, namely, yellow.
(M)
(a) pH = log 0.205 = 0.688 The indicator is red in this solution.
b
(b)
g
The total volume of the solution is 600.0 mL. We compute the amount of each
solute.
amount H 3O + = 350.0 mL 0.205 M = 71.8 mmol H 3O +
amount NO 2 = 250.0 mL 0.500 M = 125 mmol NO 2
71.8 mmol
125 mmol
H 3O + =
= 0.120 M
NO 2 =
= 0.208 M
600.0 mL
600.0 mL
The H 3O + and NO 2 react to produce a buffer solution in which
HNO 2 = 0.120 M and NO 2 = 0.208 0.120 = 0.088 M . We use the
818
Chapter 17: Additional Aspects of Acid–Base Equilibria
Henderson-Hasselbalch equation to determine the pH of this solution.
pKa = log 7.2 104 = 3.14
c
h
NO 2
= 3.14 + log 0.088 M = 3.01 The indicator is yellow in this
pH = pK a + log
HNO
0.120 M
2
solution.
(c)
The total volume of the solution is 750. mL. We compute the amount and then the
concentration of each solute. Amount OH = 150 mL 0.100 M = 15.0 mmol OH
This OH reacts with HNO 2 in the buffer solution to neutralize some of it and leave
56.8 mmol ( = 71.8 15.0 ) unneutralized.
125 +15 mmol
56.8 mmol
= 0.187 M
HNO 2 =
= 0.0757 M
NO 2 =
750. mL
750. mL
We use the Henderson-Hasselbalch equation to determine the pH of this solution.
NO 2
0.187 M
pH = pKa + log L
=
3.14
+
log
= 3.53
O
0.0757 M
NM HNO 2 QP
The indicator is yellow in this solution.
b
(d)
g
b g
We determine the OH due to the added Ba OH 2 .
OH =
b g
5.00 g Ba OH
0.750 L
This is sufficient OH
2
b g
171.34 g BabOH g
1 mol Ba OH
2
2
2 mol OH
1 mol Ba OH
b g
= 0.0778 M
2
to react with the existing HNO 2 and leave an excess
OH = 0.0778 M 0.0757 M = 0.0021M. pOH = log 0.0021 = 2.68.
pH = 14.00 2.68 = 11.32 The indicator is blue in this solution.
33.
(M) Moles of HCl = C V = 0.04050 M 0.01000 L = 4.050 104 moles
Moles of Ba(OH)2 at endpoint = C V = 0.01120 M 0.01790 L = 2.005 104 moles.
Moles of HCl that react with Ba(OH)2 = 2 moles Ba(OH)2
Moles of HCl in excess 4.050 104 moles 4.010 104 moles = 4.05 106 moles
Total volume at the equivalence point = (10.00 mL + 17.90 mL) = 27.90 mL
[HCl]excess =
4.05 106 mole HCl
= 1.45 104 M; pH = log(1.45 104) = 3.84
0.02790 L
(a)
The approximate pKHIn = 3.84 (generally + 1 pH unit)
(b)
This is a relatively good indicator (with 1 % of the equivalence point volume),
however, pKHin is not very close to the theoretical pH at the equivalence point
(pH = 7.000) For very accurate work, a better indicator is needed (i.e.,
bromthymol blue (pKHin = 7.1). Note: 2,4-dinitrophenol works relatively well
here because the pH near the equivalence point of a strong acid/strong base
819
Chapter 17: Additional Aspects of Acid–Base Equilibria
titration rises very sharply ( 6 pH units for an addition of only 2 drops (0.10
mL)).
34.
Solution (a): 100.0 mL of 0.100 M HCl, [H3O+] = 0.100 M and pH = 1.000 (yellow)
Solution (b): 150 mL of 0.100 M NaC2H3O2
Ka of HC2H3O3 = 1.8 105
Kb of C2H3O2 = 5.6 1010
C2H3O2(aq) + H2O(l)
Initial
Change
Equil.
0.100 M
x
0.100 x
K b = 5.6 10
-10
—
—
—
HC2H3O2(aq) + OH(aq)
0M
+x
x
~0M
+x
x
Assume x is small: 5.6 1011 = x2; x = 7.48 106 M (assumption valid by inspection)
[OH] = x = 7.48 106 M, pOH = 5.13 and pH = 8.87 (green-blue)
Mixture of solution (a) and (b). Total volume = 250.0 mL
nHCl = C V = 0.1000 L 0.100 M = 0.0100 mol HCl
nC2H3O2 = C V = 0.1500 L 0.100 M = 0.0150 mol C2H3O2
HCl is the limiting reagent. Assume 100% reaction.
Therefore, 0.0050 mole C2H3O2 is left unreacted, and 0.0100 moles of HC2H3O2 form.
n
0.0050 mol
n
0.0100 mol
=
=
= 0.020 M [HC2H3O2] =
= 0.0400 M
V
V
0.250 L
0.250 L
K a = 1.8 10-5
HC2H3O2(aq) + H2O(l)
+ H3O+(aq)
C2H3O2 (aq)
Initial
0.0400 M
—
0.020 M
~0M
—
+x
+x
Change
x
Equil.
—
0.020 + x
x
0.0400 x
[C2H3O2] =
1.8 10-5 =
x(0.020 x)
x(0.020)
0.0400 x
0.0400
x = 3.6 105
(proof 0.18 % < 5%, the assumption was valid)
[H3O+] = 3.6 105 pH = 4.44 Color of thymol blue at various pHs:
Red
pH
8.0
pH
2.8
pH
1.2
Orange
Solution (a)
RED
Yellow
Solution (c)
YELLOW
Green
Solution (b)
GREEN
820
pH
9.8
Blue
Chapter 17: Additional Aspects of Acid–Base Equilibria
Neutralization Reactions
35.
(E) The reaction (to second equiv. pt.) is: H 3PO 4 aq + 2 KOH aq
K 2 HPO 4 aq + 2H 2O(l) .
The molarity of the H 3 PO 4 solution is determined in the following manner.
0.2420 mmol KOH 1mmol H 3 PO 4
1mL KOH soln
2 mmol KOH
= 0.1508 M
25.00 mL H 3PO 4 soln
31.15 mL KOH soln
H 3PO 4 molarity =
36.
(E) The reaction (first to second equiv. pt.) is:
NaH 2 PO 4 aq + NaOH aq
Na 2 HPO 4 aq + H 2 O(l) . The molarity of the H 3 PO 4
solution is determined in the following manner.
0.1885mmol NaOH 1 mmol H 3 PO 4
1mL NaOH soln
1 mmol NaOH
= 0.1760 M
20.00 mL H 3 PO 4 soln
18.67 mL NaOH soln
H 3 PO 4 molarity =
37.
(M) Here we must determine the amount of H 3O + or OH in each solution, and the
amount of excess reagent.
0.0150 mmol H 2SO 4 2 mmol H 3O +
+
amount H 3O = 50.00 mL
= 1.50 mmol H 3O +
1 mL soln
1 mmol H 2SO 4
(assuming complete ionization of H2SO4 and HSO4 in the presence of OH)
amount OH = 50.00 mL
0.0385 mmol NaOH 1mmol OH
= 1.93 mmol OH
1mL soln
1mmol NaOH
Result: Titration reaction :
Initial amounts :
After reaction :
OH aq +
1.93mmol
0.43mmol
2H 2 O(l)
H 3 O + aq
1.50 mmol
0 mmol
0.43mmol OH
OH =
= 4.3 103 M
100.0 mL soln
pOH = log 4.3 103 = 2.37
821
pH = 14.00 2.37 = 11.63
Chapter 17: Additional Aspects of Acid–Base Equilibria
38.
(M) Here we must determine the amount of solute in each solution, followed by the amount of
excess reagent.
H 3O + = 102.50 = 0.0032 M
mmol HCl = 100.0 mL
0.0032 mmol H 3O + 1 mmol HCl
= 0.32 mmol HCl
1 mL soln
1 mmol H 3O +
OH = 103.00 = 1.0 103 M
0.0010 mmol OH 1 mmol NaOH
mmol NaOH = 100.0 mL
= 0.10 mmol NaOH
1 mL soln
1 mmol OH
Result:
pOH = 14.00 11.00 = 3.00
Titration reaction :
NaOH aq + HCl aq
NaCl aq
Initial amounts :
0.10 mmol
0.32 mmol
0 mmol
After reaction :
0.00 mmol
0.22 mmol
0.10 mmol
0.22 mmol HCl 1mmol H 3O +
H 3O + =
= 1.1103 M
200.0 mL soln 1mmol HCl
+ H 2 O(l)
pH = log 1.1103 = 2.96
Titration Curves
39.
(M) First we calculate the amount of HCl. The relevant titration reaction is
HCl aq + KOH aq KCl aq + H 2O(l)
amount HCl = 25.00 mL
0.160 mmol HCl
= 4.00 mmol HCl = 4.00 mmol H 3O + present
1 mL soln
Then, in each case, we calculate the amount of OH that has been added, determine
which ion, OH aq or H 3O + aq , is in excess, compute the concentration of that ion,
and determine the pH.
0.242 mmol OH
(a) amount OH = 10.00 mL
= 2.42 mmol OH ; H 3O + is in excess.
1 mL soln
b g
b g
FG
H
IJ
K
1 mmol H 3O +
1 mmol OH
= 0.0451 M
25.00 mL originally +10.00 mL titrant
4.00 mmol H 3O 2.42 mmol OH
[H 3O + ] =
b
g
pH = log 0.0451 = 1.346
822
Chapter 17: Additional Aspects of Acid–Base Equilibria
(b)
amount OH = 15.00 mL
0.242 mmol OH
= 3.63 mmol OH ; H3O+ is in excess.
1 mL soln
FG
H
IJ
K
1 mmol H 3O +
1 mmol OH
= 0.00925 M
25.00 mL originally +15.00 mL titrant
4.00 mmol H 3O 3.63 mmol OH
[H 3O + ] =
b
g
pH = log 0.00925 = 2.034
40.
(M) The relevant titration reaction is KOH aq + HCl aq KCl aq + H 2O(l)
mmol of KOH = 20.00 mL
(a)
0.275 mmol KOH
= 5.50 mmol KOH
1 mL soln
The total volume of the solution is V = 20.00 mL +15.00 mL = 35.00 mL
mmol HCl = 15.00 mL
0.350 mmol HCl
= 5.25 mmol HCl
1 mL soln
mmol excess OH - = 5.50 mmol KOH-5.25 mmol HCl ×
0.25 mmol OH
OH =
= 0.0071M
35.00 mL soln
pH = 14.00 2.15 = 11.85
(b)
1 mmol OH 1 mmol KOH
= 0.25 mmol OH -
pOH = log 0.0071 = 2.15
The total volume of solution is V = 20.00 mL + 20.00 mL = 40.00 mL
mmol HCl = 20.00 mL
0.350 mmol HCl
= 7.00 mmol HCl
1 mL soln
mmol excess H 3O + = 7.00 mmol HCl 5.50 mmol KOH ×
1 mmol H 3O +
1 mmol HCl
= 1.50 mmol H 3O +
H 3O + =
41.
1.50 mmol H 3O +
= 0.0375 M
40.00 mL
b
g
pH = log 0.0375 = 1.426
(M) The relevant titration reaction is HNO 2 aq + NaOH aq NaNO 2 aq + H 2 O(l)
amount HNO 2 = 25.00 mL
0.132 mmol HNO 2
= 3.30 mmol HNO 2
1 mL soln
823
Chapter 17: Additional Aspects of Acid–Base Equilibria
(a)
The volume of the solution is 25.00 mL +10.00 mL = 35.00 mL
amount NaOH = 10.00 mL
0.116 mmol NaOH
= 1.16 mmol NaOH
1 mL soln
1.16 mmol NaNO 2 are formed in this reaction and there is an excess of (3.30 mmol
HNO 2 1.16 mmol NaOH) = 2.14 mmol HNO 2 . We can use the
Henderson-Hasselbalch equation to determine the pH of the solution.
c
h
pKa = log 7.2 104 = 3.14
NO 2
1.16 mmol NO 2 / 35.00 mL
= 3.14 + log
= 2.87
O
2.14 mmol HNO 2 / 35.00 mL
MN HNO 2 PQ
pH = pKa + log L
(b)
The volume of the solution is 25.00 mL + 20.00 mL = 45.00 mL
amount NaOH = 20.00 mL
0.116 mmol NaOH
= 2.32 mmol NaOH
1 mL soln
2.32 mmol NaNO 2 are formed in this reaction and there is an excess of
(3.30 mmol HNO 2 2.32 mmol NaOH =) 0.98 mmol HNO 2 .
NO 2
2.32 mmol NO 2 / 45.00 mL
=
3.14
+
log
= 3.51
O
0.98 mmol HNO 2 / 45.00 mL
NM HNO 2 QP
pH = pKa + log L
42.
b g
b g
b g
bg
(M) In this case the titration reaction is NH 3 aq + HCl aq NH 4 Cl aq + H 2 O l
0.318 mmol NH 3
= 6.36 mmol NH 3
1 mL soln
The volume of the solution is 20.00 mL +10.00 mL = 30.00 mL
0.475 mmol NaOH
amount HCl = 10.00 mL
= 4.75 mmol HCl
1 mL soln
4.75 mmol NH 4 Cl is formed in this reaction and there is an excess of (6.36 mmol
NH3 – 4.75 mmol HCl =) 1.61 mmol NH3. We can use the HendersonHasselbalch equation to determine the pH of the solution.
amount NH 3 = 20.00 mL
(a)
pK b = log 1.8 105 = 4.74;
pH = pKa + log
(b)
NH 3
NH 4
+
pK a = 14.00 pK b = 14.00 4.74 = 9.26
= 9.26 + log
1.61 mmol NH 3 / 30.00 mL
= 8.79
+
4.75 mmol NH 4 / 30.00 mL
The volume of the solution is 20.00 mL +15.00 mL = 35.00 mL
amount HCl = 15.00 mL
0.475 mmol NaOH
= 7.13 mmol HCl
1 mL soln
824
Chapter 17: Additional Aspects of Acid–Base Equilibria
6.36 mmol NH 4 Cl is formed in this reaction and there is an excess of (7.13 mmol
HCl – 6.36 mmol NH3 =) 0.77 mmol HCl; this excess HCl determines the pH of
the solution.
0.77 mmol HCl 1mmol H 3O +
H 3O + =
= 0.022 M pH = log 0.022 = 1.66
35.00 mL soln 1mmol HCl
43.
(E) In each case, the volume of acid and its molarity are the same. Thus, the amount of
acid is the same in each case. The volume of titrant needed to reach the equivalence point
will also be the same in both cases, since the titrant has the same concentration in each
case, and it is the same amount of base that reacts with a given amount (in moles) of acid.
Realize that, as the titration of a weak acid proceeds, the weak acid will ionize,
replenishing the H 3O + in solution. This will occur until all of the weak acid has ionized
and all of the released H+ has subsequently reacted with the strong base. At the
equivalence point in the titration of a strong acid with a strong base, the solution contains
ions that do not hydrolyze. But the equivalence point solution of the titration of a weak
acid with a strong base contains the anion of a weak acid, which will hydrolyze to
produce a basic (alkaline) solution. (Don’t forget, however, that the inert cation of a
strong base is also present.)
44.
(E)
(a)
45.
This equivalence point solution is the result of the titration of a weak acid with a
2
strong base. The CO 3 in this solution, through its hydrolysis, will form an
alkaline, or basic solution. The other ionic species in solution, Na + , will not
hydrolyze. Thus, pH > 7.0.
+
(b)
This is the titration of a strong acid with a weak base. The NH 4 present in the
equivalence point solution hydrolyzes to form an acidic solution. Cl does not
hydrolyze. Thus, pH < 7.0.
(c)
This is the titration of a strong acid with a strong base. Two ions are present in the
solution at the equivalence point, namely, K + and Cl , neither of which
hydrolyzes. Thus the solution will have a pH of 7.00.
(D)
(a)
b
Initial OH = 0.100 M OH
g
pOH = log 0.100 = 1.000
pH = 13.00
Since this is the titration of a strong base with a strong acid, KI is the solute
present at the equivalence point, and since KI is a neutral salt, the pH = 7.00 . The
titration reaction is:
KOH aq + HI aq
KI aq + H 2 O(l)
VHI = 25.0 mL KOH soln
0.100 mmol KOH soln 1mmol HI
1mL HI soln
1mL soln
1mmol KOH 0.200 mmol HI
= 12.5 mL HI soln
825
Chapter 17: Additional Aspects of Acid–Base Equilibria
Initial amount of KOH present = 25.0 mL KOH soln 0.100 M = 2.50 mmol KOH
At the 40% titration point: 5.00 mL HI soln 0.200 M HI = 1.00 mmol HI
excess KOH = 2.50 mmol KOH 1.00 mmol HI = 1.50 mmol KOH
1.50 mmol KOH 1mmol OH
OH =
= 0.0500 M pOH = log 0.0500 = 1.30
30.0 mL total 1mmol KOH
pH = 14.00 1.30 = 12.70
At the 80% titration point: 10.00 mL HI soln 0.200 M HI = 2.00 mmol HI
excess KOH = 2.50 mmol KOH 2.00 mmol HI = 0.50 mmol KOH
0.50 mmol KOH 1mmol OH
OH =
= 0.0143M
35.0 mL total
1mmol KOH
pOH = log 0.0143 = 1.84
pH = 14.00 1.84 = 12.16
At the 110% titration point:13.75 mL HI soln 0.200 M HI = 2.75 mmol HI
excess HI = 2.75 mmol HI 2.50 mmol HI = 0.25 mmol HI
[H 3O + ]
0.25 mmol HI 1mmol H 3O
0.0064 M; pH= log(0.0064) 2.19
38.8 mL total
1mmol HI
Since the pH changes very rapidly at the equivalence point, from about pH = 10
to about pH = 4 , most of the indicators in Figure 17-8 can be used. The main
exceptions are alizarin yellow R, bromphenol blue, thymol blue (in its acid range),
and methyl violet.
(b)
Initial pH:
Since this is the titration of a weak base with a strong acid, NH4Cl is the solute
present at the equivalence point, and since NH4+ is a slightly acidic cation, the pH
should be slightly lower than 7. The titration reaction is:
Initial :
Changes :
NH 4 + aq + OH aq
NH 3 aq + H 2 O(l)
0M
1.00 M
0M
x M
+x M
+ xM
Equil :
1.00 x M
Equation:
xM
NH 4 + OH
x2
x2
5
Kb =
= 1.8 10 =
1.00 x 1.00
NH 3
(x << 1.0, thus the approximation is valid)
x = 4.2 103 M = OH , pOH = log 4.2 103 = 2.38,
pH = 14.00 2.38 = 11.62 = initial pH
Volume of titrant :
NH 3 + HCl
NH 4 Cl + H 2 O
826
xM
Chapter 17: Additional Aspects of Acid–Base Equilibria
VHCl = 10.0mL
1.00 mmol NH 3 1mmol HCl 1mL HCl soln
= 40.0 mL HCl soln
1mL soln
1mmol NH 3 0.250 mmol HCl
pH at equivalence point: The total solution volume at the equivalence point is
10.0 + 40.0 mL = 50.0 mL
+
Also at the equivalence point, all of the NH 3 has reacted to form NH 4 . It is this
+
NH 4 that hydrolyzes to determine the pH of the solution.
10.0 mL
NH 4
+
=
1.00 mmol NH 3 1 mmol NH 4+
1 mL soln
1 mmol NH 3
= 0.200 M
50.0mL total solution
NH 3 aq + H 3 O + aq
Equation : NH 4 + aq + H 2 O(l)
Initial : 0.200 M
0M
0M
Changes : x M
+ xM
+ xM
Equil :
0.200 x M
xM
xM
+
K w 1.0 1014 NH 3 H 3O
x2
x2
Ka =
=
=
=
Kb
1.8 105
NH 3
0.200 x 0.200
(x << 0.200, thus the approximation is valid)
x = 1.1105 M;
H 3O + = 1.1105 M;
pH = log 1.1105 = 4.96
Of the indicators in Figure 17-8, one that has the pH of the equivalence point
within its pH color change range is methyl red (yellow at pH = 6.2 and red at
pH = 4.5); Bromcresol green would be another choice. At the 50% titration point,
NH 3 = NH 4 + and pOH = pK b = 4.74
pH = 14.00 4.74 = 9.26
The titration curves for parts (a) and (b) follow.
827
Chapter 17: Additional Aspects of Acid–Base Equilibria
46.
(M)
(a) This part simply involves calculating the pH of a 0.275 M NH 3 solution.
+
Equation: NH 3 aq + H 2 O(l)
NH 4 aq + OH aq
Initial:
0.275 M
0M
0M
Changes: x M
+xM
+xM
Equil:
(0.275 x) M
xM
xM
NH 4
+
OH
x2
x2
LM NH 3 OP
0.275 x 0.275
N
Q
(x << 0.275, thus the approximation is valid)
pOH = log 2.2 103 = 2.66
pH = 11.34
Kb =
(b)
= 1.8 105 =
x = 2.2 10 3 M = OH
This is the volume of titrant needed to reach the equivalence point.
The relevant titration reaction is NH 3 aq + HI aq
NH 4 I aq
b g b g
VHI = 20.00 mL NH 3 aq
b g
0.275 mmol NH 3 1mmol HI
1mL HI soln
1mL NH3 soln 1mmol NH 3 0.325 mmol HI
VHI = 16.9 mL HI soln
(c)
The pOH at the half-equivalence point of the titration of a weak base with a strong
acid is equal to the pKb of the weak base.
pOH = pK b = 4.74; pH = 14.00 4.74 = 9.26
(d)
NH 4 is formed during the titration, and its hydrolysis determines the pH of the
solution. Total volume of solution = 20.00 mL +16.9 mL = 36.9 mL
+
+
mmol NH 4
NH 4
+
+
0.275 mmol NH 3 1mmol NH 4
+
= 20.00 mL NH 3 aq
= 5.50 mmol NH 4
1mL NH 3 soln
1mmol NH 3
5.50 mmol NH 4 +
=
= 0.149 M
36.9 mL soln
NH 3 aq + H 3 O + aq
Equation: NH 4 + aq + H 2 O(l)
Initial:
0.149 M
0M
0M
Changes:
+x M
+x M
xM
Equil:
(0.149 x) M
xM
xM
828
Chapter 17: Additional Aspects of Acid–Base Equilibria
+
K w 1.0 1014 NH 3 H 3 O
x2
x2
Ka =
=
=
=
Kb
0.149 x 0.149
1.8 105
NH 4 +
(x << 0.149, thus the approximation is valid)
x = 9.1 106 M = H 3 O +
47.
pH = log 9.1 106 = 5.04
(M) A pH greater than 7.00 in the titration of a strong base with a strong acid means that
the base is not completely titrated. A pH less than 7.00 means that excess acid has been
added.
(a) We can determine OH of the solution from the pH. OH is also the quotient
of the amount of hydroxide ion in excess divided by the volume of the solution:
20.00 mL base +x mL added acid.
pOH = 14.00 pH = 14.00 12.55 = 1.45
0.175 mmol OH
20.00
mL
base
1mL base
[OH ]
OH = 10 pOH = 101.45 = 0.035 M
0.200 mmol H 3 O +
x
mL
acid
1mL acid
0.035 M
20.00 mL + x mL
3.50 0.200 x = 0.70 + 0.035 x; 3.50 0.70 = 0.035 x + 0.200 x; 2.80 = 0.235 x
2.80
x=
= 11.9 mL acid added.
0.235
(b) The set-up here is the same as for part (a).
pOH = 14.00 pH = 14.00 10.80 = 3.20
OH = 10 pOH = 103.20 = 0.00063 M
0.200 mmol H 3O +
0.175 mmol OH
20.00
mL
base
mL
acid
x
1mL base
1mL acid
[OH ]
20.00 mL + x mL
mmol
= 0.00063M
mL
3.50 0.200 x = 0.0126 + 0.00063x; 3.50 0.0126 = 0.00063 x + 0.200 x; 3.49 = 0.201 x
3.49
x=
= 17.4 mL acid added. This is close to the equivalence point at 17.5 mL.
0.201
[OH ] 0.00063
829
Chapter 17: Additional Aspects of Acid–Base Equilibria
(c)
Here the acid is in excess, so we reverse the set-up of part (a). We are just slightly
beyond the equivalence point. This is close to the “mirror image” of part (b).
H3O + = 10 pH = 104.25 = 0.000056 M
FG x mLacid 0.200 mmol H O IJ FG 20.00 mL base 0.175 mmol OH IJ
1 mL acid
1mL base
H
K H
K
[H O ] =
+
3
+
3
20.00 mL+ x mL
5
= 5.6 10 M
0.200 x 3.50 = 0.0011+ 5.6 105 x; 3.50 + 0.0011 = 5.6 105 x + 0.200 x;
3.50 = 0.200 x
3.50
= 17.5 mL acid added, which is the equivalence point for this titration.
0.200
(D) In the titration of a weak acid with a strong base, the middle range of the titration,
with the pH within one unit of pKa ( = 4.74 for acetic acid), is known as the buffer
region. The Henderson-Hasselbalch equation can be used to determine the ratio of weak
acid and anion concentrations. The amount of weak acid then is used in these
calculations to determine the amount of base to be added.
x=
48.
(a)
C 2 H 3 O 2
C 2 H 3O 2
pH = pK a + log
= 3.85 = 4.74 + log
HC H O
2
3
2
HC H O
2
3
2
C 2 H 3O 2
log
= 3.85 4.74
C 2 H 3O 2
= 10 0.89 = 0.13;
HC H O
2
3
2
nHC2 H3O2 = 25.00 mL
HC H O
2
3
2
0.100 mmol HC 2 H 3 O 2
1mL acid
= 2.50 mmol
Since acetate ion and acetic acid are in the same solution, we can use their
amounts in millimoles in place of their concentrations. The amount of acetate ion
is the amount created by the addition of strong base, one millimole of acetate ion
for each millimole of strong based added. The amount of acetic acid is reduced by
the same number of millimoles. HC 2 H 3O 2 + OH C 2 H 3O 2 + H 2 O
0.200 mmol OH 1 mmol C 2 H 3O 2
x mL base
0.200 x
mL base
1 mmol OH
013
.
2.50 0.200 x
0.200 mmol OH
2.50 mmol HC 2 H 3O 2 x mL base
mL base
b
g
FG
H
IJ
K
0.200 x = 0.13 2.50 0.200 x = 0.33 0.026 x
x=
0.33
= 1.5 mL of base
0.226
830
0.33 = 0.200 x + 0.026 x = 0.226 x
Chapter 17: Additional Aspects of Acid–Base Equilibria
(b) This is the same set-up as part (a), except for a different ratio of concentrations.
C 2 H 3O 2
C 2 H 3O 2
pH = 5.25 = 4.74 + log L
log L
O
O = 5.25 4.74 = 0.51
NM HC 2 H 3O 2 QP
NM HC2 H 3O 2 QP
C 2 H 3O 2
0.51
LM HC2 H 3O 2 OP = 10 = 3.2
N
Q
3.2 =
b
g
0.200 x
2.50 0.200 x
0.200 x = 3.2 2.50 0.200 x = 8.0 0.64 x
8.0 = 0.200 x + 0.64 x = 0.84 x
8.0
= 9.5 mL base. This is close to the equivalence point, which is reached
0.84
by adding 12.5 mL of base.
x=
(c)
This is after the equivalence point, where the pH is determined by the excess added
base.
pOH = 14.00 pH = 14.00 11.10 = 2.90
OH = 10 pOH = 102.90 = 0.0013 M
0.200 mmol OH
0.200 x
1 mL base
OH = 0.0013 M =
=
x mL + 12.50 mL + 25.00 mL 37.50 + x
x mL
b
b
g
g
0.049
= 0.25 mL excess
0.200 0.0013
Total base added = 12.5 mL to equivalence point + 0.25 mL excess = 12.8 mL
0.200 x = 0.0013 37.50 + x = 0.049 + 0.0013 x
49.
x=
(D) For each of the titrations, the pH at the half-equivalence point equals the pKa of the acid.
x2
x [H 3 O ]
The initial pH is that of 0.1000 M weak acid: K a
0.1000 x
x must be found using the quadratic formula roots equation unless the approximation is
valid.
One method of determining if the approximation will be valid is to consider the ratio Ca/Ka. If
the value of Ca/Ka is greater than 1000, the assumption should be valid, however, if the value of
Ca/Ka is less than 1000, the quadratic should be solved exactly (i.e., the 5% rule will not be
satisfied).
The pH at the equivalence point is that of 0.05000 M anion of the weak acid, for which
the OH is determined as follows.
Kw
Kw
x2
x=
0.05000 [OH ]
K a 0.05000
Ka
We can determine the pH at the quarter and three quarter of the equivalence point by
using the Henderson-Hasselbalch equation (effectively + 0.48 pH unit added to the
pKa).
Kb =
831
Chapter 17: Additional Aspects of Acid–Base Equilibria
And finally, when 0.100 mL of base has been added beyond the equivalence point, the
pH is determined by the excess added base, as follows (for all three titrations).
mL
0100
.
OH =
c
0.1000 mmol NaOH 1 mmol OH
1mL NaOH soln
1 mmol NaOH
= 4.98 104 M
20.1 mL soln total
h
pOH = log 4.98 104 = 3.303
(a)
pH = 14.000 3.303 = 10.697
Ca/Ka = 14.3; thus, the approximation is not valid and the full quadratic equation
must be solved.
Initial: From the roots equation x = [H 3 O ] 0.023M pH=1.63
Half equivalence point: pH = pKa = 2.15
pH at quarter equivalence point = 2.15 - 0.48 = 1.67
pH at three quarter equivalence point = 2.15 + 0.48 = 2.63
pOH 6.57
1.0 1014
0.05000 2.7 107
Equiv: x = OH =
3
pH 14.00 6.57 7.43
7.0 10
Indicator: bromthymol blue, yellow at pH = 6.2 and blue at pH = 7.8
(b)
Ca/Ka = 333; thus, the approximation is not valid and the full quadratic equation
must be solved.
Initial:
From the roots equation x = [H 3 O ] 0.0053M
Half equivalence point: pH = pKa = 3.52
pH at quarter equivalence point = 3.52 - 0.48 = 3.04
pH at three quarter equivalence point = 3.52 + 0.48 = 4.00
Equiv:
x = OH =
pH=2.28
1.0 1014
0.05000 1.3 106
3.0 104
pOH 5.89
pH = 14.00 5.89 811
.
Indicator: thymol blue, yellow at pH = 8.0 and blue at pH = 9.6
(c)
Ca/Ka = 5×106; thus, the approximation is valid..
Initial:
[ H 3O ] 01000
.
2.0 108 0.000045 M pH = 4.35
pH at quarter equivalence point = 4.35 - 0.48 = 3.87
pH at three quarter equivalence point = 4.35 + 0.48 = 4.83
Half equivalence point: pH = pKa = 7.70
Equiv:
x = OH
10
. 1014
=
0.0500 16
. 104
8
2.0 10
.
pOH 380
pH 14.00 380
. 10.20
832
Chapter 17: Additional Aspects of Acid–Base Equilibria
Indicator: alizarin yellow R, yellow at pH = 10.0 and violet at pH = 12.0
The three titration curves are drawn with respect to the same axes in the diagram below.
12
10
pH
8
6
4
2
0
0
2
4
6
8
10
12
Volume of 0.10 M NaOH added (mL)
50.
(D) For each of the titrations, the pOH at the half-equivalence point equals the pKb of
the base. The initial pOH is that of 0.1000 M weak base, determined as follows.
x2
Kb
;
x 0.1000 K a [OH ] if Cb/Kb > 1000. For those cases
0.1000 x
where this is not the case, the approximation is invalid and the complete quadratic
equation must be solved. The pH at the equivalence point is that of 0.05000 M cation of
the weak base, for which the H 3O + is determined as follows.
Ka
Kw
x2
K b 0.05000
x
Kw
0.05000 [H 3O + ]
Kb
And finally, when 0.100 mL of acid has been added beyond the equivalence point, the
pH for all three titrations is determined by the excess added acid, as follows.
0.100 mL HCl
H 3O + =
c
0.1000 mmol HCl 1 mmol H 3O +
1 mL HCl soln
1 mmol HCl
= 4.98 104 M
20.1 mL soln total
h
pH = log 4.98 104 = 3.303
(a)
Cb/Kb = 100; thus, the approximation is not valid and the full quadratic equation must be solved.
Initial:
From the roots equation x = 0.0095. Therefore,
[OH ] 0.0095 M
pOH=2.02
pH=11.98
Half-equiv:
Equiv:
c
h
pOH = log 1 103 = 3.0
x = [H 3 O + ]
1.0 1014
0.05000 7 107
1 103
833
pH = 11.0
pH = 6.2
Chapter 17: Additional Aspects of Acid–Base Equilibria
Indicator: methyl red, yellow at pH = 6.3 and red at pH = 4.5
(b)
Cb/Kb = 33,000; thus, the approximation is valid.
Initial:
[OH ] 0.1000 3 106 5.5 104 M
Half-equiv: pOH = log 3 106 = 5.5
Equiv:
pOH=3.3
pH = 14 pOH
1.0 1014
0.05000 1 10 5
6
3 10
x = [H 3 O + ]
pH=10.7
pH = 8.5
pH -log(1 10 5 ) = 5.0
Indicator: methyl red, yellow at pH = 6.3 and red at pH = 4.5
(c)
Cb/Kb = 1.4×106; thus, the approximation is valid.
Initial:
[OH ] 01000
.
7 108 8 105 M
Half-equiv: pOH = log 7 108 = 7.2
Equiv:
pOH = 4.1
pH = 14 pOH
1.0 1014
x = [H 3O ]
0.05000 8.5 105
8
7 10
+
pH = 9.9
pH = 6.8
pH=4.1
Indicator: bromcresol green, blue at pH = 5.5 and yellow at pH = 4.0
The titration curves are drawn with respect to the same axes in the diagram below.
51.
(D) 25.00 mL of 0.100 M NaOH is titrated with 0.100 M HCl
(i) Initial pOH for 0.100 M NaOH: [OH] = 0.100 M, pOH = 1.000 or pH = 13.000
25.00 mL
= 0.0510 M
(ii) After addition of 24 mL: [NaOH] = 0.100 M
49.00 mL
24.00 mL
[HCl] = 0.100 M
= 0.0490 M
49.00 mL
834
Chapter 17: Additional Aspects of Acid–Base Equilibria
NaOH is in excess by 0.0020 M = [OH] pOH = 2.70
(iii) At the equivalence point (25.00 mL), the pOH should be 7.000 and pH = 7.000
25.00
= 0.0490 M
51.00
26.00 mL
[HCl] = 0.100 M
= 0.0510 M
51.00 mL
HCl is in excess by 0.0020 M = [H3O+] pH = 2.70 or pOH = 11.30
(iv) After addition of 26 mL: [NaOH] = 0.100M
(v) After addition of 33.00 mL HCl( xs) [NaOH] = 0.100 M
[HCl] = 0.100 M
33.00 mL
= 0.0569 M
58.00 mL
25.00 mL
= 0.0431 M
58.00 mL
[HCl]excess= 0.0138 M pH = 1.860, pOH = 12.140
The graphs look to be mirror images of one another. In reality, one must reflect about a
horizontal line centered at pH or pOH = 7 to obtain the other curve.
0.100 M NaOH(25.00 mL) + 0.100 M HCl
14
10
8
8
pH
12
10
pOH
12
6
4
6
4
2
2
0
0
52.
0.100 M NaOH(25.00 mL) + 0.100 M HCl
14
10
20
30
40
Volume of HCl (mL)
50
60
0
0
10
20
30
40
Volume of HCl (mL)
NH3(aq)
+ H2O(l)
K b = 1.8 10
-5
NH4+(aq) + OH(aq)
Initial
0.100 M
0M
Change
+x
x
Equil. 0.100 x
x
2
2
x
x
(Assume x ~0) x = 1.3 103
1.8 105 =
0.100 x
0.100
(x < 5% of 0.100, thus, the assumption is valid).
Hence, x = [OH] = 1.3 103 pOH = 2.89, pH = 11.11
(ii)
60
Kw
1 1014
=
= 5.6 1010
5
Kb
1.8 10
For initial pOH, use I.C.E.(initial, change, equilibrium) table.
(D) 25.00 mL of 0.100 M NH3 is titrated with 0.100 M HCl Ka =
(i)
50
After 2 mL of HCl is added: [HCl] = 0.100 M
2.00 mL
= 0.00741 M (after
27.00 mL
dilution)
[NH3] = 0.100 M
25.00 mL
= 0.0926 M (after dilution)
27.00 mL
835
≈0M
+x
x
Chapter 17: Additional Aspects of Acid–Base Equilibria
The equilibrium constant for the neutralization reaction is large (Kneut = Kb/Kw =
1.9×105) and thus the reaction goes to nearly 100% completion. Assume that the
limiting reagent is used up (100% reaction in the reverse direction) and re-establish
the equilibrium by a shift in the forward direction. Here H3O+ (HCl) is the limiting
reagent.
NH4+(aq) + H2O(l)
Initial
0M
Change
+x
100% rxn 0.00741
Change
y
Equil
0.00741y
5.6 1010 =
y (0.0852 y )
(0.00741 y )
Ka = 5.6 10
-10
NH3(aq) + H3O+(aq)
0.0926 M
x
0.0852
+y
re-establish equilibrium
0.0852 + y
x = 0.00741
y (0.0852)
=
(0.00741)
0.00741 M
x
0
+y
y
(set y ~ 0) y = 4.8 1011
(the approximation is clearly valid)
y = [H3O+] = 4.8 1011; pH = 10.32 and pOH = 3.68
(iii) pH at 1/2 equivalence point = pKa = log 5.6 1010 = 9.25 and pOH = 4.75
(iv) After addition of 24 mL of HCl:
[HCl] = 0.100 M
24.00 mL
25.00 mL
= 0.0490 M; [NH3] = 0.100 M
= 0.0510 M
49.00 mL
49.00 mL
The equilibrium constant for the neutralization reaction is large (see above), and thus
the reaction goes to nearly 100% completion. Assume that the limiting reagent is
used up (100% reaction in the reverse direction) and re-establish the equilibrium in
the reverse direction. Here H3O+ (HCl) is the limiting reagent.
NH4+(aq) + H2O(l)
Initial
0M
Change
+x
100% rxn 0.0490
Change
y
Equil
0.0490y
Ka = 5.6 10
-10
NH3(aq) + H3O+(aq)
0.0541 M
x = 0.0490
x
0.0020
+y
re-establish equilibrium
0.0020 + y
0.0490 M
x
0
+y
y
y (0.0020 y )
y (0.0020)
=
(Assume y ~ 0) y = 1.3 108 (valid)
(0.0490 y )
(0.0490)
+
8
y = [H3O ] = 1.3 10 ; pH = 7.89 and pOH = 6.11
5.6 1010 =
(v)
25.00 mL
=0.0500 M
50.00 mL
NH3(aq) + H3O+(aq)
Equiv. point: 100% reaction of NH3 NH4+: [NH4+] = 0.100
NH4+(aq)
Initial 0.0500 M
Change
x
Equil
0.0500x
+ H2O(l)
Ka = 5.6 10
-10
0M
+x
x
836
~0M
+x
x
Chapter 17: Additional Aspects of Acid–Base Equilibria
5.6 1010=
x2
x2
=
(Assume x ~ 0) x = 5.3 106
(0.0500 x) 0.0500
(the approximation is clearly valid)
x = [H3O+] = 5.3 106 ; pH = 5.28 and pOH = 8.72
(vi) After addition of 26 mL of HCl, HCl is in excess. The pH and pOH should be the
same as those in Exercise 51. pH = 2.70 and pOH = 11.30
(vii) After addition of 33 mL of HCl, HCl is in excess. The pH and pOH should be the
same as those in Exercise 51. pH = 1.860 and pOH = 12.140. This time the curves
are not mirror images of one another, but rather they are related through a
reflection in a horizontal line centered at pH or pOH = 7.
0.100 M NH3 (25mL) + 0.100 HCl
0.100 M NH3 (25mL) + 0.100 HCl
12
12
10
10
8
8
pH
14
pOH
14
6
6
4
4
2
2
0
0
0
10
20
30
Volume of HCl (mL)
40
50
0
10
20
30
Volume of HCl (mL)
40
Salts of Polyprotic Acids
53.
(E) We expect a solution of Na 2S to be alkaline, or basic. This alkalinity is created by the
hydrolysis of the sulfide ion, the anion of a very weak acid ( K2 = 1 1019 for H 2S ).
HS aq + OH aq
S2 aq + H 2 O(l)
54.
(E) We expect the pH of a solution of sodium dihydrogen citrate, NaH 2 Cit , to be acidic
because the pKa values of first and second ionization constants of polyprotic acids are
reasonably large. The pH of a solution of the salt is the average of pK1 and pK2 . For citric
acid, in fact, this average is 3.13 + 4.77 2 = 3.95 . Thus, NaH2Cit affords acidic
solutions.
55.
(M)
2
H 2 PO 4 aq + HCO3 aq
(a) H 3 PO 4 aq + CO3 aq
H 2 PO 4 aq + CO3
HPO 4
2
2
HPO 4 2 aq + HCO3 aq
aq
PO 43 aq + H 2 O(l)
aq + OH aq
837
50
Chapter 17: Additional Aspects of Acid–Base Equilibria
(b) The pH values of 1.00 M solutions of the three ions are;
2
1.0 M OH pH = 14.00 1.0 M CO 3 pH = 12.16
3
1.0 M PO 4 pH = 13.15
2
Thus, we see that CO 3 is not a strong enough base to remove the third proton
from H 3 PO 4 . As an alternative method of solving this problem, we can compute
the equilibrium constants of the reactions of carbonate ion with H 3 PO 4 , H2PO4and HPO42-.
H 3 PO 4 + CO 3
2
H 2 PO 4 + HCO 3 K
2
2
H 2 PO 4 + CO3
HPO 4 + HCO3 K
2
HPO 4 + CO3
2
3
PO 4 + HCO3
K
K a1{H 3 PO 4 }
K a 2 {H 2 CO 3 }
K a 2 {H 3 PO 4 }
K a 2 {H 2 CO3 }
K a 3 {H 3 PO 4 }
K a 2 {H 2 CO3 }
7.1 103
4.7 10
11
6.3 108
4.7 10 11
4.2 10 13
4.7 10
11
1.5 108
1.3 103
8.9 10 3
Since the equilibrium constant for the third reaction is much smaller than 1.00, we
conclude that it proceeds to the right to only a negligible extent and thus is not a
3
practical method of producing PO 4 . The other two reactions have large
equilibrium constants, and products are expected to strongly predominate. They
have the advantage of involving an inexpensive base and, even if they do not go to
completion, they will be drawn to completion by reaction with OH in the last step
of the process.
56.
(M) We expect CO32- to hydrolyze and the hydrolysis products to determine the pH of
the solution.
Equation:
Initial
Changes:
Equil:
HCO3 (aq)
1.00 M
–x M
(1.00 – x) M
H 2 O(l)
"H 2 CO3 "(aq)
+
0M
+x M
xM
OH (aq)
0M
+x M
xM
H 2CO3 OH x x x 2
K w 1.00 1014
8
=
= 2.3 10 =
=
Kb =
4.4 107
1.00 x 1.00
K1
HCO3
(Cb/Kb = a very large number; thus, the approximation is valid).
x 100
. 2.3 108 15
. 104 M [OH ] ; pOH = log(15
. 104 ) 382
.
pH = 10.18
For 1.00 M NaOH, OH = 1.00 pOH = log 1.00 = 0.00
pH = 14.00
Both 1.00 M NaHCO3 and 1.00 M NaOH have an equal capacity to neutralize acids
since one mole of each neutralizes one mole of strong acid.
NaOH aq + H 3O + aq Na + aq + 2H 2 O(l)
838
Chapter 17: Additional Aspects of Acid–Base Equilibria
NaHCO 3 aq + H 3O + aq Na + aq + CO 2 g + 2H 2 O(l) .
But on a per gram basis, the one with the smaller molar mass is the more effective.
Because the molar mass of NaOH is 40.0 g/mol, while that of NaHCO 3 is 84.0 g/mol,
NaOH(s) is more than twice as effective as NaHCO3(s) on a per gram basis. NaHCO3 is
preferred in laboratories for safety and expense reasons. NaOH is not a good choice
because it can cause severe burns. NaHCO3, baking soda, by comparison, is relatively
non-hazardous. It also is much cheaper than NaOH.
57.
(M) Malonic acid has the formula H2C3H2O4 MM = 104.06 g/mol
1 mol
= 0.187 mol
Moles of H2C3H2O4 = 19.5 g
104.06 g
moles
0.187 moles
Concentration of H2C3H2O4 =
=
= 0.748 M
V
0.250 L
The second proton that can dissociate has a negligible effect on pH ( K a 2 is very small).
Thus the pH is determined almost entirely by the first proton loss.
H2A(aq) + H2O(l)
Initial
0.748 M
Change
x
Equil
0.748x
+
HA(aq) + H3O (aq)
0M
≈0M
+x
+x
x
x
x2
= Ka ;
pH = 1.47, therefore, [H3O+] = 0.034 M = x ,
0.748 x
(0.034) 2
= 1.6 103
K a1 =
0.748 0.034
(1.5 103 in tables, difference owing to ionization of the second proton)
So, x =
+
A2(aq) + H3O (aq)
0M
≈0M
+x
+x
x
x
(5.5 105 ) 2
= 1.0 108
pH = 4.26, therefore, [H3O+] = 5.5 105 M = x, K a 2 =
0.300 5.5 105
HA-(aq) + H2O(l)
Initial
0.300 M
Change
x
Equil
0.300x
58.
(M) Ortho-phthalic acid. K a1 = 1.1 103, K a 2 = 3.9 106
(a)
We have a solution of HA.
HA(aq) + H2O(l)
Initial
0.350 M
Change
x
Equil
0.350x
839
+
A2(aq) + H3O (aq)
0M
~0M
+x
+x
x
x
Chapter 17: Additional Aspects of Acid–Base Equilibria
x2
x2
, x = 1.2 103
0.350 x 0.350
(x
0.350, thus, the approximation is valid)
x = [H3O+] = 1.2 103, pH = 2.92
3.9 106 =
(b)
36.35 g of potassium ortho-phthalate (MM = 242.314 g mol1)
1 mol
= 0.150 mol in 1 L
moles of potassium ortho-phthalate = 36.35 g
242.314 g
A2(aq) + H2O(l)
Initial
0.150 M
Change
x
Equil
0.150 x
K b,A2
HA(aq) +
0M
+x
x
OH(aq)
≈0M
+x
x
K w 1.0 1014
x2
x2
9
2.6
10
Ka2
0.150 x 0.150
3.9 106
x 2.0 105 (x << 0.150, so the approximation is valid) = [OH]
pOH = log 2.0105 = 4.70; pH = 9.30
General Acid–Base Equilibria
59.
(E)
(a)
b g
Ba OH
2
is a strong base.
OH = 10 2.12 = 0.0076 M
0.0076 mol OH 1 mol Ba OH 2
=
= 0.0038 M
1L
2 mol OH
pOH = 14.00 11.88 = 2.12
b g
Ba OH
(b)
2
b g
C2 H 3O 2
0.294 M
pH = 4.52 = pK a + log
= 4.74 + log
HC 2 H 3O 2
HC 2 H 3O 2
0.294 M
= 4.52 4.74
HC2 H 3O 2
log
HC2 H3O2 =
0.294 M
= 100.22 = 0.60
HC2 H 3O 2
0.294 M
= 0.49 M
0.60
840
Chapter 17: Additional Aspects of Acid–Base Equilibria
60.
(M)
(a) pOH = 14.00 8.95 = 5.05
OH = 105.05 = 8.9 106 M
Equation:
C6 H 5 NH 2 (aq)
Initial
Changes:
Equil:
xM
8.9 106 M
( x 8.9 106 ) M
Kb =
H 2 O(l)
C 6 H 5 NH 3 (aq)
0 M
8.9 106 M
8.9 106 M
0M
8.9 106 M
8.9 106 M
OH (aq)
c8.9 10 h
=
6 2
C 6 H 5 NH 3 + OH
= 7.4 10
LMC6 H 5 NH 2 OP
N
Q
c8.9 10 h
=
10
x 8.9 106
6 2
x 8.9 10
(b)
6
7.4 1010
= 0.11 M
x = 0.11 M = molarity of aniline
H 3O + = 105.12 = 7.6 106 M
Equation: NH 4 + (aq)
H 2 O(l)
Initial
xM
Changes: 7.6 106 M
Equil:
( x 7.6 106 ) M
Ka =
NH 3 H 3O +
NH 4 +
NH 3 (aq)
H 3O (aq)
0 M
7.6 106 M
7.6 106 M
0M
7.6 106 M
7.6 106 M
c
h
2
7.6 106
Kw
1.0 1014
10
=
=
= 5.6 10 =
Kb for NH 3 1.8 105
x 7.6 106
c7.6 10 h
=
6 2
x 7.6 10
61.
6
5.6 10
10
= 0.10 M
x = NH 4
+
= NH 4 Cl = 0.10 M
(M)
(a) A solution can be prepared with equal concentrations of weak acid and conjugate
base (it would be a buffer, with a buffer ratio of 1.00, where the pH = pKa = 9.26).
Clearly, this solution can be prepared, however, it would not have a pH of 6.07.
(b) These solutes can be added to the same solution, but the final solution will have an
appreciable HC 2 H 3O 2 because of the reaction of H 3O + aq with C 2 H 3O 2 aq
b g
b g
HC 2 H 3O 2 (aq) + H 2 O(l)
H 3O + (aq) + C2 H3O 2 (aq)
Initial
0.058 M
0.10 M
0M
Changes:
–0.058 M
–0.058 M
+0.058 M
Equil:
≈0.000 M
0.04 M
0.058 M
+
Of course, some H 3O will exist in the final solution, but not equivalent to 0.058 M HI.
Equation:
841
Chapter 17: Additional Aspects of Acid–Base Equilibria
(c)
Both 0.10 M KNO 2 and 0.25 M KNO 3 can exist together. Some hydrolysis of the
b g
b g
BabOH g is a strong base and will react as much as possible with the weak
conjugate acid NH , to form NH baq g .We will end up with a solution
of BaCl baq g, NH baq g , and unreacted NH Clbaq g .
NO 2 aq ion will occur, forming HNO 2 aq and OH-(aq).
(d)
2
4
2
(e)
3
62.
4
This will be a benzoic acid–benzoate ion buffer solution. Since the two
components have the same concentration, the buffer solution will have
pH = pKa = log 6.3 105 = 4.20 . This solution can indeed exist.
c
(f)
3
h
The first three components contain no ions that will hydrolyze. But C2 H3O 2 is
the anion of a weak acid and will hydrolyze to form a slightly basic solution.
Since pH = 6.4 is an acidic solution, the solution described cannot exist.
(M)
(a) When H 3O + and HC 2 H 3O 2 are high and C2 H3O 2 is very low, a common
ion H 3O + has been added to a solution of acetic acid, suppressing its ionization.
(b) When C2 H3O 2 is high and H 3O + and HC 2 H 3O 2 are very low, we are
dealing with a solution of acetate ion, which hydrolyzes to produce a small
concentration of HC2 H 3O 2 .
(c)
When HC 2 H 3O 2 is high and both H 3O + and C2 H3O 2 are low, the
solution is an acetic acid solution, in which the solute is partially ionized.
(d) When both HC 2 H 3O 2 and C2 H3O 2 are high while H 3O + is low, the
solution is a buffer solution, in which the presence of acetate ion suppresses the
ionization of acetic acid.
INTEGRATIVE AND EXERCISES
63. (M)
Na 2 SO 4 (aq) H 2 O(l)
(a) NaHSO 4 (aq) NaOH(aq)
SO 42 –(aq) H 2 O(l)
HSO 4 (aq) OH – (aq)
(b) We first determine the mass of NaHSO4.
1L
0.225 mol NaOH 1 mol NaHSO 4
1000 mL
1 L
1 mol NaOH
120.06 g NaHSO 4
0.988 g NaHSO 4
1 mol NaHSO 4
mass NaHSO 4 36.56 mL
842
Chapter 17: Additional Aspects of Acid–Base Equilibria
1.016 g sample – 0.988 g NaHSO 4
100% 2.8% NaCl
1.016 g sample
(c) At the endpoint of this titration the solution is one of predominantly SO42–, from
which the pH is determined by hydrolysis. Since Ka for HSO4– is relatively large (1.1
× 10–2), base hydrolysis of SO42– should not occur to a very great extent. The pH of a
neutralized solution should be very nearly 7, and most of the indicators represented
in Figure 17-8 would be suitable. A more exact solution follows.
% NaCl
1 mol SO 4 2
0.988 g NaHSO 4
1 mol NaHSO 4
[SO ]
0.225 M
0.03656 L
120.06 g NaHSO 4 1 mol NaHSO 4
24
Equation:
Initial:
Changes:
Equil:
Kb
HSO 4 - (aq)
SO 4 2 (aq) H 2 O(l)
0.225 M
0M
OH – (aq)
0M
x M
xM
xM
(0.225 x)M
xM
xM
[HSO 4 – ] [OH – ]
[SO 4 2 – ]
K w 1.0 10 14
xx
x2
13
9
.
1
10
0.225 x 0.225
K a2
1.1 10 2
[OH – ] 9.1 1013 0.225 4.5 107 M
(the approximation was valid since x << 0.225 M)
pOH –log(4.5 107 ) 6.35
pH 14.00 6.35 7.65
Thus, either bromthymol blue (pH color change range from pH = 6.1 to pH = 7.9) or
phenol red (pH color change range from pH = 6.4 to pH = 8.0) would be a suitable
indicator, since either changes color at pH = 7.65.
64. (D) The original solution contains 250.0 mL
pK a log(1.35 105 ) 4.87
0.100 mmol HC3 H 5O 2
25.0 mmol HC3 H 5O 2
1 mL soln
We let V be the volume added to the solution, in mL.
(a) Since we add V mL of HCl solution (and each mL adds 1.00 mmol H3O+ to the
solution), we have added V mmol H3O+ to the solution. Now the final [H3O+] = 10–1.00
= 0.100 M.
V mmol H 3 O
0.100 M
(250.0 V ) mL
25.0
27.8 mL added
V 25.0 0.100 V , therefore, V
0.900
[H 3 O ]
Now, we check our assumptions. The total solution volume is 250.0 mL+ 27.8 mL =
277.8 mL There are 25.0 mmol HC2H3O2 present before equilibrium is established,
and 27.8 mmol H3O+ also.
843
Chapter 17: Additional Aspects of Acid–Base Equilibria
[HC 2 H 3 O 2 ]
Equation:
Initial:
Changes:
Equil:
Ka
25.0 mmol
27.8 mmol
[H 3 O ]
0.0900 M
0.100 M
277.8 mL
277.8 mL
HC3 H 5O 2 (aq) H 2 O(l)
C3H 5O 2 ( aq) H 3O ( aq)
0.0900 M
-x M
(0.08999 x)M
[H 3O ][C3 H 5O 2 ]
[HC3 H 5O 2 ]
0 M
xM
xM
1.35 105
0.100 M
x M
(0.100 x)M
x (0.100 x) 0.100 x
0.0900
0.0900 x
x 1.22 105 M
The assumption used in solving this equilibrium situation, that x << 0.0900
clearly is correct. In addition, the tacit assumption that virtually all of the H3O+
comes from the HCl also is correct.
(b) The pH desired is within 1.00 pH unit of the pKa. We use the HendersonHasselbalch equation to find the required buffer ratio.
pH pK a
log
nAnHA
[A – ]
pK a log
[HA]
nA-
0.87
nHA
Vsoln 3.3 mmol C3 H5 O2
nA- / V
nHA / V
PK a log
100.87 0.13
nAnHA
4.00 4.87 log
nA25.00
nAnHA
nA- 3.3
1 mmol NaC3 H5 O 2
1 mL soln
3.3 mL added
1.00 mmol NaC3 H 5 O 2
1 mmol C3 H 5 O 2
We have assumed that all of the C3H5O2– is obtained from the NaC3H5O2 solution,
since the addition of that ion in the solution should suppress the ionization of
HC3H5O2.
(c) We let V be the final volume of the solution.
Equation:
C3 H 5 O ( aq) H 3 O ( aq)
HC3 H 5 O 2 ( aq) H 2 O(l)
2
Initial:
25.0/V
Changes:
Equil:
x/V M
(25.0/V x/V) M
Ka
[ H 3 O ] [C 3 H 5 O 2 ]
[HC 3 H 5 O 2 ]
1.35 10 5
0M
0M
x/V M
x/V M
x/V M
x/V M
(x /V )2
x2 /V
25.0 / V x / V
25.0
When V = 250.0 mL, x = 0.29 mmol H3O+
[H 3 O ]
pH = –log(1.2 × 10–3) = 2.92
An increase of 0.15 pH unit gives pH = 2.92 + 0.15 = 3.07
844
0.29 mmol
1.2 10 3 M
250.0 mL
Chapter 17: Additional Aspects of Acid–Base Equilibria
[H3O+] = 10–3.07 = 8.5 × 10–4 M
1.3 105
This is the value of x/V. Now solve for V.
(8.5 104 ) 2
25.0 / V 8.5 104
25.0 / V 8.5 104
(8.5 104 ) 2
0.056
1.3 105
25.0
4.4 102 mL
0.057
On the other hand, if we had used [H3O+] = 1.16 × 10–3 M (rather than
1.2 × 10–3 M), we would obtain V = 4.8 × 102 mL. The answer to the problem thus
is sensitive to the last significant figure that is retained. We obtain V = 4.6 × 102
mL, requiring the addition of 2.1 × 102 mL of H2O.
25.0 / V 0.056 0.00085 0.057
V
Another possibility is to recognize that [H3O+] K a Ca for a weak acid with
ionization constant Ka and initial concentration Ca if the approximation is valid.
If Ca is changed to Ca/2, [H3O+] K a C a 2 2. Since, pH = –log [H3O+], the
change in pH given by:
pH log 2 log 21 / 2 0.5 log 2 0.5 0.30103 0.15 .
This corresponds to doubling the solution volume, that is, to adding 250 mL water.
Diluted by half with water, [H3O+] goes down and pH rises, then (pH1 – pH2) < 0.
65.
(E) Carbonic acid is unstable in aqueous solution, decomposing to CO2(aq) and H2O.
The CO2(aq), in turn, escapes from the solution, to a degree determined in large part
by the partial pressure of CO2(g) in the atmosphere.
H2CO3(aq) H2O(l) + CO2(aq) CO2(g) + H2O(l)
Thus, a solution of carbonic acid in the laboratory soon will reach a low [H2CO3], since
the partial pressure of CO2(g) in the atmosphere is quite low. Thus, such a solution
would be unreliable as a buffer. In the body, however, the [H2CO3] is regulated in part
by the process of respiration. Respiration rates increase when it is necessary to decrease
[H2CO3] and respiration rates decrease when it is necessary to increase [H2CO3].
66.
(E)
(a) At a pH = 2.00, in Figure 17-9 the pH is changing gradually with added NaOH. There
would be no sudden change in color with the addition of a small volume of NaOH.
(b) At pH = 2.0 in Figure 17-9, approximately 20.5 mL have been added. Since,
equivalence required the addition of 25.0 mL, there are 4.5 mL left to add.
Therefore,
% HCl unneutralized
4.5
100% 18%
25.0
845
Chapter 17: Additional Aspects of Acid–Base Equilibria
67.
(D) Let us begin the derivation with the definition of [H3O+].
amount of excess H 3 O
volume of titrant volume of solution being titrated
Vb volume of base (titrant)
Let Va volume of acid ( solution being titrated)
[H 3 O ]
M b molarity of base
M a molarity of acid
[H 3 O ]
Va M a Vb M b
Va Vb
Now we solve this equation for Vb
V a M a Vb M b [H 3 O ] (Va Vb )
Vb
Vb ([H 3 O ] M b ) Va ( M a [H 3 O ])
Va ( M a [H 3 O ])
Vb
[H 3 O ] M b
Va ( M a 10 pH )
10 pH M b
(a) 10 pH 102.00 1.0 102 M
Vb
20.00 mL (0.1500 0.010)
25.45 mL
0.010 0.1000
(b) 10 pH 103.50 3.2 104 M
Vb
20.00 mL (0.1500 0.0003)
29.85 mL
0.0003 0.1000
(c) 10 pH 105.00 1.0 105 M
Vb
20.00 mL (0.1500 0.00001)
30.00 mL
0.00001 0.1000
Beyond the equivalence point, the situation is different.
amount of excess OH – Vb M b V a M a
–
[OH ]
total solution volume
V a Vb
–
Solve this equation for Vb. [OH ](Va + Vb) = Vb × Mb – Va × Ma
V ([OH – ] M a )
Va ([OH – ] M a ) Vb ( M b [OH – ])
Vb a
M b [OH – ]
(d) [OH ] 10 pOH 1014.00 pH 103.50 0.00032 M;
0.00032 0.1500
Vb 20.00 mL
30.16 mL
0.1000 0.00032
[OH ] 10 pOH 1014.00 pH 102.00 0.010 M;
(e)
0.010 0.1500
Vb 20.00 mL
35.56 mL
0.1000 0.010
The initial pH = –log(0.150) = 0.824. The titration curve is sketched below.
846
Chapter 17: Additional Aspects of Acid–Base Equilibria
pH
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
25
30
35
Volume NaOH (ml)
68. (D)
(a) The expressions that we obtained in Exercise 67 were for an acid being titrated by a
base. For this titration, we need to switch the a and b subscripts and exchange [OH–]
and [H3O+].
M b [OH ]
[OH ] M a
Before the equivalence point :
Va Vb
After the equivalence point :
[H 3 O ] M b
Va Vb
M a [H 3 O ]
pOH 14.00 13.00 1.00 [OH ] 101.00 1.0 101 0.10 M
0.250 0.10
9.38 mL
Va 25.00 mL
0.10 0.300
pH 12.00 pOH 14.00 12.00 2.00 [OH ] 102.00 1.0 102 0.010 M
pH 13.00
0.250 0.010
19.4 mL
0.010 0.300
pOH 14.00 10.00 4.00
Va 25.00 mL
pH 10.00
[OH ] 104.00 1.0 104 0.00010 M
0.250 0.00010
20.8 mL
0.00010 0.300
[H 3 O ] 104.00 1.0 104 0.00010 M
Va 25.00 mL
pH 4.00
0.00010 0.250
20.8 mL
0.300 0.00010
pH 3.00 [H 3 O ] 103.00 1.0 103 0.0010 M
Va 25.00
Va 25.00
0.0010 0.250
21.0 mL
0.300 0.0010
847
Chapter 17: Additional Aspects of Acid–Base Equilibria
(b) Our expression does not include the equilibrium constant, Ka. But Ka is not needed
after the equivalence point. pOH = 14.00 – 11.50 = 2.50 [OH–] = 10–2.50 = 3.2 × 10–3
= 0.0032 M
Vb
Va ( M a [OH ]) 50.00 mL (0.0100 0.0032)
14.1 mL
0.0500 0.0032
M b [OH ]
Let us use the Henderson-Hasselbalch equation as a basis to derive an expression that
incorporates Ka. Note that because the numerator and denominator of that expression
are concentrations of substances present in the same volume of solution, those
concentrations can be replaced by numbers of moles.
Amount of anion = VbMb since the added OH– reacts 1:1 with the weak acid.
Amount of acid = VaMa – VbMb the acid left unreacted
pH pK a log
Vb M b
[A ]
pK a log
V a M a Vb M b
[HA]
Rearrange and solve for Vb 10 pH pK
(V a M a Vb M b ) 10
pH pK
Vb M b
pH 4.50, 10pH pK 104.50 4.20
pH 5.50, 10pH pK 105.50 4.20
Vb M b
V a M a Vb M b
Vb
V a M a 10 pH pK
M b (1 10 pH pK )
50.00 mL 0.0100 M 2.0
2.0 Vb
6.7 mL
0.0500(1 2.0)
50.00 mL 0.0100 M 20
20. Vb
9.5 mL
0.0500(1 20.)
69. (M)
(a) We concentrate on the ratio of concentrations of which the logarithm is taken.
f initial amt. weak acid
f
equil. amount conj. base
[weak acid]eq
equil. amount weak acid (1 f ) initial amt. weak acid 1 f
The first transformation is the result of realizing that the volume in which the weak acid
and its conjugate base are dissolved is the same volume, and therefore the ratio of
equilibrium amounts is the same as the ratio of concentrations. The second
transformation is the result of realizing, for instance, that if 0.40 of the weak acid has
been titrated, 0.40 of the original amount of weak acid now is in the form of its
conjugate base, and 0.60 of that amount remains as weak acid. Equation 17.2 then is:
pH = pKa + log (f /(1 – f ))
0.27
(b) We use the equation just derived. pH 10.00 log
9.56
1 0.23
[conjugate base]eq
848
Chapter 17: Additional Aspects of Acid–Base Equilibria
70.
(M)
[HPO 42- ]
[HPO 2[HPO 42- ]
4 ]
100.20 1.6
7.20
log
7.40
[H 2 PO 4 ]
[H 2 PO 4 ]
[H 2 PO 4 ]
(b) In order for the solution to be isotonic, it must have the same concentration of ions
as does the isotonic NaCl solution.
9.2 g NaCl
1 mol NaCl
2 mol ions
[ions]
0.31 M
1 L soln
58.44 g NaCl 1 mol NaCl
Thus, 1.00 L of the buffer must contain 0.31 moles of ions. The two solutes that are
used to formulate the buffer both ionize: KH2PO4 produces 2 mol of ions (K+ and
H2PO4–) per mole of solute, while Na 2 HPO 4 12 H 2 O produces 3 mol of ions (2 Na+
and HPO42– ) per mole of solute. We let x = amount of KH2PO4 and y = amount of
Na 2 HPO 4 12 H 2 O .
y
1.6 or y 1.6 x
2x 3 y 0.31 2 x 3(1.6 x) 6.8 x
x
0.31
x
0.046 mol KH 2 PO 4 y 1.6 0.046 0.074 mol Na 2 HPO 4 12 H 2O
6.8
136.08 g KH 2 PO 4
mass KH 2 PO 4 0.046 mol KH 2 PO 4
6.3 g KH 2 PO 4
1 mol KH 2 PO 4
358.1 g Na 2 HPO 4 12H 2 O
mass of Na 2 HPO 4 12H 2O 0.074 mol Na 2 HPO 4 12H 2O
1 mol Na 2 HPO 4 12H 2O
(a) pH pK a 2 log
26 g Na 2 HPO 4 12H 2O
71. (M) A solution of NH4Cl should be acidic, hence, we should add an alkaline solution to
make it pH neutral. We base our calculation on the ionization equation for NH3(aq), and
assume that little NH4+(aq) is transformed to NH3(aq) because of the inhibition of that
reaction by the added NH3(aq), and because the added volume of NH3(aq) does not
significantly alter the Vtotal.
NH 4 (aq) OH (aq)
Equation:
NH 3 (aq) H 2 O(l)
Initial:
xM
0.500 M
1.0 10 7 M
[NH 4 ][OH ] (0.500 M)(1.0 107 M)
1.8 105
Kb
[NH 3 ]
xM
(0.500 M)(1.0 107 M)
2.8 103 M [NH 3 ]
5
1.8 10 M
2.8 103 mol NH 3 1 L conc. soln 1 drop
V 500 mL
= 2.8 drops 3 drops
1 L final soln
10.0 mol NH 3 0.05 mL
xM
849
Chapter 17: Additional Aspects of Acid–Base Equilibria
72.
(D)
(a) In order to sketch the titration curve, we need the pH at the following points.
i ) Initial point. That is the pH of 0.0100 M p-nitrophenol, which we represent by
the general formula for an indicator that also is a weak acid, HIn.
Equation:
HIn(aq)
Initial:
Changes:
Equil:
0.0100 M
–x M
(0.0100–x) M
K HIn 107.15 7.1 108
H 2 O(l)
H 3O (aq)
0M
xM
x M
–
–
–
In – (aq)
0M
x M
xM
[H 3O ] [In – ]
xx
x2
[HIn]
0.0100 x 0.0100
Ca / K a 1105 ; thus, the approximation is valid
[H 3O ] 0.0100 7.1108 2.7 105 M
pH –log(2.7 10 –5 ) 4.57
ii) At the half-equivalence point, the pH = pKHIn = 7.15
iii) At the equiv point, pH is that of In–. nIn– = 25.00 mL × 0.0100 M = 0.250 mmol In–
Vtitrant = 0.25 mmol HIn
[In ]
0.25 mmol In
6.67 103 M 0.00667 M
25.00 mL 12.5 mL
Equation:
Initial:
1 mmol NaOH
1 mL titrant
12.5 mL titrant
1 mmol HIn 0.0200 mmol NaOH
In (aq)
0.00667 M
Changes:
x M
Equil:
(0.00667 x) M
H2O(l)
HIn(aq)
0M
x M
xM
OH (aq)
0M
x M
xM
[HIn][OH ] K w 1.0 104
xx
x x
Kb
1.4 107
8
[In ]
K a 7.1 10
0.00667 x 0.00667
5
(Ca/Ka = 1.4×10 ; thus, the approximation is valid)
[OH ] 0.00667 1.4 107 3.1105
pOH log(3.1105 ) 4.51; therefore, pH 14.00 pOH 9.49
iv) Beyond the equivalence point, the pH is determined by the amount of excess OH–
in solution. After 13.0 mL of 0.0200 M NaOH is added, 0.50 mL is in excess, and
the total volume is 38.0 mL.
0.50 mL 0.0200 M
[OH ]
2.63 10 4 M pOH 3.58 pH 10.42
38.0 mL
After 14.0 mL of 0.0200 M NaOH is added, 1.50 mL is in excess, and Vtotal =
39.0 mL.
1.50 mL 0.0200 M
7.69 10 4 M pOH 3.11 pH 10.89
[OH ]
39.0 mL
v) In the buffer region, the pH is determined with the use of the HendersonHasselbalch equation.
850
Chapter 17: Additional Aspects of Acid–Base Equilibria
0.10
0.90
6.20 At f 0.90, pH 7.15 log
8.10
0.90
0.10
f = 0.10 occurs with 0.10 × 12.5 mL = 1.25 mL added titrant; f = 0.90 with 11.25
mL added. The titration curve plotted from these points follows.
At f 0.10, pH 7.15 log
12
11
10
9
pH
8
7
6
5
4
0
5
10
15
20
Volume NaOH (ml)
(b) The pH color change range of the indicator is shown on the titration curve.
(c) The equivalence point of the titration occurs at a pH of 9.49, far above the pH at which
p-nitrophenol has turned yellow. In fact, the color of the indicator changes gradually during the
course of the titration, making it unsuitable as an indicator for this titration.
Possible indicators are as follows.
Phenolphthalein: pH color change range from colorless at pH = 8.0 to red at pH = 10.0
Thymol blue: pH color change range from yellow at pH = 8.0 to blue at pH = 9.8
Thymolphthalein: pH color change range from colorless at pH 9 to blue at pH 11
The red tint of phenolphthalein will appear orange in the titrated p-nitrophenol
solution. The blue of thymol blue or thymolphthalein will appear green in the titrated
p-nitrophenol solution, producing a somewhat better yellow end point than the orange
phenolphthalein endpoint.
73.
(M)
(a) Equation (1) is the reverse of the equation for the autoionization of water. Thus, its
equilibrium constant is simply the inverse of Kw.
1
1
K
1.00 1014
K w 1.0 10 14
Equation (2) is the reverse of the hydrolysis reaction for NH4+. Thus, its equilibrium
constant is simply the inverse of the acid ionization constant for NH4+, Ka =
5.6 × 10-10
1
1
K'
1.8 10 9
10
K a 5.6 10
851
Chapter 17: Additional Aspects of Acid–Base Equilibria
(b) The extremely large size of each equilibrium constant indicates that each reaction goes
essentially to completion. In fact, a general rule of thumb suggests that a reaction is considered
essentially complete if Keq > 1000 for the reaction.
74. (D) The initial pH is that of 0.100 M HC2H3O2.
C2 H 3 O 2 (aq)
Equation:
HC2 H 3 O 2 (aq)
H 2 O(l)
Initial:
Changes:
Equil:
0.100 M
xM
(0.100 x) M
0M
x M
xM
H3 O (aq)
0M
x M
xM
[H 3 O ][C2 H 3 O 2 ]
xx
x2
1.8 105
[HC2 H 3 O 2 ]
0.100 x 0.100
3
(Ca/Ka = 5.5×10 ; thus, the approximation is valid)
Ka
[H 3 O ] 0.100 1.8 105 1.3 103 M
pH log(1.3 103 ) 2.89
At the equivalence point, we have a solution of NH4C2H3O2 which has a pH = 7.00, because
both NH4+ and C2H3O2– hydrolyze to an equivalent extent, since
K a (HC 2 H 3 O 2 ) K b (NH 3 ) and their hydrolysis constants also are virtually equal.
Total volume of titrant = 10.00 mL, since both acid and base have the same concentrations.
At the half equivalence point, which occurs when 5.00 mL of titrant have been added,
pH = pKa = 4.74. When the solution has been 90% titrated, 9.00 mL of 0.100 M NH3
has been added. We use the Henderson-Hasselbalch equation to find the pH after 90% of
the acid has been titrated
[C 2 H 3 O 2 ]
9
pH pK a log
4.74 log 5.69
[HC 2 H 3 O 2 ]
1
When the solution is 110% titrated, 11.00 mL of 0.100 M NH3 have been added.
amount NH3 added = 11.00 mL × 0.100 M = 1.10 mmol NH3
amount NH4+ produced = amount HC2H3O2 consumed = 1.00 mmol NH4+
amount NH3 unreacted = 1.10 mmol NH3 – 1.00 mmol NH4+ = 0.10 mmol NH3
We use the Henderson-Hasselbalch equation to determine the pH of the solution.
[NH 4 ]
1.00 mmol NH 4
pOH pK b log
4.74 log
5.74 pH 8.26
[ NH 3 ]
0.10 mmol NH 3
When the solution is 150% titrated, 15.00 mL of 0.100 M NH3 have been added.
amount NH3 added = 15.00 mL × 0.100 M = 1.50 mmol NH3
amount NH4+ produced = amount HC2H3O2 consumed = 1.00 mmol NH4+
amount NH3 unreacted = 1.50 mmol NH3 – 1.00 mmol NH4+ = 0.50 mmol NH3
[NH 4 ]
1.00 mmol NH 4
pOH pK b log
4.74 log
5.04 pH 8.96
[ NH 3 ]
0.50 mmol NH 3
The titration curve based on these points is sketched next. We note that the equivalence
point is not particularly sharp and thus, satisfactory results are not obtained from acetic
acid–ammonia titrations.
852
Chapter 17: Additional Aspects of Acid–Base Equilibria
10
9
8
7
pH
6
5
4
3
2
1
0
0
75.
2
4
6
8
10
Volume of NH3 (mL)
12
14
16
(D) C6H5NH3+ is a weak acid, whose acid ionization constant is determined from
Kb(C6H5NH2) = 7.4 × 10–10.
1.0 10 14
Ka
1.4 10 5 and pK a 4.85 . We first determine the initial pH.
7.4 10 10
C6 H 5 NH 2 (aq)
Equation:
C6 H 5 NH 3 (aq) H 2 O(l)
H 3O (aq)
Initial:
0.0500 M
0 M
0 M
Changes:
x M
x M
x M
Equil:
(0.0500 x) M
x M
x M
K a 1.4 10
5
[C6 H 5 NH 2 ] [H 3 O ]
xx
x2
0.0500 x 0.0500
[C6 H 5 NH 3 ]
(Since Ca /K a = 3.6×103 ; thus, the approximation is valid)
[H 3 O ] 0.0500 1.4 105 8.4 104 M
pH log (8.4 10 –4 ) 3.08
At the equivalence point, we have a solution of C6H5NH2(aq). Now find the volume of
titrant.
V 10.00 mL
0.0500 mmol C6 H 5 NH 3
1 mmol NaOH
1 mL titrant
5.00 mL
1 mL
1 mmol C6 H 5 NH 3 0.100 mmol NaOH
amount C6 H 5 NH 2 10.00 mL 0.0500 M 0.500 mmol C6 H 5 NH 2
[C6 H 5 NH 2 ]
0.500 mmol C6 H 5 NH 2
0.0333 M
10.00 mL 5.00 mL
853
Chapter 17: Additional Aspects of Acid–Base Equilibria
Initial:
0.0333 M
C6 H 5 NH 3 (aq)
H 2 O(l)
0 M
Changes:
x M
x M
x M
Equil:
(0.0333 x) M
xM
xM
Equation:
C6 H 5 NH 2 (aq)
K b 7.4 1010
OH (aq)
0M
[C6 H 5 NH 3 ][OH ]
xx
x
[C6 H 5 NH 2 ]
0.0333 x 0.0333
2
(Cb /K b = very large number; thus, the approximation is valid)
[OH ] 0.0333 7.4 1010 5.0 106
pOH 5.30
pH 8.70
At the half equivalence point, when 5.00 mL of 0.1000 M NaOH has been added,
pH = pKa = 4.85. At points in the buffer region, we use the expression derived in Exercise
69.
0.90
= 5.80
0.10
0.95
After 4.75 mL of titrant has been added, f = 0.95, pH = 4.85 + log
= 6.13
0.05
After 4.50 mL of titrant has been added, f = 0.90, pH = 4.85 + log
After the equivalence point has been reached, with 5.00 mL of titrant added, the pH of the
solution is determined by the amount of excess OH– that has been added.
0.25 mL 0.100 M
0.0016 M
15.25 mL
1.00 mL 0.100 M
At 120% titrated, [OH ]
0.00625 M
16.00 mL
At 105% titrated, [OH ]
pOH 2.80
pH 11.20
pOH 2.20 pH 11.80
Of course, one could sketch a suitable titration curve by calculating fewer points. Below is
the titration curve.
12
pH
10
8
6
4
2
0
Volume NaOH (ml)
854
5
Chapter 17: Additional Aspects of Acid–Base Equilibria
76.
In order for a diprotic acid to be titrated to two distinct equivalence points, the acid must
initially start out with two undissociated protons and the Ka values for the first and second
protons must differ by >1000. This certainly is not the case for H2SO4, which is a strong
acid (first proton is 100% dissociated). Effectively, this situation is very similar to the
titration of a monoprotic acid that has added strong acid (i.e., HCl or HNO3). With the
leveling effect of water, Ka1 = 1 and Ka2 = 0.011, there is a difference of only 100 between
Ka1 and Ka2 for H2SO4.
77.
(D) For both titration curves, we assume 10.00 mL of solution is being titrated and the
concentration of the solute is 1.00 M, the same as the concentration of the titrant. Thus,
10.00 mL of titrant is required in each case. (You may be able to sketch titration curves
based on fewer calculated points. Your titration curve will look slightly different if you
make different initial assumptions.)
(a) The initial pH is that of a solution of HCO3–(aq). This is an anion that can ionize to
CO32 (aq) or be hydrolyzed to H2CO3(aq). Thus,
pH 12 (pK a1 pK a2 ) 12 (6.35 10.33) 8.34
The final pH is that of 0.500 M H2CO3(aq). All of the NaOH(aq) has been
neutralized, as well as all of the HCO3–(aq), by the added HCl(aq).
Equation:
Initial:
Changes:
Equil:
Kb
H 2 CO3 (aq) H 2 O(l)
0.500 M
xM
(0.500 x ) M
HCO3 (aq) H 3 O (aq)
0M
xM
xM
0M
xM
xM
[HCO3 ][H 3 O ]
xx
x2
4.43 107
[H 2 CO3 ]
0.500 x 0.500
(Ca /K a = 1.1×106 ; thus, the approximation is valid)
[OH ] 0.500 4.4 107 4.7 104 M
pH 3.33
During the course of the titration, the pH is determined by the HendersonHasselbalch equation, with the numerator being the percent of bicarbonate ion
remaining, and as the denominator being the percent of bicarbonate ion that has been
transformed to H2CO3.
90% titrated: pH = 6.35 + log
10%
5.40
90%
10% titrated: pH = 6.35 + log
90%
7.30
10%
5%
95%
5.07
5% titrated: pH = 6.35 + log
7.63
95%
5%
After the equivalence point, the pH is determined by the excess H3O+.
0.50 mL 1.00 M
At 105% titrated, [H 3 O ]
0.024
pH 1.61
20.5 mL
95% titrated: pH = 6.35 + log
855
Chapter 17: Additional Aspects of Acid–Base Equilibria
2.00 mL 1.00 M
0.091 M
pH 1.04
22.0 mL
The titration curve derived from these data is sketched below.
pH
At 120% titrated, [H 3 O ]
9
8
7
6
5
4
3
2
1
0
0
2
4
6
8
Volume 1.00 M HCl
10
12
(b) The final pH of the first step of the titration is that of a solution of HCO3–(aq). This
is an anion that can be ionized to CO32–(aq) or hydrolyzed to H2CO3(aq). Thus,
pH 12 (pK a1 pK a 2 ) 12 (6.35 10.33) 8.34 . The initial pH is that of 1.000 M
CO32–(aq), which pH is the result of the hydrolysis of the anion.
Equation:
Initial:
Changes:
Equil:
Kb
CO32 (aq)
1.000 M
x M
(1.000 x) M
H 2 O(l)
HCO3 (aq) OH (aq)
0M
x M
xM
0M
x M
xM
[HCO3 ][OH ] K w 1.0 1014
xx
x2
4
2.1
10
[CO32 ]
K a2 4.7 1011
1.000 x 1.000
[OH ] 1.000 2.1104 1.4 102 M
pOH 1.85
pH 12.15
During the course of the first step of the titration, the pH is determined by the
Henderson-Hasselbalch equation, modified as in Exercise 69, but using as the
numerator the percent of carbonate ion remaining, and as the denominator the
percent of carbonate ion that has been transformed to HCO3–.
10%
9.38
90%
5%
95% titrated: pH 10.33 log
9.05
95%
90% titrated: pH 10.33 log
90%
11.28
10%
95%
5% titrated: pH 10.33 log
11.61
5%
10% titrated: pH 10.33 log
During the course of the second step of the titration, the values of pH are precisely as
they are for the titration of NaHCO3, except the titrant volume is 10.00 mL more (the
volume needed to reach the first equivalence point). The solution at the second
856
Chapter 17: Additional Aspects of Acid–Base Equilibria
equivalence point is 0.333 M H2CO3, for which the set-up is similar to that for 0.500 M
H2CO3.
[H 3 O ] 0.333 4.4 10 7 3.8 10 4 M pH 3.42
After the equivalence point, the pH is determined by the excess H3O+.
0.50 mL 1.00 M
0.0164 M pH 1.8
At 105% titrated,
[H3O ]
30.5 mL
2.00 mL 1.00 M
0.0625 M pH 1.20
At 120% titrated,
[H3O ]
32.0 mL
The titration curve from these data is sketched below.
12
10
pH
8
6
4
2
0
-3
2
(c) VHCl 1.00 g NaHCO3
119 mL 0.100 M HCl
(d) VHCl 1.00 g Na 2 CO3
189 mL 0.100 M HCl
(e)
7
12
Volume 1.00 M HCl, mL
17
22
1 mol NaHCO3
1 mol HCl
1000 mL
84.01 g NaHCO3 1 mol NaHCO3 0.100 mol HCl
1 mol Na 2 CO3
2 mol HCl
1000 mL
105.99 g Na 2 CO3 1 mol Na 2 CO3 0.100 mol HCl
The phenolphthalein endpoint occurs at pH = 8.00 and signifies that the NaOH has
been neutralized, and that Na2CO3 has been half neutralized. The methyl orange
endpoint occurs at about pH = 3.3 and is the result of the second equivalence point
of Na2CO3. The mass of Na2CO3 can be determined as follows:
857
Chapter 17: Additional Aspects of Acid–Base Equilibria
1L
0.1000 mol HCl 1 mol HCO3 1 mol Na 2 CO3
mass Na 2 CO3 0.78 mL
1000 mL
1 L soln
1 mol HCl 1 mol HCO3
105.99 g Na 2 CO3
0.0083 g Na 2 CO3
1 mol Na 2 CO3
% Na 2 CO3
0.0083g
100 8.3% Na 2 CO3 by mass
0.1000g
78. (M) We shall represent piperazine as Pip in what follows. The cation resulting from the
first ionization is HPiP+, and that resulting from the second ionization is H2Pip2+.
1.00 g C4 H10 N 2 •6 H 2 O 1mol C4 H10 N 2 •6 H 2 O
(a) [Pip] =
0.0515M
0.100 L
194.22 g C 4 H10 N 2 •6 H 2 O
HPip (aq)
H 2O(l)
OH (aq)
Initial:
0.0515 M
0M
0 M
Changes:
x M
x M
xM
Equil:
(0.0515 x) M
xM
xM
Ca/Ka = 858; thus, the approximation is not valid. The full quadratic equation must be
solved.
[HPip ][OH ]
xx
K b1 104.22 6.0 105
[ Pip ]
0.0515 x
Equation:
Pip (aq)
From the roots of the equation, x = [OH ] 1.7 103 M
pOH 2.77
pH 11.23
(b) At the half-equivalence point of the first step in the titration, pOH = pKb1 = 4.22
pH = 14.00 – pOH = 14.00 – 4.22 = 9.78
(c) Volume of HCl = 100. mL ×
0.0515 mmol Pip 1mmol HCl
1mL titrant
×
×
= 10.3mL
1mL
1mmol Pip 0.500 mmol HCl
(d) At the first equivalence point we have a solution of HPip+. This ion can react as a base
with H2O to form H2Pip2+ or it can react as an acid with water, forming Pip (i.e. HPip+
is amphoteric). The solution’s pH is determined as follows (base hydrolysis
predominates):
pOH = ½ (pKb1 + pKb2) = ½ (4.22 + 8.67) = 6.45, hence pH = 14.00 – 6.45 = 7.55
(e) The pOH at the half-equivalence point in the second step of the titration equals pKb2.
pH = 14.00 – 8.67 = 5.33
pOH = pKb2 = 8.67
(f) The volume needed to reach the second equivalence point is twice the volume
needed to reach the first equivalence point, that is 2 × 10.3 mL = 20.6 mL.
(g) The pH at the second equivalence point is determined by the hydrolysis of the
H2Pip2+ cation, of which there is 5.15 mmol in solution, resulting from the reaction
858
Chapter 17: Additional Aspects of Acid–Base Equilibria
of HPip+ with HCl. The total solution volume is 100. mL + 20.6 mL = 120.6 mL
5.15 mmol
0.0427 M
[H2Pip2+] =
K b 2 108.67 2.1 109
120.6 mL
Equation:
Initial:
Changes:
Equil:
HPip (aq) H 3O (aq)
H 2 Pip 2+ (aq) H 2 O(l)
0.0427 M
0M
0 M
x M
x M
x M
xM
xM
(0.0427 x) M
K w 1.00 1014 [HPip ][H 3O ]
x x
x2
6
Ka
4.8 10
Kb 2
2.1 109
[H 2 Pip 2+ ]
0.0427 x 0.0427
x [H 3O ] 0.0427 4.8 106 4.5 104 M
(x << 0.0427; thus, the approximation is valid).
79. (M) Consider the two equilibria shown below.
HPO42-(aq) + H3O+(aq)
H2PO4 (aq) + H2O(l)
-
pH 3.347
[HPO 4 2- ][H 3 O + ]
Ka2 =
[H 2 PO 4- ]
[H 3 PO 4 ][OH - ]
Kb = Kw/Ka1 =
[H 2 PO 4- ]
All of the phosphorus containing species must add up to the initial molarity M.
Hence, mass balance: [HPO42- ] + [H2PO4- ] + [H3PO4] = M
charge balance: [H3O+] + [Na+] = [H2PO4- ] + 2×[HPO42- ]
Note: [Na+] for a solution of NaH2PO4 = M
H3PO4(aq) + OH-(aq)
H2PO4 (aq) + H2O(l)
-
Thus,
[H3O+ ] + [Na+ ] = [H2PO4- ] + 2×[HPO42- ] = [H3O+ ] + M (substitute mass balance equation)
[H3O+ ] + [HPO42- ] + [H2PO4- ] + [H3PO4] = [H2PO4- ] + 2×[HPO42- ] (cancel terms)
[H3O+ ] = [HPO42- ] - [H3PO4]
(Note: [HPO42- ] = initial [H3O+ ] and [H3PO4] = initial [OH- ])
Thus: [H3O+ ]equil = [H3O+ ]initial ]initial (excess H3O+ reacts with OH- to form H2O)
Rearrange the expression for Ka2 to solve for [HPO42-], and the expression for Kb to solve
for [H3PO4].
[H 3 O + ] = [HPO 4 2- ] - [H 3 PO 4 ] =
K a2 [H 2 PO 4 - ] K b [H 2 PO 4- ]
[H 3 O + ]
[OH - ]
Kw
[H 2 PO 4 - ]
K
[H
PO
]
K [H PO - ] [H O + ][H 2 PO 4 - ]
K
[H 3 O + ] a2 2 + 4 a1
a2 2 + 4 3
Kw
K a1
[H 3 O ]
[H 3 O ]
+
[H 3 O ]
-
Multiply through by [H3O+] Ka1 and solve for [H3O+].
859
Chapter 17: Additional Aspects of Acid–Base Equilibria
K a1 [H 3 O + ][H 3 O + ] K a1 K a2 [H 2 PO 4- ] [H 3 O + ][H 3 O + ][H 2 PO 4- ]
K a1 [H 3 O + ]2 K a1 K a2 [H 2 PO 4 - ] [H 3 O + ]2 [H 2 PO 4 - ]
K a1 [H 3 O + ]2 [H 3 O + ]2 [H 2 PO 4- ] K a1 K a2 [H 2 PO 4- ] = [H 3 O + ]2 (K a1 [H 2 PO 4- ])
K a1 [H 3 O + ]2 [H 3 O + ]2 [H 2 PO 4- ] K a1 K a2 [H 2 PO 4- ] = [H 3 O + ]2 (K a1 [H 2 PO 4- ])
K a1 K a2 [H 2 PO 4 - ]
For moderate concentrations of [H 2 PO 4 - ], K a1 << [H 2 PO4 - ]
(K a1 [H 2 PO 4 ])
[H 3 O + ]2 =
This simplifies our expression to:
[H 3O + ]2 =
K a1K a2 [H 2 PO 4 - ]
= K a1K a2
[H 2 PO 4 - ]
Take the square root of both sides: [H 3O + ]2 = K a1K a2
[H3O+ ] = K a1K a2 (K a1K a2 )1 2
Take the -log of both sides and simplify:
-log[H 3O + ] = log(K a1K a2 )1 2 1 2 log(K a1K a2 ) 1 2 (log K a1 +logK a2 )
-log[H 3O + ] 1 2 ( log K a1 logK a2 ) Use -log[H 3O + ] = pH and log K a1 pK a1 logK a2 pK a2
Hence, -log[H 3O + ] = 1 2 ( log K a1 logK a2 ) becomes pH =1 2 (pK a1 +pK a2 ) (Equation 17.5)
Equation 17.6 can be similarly answered.
80.
(D) H2PO4– can react with H2O by both ionization and hydrolysis.
HPO 4 2 (aq) H 3 O (aq)
H 2 PO 4 - (aq) H 2 O(l)
PO 4 3 (aq) H3 O (aq)
HPO 4 2- (aq) H 2 O(l)
The solution cannot have a large concentration of both H3O+ and OH– (cannot be
simultaneously an acidic and a basic solution). Since the solution is acidic, some of the
H3O+ produced in the first reaction reacts with virtually all of the OH– produced in the
second. In the first reaction [H3O+] = [HPO42–] and in the second reaction [OH–] =
[H3PO4]. Thus, following the neutralization of OH– by H3O+, we have the following.
Kw
[H 2 PO-4 ]
K
K
K
[H
PO
]
[H
PO
]
K [H PO ]
2
2
a
a
a
b 2 4 2
1
[H 3O ] [HPO 42- ] [H 2 PO -4 ] 2
Kw
[H 3O ]
[OH ]
[H 3O ]
[H 3O ]
4
K a2 [H 2 PO-4 ] [H 3O ][H 2 PO-4 ]
K a1
[H 3O ]
2
4
Then multiply through by [H3O ]Ka
1
and solve for [H3O ].
2
[H 3 O ] K a1 K a1 K a2 [H 2 PO ] [H 3 O ] [H 2 PO ]
4
4
[H 3 O ]
K a1 K a2 [H 2 PO -4 ]
[ K a1 [H 2 PO-4 ]
But, for a moderate [H2PO -4 ], from Table 16-6, K a1 = 7.1 × 10–3 < [H2PO -4 ] and thus
K a1 [H 2 PO 4 ] [H 2 PO 4 ] . Then we have the following expressions for [H3O+] and pH.
860
Chapter 17: Additional Aspects of Acid–Base Equilibria
[H 3O ] K a1 K a2
pH log[H 3O ] log(K a1 K a2 )1/2 12 [ log(K a1 K a2 )] 12 ( log K a1 log K a2 )
pH
1
2
(pK a1 pK a2 )
Notice that we assumed Ka < [H2PO4–]. This assumption is not valid in quite dilute
solutions because Ka = 0.0071.
81. (D)
(a) A buffer solution is able to react with small amounts of added acid or base. When strong acid is
added, it reacts with formate ion.
CHO 2 (aq) H 3 O (aq)
HCHO 2 (aq) H 2 O
Added strong base reacts with acetic acid.
HC 2 H 3 O 2 (aq) OH (aq)
H 2 O(l) C 2 H 3 O 2 (aq)
Therefore neither added strong acid nor added strong base alters the pH of the solution very much.
Mixtures of this type are referred to as buffer solutions.
(b) We begin with the two ionization reactions.
CHO 2 (aq) H 3 O (aq)
HCHO 2 (aq) H 2 O(l)
C2 H 3 O 2 (aq) H 3 O (aq)
HC2 H 3 O 2 (aq) H 2 O(l)
[Na ] 0.250
[OH ] 0
[H 3O ] x
0.150 [HC 2 H 3 O 2 ] [C 2 H 3 O 2 ]
[C2 H 3O 2 ] y
[CHO 2 ] z
[HC 2 H 3 O 2 ] 0.150 [C 2 H 3 O 2 ] 0.150 y
0.250 [HCHO 2 ] [CHO 2 ]
[HCHO 2 ] 0.250 [CHO 2 ] 0.250 z
[ Na ] [H 3 O ] [C 2 H 3 O 2 ] [CHO 2 ] [OH ] (electroneutrality)
0.250 x [C 2 H 3 O 2 ] [CHO 2 ] y z
(1)
[H 3 O ][C 2 H 3 O 2 ]
x y
K A 1.8 10 5
[HC 2 H 3 O 2 ]
0.150 y
(2)
[H 3 O ][CHO 2 ]
xz
K F 1.8 10 4
[HCHO 2 ]
0.250 z
(3)
There now are three equations—(1), (2), and (3)—in three unknowns—x, y, and z.
We solve equations (2) and (3), respectively, for y and z in terms of x.
861
Chapter 17: Additional Aspects of Acid–Base Equilibria
0.150 K A y K A xy
0.250 K F z K F xz
y
0.150 K A
KA x
z
0.250 K F
KF x
Then we substitute these expressions into equation (1) and solve for x.
0.250 x
0.150 K A 0.250 K F
0.250
KA x
KF x
since x 0.250
0.250 ( K F x) ( K A x) 0.150 K A ( K F x) 0.250 K F ( K A x)
K A K F ( K A K F ) x x 2 1.60 K A K F x(0.600 K A 1.00 K F )
x 2 0.400 K A x 0.600 K A K F 0 x 2 7.2 10 6 x 1.9 10 9
7.2 10 6 5.2 10 11 7.6 10 9
4.0 10 5 M [H 3 O ]
2
pH 4.40
(c) Adding 1.00 L of 0.100 M HCl to 1.00 L of buffer of course dilutes the
concentrations of all components by a factor of 2. Thus, [Na+] = 0.125 M; total
acetate concentration = 0.0750 M; total formate concentration = 0.125 M. Also, a
new ion is added to the solution, namely, [Cl–] = 0.0500 M.
x
[ Na ] 0.125 [OH ] 0 [H 3 O ] x [Cl ] 0.0500 M
0.0750 [HC 2 H 3 O 2 ] [C 2 H 3 O 2 ]
[C 2 H 3 O 2 ] y [CHO 2 ] z
[HC 2 H 3 O 2 ] 0.0750 – [C 2 H 3 O 2 ] 0.0750 y
0.125 [HCHO 2 ] [CHO 2 ]
[HCHO 2 ] 0.125 [CHO 2 ] 0.125 z
[ Na ] [H 3 O ] [C 2 H 3 O 2 ] [OH ] [CHO 2 ] [Cl ] (electroneutrality)
0.125 x [C 2 H 3 O 2 ] [CHO 2 ] [Cl ] y z 0 0.0500 0.075 x y z
[H 3 O ][C 2 H 3 O 2 ]
x y
K A 1.8 10 5
[ HC 2 H 3 O 2 ]
0.0750 y
[H 3 O ][CHO 2 ]
[HCHO 2 ]
K F 1.8 10 4
xz
0.125 z
Again, we solve the last two equations for y and z in terms of x.
0.0750 K A y K A xy
0.125 K F z K F xz
y
0.0750 K A
z
KA x
0.125 K F
KF x
862
Chapter 17: Additional Aspects of Acid–Base Equilibria
Then we substitute these expressions into equation (1) and solve for x.
0.0750 x
0.0750 K A
KA x
0.125 K F
KF x
0.0750
since x 0.0750
0.0750 (K F x) (K A x) 0.0750 K A (K F x) 0.125 K F (K A x)
K A K F (K A K F )x x 2 2.67 K A K F x(1.67 K F 1.00 K A )
x 2 0.67 K F x 1.67 K A K F 0 x 2 1.2 10 4 x 5.4 10 9
1.2 104 1.4 108 2.2 10 8
x
2
1.55 10 4 M [H 3O ]
pH 3.81 3.8
As expected, the addition of HCl(aq), a strong acid, caused the pH to drop. The
decrease in pH was relatively small, nonetheless, because the H3O+(aq) was
converted to the much weaker acid HCHO2 via the neutralization reaction:
CHO2-(aq) + H3O+(aq) → HCHO2(aq) + H2O(l)
(buffering action)
82. (M) First we find the pH at the equivalence point:
CH 3 CH(OH)COO - (aq) H 2 O(l)
CH 3 CH(OH)COOH(aq) OH ( aq)
1 mmol
1 mmol
1 mmol
The concentration of the salt is 1 103 mol/0.1 L = 0.01 M
The lactate anion undergoes hydrolysis thus:
Initial
CH 3 CH(OH)COOH(aq) OH (aq )
CH 3 CH(OH)COO - ( aq ) H 2 O(l)
0 M
0.01 M
0M
Change
Equilibrium
x
+x
(0.01 x) M
x
+x
x
-
Where x is the [hydrolyzed lactate ion], as well as that of the [OH ] produced by hydrolysis
K for the above reaction
[CH 3 CH(OH)COOH][OH ]
K w 1.0 1014
7.2 1011
[CH 3 CH(OH)COO- ]
Ka
103.86
x2
x2
7.2 1011 and x [OH - ] 8.49 107 M
0.01 x 0.01
x 0.01, thus, the assumption is valid
so
pOH log (8.49 107 M ) 6.07 pH 14 pOH 14.00 6.07 7.93
(a) Bromthymol blue or phenol red would be good indicators for this titration since they change
color over this pH range.
863
Chapter 17: Additional Aspects of Acid–Base Equilibria
(b) The H2PO4–/HPO42- system would be suitable because the pKa for the acid (namely, H2PO4–) is
close to the equivalence point pH of 7.93. An acetate buffer would be too acidic, an ammonia
buffer too basic.
H 3 O HPO 4 2 K a2 6.3 10 8
(c) H 2 PO 4 - H 2 O
K a2
[HPO 4 2 ]
6.3 108
5.4 (buffer ratio required)
Solving for
[H 2 PO 4 - ] [H 3 O ]
107.93
83.
(M)
H 3 O (aq) HO 2 (aq) K a
H 2 O 2 (aq) H 2 O(l)
[H 3 O ][HO 2 ]
[H 2 O 2 ]
Data taken from experiments 1 and 2:
[H 2 O 2 ] [HO 2 ] 0.259 M
[HO 2 ] 0.235 M
(6.78) (0.00357 M) [HO 2 ] 0.259 M
[H 2 O 2 ] (6.78)(0.00357 M) 0.0242 M
[H 3 O ] 10 ( pKw pOH )
log 0.250 - 0.235) M NaOH log (0.015 M NaOH) = 1.824
[H 3 O ] 10 (14.941.824) 7.7 1014
Ka
[H 3 O ][HO 2 ] (7.7 1014 M)(0.235 M)
7.4 1013
[H 2 O 2 ]
(0.0242 M)
pK a 12.14
From data taken from experiments 1 and 3:
[H 2 O 2 ] [HO 2 ] 0.123 M
(6.78)(0.00198 M) [HO 2 ] 0.123 M
[HO 2 ] 0.1096 M
[H 2 O 2 ] (6.78)(0.00198 M) 0.0134 M
[H 3 O ] 10 ( pKw pOH ) , pOH = -log[(0.125- 0.1096 ) M NaOH] 1.81
[H 3 O ] 10 (14.941.81) 7.41 1014
[H 3 O ][HO 2 ] (7.41 1014 M)(0.1096 M)
Ka
6.06 1013
[H 2 O 2 ]
(0.0134 M)
pK a = -log (6.06 1013 ) = 12.22
Average value for pK a = 12.17
864
Chapter 17: Additional Aspects of Acid–Base Equilibria
84.
(D) Let's consider some of the important processes occurring in the solution.
H 2 PO 4 (aq) OH (aq)
(1) HPO 4 2 (aq) H 2 O(l)
K (1) = K a2 4.2 10-13
1.0 1014
K (2) = K b
(2) HPO 4
1.6 107
8
6.3 10
1.0 1014
H 3 O (aq) NH 3 (aq)
K (3) = K a
(3) NH 4 (aq) H 2 O(l)
5.6 1010
1.8 105
1
H 2 O(l) NH 3 (aq)
K (4)
(4) NH 4 (aq) OH - (aq)
5.6 104
1.8 105
May have interaction between NH 4 and OH - formed from the hydrolysis of HPO 4 2 .
2
H 3 O (aq) PO 43 (aq)
(aq) H 2 O(l)
H 2 PO 4 - (aq) NH 3 (aq)
NH 4 (aq) HPO 4 2 (aq)
Initial 0.10M
0.10M
0 M
0 M
x
x
x
Change x
Equil. 0.100 x
0.100 x
x
x
(where x is the molar concentration of NH 4 that hydrolyzes)
K K (2) K (4) (1.6 107 ) (5.6 104 ) 9.0 103
K
[H 2 PO 4 - ] [NH 3 ]
[x]2
9.0 103 and x 8.7 103M
2
2
[NH 4 ] [HPO 4 ] [0.10-x]
Finding the pH of this buffer system:
pH pK a log
0.100-8.7 103 M
[HPO 4 2 ]
8
log(6.3
10
)
log
8.2
3
[H 2 PO 4 - ]
8.7 10 M
As expected, we get the same result using the NH 3 /NH 4 buffer system.
85. Consider the two weak acids and the equilibrium for water (autodissociation)
H3 O+ (aq) + A- (aq) K HA =
HA(aq) + H 2 O(l)
[H 3 O + ][A - ]
[HA]
[H 3 O + ][A - ]
[HA]=
K HA
H 3 O + (aq) + B- (aq) K HB =
HB(aq) + H 2 O(l)
[H 3 O + ][ B- ]
[HB]
[HB]=
H3 O+ (aq) + OH - (aq) K w =[H 3 O + ][OH - ]
H 2 O(l) + H 2 O(l)
865
[H 3 O + ][ B- ]
K HB
[OH - ] =
Kw
[H 3 O + ]
Chapter 17: Additional Aspects of Acid–Base Equilibria
Mass Balance: [HA]initial =[HA]+[A - ] =
From which: [A - ] =
[HA]initial
[H 3O + ]
+1
K HA
[HB]initial = [HB] + [B- ] =
From which: [B- ] =
[H O + ]
[H 3O + ][A - ]
+[A - ] = [A - ] 3
+1
K HA
K HA
[H O + ]
[H 3O + ][B- ]
+ [B- ] = [B- ] 3
+1
K HB
K
HB
[HB]initial
[H 3O + ]
+1
K HB
Charge Balance: [H3O+ ] = [A - ] + [B- ] + [OH - ] (substitute above expressions)
[H 3O + ] =
86.
[HA]initial
[HB]initial
Kw
+
+
[H3O ] [H 3O ] [H 3O + ]
+1
+1
K HA
K HB
(D) We are told that the solution is 0.050 M in acetic acid (Ka = 1.8 ×10-5) and 0.010 M in phenyl
acetic acid (Ka = 4.9 ×10-5). Because the Ka values are close, both equilibria must be satisfied
simultaneously. Note: [H3O+] is common to both equilibria (assume z is the concentration of
[H3O+]).
C2H3O2-(aq) + H3O+(aq)
HC2H3O2(aq) + H2O(l)
0.050 M
—
0M
≈0M
-x M
—
+xM
+zM
(0.050 – x)M
—
xM
zM
+
[H3 O ][C2 H 3 O 2 ]
1.8 105 (0.050 x)
xz
1.8 105 or x
For acetic acid: K a =
[HC2 H3 O 2 ]
(0.050 x)
z
Reaction:
Initial:
Change:
Equilibrium
Reaction:
H3O+(aq)
Initial:
Change:
Equilibrium
HC8H7O2(aq)
0.010 M
-y M
(0.010 – y)M
+ H2O(l)
—
—
—
C8H7O2-(aq)
0M
+yM
yM
[H3O + ][C8 H 7 O 2- ]
yz
For phenylacetic acid: K a =
4.9 105
[HC8 H 7 O 2 ]
(0.010 y )
or y
4.9 105 (0.010 y )
z
866
≈0M
+zM
zM
+
Chapter 17: Additional Aspects of Acid–Base Equilibria
There are now three variables: x = [C2H3O2- ], y = [C8H7O2- ], and z = [H3O+ ]
These are the only charged species, hence, x + y = z.
(This equation neglects contribution from the [H3O+] from water.)
Next we substitute in the values of x and y from the rearranged Ka expressions above.
1.8 105 (0.050 x) 4.9 105 (0.010 y )
z
z
x+y=z=
+
Since these are weak acids, we can simplify this expression by assuming that x << 0.050 and y <<
0.010.
1.8 105 (0.050) 4.9 105 (0.010)
z
z
z=
+
Simplify even further by multiplying through by z.
z2 = 9.0×10-7 + 4.9×10-7 = 1.4 ×10-6
z = 1.18×10-3 = [H3O+]
From which we find that x = 7.6 × 10-4 and y = 4.2×10-4
(as a quick check, we do see that x + y = z).
We finish up by checking to see if the approximation is valid (5% rule)..
For x :
7.6 104
100% 1.5%
0.050
For y :
4.2 104
100% 4.2%
0.010
Since both are less than 5%, we can be assured that the assumption is valid.
The pH of the solution = -log(1.18×10-3) = 2.93.
(If we simply plug the appropriate values into the equation developed in the previous question we
get the exact answer 2.933715 (via the method of successive approximations), but the final answers
can only be reported to 2 significant figures. Hence the best we can do is say that the pH is
expected to be 2.93 (which is the same level of precision as that for the result obtained following
the 5% rule).
87.
(D)
(a) By using dilution, there are an infinite number of ways of preparing the pH = 7.79
buffer. We will consider a method that does not use dilution.
Let TRIS = weak base and TRISH+ be the conjugate acid.
pKb = 5.91, pKa = 14 – pKb = 8.09
Use the Henderson Hasselbalch equation.
pH = pKa + log (TRIS/TRISH+) = 7.79 = 8.09+ log (TRIS/TRISH+)
log (TRIS/TRISH+) = 7.79 – 8.09 = -0.30 Take antilog of both sides
(TRIS/TRISH+) = 10-0.30 = 0.50; Therefore, nTRIS = 0.50(nTRISH+)
867
Chapter 17: Additional Aspects of Acid–Base Equilibria
Since there is no guidance given on the capacity of the buffer, we will choose an arbitrary
starting point. We start with 1.00 L of 0.200 M TRIS (0.200 moles). We need to convert 2/3 of
this to the corresponding acid (TRISH+) using 10.0 M HCl, in order for the TRIS/TRISH+ ratio
to be 0.5. In all, we need (2/3)×0.200 mol of HCl, or a total of 0.133 moles of HCl.
Volume of HCl required = 0.133 mol 10.0 mol/L = 0.0133 L or 13.3 mL.
This would give a total volume of 1013.3 mL, which is almost a liter (within 1.3%).
If we wish to make up exactly one liter, we should only use 987 mL of 0.200 M TRIS.
This would require 13.2 mL of HCl, resulting in a final volume of 1.0002 L.
Let’s do a quick double check of our calculations.
nHCl = 0.0132 L×10.0 M = 0.132 mol = nTRISH+
nTRIS = ninitial – nreacted = 0.200 M×0.987 L – 0.132 mol = 0.0654 mol
pH = pKa + log (TRIS/TRISH+) = 8.09+ log (0.0654 mol/0.132 mol) = 7.785
We have prepared the desired buffer (realize that this is just one way of preparing the buffer).
(b) To 500 mL of the buffer prepared above, is added 0.030 mol H3O+.
In the 500 mL of solution we have 0.132 mol 2 mol TRISH+ = 0.0660 mol TRISH+
and 0.0654 mol 2 mol TRIS = 0.0327 mol TRIS (we will assume no change in
volume).
HCl will completely react (≈ 100%) with TRIS, converting it to TRISH+.
After complete reaction, there will be no excess HCl, 0.0327 mol – 0.0300 mol =
0.0027 mol TRIS and 0.0660 mol + 0.0300 mol = 0.0960 mol TRISH+.
By employing the Henderson Hasselbalch equation, we can estimate the pH of the
resulting solution:
pH = pKa + log (TRIS/TRISH+) = 8.09 + log (0.0027 mol/0.0960 mol) = 6.54
This represents a pH change of 1.25 units. The buffer is nearly exhausted owing to
the fact that almost all of the TRIS has been converted to TRISH+. Generally, a pH
change of 1 unit suggests that the capacity of the buffer has been pretty much
completely expended. This is the case here.
(Alternatively, you may solve this question using an I.C.E. table).
(c) Addition of 20.0 mL of 10.0 M HCl will complete exhaust the buffer (in part (b) we saw
that addition of 3 mL of HCl used up most of the TRIS in solution). The buffer may be
regenerated by adding 20.0 mL of 10.0 M NaOH. The buffer will be only slightly
diluted after the addition of HCl and NaOH (500 mL → 540 mL). Through the slow and
careful addition of NaOH, one can regenerate the pH = 7.79 buffer in this way (if a pH
meter is used to monitor the addition).
868
Chapter 17: Additional Aspects of Acid–Base Equilibria
88.
(M)
(a) The formula given needs to be rearranged to isolate α, as shown below.
1
pH pK a log 1
H 3O
log 1 1
pH pKa log
Ka
H 3O 1
1
Ka
=
Ka
K
pH a
H 3O K a 10 K a
Now, use the given Ka (6.3×10-8) and calculate α for a range of pH values.
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
(c)
0
0
1
2
3
4
5
6
7
8
9
10
11
12
pH
(b) When pH = pKa, α = 0.5 or 50%.
(c) At a pH of 6.0, =
Ka
6.3 108
0.059 or 5.9%.
10 pH K a 106.0 6.3 108
869
13
14
Chapter 17: Additional Aspects of Acid–Base Equilibria
89.
(D)
(a) We start by writing the equilibrium expression for all reactions:
CO 2 aq
K1
CO 2 g
HCO3
K3
H 3O CO32
K 2 Ca 2 CO32
K4
CO 2 aq
HCO3 H 3O
First, we have to express [H3O+] using the available expressions:
HCO3
H 3O
K 3 CO32
CO aq
H 3O 2
K 4 HCO3
HCO3
CO 2 aq
CO 2 aq
H 3O
2
K 3 CO3 K 4 HCO3 K 3 K 4 CO32
2
From the expression for K1, we know that [CO2(aq)] = K1[CO2(g)]. Therefore,
K CO g
2
H 3O 1 2 2
K 3 K 4 CO3
Now, the expression for K2 can be plugged into the above expression as follows:
K1 CO 2 g Ca 2
K1 CO 2 g
2
H 3O
K 2 K3 K 4
K 3 K 4 CO32
H 3O
K1 CO 2 g Ca 2
K 2 K3 K 4
(b) The K values for the reactions are as follows:
K1 = 0.8317 (given in the problem)
K2 = Ksp (CaCO3) = 2.8×10-9
K3 = 1/Ka of HCO3– = 1/(4.7×10-11) = 2.13×1010
K4 = 1/Ka of H2CO3 = 1/(4.4×10-7) = 2.27×106
280 106 L CO 2 1 mol CO 2
mol CO 2
1.145 105
CO2 (g)
L air
24.45 L CO 2
L air
870
Chapter 17: Additional Aspects of Acid–Base Equilibria
H 3O
K1 CO 2 g Ca 2
K 2 K3 K 4
0.8317 1.145 105 10.24 103
2.8 10 2.13 10 2.27 10
9
10
6
2.684 108 M
pH log 2.684 108 7.57
90.
(M) To determine pH, we must first determine what is the final charge balance of the solution
without adding any H3O+ or OH¯ ions. As such, we have to calculate the number of moles of each
ion (assume 1 L):
Moles of Na+: 23.0 g Na+ × (1 mol/23.0 g) = 1.00 mol Na+
Moles of Ca2+: 10.0 g Ca2+ × (1 mol/40.0 g) = 0.250 mol Ca2+
Moles of CO32-: 40.02 g CO32- × (1 mol/60.01 g) = 0.670 mol CO32Moles of SO42-: 9.6 g SO42- × (1 mol/96.056 g) = 0.100 mol
Looking at the list of ions and consulting the solubility guide, we note that Ca2+ will precipitate
with both SO42- and CO32-. Since both of these anions have a 2– charge, it doesn’t matter with
which anion the precipitation occurs (in reality, it does a little, but the effects are small compared to
the effects of adding H3O+ or OH¯ ions).
Moles of anions left = (0.670 + 0.100) – 0.250 = 0.520 moles
Since we have 0.520 moles of 2– ions, we must have 1.04 moles of 1+ ions to balance. However,
we only have 1 mole of Na+. The difference in charge is:
1.04(-) – 1.0(+) = 0.04(-) moles of ions
To balance, we need 0.04 moles of H3O+. The pH = –log (0.04) = 1.40.
91.
(M)
(a) We note that the pH is 5.0. Therefore, H 3O 105.0 1.0 105 M
C K a H 3O 2.0 102 1.8 105 1.0 105
4.6 103
2
5
5 2
K a H 3O
1.8 10 1.0 10
(b)
dCA d pH d pH dCA
d pH 1.0 103 4.6 103 0.22
pH 5 0.22 4.78
(c) At an acetic acid concentration of 0.1 M, C is also 0.1, because C is the total concentration of
the acetic acid and acetate. The maximum buffer index β (2.50×10-2) happens at a pH of 4.75,
871
Chapter 17: Additional Aspects of Acid–Base Equilibria
where [HAc] = [Ac–]. The minima are located at pH values of 8.87 (which correspond to pH of a
solution of 0.1 M acetic acid and 0.1 M acetate, respectively).
3.0E-02
2.5E-02
β
2.0E-02
1.5E-02
1.0E-02
5.0E-03
0.0E+00
0
1
2
3
4
5
6
7
8
9
pH
FEATURE PROBLEMS
92.
(D)
(a)
(b)
The two curves cross the point at which half of the total acetate is present as
acetic acid and half is present as acetate ion. This is the half equivalence point in a
titration, where pH = pKa = 4.74 .
For carbonic acid, there are three carbonate containing species: “ H 2 CO 3 ” which
predominates at low pH, HCO3 , and CO 2
3 , which predominates in alkaline
solution. The points of intersection should occur at the half-equivalence points in
each step-wise titration: at pH = pKa1 = log 4.4 107 = 6.36 and at
c
c
h
h
pH = pKa 2 = log 4.7 10 11 = 10.33 . The following graph was computercalculated (and then drawn) from these equations. f in each instance represents
the fraction of the species whose formula is in parentheses.
1
f H2A
= 1+
K1
H
+
+
K1 K 2
H +
2
H
K2
= +1+
K1
H +
+
1
f HA
2
1
f A 2
H + H +
=
+
+1
K1 K 2
K2
872
Chapter 17: Additional Aspects of Acid–Base Equilibria
f H 2 CO3
f HCO3
f CO3 2
(c) For phosphoric acid, there are four phosphate containing species: H 3 PO 4 under acidic
2
3
conditions, H 2 PO 4 , HPO 4 , and PO 4 , which predominates in alkaline solution. The points
c
h
= logc4.2 10 h = 12.38 , a quite alkaline
of intersection should occur at pH = pKa1 = log 7.1 103 = 2.15,
c
h
pH = pKa2 = log 6.3 10 8 = 7.20 , and pH = pKa3
13
solution. The graph that follows was computer-calculated and drawn.
f Hf 3(H
PO34 PO
4)
-
f (H PO )
f H 2 PO24 4
f (HPO42-)
f (PO4 )
f HPO4 2 3f PO43
873
Chapter 17: Additional Aspects of Acid–Base Equilibria
93. (D)
(a) This is exactly the same titration curve we would obtain for the titration of 25.00 mL of
0.200 M HCl with 0.200 M NaOH, because the acid species being titrated is H 3O + . Both acids
are strong acids and have ionized completely before titration begins. The initial pH is that of
0.200 M H3O+ = [HCl] + [HNO3]; pH = log 0.200 = 0.70. At the equivalence point,
pH = 7.000 . We treat this problem as we would for the titration of a single strong acid with a
strong base.
14
12
pH
10
8
6
4
2
0
0
5
10
15
20
25
Volume of 0.200 M NaOH (ml)
(b) In Figure 17-9, we note that the equivalence point of the titration of a strong acid occurs at
pH = 7.00 , but that the strong acid is essentially completely neutralized at pH = 4. In Figure 1713, we see that the first equivalence point of H 3 PO 4 occurs at about pH = 4.6. Thus, the first
equivalence point represents the complete neutralization of HCl and the neutralization of
H 3 PO 4 to H 2 PO 4 . Then, the second equivalence point represents the neutralization of
2
H 2 PO 4 to HPO 4 . To reach the first equivalence point requires about 20.0 mL of 0.216 M
NaOH, while to reach the second one requires a total of 30.0 mL of 0.216 M NaOH, or an
additional 10.0 mL of base beyond the first equivalence point. The equations for the two
titration reactions are as follows.
NaOH + H 3PO 4
NaH 2 PO 4 + H 2 O
To the first equivalence point:
NaOH + HCl
NaCl + H 2 O
To the second equivalence point:
NaOH + NaH 2 PO 4
Na 2 HPO 4 + H 2O
There is a third equivalence point, not shown in the figure, which would require an additional
10.0 mL of base to reach. Its titration reaction is represented by the following equation.
To the third equivalence point:
NaOH + Na 2 HPO 4
Na 3 PO 4 + H 2 O
874
Chapter 17: Additional Aspects of Acid–Base Equilibria
We determine the molar concentration of H 3 PO 4 and then of HCl. Notice that only 10.0 mL of
the NaOH needed to reach the first equivalence point reacts with the HCl(aq); the rest reacts
with H 3 PO 4 .
(30.0 20.0) mL NaOH(aq)
0.216 mmol NaOH 1 mmol H 3 PO 4
1 mL NaOH soln 1 mmol NaOH
10.00 mL acid soln
(20.0 10.0) mL NaOH(aq)
0.216 mmol NaOH
1 mL NaOH soln
0.216 M H 3 PO 4
1 mmol HCl
1 mmol NaOH
10.00 mL acid soln
0.216 M HCl
(c) We start with a phosphoric acid-dihydrogen phosphate buffer solution and titrate until all of
the H 3 PO 4 is consumed. We begin with
10.00 mL
0.0400 mmol H 3PO 4
= 0.400 mmol H 3PO 4
1 mL
and the diprotic anion,
0.0150 mmol H 2 PO 4
2
10.00 mL
= 0.150 mmol H 2 PO 4
1 mL
. The volume of 0.0200 M
NaOH needed is: 0.400 mmol H 3 PO 4
1 mmol NaOH
1 mmol H 3 PO 4
1 mL NaOH
0.0200 mmol NaOH
= 20.00 mL
To reach the first equivalence point. The pH value of points during this titration are computed
with the Henderson-Hasselbalch equation.
H 2 PO 4
0.0150
3
Initially : pH = pK1 + log
=
log
7.1
10
+
log
= 2.15 0.43 = 1.72
0.0400
H 3 PO 4
0.150 + 0.100
At 5.00 mL : pH = 2.15 + log
= 2.15 0.08 2.07
0.400 0.100
At 10.0 mL, pH = 2.15 log
0.150 0.200
2.15 0.24 2.39
0.400 0.200
At 15.0 mL, pH = 2.15 log
0.150 0.300
2.15 0.65 2.80
0.400 0.300
This is the first equivalence point, a solution of 30.00 mL (= 10.00 mL originally + 20.00 mL
titrant), containing 0.400 mmol H 2 PO 4 from the titration and the 0.150 mmol H 2 PO 4 2
originally present.
This is a solution with
875
Chapter 17: Additional Aspects of Acid–Base Equilibria
H 2 PO 4 =
pH =
(0.400 + 0.150) mmol H 2 PO 4
= 0.0183 M , which has
30.00 mL
1
pK1 + pK 2 = 0.50 2.15 log 6.3 108 = 0.50 2.15 + 7.20 = 4.68
2
To reach the second equivalence point means titrating 0.550 mmol H 2 PO 4 , which
requires an additional volume of titrant given by
0.550 mmol H 2 PO 4
1 mmol NaOH
1 mL NaOH
= 27.5 mL .
1 mmol H 2 PO 4 0.0200 mmol NaOH
To determine pH during this titration, we divide the region into five equal portions
of 5.5 mL and use the Henderson-Hasselbalch equation.
At 20.0 + 5.5 mL,
2
HPO 4 2
0.20 0.550 mmol HPO 4 formed
pH = pK 2 + log
= 7.20 + log
H 2 PO 4
0.80 0.550 mmol H 2 PO 4 remaining
pH = 7.20 0.60 = 6.60
At 20.0 +11.0 mL = 31.0 mL, pH = 7.20 + log
At 36.5 mL, pH = 7.38
0.40 0.550
= 7.02
0.60 0.550
At 42.0 mL, pH = 7.80
The pH at the second equivalence point is given by
pH =
b
g
e
c
hj
b
g
1
pK 2 + pK3 = 0.50 7.20 log 4.2 1013 = 0.50 7.20 + 12.38 = 9.79 .
2
Another 27.50 mL of 0.020 M NaOH would be required to reach the third
equivalence point. pH values at each of four equally spaced volumes of 5.50 mL
additional 0.0200 M NaOH are computed as before, assuming the HendersonHasselbalch equation is valid.
PO 43
= 12.38 + log 0.20 0.550
At 47.50 + 5.50 mL = 53.00 mL, pH = pK 3 + log
2
0.80 0.550
HPO 4
= 12.38 0.60 = 11.78
876
Chapter 17: Additional Aspects of Acid–Base Equilibria
At 58.50 mL, pH = 12.20
At 64.50 mL, pH = 12.56
At 70.00 mL, pH = 12.98
But at infinite dilution with 0.0200 M NaOH, the pH = 12.30, so this point can’t be
reached.
At the last equivalence point, the solution will contain 0.550 mmol PO 43 in a total of
10.00 + 20.00 + 27.50 + 27.50 mL = 85.00 mL of solution, with
0.550 mmol
PO 43 =
= 0.00647 M. But we can never reach this point, because the
85.00 mL
pH of the 0.0200 M NaOH titrant is 12.30. Moreover, the titrant is diluted by its
addition to the solution. Thus, our titration will cease sometime shortly after the
second equivalence point. We never will see the third equivalence point, largely
because the titrant is too dilute. Our results are plotted below.
877
Chapter 17: Additional Aspects of Acid–Base Equilibria
94.
(D) p K a1 = 2.34; K a1 = 4.6 103 and p K a 2 = 9.69; K a 2 = 2.0 1010
(a)
Since the Ka values are so different, we can treat alanine (H2A+) as a monoprotic acid
with K a1 = 4.6 103. Hence:
H2A+(aq)
+ H2O(l)
Initial
0.500 M
—
Change
—
x M
—
Equilibrium (0.500 x) M
HA(aq) + H3O+(aq)
0M
0M
+x M
+x M
xM
xM
[HA][H 3O + ]
( x)( x)
x2
3
=
4.6
10
=
0.500
(0.500 x)
[H 2 A + ]
+
x = 0.048 M = [H3O ] (x = 0.0457, solving the quadratic equation)
K a1 =
pH = log[H3O+] = log(0.046) = 1.34
(b)
At the first half-neutralization point a buffer made up of H2A+/HA is formed, where
[H2A+] = [HA]. The Henderson-Hasselbalch equation gives pH = pKa = 2.34.
(c)
At the first equivalence point, all of the H2A+(aq) is converted to HA(aq).
HA(aq) is involved in both K a1 and K a 2 , so both ionizations must be considered.
If we assume that the solution is converted to 100% HA, we must consider two
reactions. HA may act as a weak acid (HAA + H+ ) or HA may act as a base
(HA + H+ H2A+). See the following diagram.
Some HA
unreacted
100 %
HA
amphoteric nature
(reactions occur)
HA
reacts as
a base
H
acid
A as
HA a s
+
+ H3O
A-
base
H2A+
Some HA
reacts as
an acid
Using the diagram above, we see that the following relations must hold true.
[A] = [H3O+] + [H2A+]
Ka2 =
K a [HA]
[HA][H 3O + ]
[A ][H 3O + ]
[H 3O + ][HA]
+
or [A] = 2 + & K a1 =
A
]
=
or
[H
2
[H3O ]
[HA]
K a1
[H 2 A + ]
Substitute for [A] and [H2A+] in [A] = [H3O+] + [H2A+]
878
Chapter 17: Additional Aspects of Acid–Base Equilibria
K a 2 [HA]
[H 3O + ]
= [H3O+] +
[H 3O + ][HA]
(multiply both sides by K a1 [H3O+])
K a1
K a1 K a 2 [HA] = K a1 [H3O+][H3O+] + [H 3O + ][H 3O + ][HA]
K a1 K a 2 [HA] = [H3O+]2( K a1 + [HA])
[H3O+]2 =
K a1 K a 2 [HA]
(K a1 + [HA])
Usually, [HA] K a1 (Here, 0.500 4.6 103)
Make the assumption that K a1 + [HA] [HA]
[H3O+]2 =
K a1 K a 2 [HA]
[HA]
= K a1 K a 2
Take log of both sides
log[H3O+]2 = 2log [H3O+] = 2(pH) = log K a1 K a 2 = log K a1 log K a 2 = p K a1 + p K a 2
2(pH) = p K a1 + p K a 2
pH =
pK a1 + pK a 2
2
=
2.34 + 9.69
= 6.02
2
(d)
Half-way between the first and second equivalence points, half of the HA(aq) is
converted to A(aq). We have a HA/A buffer solution where [HA] = [A]. The
Henderson-Hasselbalch equation yields pH = p K a 2 = 9.69.
(e)
At the second equivalence point, all of the H2A+(aq) is converted to A(aq). We can
treat this simply as a weak base in water having:
K
11014
Kb = w =
= 5.0 105
10
Ka2
2.0 10
1V
= 0.167 M
3V
HA(aq) + OH(aq)
0M
0M
+x M
+x M
xM
xM
Note: There has been a 1:3 dilution, hence the [A] = 0.500 M
+ H2O(l)
A(aq)
Initial
0.167 M
—
—
Change
x M
—
Equilibrium (0.167 x) M
Kb =
[HA][OH ]
( x)( x)
x2
5
=
=
5.0
10
[A ]
0.167
(0.167 x)
x = 0.0029 M = [OH]; pOH = log[OH] = 2.54;
pH = 14.00 pOH = 14.00 2.54 = 11.46
879
Chapter 17: Additional Aspects of Acid–Base Equilibria
mL NaOH
0.0
10.0
20.0
30.0
40.0
50.0
60.0
70.0
80.0
90.0
100.0 110.0
%H2A+
%HA
%A-
100
0
0
80
20
0
60
40
0
40
60
0
20
80
0
0
100
0
0
80
20
0
60
40
0
40
60
0
20
80
0
0
100
0
0
100
0
0.25
0.67
1.5
4.0
0.25
0.67
1.5
4.0
8
8
All of the points required in (f) can be obtained using the Henderson-Hasselbalch
equation (the chart below shows that the buffer ratio for each point is within the
acceptable range (0.25 to 4.0))
8
(f)
buffer base
= acid
ratio
May use Henderson-Hasselbalch equation
May use Henderson-Hasselbalch equation
(i) After 10.0 mL
Here we will show how to obtain the answer using both the Henderson-Hasselbalch
equation and setting up the I. C. E. (Initial, Change, Equilibrium) table. The results
will differ within accepted experimental limitation of the experiment (+ 0.01 pH units)
nH2A+ = (C V) = (0.500 M)(0.0500 L) = 0.0250 moles H2A
nOH- = (C V) = (0.500 M)(0.0100 L) = 0.00500 moles OH
Vtotal = (50.0 + 10.0) mL = 60.0 mL or 0.0600 L
nH A+ 0.0250 mol
n 0.00500 mol
= 0.417 M
[H2A+] = 2 =
[OH] = OH =
= 0.0833M
Vtotal
0.0600 L
Vtotal
0.0600 L
K a1 4.6 103
1
1
=
=
4.6 1011
Keq for titration reaction =
14
K b(HA) K w K w 1.00 10
K a1
H2A+(aq) + OH(aq)
HA(aq)
+ H2O(l)
Initial:
0.417 M
100% rxn:
-0.0833
New initial:
0.334 M
Change:
+x M
Equilibrium: 0.334 M
4.6 1011 =
0.0833 M
-0.0833 M
0M
+x M
xM
0M
+0.0833 M
0.0833 M
x M
0.0833 M
—
—
—
—
—
(0.0833)
(0.0833)
; x=
= 5.4 1013 (valid assumption)
11
(0.334)( x)
(0.334)(4.6 10 )
x = 5.4 1013 = [OH]; pOH = log(5.4 1013) = 12.27;
pH = 14.00 – pOH = 14.00 – 12.27 = 1.73
Alternative method using the Henderson-Hasselbalch equation:
880
Chapter 17: Additional Aspects of Acid–Base Equilibria
(i)
After 10.0 mL, 20% of H2A+ reacts, forming the conjugate base HA.
Hence the buffer solution is 80% H2A+ (acid) and 20% HA (base).
pH =
pK a1
base
20.0
+ log acid = 2.34 + log 80.0 = 2.34 + (0.602) = 1.74 (within + 0.01)
For the remainder of the calculations we will employ the Henderson-Hasselbalch equation
with the understanding that using the method that employs the I.C.E. table gives the same
result within the limitation of the data.
(ii)
After 20.0 mL, 40% of H2A+ reacts, forming the conjugate base HA.
Hence the buffer solution is 60% H2A+ (acid) and 40% HA (base).
pH =
pK a1
base
40.0
+ log acid = 2.34 + log 60.0 = 2.34 + (0.176) = 2.16
(iii) After 30.0 mL, 60% of H2A+ reacts, forming the conjugate base HA.
Hence the buffer solution is 40% H2A+ (acid) and 60% HA (base).
pH =
pK a1
base
60.0
+ log acid = 2.34 + log 40.0 = 2.34 + (+0.176) = 2.52
(iv) After 40.0 mL, 80% of H2A+ reacts, forming the conjugate base HA.
Hence the buffer solution is 20% H2A+ (acid) and 80% HA (base).
base
80.0
= 2.34 + log
= 2.34 + (0.602) = 2.94
pH = pK a1 + log
acid
20.0
(v) After 50 mL, all of the H2A+(aq) has reacted, and we begin with essentially 100%
HA(aq), which is a weak acid. Addition of base results in the formation of the
conjugate base (buffer system) A(aq). We employ a similar solution, however,
now we must use p K a 2 = 9.69.
(vi) After 60.0 mL, 20% of HA reacts, forming the conjugate base A.
Hence the buffer solution is 80% HA (acid) and 20% A (base)
base
20.0
= 9.69 + log
= 9.69 + (0.602) = 9.09
pH = pK a 2 + log
acid
80.0
(vii) After 70.0 mL, 40% of HA reacts, forming the conjugate base A.
Hence the buffer solution is 60% HA (acid) and 40% A (base).
base
40.0
= 9.69 + log
= 9.69 + (0.176) = 9.51
pH = pK a 2 + log
acid
60.0
881
Chapter 17: Additional Aspects of Acid–Base Equilibria
(viii) After 80.0 mL, 60% of HA reacts, forming the conjugate base A.
Hence the buffer solution is 40% HA (acid) and 60% A (base).
base
60.0
= 9.69 + log
= 9.69 + (+0.176) = 9.87
pH = pK a 2 + log
acid
40.0
(ix) After 90.0 mL, 80% of HA reacts, forming the conjugate base A.
Hence the buffer solution is 20% HA (acid) and 80% A (base).
base
80.0
= 9.69 + log
= 9.69 + (0.602) = 10.29
pH = pK a 2 + log
acid
20.0
(x) After the addition of 110.0 mL, NaOH is in excess. (10.0 mL of 0.500 M NaOH is in excess,
or, 0.00500 moles of NaOH remains unreacted). The pH of a solution that has NaOH in
excess is determined by the [OH] that is in excess.
(For a diprotic acid, this occurs after the second equivalence point.)
n 0.00500 mol
= 0.0312 M ; pOH = log(0.03125) = 1.51
[OH]excess = OH =
Vtotal
0.1600 L
pH = 14.00 – pOH = 14.00 –1.51 = 12.49
(g) A sketch of the titration curve for the 0.500 M solution of alanine hydrochloride,
with some significant points labeled on the plot, is shown below.
14
Plot of pH versus Volume of NaOH Added
HA/Abuffer region
12
NaOH
in excess
10
8
pH
pH = pKa2
isoelectric point
6
+
H2A /HA
buffer region
4
2
pH = pKa1
0
0
20
40
60
NaOH volume (mL)
882
80
100
120
Chapter 17: Additional Aspects of Acid–Base Equilibria
SELF-ASSESSMENT EXERCISES
95.
(E)
(a) mmol: millimoles, or 1×10-3 mol
(b) HIn: An indicator, which is a weak acid
(c) Equivalence point of a titration: When the moles of titrant equals the moles of the substance
being titrated
(d) Titration curve: A curve of pH of the solution being titrated versus the pH of the titrating
solution
96.
(E)
(a) The common-ion effect: A process by which ionization of a compound is suppressed by having
present one of the product ions (from another source) in the solution
(b) use of buffer to maintain constant pH: A buffer is a solution of a weak acid and its conjugate
base, and it resists large changes in pH when small amounts of an acid or base are added.
(c) determination of pKa from titration curve: At the half-way point (when half of the species being
titrated is consumed, the concentration of that species and its conjugate is the same, and the
equilibrium expression simplifies to pKa = pH
(d) measurement of pH with an indicator: An approximate method of measuring the pH, where the
color of an ionizable organic dye changes based on the pH
97.
(E)
(a) Buffer capacity and buffer range: Buffer capacity is a measure of how much acid or base can be
added to the buffer without an appreciable change in the pH, and is determined by the
concentration of the weak acid and conjugate base in the buffer solution. The buffer range,
however, refers to the range over which a buffer effectively neutralizes added acids and bases
and maintains a fairly constant pH.
(b) Hydrolysis and neutralization: Hydrolysis is reaction of an acid or base with water molecules,
which causes water to split into hydronium and hydroxide ions. Neutralization is the reaction of
H3O+ and OH– together to make water.
(c) First and second equivalence points in the titration of a weak diprotic acid: First equivalence
point is when the first proton of a weak diprotic acid is completely abstracted and the resulting
acid salt and base have been consumed and neither is in excess. Second equivalence point is the
equivalence point at which all protons are abstracted from the acid and what remains is the (2-)
anion.
(d) Equivalence point of a titration and end point of an indicator: Equivalence point of a titration is
when the moles of titrant added are the same as the moles of acid or base in the solution being
titrated. The endpoint of an indicator is when the indicator changes color because of abstraction
or gain of a proton, and is ideally as close as possible to the pH of the equivalence point.
883
Chapter 17: Additional Aspects of Acid–Base Equilibria
98.
(E)
(a) HCHO2 + OH¯ → CHO2¯ + H2O
CHO2¯ + H3O+ → HCHO2 + H2O
(b) C6H5NH3+ + OH¯ → C6H5NH2 + H2O
C6H5NH2 + H3O+ → C6H5NH3+ + H2O
(c) H2PO4¯ + OH¯ → HPO42- + H2O
HPO42- + H3O+ → H2PO4¯ + H2O
pH
(M)
(a) The pH at theequivalence point is 7. Use bromthymol blue.
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Vol of Titrant (mL)
(b) The pH at the equivalence point is ~5.3 for a 0.1 M solution. Use methyl red.
pH
99.
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Vol of Titrant (mL)
884
Chapter 17: Additional Aspects of Acid–Base Equilibria
pH
(c) The pH at the equivalence point is ~8.7 for a 0.1 M solution. Use phenolphthalein, because it
just begins to get from clear to pink around the equivalence point.
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Vol of Titrant (mL)
pH
(d) The pH for the first equivalence point (NaH2PO4– to Na2HPO42–) for a 0.1 M solution is right
around ~7, so use bromthymol blue.
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Vol of Titrant (mL)
885
Chapter 17: Additional Aspects of Acid–Base Equilibria
100. (D)
(a) This is the initial equilibrium before any base has reacted with the acid. The reaction that
dominates, along with changes in concentration, is shown below:
HC7H5O2
0.0100
-x
0.0100 − x
+ H2O
C7H5O2¯
0
+x
x
+ H3O+
0
+x
x
H 3O + C7 H 5O-2
Ka
HC7 H5O2
6.3 10-5
x x
0.0100 - x
Solving for x using the quadratic formula, x = 7.63 104 M.
pH = log H3O+ log 7.63 104 3.12
(b) In this case, we titrate the base with 0.00625 L of Ba(OH)2. Therefore, we have to calculate the
final moles of the base and the total volume to determine the concentration.
mol HC7 H 5O 2 0.02500 L 0.0100 M HC7 H 5O 2 2.5 104 mol
2 mol OH
mol OH 0.00625 L 0.0100 M Ba OH 2
= 1.25 104 mol
1 mol Ba OH 2
Since the amount of OH¯ is half of the initial amount of HC7H5O2, the moles of HC7H5O2 and
C7H5O2¯ are equal. Therefore, Ka = [H3O+], and pH = −log(6.3×10-5) = 4.20.
(c) At the equivalence point, there is initially no HC7H5O2. The equilibrium is dominated by
C7H5O2¯ hydrolyzing water. The concentration of C7H5O2¯ is:
Total moles of HC7H5O2 = 2.5×10-4 mol as shown previously. At the equivalence point, the
moles of acid equal moles of OH¯.
1 mol Ba OH 2
mol Ba OH 2 2.5 104 mol OH
1.25 104 mol
2 mol OH
Vol of Ba(OH)2 = 1.25×10-4 mol / 0.0100 M = 0.0125 L
Total volume of the solution at the equivalence point is the sum of the initial volume plus the
volume of Ba(OH)2 added. That is,
VTOT = 0.02500 L + 0.0125 L = 0.0375 L
Therefore, the concentration of C7H5O2¯ = 2.5×10-4/0.0375 L = 0.00667 M.
886
Chapter 17: Additional Aspects of Acid–Base Equilibria
C7H5O2¯
0.00667
-x
0.00667 − x
+ H2O
HC7H5O2
0
+x
x
+ OH¯
0
+x
x
Since this is a base reaction, K b K w K a 1.00 104 6.3 105 1.587 1010.
OH - HC7 H 5O 2
Kb =
C7 H 5O-2
1.59 1010
x x
0.00667 x
solving for x (by simplifying the formula above) yields x = 1.03 10 6 M.
pH = 14 pOH 14 log 1.03 106 14 6.00 8.00
(d) In this part, we have an excess of a strong base. As such, we have to determine how much
excess base there is and what is the final volume of the solution.
mol Ba OH 2
2 mol OH
mol OH 0.01500 L 0.0100
L
1 mol Ba OH 2
3.000 104 mol OH
Excess mol OH mol HC7 H5 O 2 mol OH 3.000 104 2.500 104
5.000 105 mol
5.0 105 mol
OH
0.00125 M
0.02500 L+0.01500 L
pH = 14 pOH 14 log 0.00125 14 2.903 11.1
101. (E) The answer is (c); because of the common ion-effect, the presence of HCO2¯ will repress
ionization of formic acid.
102. (E) The answer is (d), because NaHCO3 is a weak base and will react with protons in water, shifting
the formic acid ionization equilibrium to the right.
103. (E) The answer is (b), raise the pH. NH4+ is an acid, and to be converted to its conjugate base, it
must react with a base to abstract its proton.
104. (E) The answer is (b), because at that point, the number of moles of weak base remaining is the
same as its conjugate acid, and the equilibrium expression simplifies to Ka = [H3O+].
887
Chapter 17: Additional Aspects of Acid–Base Equilibria
105. (M) The base, C2H5NH2, is reacted with HClO4. The reaction is:
C2 H 5 NH 2 HClO 4 C2 H 5 NH 3 ClO 4
Assuming a volume of 1 L for each solution,
1.49 mol – 1.001 mol
0.489 mol
0.2445 M
2L
2L
1.001 mol
C2 H 5 NH 3
0.5005 M
2L
C2 H5 NH 2
C2 H 5 NH 3 OH 0.5005 +x x
4.3 104
0.2445 x
C2 H5 NH 2
x 2.10 104
pOH log 2.10 104 3.68
pH 14 3.68 10.32
106. (D) We assume that all of Ca(HSe)2 dissociates in water. The concentration of HSe– is therefore:
2 mol HSe
1.0 M HSe
1 mol Ca HSe 2
We note that HSe– is amphoteric; that is, it can act either as an acid or a base. The acid reaction of
HSe– and the concentration of [H3O+] generated, are as follows:
0.5 M Ca HSe 2
HSe– + H2O Se2– + H3O+
1.00 10
11
Se 2 H 3O
x x
HSe
1.00 x
x = 1.00 1011 3.16 106
The basic reaction of HSe– and the concentration of [OH–] generated, are as follows:
HSe– + H2O H2Se + OH–
Kb = 1.00×10-14/1.3×10-4 = 7.69×10-11
7.69 10
11
H 2Se OH
HSe
x x
1.00 x
x = 7.69 1011 8.77 106
888
Chapter 17: Additional Aspects of Acid–Base Equilibria
Therefore, we have [H3O+] = 3.16×10-6 and [OH–] = 8.77×10-6. Since these two react to give H2O,
the result is 8.77×10-6 – 3.16×10-6 = 5.61×10-6 M [OH–]. The pH of the solution is:
pH = 14 – pOH = 14 – 5.25 = 8.75
107. (E) The answer is (a). The solution system described is a buffer, and will resist large changes in
pH. Adding KOH should raise the pH slightly.
108. (E) The answer is (b), because HSO3¯ is a much stronger acid (Ka = 1.3×10-2) than H2PO4¯
(Ka = 6.3×10-8).
109. (E) The answer is (b). The pKa of the acid is 9, which puts it squarely in the middle of the 8–10 pH
range for the equivalence point.
110. (E)
(a) NaHCO3 titrated with NaOH: pH > 7, because HCO3¯ is itself slightly basic, and is being
titrated with NaOH to yield CO32- at the equivalence point, which is even more basic.
(b) HCl titrated with NH3: pH < 7, because the resulting NH4+ at the equivalence point is acidic.
(c) KOH titrated with HI: pH = 7, because a strong base is being titrated by a strong acid, and the
resulting anions and cations are all non-basic and non-acidic.
111. (M) The concepts that define Sections 17-2, 17-3, and 17-4 are buffers, indicators, and titrations.
For buffers, after definition, there are a number of concepts, such as composition, application, the
equilibrium expression (the Henderson-Hasselbalch equation is a subtopic of the equilibrium
expression). Under composition, there are other subtopics such as buffer range and capacity. For
indicators, after defining it, there are subtopics such as equilibrium expression and usage. With
titration, topics such as form (weak acid titrated by strong base, and weak base titrated by strong
acid), the titration curve, and the equivalence point. Look at these sections to find inter-related
terminology and other concepts.
889
CHAPTER 18
SOLUBILITY AND COMPLEX-ION EQUILIBRIA
PRACTICE EXAMPLES
1A
1B
(E) In each case, we first write the balanced equation for the solubility equilibrium and then
the equilibrium constant expression for that equilibrium, the Ksp expression:
(a)
Mg 2+ aq + CO3
MgCO3 s
(b)
3Ag + aq + PO 4
Ag 3 PO 4 s
(E)
(a)
(b)
2A
2
aq
Ksp = [Mg2+][CO32 ]
3
aq
Ksp = [Ag+]3[PO43 ]
Provided the [OH] is not too high, the hydrogen phosphate ion is not expected to ionize
in aqueous solution to a significant extent because of the quite small values for the
second and third ionization constants of phosphoric acid.
Ca 2+ aq + HPO 4 2 aq
CaHPO 4 s
The solubility product constant is written in the manner of a Kc expression:
Ksp = [Ca2+][HPO42] = 1. 107
(M) We calculate the solubility of silver cyanate, s , as a molarity. We then use the solubility
equation to (1) relate the concentrations of the ions and (2) write the Ksp expression.
s=
7 mg AgOCN 1000 mL
1g
1 mol AgOCN
= 5 104 moles/L
100 mL
1L
1000 mg 149.9 g AgOCN
Ag + aq + OCN aq
AgOCN s
Equation :
Solubility Product :
s
s
K sp = Ag + OCN = s s = s 2 = 5 104 = 3 107
2
2B
(E) We calculate the solubility of lithium phosphate, s , as a molarity. We then use the
solubility equation to (1) relate the concentrations of the ions and (2) write the Ksp expression.
s=
0.034 g Li3 PO 4 1000 mL 1 mol Li3 PO 4
= 0.0029 moles/L
100 mL soln
1L
115.79 g Li3 PO 4
3Li + aq + PO 4
Equation: Li3 PO 4 s
Solubility Product:
(3 s)3
3
aq
s
3
3
K sp = Li + PO 4 = 3s s = 27 s 4 27 (0.0029) 4 1.9 109
3
890
Chapter 18: Solubility and Complex-Ion Equilibria
3A
(E) We use the solubility equilibrium to write the Ksp expression, which we then solve to
obtain the molar solubility, s , of Cu3(AsO4)2.
3 Cu 2+ aq + 2 AsO 4 aq
Cu 3 (AsO 4 ) 2 . s
K sp = Cu 2+ AsO 4 = 3s 2 s = 108s 5 = 7.6 1036
3
2
s
Solubility :
3B
5
3
2
7.6 1036
3.7 108 M
108
(E) First we determine the solubility of BaSO 4 , and then find the mass dissolved.
Ba 2+ aq + SO 4 2 aq
BaSO 4 aq
Ksp = Ba 2+ SO 4 2 = s 2
The last relationship is true because Ba 2+ = SO 4 2 in a solution produced by dissolving
BaSO 4 in pure water. Thus, s = K sp 1.1 1010 1.05 105 M.
mass BaSO 4 = 225 mL
4A
1.05 105 mmol BaSO 4 233.39 mg BaSO 4
= 0.55 mg BaSO 4
1 mL sat’d soln
1mmol BaSO4
(M) For PbI 2 , Ksp = Pb 2+ I
calculation.
Equation:
Initial:
Changes:
Equil:
Ksp = Pb
2+
bg
PbI 2 s
—
—
—
I
2
9
2
= 7.1 109 . The solubility equilibrium is the basis of the
b g
b g
2I aq
0M
+2s M
2s M
+
Pb 2+ aq
0.10 M
+s M
(0.10 + s) M
b
gb g
= 7.1 10 = 0.10 + s 2 s 0.40 s
2
7.1 109
. 104 M
s
13
0.40
2
(assumption 0.10 s is valid)
This value of s is the solubility of PbI 2 in 0.10 M Pb NO 3
b g baqg .
4B
2
(E) We find pOH from the given pH:
pOH = 14.00 – 8.20 = 5.80; [OH–] = 10-pOH = 10-5.80 = 1.6×10-6 M.
We assume that pOH remains constant, and use the Ksp expression for Fe OH 3 .
b g
K sp = Fe3+ OH = 4 1038 = Fe3+ 1.6 106
3
3
Fe3+ =
Therefore, the molar solubility of Fe(OH)3 is 9.81021 M.
b g
The dissolved Fe OH
3
does not significantly affect OH .
891
4 1038
1.6 10
6 3
= 9.8 1021 M
Chapter 18: Solubility and Complex-Ion Equilibria
5A
(M) First determine I as altered by dilution. We then compute Qsp and compare it with Ksp .
3drops
I =
0.05 mL 0.20 mmol KI 1mmol I
1drop
1mL
1mmol KI
= 3 104 M
100.0 mL soln
Qsp = Ag + I = 0.010 3 104 = 3 106
Qsp 8.5 1017 = K sp Thus, precipitation should occur.
5B
(M) We first use the solubility product constant expression for PbI 2 to determine the I
needed in solution to just form a precipitate when Pb 2+ = 0.010 M. We assume that the
volume of solution added is small and that Pb 2+ remains at 0.010 M throughout.
Ksp = Pb
2+
I
2
b
9
g
= 7.1 10 = 0.010 I
2
I
7.1 109
= 8.4 104 M
=
0.010
We determine the volume of 0.20 M KI needed.
b g
8.4 104 mmol I 1 mmol KI 1 mL KI aq
1 drop
1 mL
1 mmol I
0.20 mmol KI 0.050 mL
= 8.4 drops = 9 drops
Since one additional drop is needed, 10 drops will be required. This is an insignificant volume
compared to the original solution, so Pb 2+ remains constant.
b g
volume of KI aq = 100.0 mL
6A
(E) Here we must find the maximum Ca 2+ that can coexist with OH = 0.040 M.
2
5.5 106
2
K sp = 5.5 106 = Ca 2+ OH = Ca 2+ 0.040 ; Ca 2+ =
= 3.4 103 M
2
0.040
For precipitation to be considered complete, Ca 2+ should be less than 0.1% of its original
b g
value. 3.4 103 M is 34% of 0.010 M and therefore precipitation of Ca OH
2
is not complete
under these conditions.
6B
(E) We begin by finding Mg 2+ that corresponds to1 g Mg 2+ / L .
1 g Mg 2+
1g
1 mol Mg 2+
Mg =
6
= 4 108 M
2+
1 L soln 10 g 24.3 g Mg
2+
b g to determine OH .
18
. 10
= c4 10 h OH
OH =
4 10
Now we use the Ksp expression for Mg OH
Ksp = 1.8 10
11
= Mg
2+
OH
2
2
8
2
11
8
892
0.02 M
Chapter 18: Solubility and Complex-Ion Equilibria
7A
(M) Let us first determine Ag + when AgCl(s) just begins to precipitate. At this point,
Qsp and Ksp are equal.
Ksp = 1.8 1010 = Ag + Cl = Qsp = Ag + 0.115M
Ag + =
1.8 1010
= 1.6 109 M
0.115
Now let us determine the maximum Br that can coexist with this Ag + .
Ksp = 5.0 1013 = Ag + Br = 1.6 109 M Br ; Br =
5.0 1013
= 3.1 10 4 M
1.6 109
b g
The remaining bromide ion has precipitated as AgBr(s) with the addition of AgNO 3 aq .
Percent of Br remaining
7B
[Br ]final
3.1 104 M
100% 0.12%
100%
[Br ]initial
0.264 M
(M) Since the ions have the same charge and the same concentrations, we look for two Ksp
values for the salt with the same anion that are as far apart as possible. The Ksp values for the
carbonates are very close, while those for the sulfates and fluorides are quite different.
However, the difference in the Ksp values is greatest for the chromates; Ksp for
BaCrO 4 1.2 1010 is so much smaller than Ksp for SrCrO 4 2.2 105 , BaCrO 4 will
2
precipitate first and SrCrO 4 will begin to precipitate when CrO 4 has the value:
K sp
2.2 105
CrO 4 2 =
=
= 2.2 104 M .
Sr 2+
0.10
At this point Ba 2+ is found as follows.
K sp
1.2 1010
Ba 2+ =
=
= 5.5 107 M ;
4
2
CrO 4 2.2 10
[Ba2+] has dropped to 0.00055% of its initial value and therefore is considered to be
completely precipitated, before SrCrO 4 begins to precipitate. The two ions are thus effectively
separated as chromates. The best precipitating agent is a group 1 chromate salt.
8A
(M) First determine OH resulting from the hydrolysis of acetate ion.
Equation:
Initial:
Changes:
Equil:
C2 H 3 O-2 aq + H 2 O(l)
0.10 M
x M
0.10 x M
b
g
—
—
—
b g
HC 2 H 3O 2 aq
0M
+x M
xM
893
+
b g
OH aq
0M
+x M
xM
Chapter 18: Solubility and Complex-Ion Equilibria
Kb
HC2 H3O2 OH
1.0 1014
xx
x2
10
5.6
10
=
1.8 105
0.10 x 0.10
C 2 H 3O 2
x [OH ] 010
. 5.6 1010 7.5 106 M (the assumption x 0.10 was valid)
Now compute the value of the ion product in this solution and compare it with the value of
Ksp for Mg OH 2 .
b g
Qsp = Mg 2+ OH
2
b
gc
b g
, this solution is unsaturated and precipitation of MgbOH g bsg
= 0.010M 7.5 106 M
Because Qsp is smaller than Ksp
h
2
= 5.6 1013 1.8 1011 = Ksp Mg OH
2
2
will not occur.
8B
(M) Here we can use the Henderson–Hasselbalch equation to determine the pH of the buffer.
C 2 H 3O 2
c
= log 1.8 10
MN HC2 H 3O 2 OPQ
pH = pKa + log L
5
M
= 4.74 + 0.22 = 4.96
h + log 0.250
0.150 M
OH = 10 pOH = 10 9.04 = 9.1 1010 M
pOH = 14.00 pH = 14.00 4.96 = 9.04
Now we determine Qsp to see if precipitation will occur.
Qsp = Fe3+ OH = 0.013M 9.1 1010 = 9.8 1030
Qsp 4 1038 = K sp ; Thus, Fe(OH)3 precipitation should occur.
3
9A
3
b g
(M) Determine OH , and then the pH necessary to prevent the precipitation of Mn OH 2 .
K sp = 1.9 10
13
= Mn 2+ OH = 0.0050M OH OH =
2
c
h
pOH = log 6.2 106 = 5.21
2
1.9 1013
6.2 106 M
0.0050
pH = 14.00 5.21 = 8.79
+
We will use this pH in the Henderson–Hasselbalch equation to determine NH 4 .
pK b = 4.74 for NH 3 .
pH = pK a + log
log
NH3
NH 4
+
= 8.79 = 14.00 4.74 + log
0.025 M
= 8.79 14.00 4.74 = 0.47
NH 4 +
0.025
NH 4 + =
0.34 = 0.074M
894
0.025 M
NH 4 +
0.025
NH 4
+
= 100.47 = 0.34
Chapter 18: Solubility and Complex-Ion Equilibria
9B
(M) First we must calculate the [H3O+] in the buffer solution that is being employed to
dissolve the magnesium hydroxide:
NH4+(aq) + OH(aq) ; Kb = 1.8 105
NH3(aq) + H2O(l)
Kb
[NH 4 ][OH ] [0.100M][OH ]
1.8 105
[NH 3 ]
[0.250M]
1.00 1014 M 2
2.22 1010 M
4.5 105 M
Now we can employ Equation 18.4 to calculate the molar solubility of Mg(OH)2 in the buffer
solution; molar solubility Mg(OH)2 = [Mg2+]equil
[OH ] 4.5 105 M ; [H 3O ]
K = 1.8 10
2+
Mg(OH)2(s) + 2 H3O+(aq)
Mg (aq) + 4H2O(l) ;
Equilibrium
2.2 1010
x
17
K
x
[Mg 2 ]
1.8 1017
2
10
2
[H 3O ] [2.2 10 M]
x 8.7 103 M [Mg 2 ]equil
So, the molar solubility for Mg(OH) 2 = 8.7 10-3 M.
10A (M)
(a) In solution are Cu 2+ aq , SO 4 2 aq , Na + aq , and OH aq .
b g
Cu
(b)
2+
b g
aq + 2 OH aq Cu OH 2 s
b g
b g bsg is Cu baqg :
In the solution above, Cu OH
2+
2
Cu 2+ aq + 2 OH aq
Cu OH 2 s
b g
b g
This Cu 2+ aq reacts with the added NH 3 aq :
Cu NH 3
Cu 2+ aq + 4 NH 3 aq
4
2+
aq
Cu NH 3
The overall result is: Cu OH 2 s + 4NH 3 aq
4
(c)
b g
b g
2+
aq + 2 OH aq
b g
b g
HNO 3 aq (a strong acid), forms H 3O + aq , which reacts with OH aq and NH 3 aq .
OH aq + H 3O + aq 2 H 2 O(l);
b g
NH 3 aq + H 3O + aq NH 4 aq + H 2 O(l)
+
As NH 3 aq is consumed, the reaction below shifts to the left.
Cu NH 3
Cu OH 2 s + 4 NH 3 aq
4
2+
b g
aq + 2 OH aq
But as OH aq is consumed, the dissociation reactions shift to the side with the
Cu 2+ aq + 2 OH aq
dissolved ions: Cu OH s
2
The species in solution at the end of all this are
895
Chapter 18: Solubility and Complex-Ion Equilibria
Cu 2+ aq , NO3 aq , NH 4 + aq , excess H 3O + aq , Na + aq , and SO 4 2 aq
(probably HSO4(aq) as well).
10B (M)
2
(a) In solution are Zn 2+ (aq), SO 4 (aq), and NH 3 (aq),
Zn NH 3
Zn 2+ aq + 4 NH 3 aq
4
(b)
b g
2+
aq
b g
b g
HNO 3 aq , a strong acid, produces H 3O + aq , which reacts with NH 3 aq .
b g
NH 3 aq + H 3O + aq NH 4 aq + H 2 O(l) As NH 3 aq is consumed, the tetrammine
+
complex ion is destroyed.
Zn NH 3 4
(c)
2+
Zn H 2 O aq + 4NH 4 + aq
aq + 4H3O+ aq
4
2+
b g
NaOH(aq) is a strong base that produces OH aq , forming a hydroxide precipitate.
Zn OH s + 4 H 2O l
Zn H 2 O 4 aq + 2 OH aq
2
Another possibility is a reversal of the reaction of part (b).
2+
2+
Zn NH 3 aq + 8 H 2O l
Zn H 2 O 4 aq + 4 NH 4 + aq + 4 OH aq
4
2+
(d)
The precipitate dissolves in excess base.
2
Zn OH aq
Zn OH 2 s + 2 OH aq
4
11A (M) We first determine Ag + in a solution that is 0.100 M Ag + (aq) (from AgNO 3 ) and
b g
0.225 M NH 3 aq . Because of the large value of Kf = 1.6 107 , we start by having the
reagents form as much complex ion as possible, and approach equilibrium from this point.
b g
Ag + aq
Equation:
b g
+
2 NH 3 aq
In soln
0.100 M
Form complex –0.100 M
Initial
0M
Changes
+x M
Equil
xM
Kf = 1.6 107 =
x=
b g
Ag NH 3
0.100
b0.025g 1.6 10
2
7
Ag NH 3
0.225 M
–0.200 M
0.025 M
+2x M
(0.025 + x) M
+
2
0M
+0.100 M
0.100 M
–x M
(0.100 – x) M
+
2
Ag LNM NH 3 OQP
+
b g baqg
2
=
0.100 x
x 0.025 + 2 x
2
0.100
x 0.025
2
= 1.0 105 M = Ag + = concentration of free silver ion
(x << 0.025 M, so the approximation was valid)
The Cl is diluted:
1.00 mLinitial
Cl = Cl
= 3.50 M 1500 = 0.00233M
final
initial 1,500 mL
final
896
Chapter 18: Solubility and Complex-Ion Equilibria
Finally we compare Qsp with Ksp to determine if precipitation of AgCl(s) will occur.
c
g
hb
Qsp = Ag + Cl = 1.0 105 M 0.00233 M = 2.3 108 1.8 10 10 = Ksp
Because the value of the Qsp is larger than the value of the Ksp, precipitation of AgCl(s) should
occur.
11B (M) We organize the solution around the balanced equation of the formation reaction.
b g
Pb 2+ aq
Equation:
b g
EDTA 4 aq
+
Initial
0.100 M
Form Complex: –0.100 M
Equil
xM
0.250 M
(0.250 – 0.100) M
0.150 + x M
b
g
PbEDTA
b
2
baqg
0M
0.100 M
0.100 x M
g
PbEDTA 2
= 2 1018 = 0.100 x 0.100
K f = 2+
4
x 0.150 + x 0.150 x
Pb EDTA
x=
0.100
= 3 1019 M
0.150 2 1018
(x << 0.100 M, thus the approximation was valid.)
We calculate Qsp and compare it to Ksp to determine if precipitation will occur.
Qsp = Pb 2+ I = 3 1019 M 0.10 M = 3 1021 .
Qsp 7.1109 = K sp Thus precipitation will not occur.
2
2
12A (M) We first determine the maximum concentration of free Ag + .
1.8 1010
= 2.4 108 M.
0.0075
+
This is so small that we assume that all the Ag in solution is present as complex ion:
K sp = Ag + Cl = 1.8 1010
Ag + =
b g = 0.13 M. We use K to determine the concentration of free NH .
Agb NH g
0.13M
Ag NH 3
+
f
2
3
+
Kf =
3 2
2
Ag LNM NH 3 OQP
NH 3 =
+
= 1.6 107 =
2.4 108 LNM NH 3 OQP
2
013
.
= 0.58 M.
2.4 10 16
. 107
8
If we combine this with the ammonia present in the complex ion, the total ammonia
concentration is 0.58 M + 2 0.13 M = 0.84 M . Thus, the minimum concentration of
ammonia necessary to keep AgCl(s) from forming is 0.84 M.
b
g
897
Chapter 18: Solubility and Complex-Ion Equilibria
12B (M) We use the solubility product constant expression to determine the maximum Ag +
that can be present in 0.010 M Cl without precipitation occurring.
1.8 1010
Ag + =
K sp = 1.8 1010 = Ag + Cl = Ag + 0.010 M
= 1.8 108 M
0.010
This is also the concentration of free silver ion in the Kf expression. Because of the large
value of Kf , practically all of the silver ion in solution is present as the complex ion,
[Ag(S2O3)2]3]. We solve the expression for [S2O32] and then add the [S2O32] “tied up” in
the complex ion.
Ag S O 3
2 3 2
0.10 M
K f 1.7 1013
2
2
8
S2 O32
Ag + S2 O32
1.8
10
M
0.10
2
S2 O32 =
= 5.7 104 M = concentration of free S2O3
8
13
1.8 10 1.7 10
2 mol S2 O32
3
total S2 O32 = 5.7 104 M + 0.10 M Ag S2 O3 2
1mol Ag S2 O3 2
3
= 0.20 M + 0.00057 M = 0.20 M
13A (M) We must combine the two equilibrium expressions, for Kf and for Ksp, to find Koverall.
Fe3+ aq + 3OH aq
Fe OH 3 s
Fe3+ aq + 3C2 O 4
Fe OH 3 s + 3C2 O 4
Initial
Changes
Equil
2
2
3
Fe C2 O 4 aq
aq
3
3
Fe C2 O 4 aq + 3OH aq
aq
3
0.100 M
3x M
0.100 3x M
b
0M
+x M
xM
g
Kf = 2 1020
K overall = 8 1018
0M
+3x M
3x M
3
4
Fe C 2 O 4 3 OH
27 x
x 3 x
18
=
= 8 10
=
3
3
2 3
0.100
3
0.100
x
C
O
2 4
3
K overall
Ksp = 4 1038
3
(3x << 0.100 M, thus the approximation was valid.)
(0100
. ) 3 8 1018
x
4 106 M The assumption is valid.
27
2
Thus the solubility of Fe OH 3 in 0.100 M C 2 O 4 is 4 10 6 M .
4
b g
898
Chapter 18: Solubility and Complex-Ion Equilibria
13B (M) In Example 18-13 we saw that the expression for the solubility, s , of a silver halide in an
aqueous ammonia solution, where NH 3 is the concentration of aqueous ammonia, is given
by:
s
K sp K f =
NH 3 2 s
2
K sp K f
or
s
[NH 3 ] 2 s
For all scenarios, the NH 3 stays fixed at 0.1000 M and K f is always 1.6 107 .
We see that s will decrease as does Ksp . The relevant values are:
K sp AgCl = 1.8 1010 , K sp AgBr = 5.0 1013 , K sp AgI = 8.5 1017 .
Thus, the order of decreasing solubility must be: AgI > AgBr > AgCl.
14A (M) For FeS, we know that Kspa = 6 102 ; for Ag 2S, Kspa = 6 1030 .
We compute the value of Qspa in each case, with H 2S = 0.10 M and H 3O + = 0.30 M .
For FeS, Qspa =
0.020 0.10
0.30
2
= 0.022 6 102 = K spa
Thus, precipitation of FeS should not occur.
2
0.010 0.10
For Ag 2S, Qspa =
= 1.1 104
2
0.30
Qspa 6 1030 = K spa ; thus, precipitation of Ag 2S should occur.
14B (M) The H 3O + needed to just form a precipitate can be obtained by direct substitution of the
provided concentrations into the Kspa expression. When that expression is satisfied, a
precipitate will just form.
Fe 2+ H 2S
0.015 M Fe 2+ 0.10 M H 2S
2
=
6
10
=
, H 3 O + =
K spa =
+ 2
+ 2
H 3O
H 3O
b
g
pH = log H 3O + = log 0.002 = 2.7
899
0.015 0.10
6 102
= 0.002 M
Chapter 18: Solubility and Complex-Ion Equilibria
INTEGRATIVE EXAMPLES
A.
(D) To determine the amount of Ca(NO3)2 needed, one has to calculate the amount of Ca2+ that
will result in only 1.00×10-12 M of PO43Ca 3 PO 4 2 3 Ca 2+ +2 PO3-4
K sp 3s 2s
3
2
Using the common–ion effect, we will try to determine what concentration of Ca2+ ions forces
the equilibrium in the lake to have only 1.00×10-12 M of phosphate, noting that (2s) is the
equilibrium concentration of phosphate.
1.30 1032 Ca 2 1.00 1012
Solving for [Ca2+] yields a concentration of 0.00235 M.
3
2
The volume of the lake: V = 300 m × 150 m × 5 m = 225000 m3 or 2.25×108 L.
Mass of Ca(NO3)2 is determined as follows:
0.00235 mol Ca 2 1 mol Ca NO3 2 164.1 g Ca NO3 2
mass Ca NO3 2 2.25 108 L
L
1 mol Ca 2
1 mol Ca NO3 2
87 106 g Ca NO3 2
B.
(M) The reaction of AgNO3 and Na2SO4 is as follows:
2AgNO3 Na 2SO 4 2NaNO3 Ag 2SO 4
mol Ag 2SO 4 0.350 L 0.200 M AgNO3
1 mol Ag 2SO 4
3.5 102 mol
2 mol AgNO3
2 mol NaNO3
0.12 mol
1 mol Na 2SO 4
Ag2SO4 is the precipitate. Since it is also the limiting reagent, there are 3.5×10-2 moles of
Ag2SO4 produced.
mol NaNO3 0.250 L 0.240 M Na 2SO 4
The reaction of Ag2SO4 with Na2S2O3 is as follows:
Na 2S2 O3 Ag 2SO 4 Na 2SO 4 Ag 2S2 O3
mol Na 2S2 O3 0.400 L 0.500 M Na 2S2 O3 = 0.200 mol
Ag2SO4 is the limiting reagent, so no Ag2SO4 precipitate is left.
900
Chapter 18: Solubility and Complex-Ion Equilibria
EXERCISES
Ksp and Solubility
1.
2.
3.
4.
(E)
(a)
2 Ag + aq + SO 4 2 aq
Ag 2SO 4 s
2
K sp = Ag + SO 4
(b)
Ra 2+ aq + 2 IO3 aq
Ra IO3 2 s
K sp = Ra 2+ IO3
(c)
3 Ni 2+ aq + 2 PO 43 aq
Ni3 PO 4 2 s
3
K sp = Ni 2+ PO 4
(d)
PuO 2 2+ aq + CO32 aq
PuO 2 CO3 s
2+
2
K sp = PuO 2 CO3
2
2
3
2
(E)
Fe3+ aq + 3OH aq
Fe OH 3 s
3
(a)
K sp = Fe3+ OH
(b)
K sp = BiO + OH
BiO + aq + OH aq
BiOOH s
(c)
2+
K sp = Hg 2 I
Hg 2 2+ aq + 2 I aq
Hg 2 I 2 s
(d)
3
K sp = Pb 2+ AsO 4
2
3
3Pb 2+ aq + 2 AsO 43 aq
Pb3 AsO 4 2 s
2
(E)
(a)
Cr 3+ aq + 3F aq
CrF3 s
(b)
2 Au 3+ aq + 3C 2O 4 2 aq K sp = Au 3+ C 2 O 4 2 = 11010
Au 2 C2O 4 3 s
(c)
3Cd 2+ aq + 2 PO 4
Cd 3 PO 4 2 s
(d)
Sr 2+ aq + 2 F aq
SrF2 s
3
K sp = Cr 3+ F = 6.6 1011
3
2
3
aq
Ksp = Cd 2+
3
PO 4
Ksp = Sr 2+ F
2
3 2
= 2.1 1033
= 2.5 109
(E) Let s = solubility of each compound in moles of compound per liter of solution.
b gb g
Br = b sgb2 sg = 4 s = 4.0 10
F = b sgb3sg = 27 s = 8 10
AsO
= b3sg b2 sg = 108s = 2.1 10
(a)
Ksp = Ba 2+ CrO 4 2 = s s = s 2 = 1.2 1010
(b)
Ksp = Pb 2+
(c)
Ksp = Ce 3+
(d)
Ksp = Mg 2+
2
2
3
3 2
3
5
3
3
3
2
901
s = 2.2 102 M
16
4
4
s = 1.1 105 M
5
s = 7 105 M
20
s = 4.5 105 M
Chapter 18: Solubility and Complex-Ion Equilibria
5.
(E) We use the value of Ksp for each compound in determining the molar solubility in a
saturated solution. In each case, s represents the molar solubility of the compound.
AgCN
K sp = Ag + CN = s s = s 2 = 1.2 1016
s = 1.1108 M
K sp = Ag + IO3
K sp = Ag + I
AgIO3
AgI
= s s = s 2 = 3.0 108
s = 1.7 104 M
= s s = s 2 = 8.5 1017
s = 9.2 109 M
K sp = Ag + NO 2 = s s = s 2 = 6.0 104
AgNO 2
s = 2.4 102 M
K sp = Ag + SO 42 = 2s s = 4s 3 = 1.4 105 s = 1.5 102 M
Thus, in order of increasing molar solubility, from smallest to largest:
2
Ag 2SO 4
2
AgI AgCN AgIO 3 Ag 2SO 4 AgNO 2
6.
(E) We use the value of Ksp for each compound in determining Mg 2+ in its saturated
solution. In each case, s represents the molar solubility of the compound.
(a)
b gb g
Ksp = Mg 2+ CO 32 = s s = s2 = 3.5 108
MgCO 3
s = 1.9 104 M
(b)
Ksp = Mg 2+ F
MgF2
Mg 2+ = 1.9 104 M
2
b gb g
= s 2 s = 4 s 3 = 3.7 108
s = 2.1 103 M
(c)
2
Mg 2+ = 2.1 103 M
3
2
3
K sp = Mg 2+ PO 4 = 3s 2 s = 108s 5 = 2.11025
Mg 2+ = 1.4 105 M
s = 4.5 106 M
3
Mg 3 (PO 4 ) 2
2
Thus a saturated solution of MgF2 has the highest Mg 2+ .
7.
(M) We determine F in saturated CaF2 , and from that value the concentration of F- in ppm.
For CaF2
Ksp = Ca 2+ F
2
b gb g
= s 2 s = 4 s 3 = 5.3 109
2
s = 1.1 103 M
The solubility in ppm is the number of grams of CaF2 in 106 g of solution. We assume a
solution density of 1.00 g/mL.
mass of F 106 g soln
1mL
1L soln 1.1103 mol CaF2
1.00 g soln 1000 mL
1L soln
2 mol F 19.0 g F
42 g F
1mol CaF2 1mol F
This is 42 times more concentrated than the optimum concentration of fluoride ion for
fluoridation. CaF2 is, in fact, more soluble than is necessary. Its uncontrolled use might lead to
excessive F in solution.
902
Chapter 18: Solubility and Complex-Ion Equilibria
8.
(E) We determine OH in a saturated solution. From this OH we determine the pH.
Ksp = BiO + OH = 4 1010 = s 2
c
s = 2 105 M = OH = BiO +
h
pOH = log 2 105 = 4.7
9.
pH = 9.3
(M) We first assume that the volume of the solution does not change appreciably when its
temperature is lowered. Then we determine the mass of Mg C16 H 31O 2 2 dissolved in each
b
b
solution, recognizing that the molar solubility of Mg C16 H 31O 2
g
g
2
equals the cube root of one
fourth of its solubility product constant, since it is the only solute in the solution.
Ksp 4 s 3
s 3 Ksp / 4
3
3
At 50 C : s = 4.8 1012 / 4 1.1 104 M; At 25C: s = 3.3 1012 / 4 9.4 105 M
b
g
b
g
amount of Mg C16 H 31O 2 2 (50C) = 0.965 L
1.1104 mol Mg C16 H 31O 2 2
amount of Mg C16 H 31O 2 2 (25C) = 0.965 L
1L soln
9.4 105 mol Mg C16 H 31O 2 2
1L soln
= 1.1104 mol
= 0.91104 mol
mass of Mg C16 H 31O 2 2 precipitated:
535.15g Mg C16 H 31O 2 2 1000 mg
= 11mg
1mol Mg C16 H 31O 2 2
1g
(M) We first assume that the volume of the solution does not change appreciably when its
temperature is lowered. Then we determine the mass of CaC 2 O 4 dissolved in each solution,
recognizing that the molar solubility of CaC 2 O 4 equals the square root of its solubility product
constant, since it is the only solute in the solution.
= 1.1 0.91 104 mol
10.
At 95 C s = 1.2 108 1.1 104 M; At 13 C:
mass of CaC 2 O 4 (95 C) = 0.725 L
mass of CaC 2 O 4 (13 C) = 0.725 L
s 2.7 109 5.2 105 M
1.1 10 4 mol CaC 2 O 4
1 L soln
5.2 10 5 mol PbSO 4
1 L soln
mass of CaC2 O 4 precipitated = 0.010 g 0.0048 g
903
128.1g CaC 2 O 4
1 mol CaC 2 O 4
128.1g CaC 2 O 4
1mol CaC 2 O 4
1000 mg
= 5.2 mg
1g
= 0.010 g CaC 2 O 4
= 0.0048 g CaC 2 O 4
Chapter 18: Solubility and Complex-Ion Equilibria
11.
(M) First we determine I in the saturated solution.
Ksp = Pb 2+ I
2
b gb g
= 7.1 109 = s 2 s = 4 s 3
s = 1.2 103 M
2
The AgNO 3 reacts with the I in this saturated solution in the titration.
Ag + aq + I aq AgI s We determine the amount of Ag + needed for this titration, and
b g b g
bg
then AgNO 3 in the titrant.
moles Ag + = 0.02500 L
AgNO 3 molarity =
12.
1.2 103 mol PbI 2
2 mol I
1 mol Ag +
= 6.0 105 mol Ag +
1 L soln
1 mol PbI 2
1 mol I
6.0 105 mol Ag + 1 mol AgNO 3
= 4.5 103 M
+
0.0133 L soln
1 mol Ag
(M) We first determine C 2 O 4
2
= s, the solubility of the saturated solution.
2
C2O4
2
0.00134 mmol KMnO 4 5 mmol C 2 O 4
4.8 mL
1 mL soln
2 mmol MnO 4
=
= 6.4 105 M = s = Ca 2+
250.0 mL
Ksp = Ca 2+ C 2 O 4
13.
2
b gb g
c
= s s = s2 = 6.4 105
h
2
= 4.1 109
(M) We first use the ideal gas law to determine the moles of H 2S gas used.
FG 748 mmHg 1atm IJ FG 30.4 mL 1 L IJ
1000 mL K
760 mmHg K H
PV H
. 10
123
n
RT
1
3
1
0.08206 L atm mol K (23 273) K
moles
If we assume that all the H 2S is consumed in forming Ag 2S , we can compute the Ag + in
the AgBrO 3 solution. This assumption is valid if the equilibrium constant for the cited reaction
is large, which is the case, as shown below:
1.0 1019
3.8 1031
51
2.6 10
Ag 2S s + H 2 O(l)
2Ag + aq + HS aq + OH aq
K a 2 /K sp =
HS aq + H 3O + aq
H 2S aq + H 2 O(l)
K1 = 1.0 107
H 3O + aq + OH aq
2H 2 O(l)
K w = 1.0 1014
Ag 2 S s + 2H 3 O + aq
2Ag + aq + H 2 S aq + 2H 2 O(l)
K overall = ( K a 2 / K sp )( K1 )( K w ) (3.8 1031 )(1.0 107 )(1.0 1014 ) = 3.8 1010
Ag + =
1.23 103 mol H 2S 1000 mL 2 mol Ag +
= 7.28 103 M
338 mL soln
1 L soln 1 mol H 2S
Then, for AgBrO3 , K sp = Ag + BrO3 = 7.28 103 = 5.30 105
2
904
Chapter 18: Solubility and Complex-Ion Equilibria
14.
(M) The titration reaction is Ca OH 2 aq + 2HCl aq
CaCl2 aq + 2H 2 O(l)
1L
0.1032 mol HCl 1 mol OH
10.7 mL HCl
1000 mL
1L
1 mol HCl
OH
0.0221 M
1L
50.00 mL Ca(OH) 2 soln
1000 mL
In a saturated solution of Ca OH 2 , Ca 2+ = OH 2
K sp = Ca 2+ OH = 0.0221 2 0.0221 = 5.40 106 ( 5.5 106 in Appendix D).
2
2
The Common-Ion Effect
15.
b g
(E) We let s = molar solubility of Mg OH
(a)
Ksp = Mg 2+ OH
(b)
Equation :
Initial :
Changes :
Equil :
2
b gb g
2
in moles solute per liter of solution.
= s 2 s = 4 s 3 = 1.8 1011
s = 1.7 104 M
2
Mg 2+ aq
Mg OH 2 s
+
0.0862 M
+s M
0.0862 + s M
2OH aq
0M
+2s M
2s M
K sp = 0.0862 + s 2 s = 1.8 1011 0.0862 2s = 0.34 s 2 s = 7.3 106 M
(s << 0.0802 M, thus, the approximation was valid.)
2
(c)
2
OH = KOH = 0.0355 M
Equation :
Initial :
Changes :
Equil :
Mg 2+ aq
Mg OH 2 s
0M
+ sM
sM
+
2OH aq
0.0355 M
+ 2s M
0.0355 + 2s M
K sp = s 0.0355 + 2s = 1.8 1011 s 0.0355 = 0.0013 s
2
16.
s = 1.4 108 M
2
(E) The solubility equilibrium is
Ca 2+ aq + CO3
CaCO3 s
b g
2
aq
b g
(a)
The addition of Na 2 CO 3 aq produces CO 32 aq in solution. This common ion
suppresses the dissolution of CaCO 3 s .
(b)
HCl(aq) is a strong acid that reacts with carbonate ion:
2
CO3 aq + 2H 3O + aq CO 2 g + 3H 2 O(l) .
bg
bg
This decreases CO 32 in the solution, allowing more CaCO 3 s to dissolve.
905
Chapter 18: Solubility and Complex-Ion Equilibria
b g
HSO 4 aq is a moderately weak acid. It is strong enough to protonate appreciable
(c)
concentrations of carbonate ion, thereby decreasing CO 32 and enhancing the solubility
bg
of CaCO 3 s , as the value of Kc indicates.
HSO 4 aq + CO3
2
SO 4 2 aq + HCO3 aq
aq
Kc
17.
K a (HSO 4 )
0.011
2.3 108
11
K a (HCO3 ) 4.7 10
(E) The presence of KI in a solution produces a significant I in the solution. Not as much AgI
can dissolve in such a solution as in pure water, since the ion product, Ag + I , cannot exceed
the value of K sp (i.e., the I- from the KI that dissolves represses the dissociation of AgI(s). In
similar fashion, AgNO 3 produces a significant Ag + in solution, again influencing the value of
the ion product; not as much AgI can dissolve as in pure water.
18.
(E) If the solution contains KNO 3 , more AgI will end up dissolving than in pure water, because
the activity of each ion will be less than its molarity. On an ionic level, the reason is that ion
pairs, such as Ag+NO3– (aq) and K+I– (aq) form in the solution, preventing Ag+ and I– ions from
coming together and precipitating.
19.
(E) Equation:
Ag 2SO 4 s
Original:
Add solid:
Equil:
—
—
—
2Ag + aq
0M
+2x M
2x M
SO 24 aq
+
0.150 M
+x M
0.150 + x M
2 x = Ag + = 9.7 103 M; x = 0.00485 M
2
2
K sp = Ag + SO 4 = 2 x 0.150 + x = 9.7 103 0.150 + 0.00485 = 1.5 105
2
2
20.
(M) Equation: CaSO 4 s
Ca 2+ aq +
SO 4
0.0025 M
+x M
Soln:
Add CaSO 4 s
—
—
0M
+x M
Equil:
—
xM
2
aq .
0.0025 + x
M
If we use the approximation that x << 0.0025, we find x = 0.0036. Clearly, x is larger than
0.0025 and thus the approximation is not valid. The quadratic equation must be solved.
2
K sp = Ca 2+ SO 4 = 9.1106 = x 0.0025 + x = 0.0025 x + x 2
x 2 + 0.0025 x 9.1106 = 0
906
Chapter 18: Solubility and Complex-Ion Equilibria
x=
b b 2 4ac 0.0025 6.3 106 + 3.6 105
= 2.0 10 3 M = CaSO 4
=
2
2a
mass CaSO 4 = 0.1000 L
21.
2.0 10 3 mol CaSO 4 136.1 g CaSO 4
= 0.027 g CaSO 4
1 L soln
1 mol CaSO 4
(M) For PbI 2 , Ksp = 7.1 109 = Pb 2+ I
2
In a solution where 1.5 104 mol PbI 2 is dissolved, Pb 2 1.5 104 M , and
I = 2 Pb 2+ = 3.0 104 M
Equation:
Initial:
Add lead(II):
Equil:
PbI 2 s
Pb 2+ aq
+
1.5 104 M
+x M
0.00015 + x M
—
—
—
2I aq
3.0 104 M
0.00030 M
K sp = 7.1109 = 0.00015 + x 0.00030 ; 0.00015 + x = 0.079 ; x = 0.079 M = Pb 2+
2
22.
(M) For PbI 2 , Ksp = 7.1 109 = Pb 2+ I
2
In a solution where 1.5 105 mol PbI 2 /L is dissolved, Pb 2 1.5 105 M , and
I = 2 Pb 2+ = 3.0 105 M
Equation:
Initial:
Add KI:
Equil:
PbI 2 s
—
—
—
Pb 2+ aq +
2I aq
1.5 105 M
3.0 105 M
+x M
(3.0 105 x) M
1.5 105 M
Ksp = 7.1 × 10-9 = (1.5 × 10-5) × (3.0 × 10-5 + x)2
(3.0 × 10-5 + x) = (7.1 × 10-9 ÷ 1.5 × 10-5)1/2
x = (2.2 × 10-2) – (3.0 × 10-5) = 2.1 × 10-2 M = [I-]
23.
2
(D) For Ag 2 CrO 4 , K sp = 1.11012 = Ag + CrO 4
In a 5.0 108 M solution of Ag 2 CrO 4 , CrO 4 2 = 5.0 108 M and Ag + = 1.0 107 M
2
Equation:
Initial:
Add chromate:
Equil:
Ag 2 CrO 4 s
2Ag + aq +
1.0 107 M
—
—
—
1.0 107 M
K sp = 1.1 1012 = 1.0 107 5.0 108 + x ;
2
907
CrO 4
2
aq
5.0 108 M
+x M
5.0 108 + x M
c
5.0 10
h
8
+ x = 1.1 102 = CrO 4 2 .
Chapter 18: Solubility and Complex-Ion Equilibria
This is an impossibly high concentration to reach. Thus, we cannot lower the solubility of
2
Ag 2 CrO 4 to 5.0 108 M with CrO 4 as the common ion. Let’s consider using Ag+ as the
common ion.
2Ag + aq + CrO 4 2 aq
Ag 2CrO 4 s
Equation:
Initial:
Add silver(I) ion:
Equil:
1.0 107 M
+x M
1.0 107 + x M
—
—
—
5.0 108 M
5.0 108 M
1.1 10-12
= 4.7 10-3
-8
5.0 10
3
7
3
x = 4.7 10 1.0 10 = 4.7 10 M = I ; this is an easy-to-reach concentration. Thus,
K sp = 1.1 10-12 = 1.0 + x 5.0 10-8
2
1.0 10-7 + x =
the solubility can be lowered to 5.0×10-8 M by carefully adding Ag+(aq).
24.
(M) Even though BaCO 3 is more soluble than BaSO 4 , it will still precipitate when 0.50 M
b g
Na 2 CO 3 aq is added to a saturated solution of BaSO 4 because there is a sufficient Ba 2+
in such a solution for the ion product Ba 2+ CO 32 to exceed the value of Ksp for the compound.
An example will demonstrate this phenomenon. Let us assume that the two solutions being mixed
are of equal volume and, to make the situation even more unfavorable, that the saturated
BaSO 4 solution is not in contact with solid BaSO 4 , meaning that it does not maintain its
saturation when it is diluted. First we determine Ba 2+ in saturated BaSO 4 (aq) .
Ksp = Ba 2+ SO 4
2
= 1.1 1010 = s 2
s = 11
. 1010 10
. 105 M
Mixing solutions of equal volumes means that the concentrations of solutes not common to the
two solutions are halved by dilution.
1 1.0 105 mol BaSO 4 1 mol Ba 2+
Ba 2+ =
= 5.0 106 M
2
1L
1 mol BaSO 4
2
1 0.50 mol Na 2 CO3 1 mol CO3
CO32 =
= 0.25 M
2
1L
1 mol Na 2 CO3
2
Qsp {BaCO3 } = Ba 2+ CO3 = 5.0 106 0.25 = 1.3 106 5.0 109 = K sp {BaCO3}
Thus, precipitation of BaCO 3 indeed should occur under the conditions described.
25.
115g Ca 2+ 1 mol Ca 2+ 1000 g soln
= 2.87 103 M
(E) Ca 2+ = 6
2+
10 g soln 40.08 g Ca
1 L soln
2+
9
3
3
Ca F = K sp = 5.3 10 = 2.87 10 F
F = 1.4 10 M
1.4 103 mol F 19.00 g F 1 L soln
ppm F =
106 g soln = 27 ppm
1 L soln
1 mol F
1000 g
2
2
908
Chapter 18: Solubility and Complex-Ion Equilibria
26.
(M) We first calculate the Ag + and the Cl in the saturated solution.
b gb g
Ksp = Ag + Cl = 1.8 1010 = s s = s 2
s = 1.3 105 M = Ag + = Cl
Both of these concentrations are marginally diluted by the addition of 1 mL of NaCl(aq)
Ag + = Cl = 1.3 105 M
100.0 mL
= 1.3 105 M
100.0 mL +1.0 mL
1.0 mL
= 9.9 103 M
The Cl in the NaCl(aq) also is diluted. Cl = 1.0 M
100.0 mL +1.0 mL
Let us use this Cl to determine the Ag + that can exist in this solution.
1.8 1010
Ag + =
= 1.8 108 M
9.9 103
We compute the amount of AgCl in this final solution, and in the initial solution.
Ag + Cl = 1.8 1010 = Ag + 9.9 103 M
mmol AgCl final = 101.0 mL
1.8 108 mol Ag + 1 mmol AgCl
= 1.8 106 mmol AgCl
+
1 L soln
1 mmol Ag
1.3 105 mol Ag + 1 mmol AgCl
= 1.3 103 mmol AgCl
+
1 L soln
1 mmol Ag
The difference between these two amounts is the amount of AgCl that precipitates. Next we
compute its mass.
143.3 mg AgCl
mass AgCl = 1.3 103 1.8 106 mmol AgCl
= 0.19 mg
1 mmol AgCl
We conclude that the precipitate will not be visible to the unaided eye, since its mass is less
than 1 mg.
mmol AgCl initial = 100.0 mL
Criteria for Precipitation from Solution
27.
(E) We first determine Mg 2+ , and then the value of Qsp in order to compare it to the value of
Ksp . We express molarity in millimoles per milliliter, entirely equivalent to moles per liter.
[Mg 2+ ]
22.5 mg MgCl2 1mmol MgCl2 6H 2O
1mmol Mg 2+
3.41 104 M
325 mL soln
203.3mg MgCl2 6H 2O 1mmol MgCl2
Qsp [Mg 2+ ][F ]2 (3.41 104 )(0.035) 2 4.2 107 3.7 108 K sp
bg
Thus, precipitation of MgF2 s should occur from this solution.
28.
(E) The solutions mutually dilute each other.
155 mL
Cl = 0.016 M
= 6.2 103 M
155 mL + 245 mL
245 mL
Pb 2+ = 0.175 M
= 0.107 M
245 mL +155 mL
909
Chapter 18: Solubility and Complex-Ion Equilibria
Then we compute the value of the ion product and compare it to the solubility product constant
value.
Qsp = Pb 2+ Cl = 0.107 6.2 103 = 4.1106 1.6 105 = K sp
2
2
bg
Thus, precipitation of PbCl 2 s will not occur from these mixed solutions.
29.
(E) We determine the OH needed to just initiate precipitation of Cd(OH)2.
K sp = Cd 2+ OH
c
2
b
h
pOH = log 2.1 106 = 5.68
b g
Thus, Cd OH
30.
2
g
= 2.5 10 14 = 0.0055M OH
2
pH = 14.00 5.68 = 8.32
will precipitate from a solution with pH 8.32.
(E) We determine the OH needed to just initiate precipitation of Cr(OH)3.
K sp [Cr 3+ ][OH ]3 6.3 1031 (0.086 M) [OH ]3 ; [OH ]
c
Thus, Cr bOH g
h
pOH = log 1.9 1010 = 9.72
31.
2.5 10 14
= 2.1 10 6 M
0.0055
OH =
(D)
(a)
3
6.3 1031
1.9 1010 M
0.086
pH = 14.00 9.72 = 4.28
will precipitate from a solution with pH > 4.28.
First we determine Cl due to the added NaCl.
Cl =
0.10 mg NaCl
1g
1 mol NaCl 1 mol Cl
= 1.7 106 M
1.0 L soln
1000 mg 58.4 g NaCl 1 mol NaCl
Then we determine the value of the ion product and compare it to the solubility product
constant value.
Qsp = Ag + Cl = 0.10 1.7 106 = 1.7 107 1.8 1010 = K sp for AgCl
Thus, precipitation of AgCl(s) should occur.
(b)
The KBr(aq) is diluted on mixing, but the Ag + and Cl are barely affected by
dilution.
Br = 0.10 M
0.05 mL
= 2 105 M
0.05 mL + 250 mL
Now we determine Ag + in a saturated AgCl solution.
b gb g
Ksp = Ag + Cl = s s = s 2 = 1.8 1010
910
s = 1.3 105 M
Chapter 18: Solubility and Complex-Ion Equilibria
Then we determine the value of the ion product for AgBr and compare it to the solubility
product constant value.
Qsp = Ag + Br = 1.3 105 2 105 = 3 1010 5.0 1013 = K sp for AgBr
Thus, precipitation of AgBr(s) should occur.
(c)
The hydroxide ion is diluted by mixing the two solutions.
0.05 mL
OH = 0.0150 M
= 2.5 107 M
0.05 mL + 3000 mL
But the Mg 2+ does not change significantly.
2.0 mg Mg 2+
1g
1 mol Mg 2+
Mg 2+ =
= 8.2 105 M
1.0 L soln
1000 mg 24.3 g Mg
Then we determine the value of the ion product and compare it to the solubility product
constant value.
Qsp = Mg 2+ OH = 2.5 107 8.2 105 = 5.1 1018
2
2
Qsp 1.8 1011 = K sp for Mg OH 2 Thus, no precipitate forms.
32.
First we determine the moles of H 2 produced during the electrolysis, then determine [OH–].
1 atm
752 mmHg
0.652 L
PV
760 mmHg
moles H 2 =
=
= 0.0267 mol H 2
RT
0.08206 L atm mol1 K 1 295 K
OH =
2 mol OH
1 mol H 2
= 0.170 M
0.315 L sample
0.0267 mol H 2
Qsp = Mg 2+ OH = 0.185 0.170 = 5.35 103 1.8 1011 = K sp
2
2
b g bsg should occur during the electrolysis.
Thus, yes, precipitation of Mg OH
33.
2
(D) First we must calculate the initial [H2C2O4] upon dissolution of the solid acid:
1 mol H 2 C 2 O 4
1
[H2C2O4]initial = 1.50 g H2C2O4
= 0.0833 M
90.036 g H 2 C2 O 4 0.200 L
(We assume the volume of the solution stays at 0.200 L.)
Next we need to determine the [C2O42-] in solution after the hydrolysis reaction between oxalic
acid and water reaches equilibrium. To accomplish this we will need to solve two I.C.E. tables:
911
Chapter 18: Solubility and Complex-Ion Equilibria
K 5.2 102
+
a1
H2C2O4(aq) + H2O(l)
HC2O4 (aq) + H3O (aq)
0.0833 M
—
0M
0M
–x
—
+x M
+x M
0.0833 – x M
—
xM
xM
Table 1:
Initial:
Change:
Equilibrium:
Since Ca/Ka = 1.6, the approximation cannot be used, and thus the full quadratic equation must
be solved: x2/(0.0833 – x) = 5.2×10-2;
x2 + 5.2×10-2x – 4.33×10-3
-5.2×10-2
2.7×10-3 + 0.0173
x = 0.045 M = [HC 2 O 4- ] [H 3 O ]
2
Now we can solve the I.C.E. table for the second proton loss:
x=
K 5.4 105
2+
a1
HC2O4-(aq) + H2O(l)
C2O4 (aq) + H3O (aq)
0.045 M
—
0M
≈ 0.045 M
–y
—
+y M
+yM
(0.045 – y) M
—
yM
≈ (0.045 +y) M
Table 2:
Initial:
Change:
Equilibrium:
Since Ca/Ka = 833, the approximation may not be valid and we yet again should solve the full
quadratic equation:
y (0.045 y )
= 5.4 10-5 ;
y2 + 0.045y = 2.43×10-6 – 5.4×10-5y
(0.045 y )
-0.045
2.03×10-3 + 9.72×10-6
y = 8.2 10-5 M = [C2 O 4 2- ]
2
Now we can calculate the Qsp for the calcium oxalate system:
Qsp = [Ca2+]initial×[ C2O42-]initial = (0.150)(8.2×10-5) = 1.2×10-5 > 1.3 ×10-9 (Ksp for CaC2O4)
Thus, CaC2O4 should precipitate from this solution.
y=
34.
(D) The solutions mutually dilute each other. We first determine the solubility of each
compound in its saturated solution and then its concentration after dilution.
2
2
K sp = Ag + SO 4 = 1.4 105 = 2 s s = 4s 3
2
3
s=
100.0 mL
SO 4 2 = 0.015 M
= 0.0043 M
100.0 mL + 250.0 mL
Ksp [ Pb 2+ ][CrO 24 ] 2.8 1013 ( s)( s) s2
1.4 105
1.5 102 M
4
Ag + = 0.0086 M
s 2.8 1013 5.3 107 M
250.0 mL
Pb 2+ = CrO 4 2 = 5.3 107
= 3.8 107 M
250.0 mL +100.0 mL
From the balanced chemical equation, we see that the two possible precipitates are PbSO 4 and
Ag 2 CrO 4 . (Neither PbCrO 4 nor Ag 2SO 4 can precipitate because they have been diluted
PbSO 4 + Ag 2CrO 4
below their saturated concentrations.) PbCrO 4 + Ag 2SO 4
Thus, we compute the value of Qsp for each of these compounds and compare those values
with the solubility constant product value.
912
Chapter 18: Solubility and Complex-Ion Equilibria
2
Qsp = Pb 2+ SO 4 = 3.8 107 0.0043 = 1.6 109 1.6 108 = K sp for PbSO 4
Thus, PbSO 4 (s) will not precipitate.
2
2
Qsp = Ag + CrO 4 = 0.0086 3.8 107 = 2.8 1011 1.11012 = K sp for Ag 2 CrO 4
2
bg
Thus, Ag 2 CrO 4 s should precipitate.
Completeness of Precipitation
35.
(M) First determine that a precipitate forms. The solutions mutually dilute each other.
CrO 4 2 = 0.350 M
Ag + = 0.0100 M
200.0 mL
= 0.175 M
200.0 mL + 200.0 mL
200.0 mL
= 0.00500 M
200.0 mL + 200.0 mL
We determine the value of the ion product and compare it to the solubility product constant
value.
2
2
Qsp = Ag + CrO 4 = 0.00500 0.175 = 4.4 106 1.110 12 = K sp for Ag 2 CrO 4
2
Ag 2 CrO 4 should precipitate.
Now, we assume that as much solid forms as possible, and then we approach equilibrium by
dissolving that solid in a solution that contains the ion in excess.
12
1.1 10
+
2
Equation:
Ag 2 CrO 4 s
2Ag aq + CrO 4 aq
Orig. soln :
0.00500 M
0.175 M
Form solid :
0.00500 M
0.00250 M
0M
0.173 M
Not at equilibrium
Changes :
+2 x M
+x M
Equil :
2x M
0.173 + x M
K sp Ag CrO 4 2 1.1 1012 2 x 0.173 x 4 x 2 0.173
2
x = 1.11012 4 0.173 = 1.3 106 M
% Ag + unprecipitated
36.
2.6 106 M final
100% 0.052% unprecipitated
0.00500 M initial
(M) [Ag+]diluted = 0.0208 M 175 mL = 0.008565 M
425 mL
[CrO42]diluted = 0.0380 M 250 mL = 0.02235 M
425 mL
Qsp Ag CrO 4 1.6 106 K sp
2
Ag + = 2 x = 2.6 106 M
2
913
Chapter 18: Solubility and Complex-Ion Equilibria
Because Qsp > Ksp, then more Ag2Cr2O4 precipitates out. Assume that the limiting reagent is
used up (100% reaction in the reverse direction) and re-establish the equilibrium in the reverse
direction. Here Ag+ is the limiting reagent.
Ag2CrO4(s)
Initial
Change
100% rxn
Change
Equil
Ksp(Ag
2 Ag+(aq)
=1.1 10-12
2CrO4 )
+ CrO42(aq)
0.008565 M
2x
0
+2y
2y
x = 0.00428 M
re-establish equil
(assume y ~ 0)
0.02235 M
x
0.0181
+y
0.0181 + y
1.1 1012 = (2y)2(0.0181+ y) (2y)2(0.0181)
y = 3.9 106 M
y << 0.0181, so this assumption is valid.
2y = [Ag+] = 7.8 106 M after precipitation is complete.
7.8 106
+
100% = 0.091% (precipitation is essentially quantitative)
% [Ag ]unprecipitated =
0.00856
37.
(M) We first use the solubility product constant expression to determine Pb 2+ in a solution
with 0.100 M Cl .
K sp = Pb 2+ Cl = 1.6 105 = Pb 2+ 0.100
2
Thus, % unprecipitated
2
Pb 2+ =
1.6 105
0.100
2
= 1.6 103 M
3
1.6 10 M
100% 2.5%
0.065 M
Now, we want to determine what [Cl-] must be maintained to keep [Pb 2+ ]final 1%;
[Pb 2+ ]initial 0.010 0.065 M = 6.5 104 M
Ksp = Pb 2+ Cl
2
c
h
= 1.6 105 = 6.5 10 4 Cl
2
Cl =
6 105
= 0.16 M
6.5 10 4
38. (M) Let’s start by assuming that the concentration of Pb2+ in the untreated wine is no higher
than 1.5 104M (this assumption is not unreasonable.) As long as the Pb2+ concentration is
less than 1.5 104 M, then the final sulfate ion concentration in the CaSO4 treated wine
should be virtually the same as the sulfate ion concentration in a saturated solution of CaSO4
formed by dissolving solid CaSO4 in pure water ( i.e., with [Pb2+] less than or equal to 1.5
104 M, the [SO42] will not drop significantly below that for a saturated solution, 3.0 103
M.) Thus, the addition of CaSO4 to the wine would result in the precipitation of solid PbSO4,
which would continue until the concentration of Pb2+ was equal to the Ksp for PbSO4 divided
by the concentration of dissolved sulfate ion, i.e.,
[Pb2+]max = 1.6 108M2/3.0 103 M = 5.3 106 M.
914
Chapter 18: Solubility and Complex-Ion Equilibria
Fractional Precipitation
39.
(M) First, assemble all of the data. Ksp for Ca(OH)2 = 5.5×10-6, Ksp for Mg(OH)2 = 1.8×10-11
440 g Ca 2+
1 mol Ca 2+
1 kg seawater 1.03 kg seawater
2+
[Ca ]
0.0113 M
2+
1000 kg seawater 40.078 g Ca
1000 g seawater
1 L seawater
[Mg2+] = 0.059 M, obtained from Example 18-6. [OH-] = 0.0020 M (maintained)
(a)
(b)
40.
Qsp = [Ca2+]×[OH-]2 = (0.0113)( 0.0020)2 = 4.5×10-8
Qsp < Ksp no precipitate forms.
2+
For the separation to be complete, >>99.9 % of the Mg must be removed before Ca2+
begins to precipitate. We have already shown that Ca2+ will not precipitate if the [OH-] =
0.0020 M and is maintained at this level. Let us determine how much of the 0.059 M
Mg2+ will still be in solution when [OH-] = 0.0020 M.
Ksp = [Mg2+]×[OH-]2 = (x)( 0.0020)2 = 1.8×10-11
x = 4.5×10-6 M
4.5 106
100% = 0.0076 %
The percent Mg2+ ion left in solution =
0.059
This means that 100% - 0.0076 % Mg = 99.992 % has precipitated.
Clearly, the magnesium ion has been separated from the calcium ion (i.e., >> 99.9% of the
Mg2+ ions have precipitated and virtually all of the Ca2+ ions are still in solution.)
(D)
(a)
0.10 M NaCl will not work at all, since both BaCl 2 and CaCl 2 are soluble in water.
(b)
Ksp = 1.1 1010 for BaSO 4 and Ksp = 9.1 106 for CaSO 4 . Since these values differ by
more than 1000, 0.05 M Na 2SO 4 would effectively separate Ba 2+ from Ca 2+ .
We first compute SO 4 2 when BaSO 4 begins to precipitate.
Ba 2+ SO 4 2 = 0.050 SO 4 2 = 1.11010 ;
SO 4 2 =
1.11010
= 2.2 109 M
0.050
And then we calculate SO 4 2 when Ba 2+ has decreased to 0.1% of its initial value,
that is, to 5.0 105 M.
1.11010
6
Ba 2+ SO 4 2 = 5.0 105 SO 4 2 = 1.11010 ; SO 4 2 =
5.0 105 = 2.2 10 M
And finally, SO 4 2 when CaSO 4 begins to precipitate.
9.1 106
Ca 2+ SO 4 2 = 0.050 SO 4 2 = 9.1106 ; SO 4 2 =
= 1.8 104 M
0.050
915
Chapter 18: Solubility and Complex-Ion Equilibria
(c)
b g
Now, Ksp = 5 103 for Ba OH
2
and Ksp = 5.5 106 for Ca OH 2 . The fact that these
two Ksp values differ by almost a factor of 1000 does not tell the entire story, because
b g
OH is squared in both Ksp expressions. We compute OH when Ca OH
2
begins
to precipitate.
. 106
55
= 1.0 102 M
0.050
Precipitation will not proceed, as we only have 0.001 M NaOH, which has
OH = 1 103 M.
Ca 2+ OH
(d)
2
b
g
= 5.5 106 = 0.050 OH
2
OH =
Ksp = 5.1 109 for BaCO 3 and Ksp = 2.8 109 for CaCO3 . Since these two values
differ by less than a factor of 2, 0.50 M Na 2 CO 3 would not effectively separate Ba 2+
from Ca 2+ .
41.
(M)
(a) Here we need to determine I when AgI just begins to precipitate, and I when
PbI 2 just begins to precipitate.
b g
Ksp = Ag + I = 8.5 1017 = 0.10 I
I = 8.5 1016 M
K sp = Pb 2+ I = 7.1 109 = 0.10 I
2
2
I
7.1 109
= 2.7 104 M
0.10
Since 8.5 1016 M is less than 2.7 104 M, AgI will precipitate before PbI 2 .
(b)
I must be equal to 2.7 104 M before the second cation, Pb 2+ , begins to precipitate.
(c)
Ksp = Ag + I = 8.5 1017 = Ag + 2.7 104
(d)
Since Ag + has decreased to much less than 0.1% of its initial value before any PbI 2
c
h
Ag + = 3.1 10 13 M
begins to precipitate, we conclude that Ag + and Pb 2+ can be separated by precipitation
with iodide ion.
42.
(D) Normally we would worry about the mutual dilution of the two solutions, but the values
of the solubility product constants are so small that only a very small volume of 0.50 M
Pb NO 3 2 solution needs to be added, as we shall see.
b g
(a)
Since the two anions are present at the same concentration and they have the same type
of formula (one anion per cation), the one forming the compound with the smallest Ksp
value will precipitate first. Thus, CrO 4
2
916
is the first anion to precipitate.
Chapter 18: Solubility and Complex-Ion Equilibria
(b)
At the point where SO 4
2
begins to precipitate, we have
1.6 108
2
8
2+
2+
K sp = Pb SO 4 = 1.6 10 = Pb 0.010M ; Pb =
= 1.6 106 M
0.010
Now we can test our original assumption, that only a very small volume of 0.50 M
Pb NO 3 2 solution has been added. We assume that we have 1.00 L of the original
2+
b g
solution, the one with the two anions dissolved in it, and compute the volume of 0.50 M
Pbb NO g that has to be added to achieve Pb = 1.6 10 M.
6
2+
3 2
Vadded
1.6 106 mol Pb 2+ 1 mol Pb NO3 2
1 L Pb 2+soln
= 1.00 L
1 L soln
1 mol Pb 2+
0.50 mol Pb NO3 2
Vadded = 3.2 105 L Pb 2+soln = 0.0032mL Pb 2+soln
This is less than one drop (0.05 mL) of the Pb 2+ solution, clearly a very small volume.
(c)
The two anions are effectively separated if Pb 2+ has not reached 1.6 106 M when
CrO 4 2 is reduced to 0.1% of its original value, that is, to
2
0.010 103 M = 1.0 105 M = CrO 4
Ksp = Pb 2+ CrO 4
2
c
= 2.8 10 13 = Pb 2+ 1.0 105
h
2.8 1013
Pb 2+ =
= 2.8 108 M
5
1.0 10
Thus, the two anions can be effectively separated by fractional precipitation.
43.
(M) First, let’s assemble all of the data. Ksp for AgCl = 1.8×10-10
[Ag+] = 2.00 M
[Cl-] = 0.0100 M
[I-] = 0.250 M
Ksp for AgI = 8.5×10-17
(a)
AgI(s) will be the first to precipitate by virtue of the fact that the Ksp value for AgI is
about 2 million times smaller than that for AgCl.
(b)
AgCl(s) will begin to precipitate when the Qsp for AgCl(s) > Ksp for AgCl(s). The
concentration of Ag+ required is: Ksp = [Ag+][Cl-] = 1.8×10-10 = (0.0100)×(x)
1.8×10-8 M
Using this data, we can determine the remaining concentration of I- using the Ksp.
Ksp = [Ag+][I-] = 8.5×10-17 = (x)×( 1.8×10-8)
x = 4.7×10-9 M
(c)
x=
In part (b) we saw that the [I-] drops from 0.250 M → 4.7×10-9 M. Only a small
4.7 109
100% = 0.0000019 %
percentage of the ion remains in solution.
0.250
This means that 99.999998% of the I- ion has been precipitated before any of the Cl- ion
has precipitated. Clearly, the fractional separation of Cl- from I- is feasible.
917
Chapter 18: Solubility and Complex-Ion Equilibria
44.
(M)
(a) We first determine the Ag + needed to initiate precipitation of each compound.
AgCl : K sp = Ag + Cl = 1.8 1010 = Ag + 0.250 ;
1.8 1010
= 7.2 1010 M
0.250
5.0 1013
Ag + =
= 2.3 1010 M
0.0022
Ag + =
AgBr : Ksp = Ag + Br = 5.0 1013 = Ag + 0.0022 ;
Thus, Br precipitates first, as AgBr, because it requires a lower Ag + .
Ag + = 7.2 1010 M when chloride ion, the second anion, begins to precipitate.
(b)
(c)
Cl and Br cannot be separated by this fractional precipitation. Ag + will have to
rise to 1000 times its initial value, to 2.3 107 M, before AgBr is completely precipitated.
But as soon as Ag + reaches 7.2 1010 M, AgCl will begin to precipitate.
Solubility and pH
45.
(E) In each case we indicate whether the compound is more soluble in water. We write the net
ionic equation for the reaction in which the solid dissolves in acid. Substances are more
soluble in acid if either (1) an acid-base reaction occurs or (2) a gas is produced, since escape
of the gas from the reaction mixture causes the reaction to shift to the right.
Same: KCl (K+ and Cl- do not react appreciably with H2O)
Acid: MgCO3 s + 2H + aq
Mg 2+ aq + H 2O(l) + CO 2 g
Fe 2+ aq + H 2S g
Acid: FeS s + 2H + aq
Acid: Ca OH 2 s + 2H + aq
Ca 2+ aq + 2H 2 O(l)
Water:
46.
C6 H 5COOH is less soluble in acid, because of the H 3O + common ion.
(E) In each case we indicate whether the compound is more soluble in base than in water. We
write the net ionic equation for the reaction in which the solid dissolves in base. Substances are
soluble in base if either (1) acid-base reaction occurs [as in (b)] or (2) a gas is produced, since
escape of the gas from the reaction mixture causes the reaction to shift to the right.
Water:
BaSO 4 is less soluble in base, hydrolysis of SO 4
Base: H 2C 2 O 4 s + 2OH aq
C2O4
b g
2
2
will be repressed.
aq + 2H 2O(l)
is less soluble in base because of the OH common ion.
Water:
Fe OH
Same:
NaNO 3 (neither Na+ nor NO3- react with H2O to a measurable extent).
3
Water: MnS is less soluble in base because hydrolysis of S2 will be repressed.
918
Chapter 18: Solubility and Complex-Ion Equilibria
47.
(E) We determine Mg 2+ in the solution.
Mg 2+ =
b g
0.65 g Mg OH
1 L soln
2
b g
58.3 g MgbOH g
1 mol Mg OH
2
2
1 mol Mg 2+
1 mol Mg OH
b g
= 0.011 M
2
Then we determine OH in the solution, and its pH.
1.8 1011
4.0 105 M
0.011
K sp Mg 2 OH 1.8 1011 0.011 OH ; OH
2
2
pOH = log 4.0 105 = 4.40
48.
pH = 14.00 4.40 = 9.60
(M) First we determine the [Mg2+] and [NH3] that result from dilution to a total volume of
0.500 L.
0.150 Linitial
Mg 2+ = 0.100 M
= 0.0300 M;
0.500 Lfinal
NH3 = 0.150 M
0.350 Linitial
= 0.105 M
0.500 L final
Then determine the OH that will allow Mg 2+ = 0.0300 M in this solution.
1.8 10 11
K sp 1.8 10 11 = Mg 2+ OH = 0.0300 OH ; OH =
2
2
0.0300
= 2.4 10 5 M
This OH is maintained by the NH3 / NH +4 buffer; since it is a buffer, we can use the
Henderson–Hasselbalch equation to find the [NH4+].
pH = 14.00 pOH = 14.00 + log 2.4 105 = 9.38 = pK a + log
NH3
log
= 9.38 9.26 = +0.12;
NH 4 +
mass NH 4 2 SO 4 = 0.500 L
49.
NH3
= 10+0.12 = 1.3;
NH 4 +
0.081 mol NH 4
L soln
+
NH3
NH 4
+
NH 4 + =
1 mol NH 4 2 SO 4
2 mol NH 4
+
= 9.26 + log
132.1 g NH 4 2 SO 4
1 mol NH 4 2 SO 4
K sp [Al3+ ][OH ]3 1.3 1033 (0.075 M)[OH ]3
1.3 1033
2.6 1011 pOH = log 2.6 1011 = 10.59
0.075
pH = 14.00 10.59 = 3.41
3
919
NH 4 +
0.105 M NH3
= 0.081 M
1.3
(M)
(a) Here we calculate OH needed for precipitation.
[OH ]
NH3
= 2.7 g
Chapter 18: Solubility and Complex-Ion Equilibria
(b)
We can use the Henderson–Hasselbalch equation to determine C 2 H 3O 2 .
pH = 3.41 = pKa + log L
C 2 H 3O 2
C 2 H 3O 2
O = 4.74 + log 1.00 M
NM HC2 H 3O 2 QP
C2 H3O2
C2 H3O 2
= 101.33 = 0.047; C H O = 0.047 M
log
= 3.41 4.74 = 1.33;
2 3 2
1.00 M
1.00 M
This situation does not quite obey the guideline that the ratio of concentrations must fall
in the range 0.10 to 10.0, but the resulting error is a small one in this circumstance.
0.047 mol C2 H3O 2 1 mol NaC2 H3O 2 82.03 g NaC2 H3O2
1 L soln
1 mol NaC2 H3O2
1 mol C2 H3O2
0.96 g NaC2 H3O2
mass NaC2 H3O2 = 0.2500 L
50.
(D)
(a)
Since HI is a strong acid, I = 1.05 103 M +1.05 103 M = 2.10 103 M.
We determine the value of the ion product and compare it to the solubility product
constant value.
Qsp = Pb 2+ I = 1.1103 2.10 103 = 4.9 109 7.1109 = K sp for PbI 2
2
2
Thus a precipitate of PbI 2 will not form under these conditions.
(b) We compute the OH needed for precipitation.
Ksp Mg 2 OH 1.8 1011 0.0150 OH ; OH
2
2
1.8 1011
3.5 105 M
0.0150
Then we compute OH in this solution, resulting from the ionization of NH3.
0.05 103 L
= 1.2 104 M
NH3 = 6.00 M
2.50 L
Even though NH 3 is a weak base, the OH produced from the NH 3 hydrolysis reaction
will approximate 4 105 M (3.85 if you solve the quadratic) in this very dilute solution.
(Recall that degree of ionization is high in dilute solution.) And since OH = 3.5 105
b g
M is needed for precipitation to occur, we conclude that Mg OH
this solution (note: not much of a precipitate is expected).
(c)
2
will precipitate from
0.010 M HC2 H 3O 2 and 0.010 M NaC2 H 3O 2 is a buffer solution with pH = pKa of
acetic acid (the acid and its anion are present in equal concentrations.) From this, we
determine the OH .
920
Chapter 18: Solubility and Complex-Ion Equilibria
OH = 109.26 = 5.5 1010
pOH = 14.00 4.74 = 9.26
pH = 4.74
Q = Al3+ OH = 0.010 5.5 10 10 = 1.7 10 30 1.3 10 33 = K sp of Al OH 3
3
3
b g bsg should precipitate from this solution.
Thus, Al OH
3
Complex-Ion Equilibria
51.
(E) Lead(II) ion forms a complex ion with chloride ion. It forms no such complex ion with
nitrate ion. The formation of this complex ion decreases the concentrations of free Pb 2+ aq
b g
b g
and free Cl aq . Thus, PbCl 2 will dissolve in the HCl(aq) up until the value of the solubility
PbCl3 aq
product is exceeded. Pb 2+ aq + 3Cl aq
52.
Zn NH 3
(E) Zn 2+ aq + 4NH 3 aq
4
b g
2+
aq
K f = 4.1108
NH 3 aq will be least effective in reducing the concentration of the complex ion. In fact, the
addition of NH 3 aq will increase the concentration of the complex ion by favoring a shift of
b g
b g
baqg and, thus, increasing
b g
the equilibrium to the right. NH 4 + aq will have a similar effect, but not as direct. NH 3 aq is
formed by the hydrolysis of NH 4 +
NH 4 + will eventually increase
b g
cause the greatest decrease in the concentration of the complex ion. HCl(aq) will react with
NH baq g to decrease its concentration (by forming NH ) and this will cause the complex ion
NH 3 aq + H 3O + aq . The addition of HCl(aq) will
NH 3 aq : NH 4 aq + H 2 O(l)
+
+
3
4
equilibrium reaction to shift left toward free aqueous ammonia and Zn2+(aq).
53.
(E) We substitute the given concentrations directly into the Kf expression to calculate Kf.
Kf
54.
[[Cu(CN) 43]
0.0500
2.0 1030
4
+
[Cu ][CN ]
(6.1 1032 )(0.80) 4
(M) The solution to this problem is organized around the balanced chemical equation. Free
NH 3 is 6.0 M at equilibrium. The size of the equilibrium constant indicates that most
copper(II) is present as the complex ion at equilibrium.
2+
Equation:
Cu 2+ aq + 4NH 3 aq
Cu NH 3 4
aq
b g b g
Initial:
0.10 M
Change(100 % rxn): 0.10 M
Completion:
0M
Changes:
+x M
Equil:
x M
6.00 M
0.40 M
5.60 M
+4x M
5.60 + 4x M
921
b
0M
+0.10 M
0.10 M
x M
0.10 x M
g
Chapter 18: Solubility and Complex-Ion Equilibria
Let’s assume (5.60 + 4x) M ≈ 5.60 M and (0.10 –x) M ≈ 0.10 M
Cu NH 2+
3 4
0.10 x
0.10
0.10
= 1.1 1013
Kf =
4
4
4
2+
(5.60) x 983.4x
Cu NH 3
x 5.60 4 x
x
55.
0.10
9.2 1018 M Cu 2
983.4 (1.1 1013 )
( x << 0.10, thus the approximation was valid)
(M) We first find the concentration of free metal ion. Then we determine the value of Qsp for
the precipitation reaction, and compare its value with the value of Ksp to determine whether
precipitation should occur.
Equation:
Initial:
Changes:
Equil:
b g
Ag + aq +
0M
+x M
x M
2 S2 O3
2
aq
b
0.76 M
+2 x M
Ag S2 O 3
g baqg
3
2
0.048 M
x M
0.048 x M
0.76 + 2 x M
Ag S2 O3 3
2
= 1.7 1013 = 0.048 x 0.048 ; x = 4.9 1015 M = Ag +
Kf =
2
2 2
+
(0.76) 2 x
x 0.76 + 2 x
Ag S2 O3
(x << 0.048 M, thus the approximation was valid.)
Qsp = Ag + I = 4.9 1015 2.0 = 9.8 1015 8.5 1017 = K sp .
Because Qsp K sp , precipitation of AgI(s) should occur.
56.
(M) We need to determine OH in this solution, and also the free Cu 2+ .
pH = pK a + log
NH3
NH 4
+
= 9.26 + log
OH = 104.74 = 1.8 105 M
0.10 M
= 9.26 pOH = 14.00 9.26 = 4.74
0.10 M
Cu NH 3
Cu 2+ aq + 4NH 3 aq
4
2+
aq
Cu NH 3 2+
4
0.015
0.015
= 1.1 1013 =
Kf =
; Cu 2+ =
= 1.4 1011 M
4
13
4
2+
4
2+
1.1
10
0.10
Cu
0.10
Cu NH 3
b g
Now we determine the value of Qsp and compare it with the value of Ksp for Cu OH 2 .
Qsp = Cu 2+ OH = 1.4×1011 1.8×10 = 4.5×1021 < 2.2×1020 (K sp for Cu OH 2 )
Precipitation of Cu OH 2 s from this solution should not occur.
2
5 2
b gbg
922
Chapter 18: Solubility and Complex-Ion Equilibria
57.
(M) We first compute the free Ag + in the original solution. The size of the complex ion
formation equilibrium constant indicates that the reaction lies far to the right, so we form as
much complex ion as possible stoichiometrically.
b g
b g
Equation:
Ag + aq +
2 NH 3 aq
In soln:
Form complex:
0.10 M
0.10 M
0M
+x M
x M
1.00 M
0.20 M
0.80 M
+2x M
0.80 + 2x M
Changes:
Equil:
b
b g baqg
Ag NH 3
b
g
Ag NH
3 2
0.10 x
0.10
=
K f = 1.6 10 =
2
2
2
+
Ag NH3
x 0.80 + 2 x
x 0.80
+
2
0M
+0.10 M
0.10 M
x M
0.10 x M
g
+
7
x=
0.10
= 9.8 109 M .
2
7
1.6 10 0.80
(x << 0.80 M, thus the approximation was valid.)
Thus, Ag + = 9.8 109 M. We next determine the I that can coexist in this solution
without precipitation.
8.5 1017
+
17
9
I =
K sp = Ag I = 8.5 10 = 9.8 10 I ;
= 8.7 109 M
9
9.8 10
Finally, we determine the mass of KI needed to produce this I
mass KI = 1.00 L soln
58.
8.7 109 mol I 1mol KI 166.0 g KI
= 1.4 106 g KI
1 L soln
1mol I
1mol KI
(M) First we determine Ag + that can exist with this Cl . We know that Cl will be
unchanged because precipitation will not be allowed to occur.
1.8 1010
= 1.8 109 M
0.100
We now consider the complex ion equilibrium. If the complex ion’s final concentration is x ,
then the decrease in NH 3 is 2x , because 2 mol NH 3 react to form each mole of complex
K sp = Ag + Cl = 1.8 1010 = Ag + 0.100 M;
Ag + =
Ag NH 3 aq We can solve the Kf
ion, as follows. Ag + aq + 2NH 3 aq
2
+
expression for x.
Kf = 1.6 107 =
c
hc
b g
Ag NH 3
+
2
Ag + LMN NH 3 OPQ
hb
2
=
x
b
1.8 109 1.00 2 x
x = 1.6 107 1.8 109 1.00 2 x
g
2
c
g
2
h
= 0.029 1.00 4.00 x + 4.00 x 2 = 0.029 0.12 x + 0.12 x 2
0 = 0.029 1.12 x + 0.12 x 2 We use the quadratic formula roots equation to solve for x.
x=
b b 2 4ac 1.12 (1.12) 2 4 0.029 0.12 1.12 1.114
=
=
= 9.3, 0.025
2a
2 0.12
0.24
923
Chapter 18: Solubility and Complex-Ion Equilibria
Thus, we can add 0.025 mol AgNO 3 (~ 4.4 g AgNO 3 ) to this solution before we see a
precipitate of AgCl(s) form.
Precipitation and Solubilities of Metal Sulfides
59.
(M) We know that Kspa = 3 107 for MnS and Kspa = 6 102 for FeS. The metal sulfide will
begin to precipitate when Qspa = Kspa . Let us determine the H 3O + just necessary to form each
precipitate. We assume that the solution is saturated with H 2S, H 2S = 0.10 M .
K spa
[M 2 ][H 2S]
[H 3O + ]2
H 3O + =
[M 2 ][H 2S]
(0.10 M)(0.10 M)
1.8 105 M for MnS
[H 3O ]
7
3 10
K spa
+
(0.10 M)(0.10 M)
4.1103 M for FeS
2
6 10
Thus, if the [H3O+] is maintained just a bit higher than 1.8 105 M , FeS will precipitate and
Mn 2+ aq will remain in solution. To determine if the separation is complete, we see whether
b g
Fe
2+
has decreased to 0.1% or less of its original value when the solution is held at the
aforementioned acidity. Let H 3O + = 2.0 105 M and calculate Fe 2+ .
Fe 2+ H 2S
Fe 2+ 0.10 M
6 102 2.0 105 = 2.4 106 M
2
2+
=
=
6
10
=
;
Fe
=
2
2
0.10
H 3O +
2.0 105 M
2
K spa
2.4 106 M
% Fe aq remaining =
100% = 0.0024%
0.10 M
2+
60.
Separation is complete.
(M) Since the cation concentrations are identical, the value of Qspa is the same for each one. It
is this value of Qspa that we compare with Kspa to determine if precipitation occurs.
Qspa
M 2+ H 2S 0.05 M 0.10 M
=
=
= 5 101
2
+ 2
0.010 M
H 3O
If Qspa Kspa , precipitation of the metal sulfide should occur. But, if Qspa Kspa , precipitation
will not occur.
For CuS, Kspa = 6 10 16 Qspa = 5 101
Precipitation of CuS(s) should occur.
For HgS, Kspa = 2 1032 Qspa = 5 101
Precipitation of HgS(s) should occur.
For MnS, Kspa = 3 107 Qspa = 5 101
Precipitation of MnS(s) will not occur.
924
Chapter 18: Solubility and Complex-Ion Equilibria
61.
(M)
(a) We can calculate H 3O + in the buffer with the Henderson–Hasselbalch equation.
C 2 H 3O 2
0.15 M
pH = pK a + log
= 4.74 + log
= 4.52 H 3O + = 10 4.52 = 3.0 10 5 M
HC H O
0.25 M
2
3
2
We use this information to calculate a value of Qspa for MnS in this solution and then
comparison of Qspa with Kspa will allow us to decide if a precipitate will form.
Qspa
Mn 2+ H 2S 0.15 0.10
=
=
= 1.7 107 3 107 = K spa for MnS
+ 2
5 2
H 3O
3.0 10
Thus, precipitation of MnS(s) will not occur.
(b)
We need to change H 3O + so that
Qspa = 3 107 =
0.15 0.10
H 3 O
+
2
; H 3 O + =
(0.15) 0.10
3 10
7
[H 3 O + ] = 2.2 105 M
pH = 4.66
This is a more basic solution, which we can produce by increasing the basic component
of the buffer solution, namely, the acetate ion. We can find out the necessary acetate ion
concentration with the Henderson–Hasselbalch equation.
[C 2 H 3O 2 ]
[C2 H 3O 2 ]
4.66 4.74 log
pH pK a log
[HC2 H 3O 2 ]
0.25 M
log
[C2 H 3O 2 ]
4.66 4.74 0.08
0.25 M
C 2 H 3O 2
= 100.08 = 0.83
0.25 M
62.
C 2 H 3O 2 = 0.83 0.25 M = 0.21M
(M)
(a) CuS is in the hydrogen sulfide group of qualitative analysis. Its precipitation occurs
when 0.3 M HCl is saturated with H 2S. It will certainly precipitate from a (non-acidic)
saturated solution of H2S which has a much higher S2 .
Cu 2+ aq + H 2S satd aq
CuS s + 2H + aq
This reaction proceeds to an essentially quantitative extent in the forward direction.
(b)
MgS is soluble, according to the solubility rules listed in Chapter 5.
0.3 M HCl
Mg 2+ aq + H 2S satd aq
no reaction
(c)
As in part (a), PbS is in the qualitative analysis hydrogen sulfide group, which
precipitates from a 0.3 M HCl solution saturated with H 2S. Therefore, PbS does not
dissolve appreciably in 0.3 M HCl. PbS s + HCl 0.3M
no reaction
925
Chapter 18: Solubility and Complex-Ion Equilibria
(d)
Since ZnS(s) does not precipitate in the hydrogen sulfide group, we conclude that it is
soluble in acidic solution.
ZnS s + 2HNO3 aq
Zn NO3 2 aq + H 2S g
Qualitative Cation Analysis
63.
(E) The purpose of adding hot water is to separate Pb 2+ from AgCl and Hg 2 Cl2 . Thus, the
most important consequence would be the absence of a valid test for the presence or absence
of Pb 2+ . In addition, if we add NH 3 first, PbCl 2 may form Pb OH 2 . If Pb OH 2 does
b g
form, it will be present with Hg 2 Cl 2 in the solid, although Pb OH
b g
2
will not darken with
+
added NH 3 . Thus, we might falsely conclude that Ag is present.
64.
(M) For PbCl2 aq , 2 Pb 2+ = Cl where s = molar solubility of PbCl 2 .
Thus s = Pb 2+ .
3
Ksp [Pb 2+ ][Cl ]2 ( s )(2s ) 2 4s 3 1.6 105 ; s 1.6 105 4 1.6 102 M = [Pb 2+ ]
Both Pb 2+ and CrO 4 2 are diluted by mixing the two solutions.
1.00 mL
Pb 2+ = 0.016 M
= 0.015 M
1.05 mL
0.05 mL
CrO 4 2 = 1.0 M
= 0.048 M
1.05 mL
2
Qsp = Pb 2+ CrO 4 = 0.015 M 0.048 M = 7.2 104 2.8 1013 = K sp
Thus, precipitation should occur from the solution described.
65.
(E)
(a)
2+
Ag + and/or Hg 2 are probably present. Both of these cations form chloride precipitates
from acidic solutions of chloride ion.
(b)
We cannot tell whether Mg 2+ is present or not. Both MgS and MgCl 2 are water soluble.
(c)
Pb 2+ possibly is absent; it is the only cation of those given which forms a precipitate in
an acidic solution that is treated with H 2S, and no sulfide precipitate was formed.
(d)
We cannot tell whether Fe2+ is present. FeS will not precipitate from an acidic solution that
is treated with H 2S; the solution must be alkaline for a FeS precipitate to form.
(a) and (c) are the valid conclusions.
66.
(E)
(a)
Pb 2+ aq + 2Cl aq
PbCl2 s
(b)
Zn OH 2 s + 2OH aq
Zn OH 4
(c)
Fe OH 3 s + 3H 3O + aq
Fe3+ aq + 6H 2 O(l) or Fe H 2O 6
926
2
aq
3+
aq
Chapter 18: Solubility and Complex-Ion Equilibria
(d)
Cu 2+ aq + H 2S aq
CuS s + 2H + aq
INTEGRATIVE AND ADVANCED EXERCISES
67. (M) We determine s, the solubility of CaSO4 in a saturated solution, and then the concentration of
CaSO4 in ppm in this saturated solution, assuming that the solution’s density is 1.00 g/mL.
Ksp [Ca 2 ][SO 4 2 ] ( s )( s ) s 2 9.1 106
s 3.0 103 M
1 mL 1 L soln 0.0030 mol CaSO4 136.1 g CaSO 4
4.1 102 ppm
1.00 g 1000 mL
1 L soln
1 mol CaSO 4
Now we determine the volume of solution remaining after we evaporate the 131 ppm CaSO4
down to a saturated solution (assuming that both solutions have a density of 1.00 g/mL.)
10 6 g sat' d soln
1 mL
volume sat' d soln 131 g CaSO 4
3.2 10 5 mL
2
4.1 10 g CaSO 4 1.00 g
Thus, we must evaporate 6.8 × 105 mL of the original 1.000 × 106 mL of solution, or 68% of the
water sample.
ppm CaSO 4 106 g soln
68. (M)
(a) First we compute the molar solubility, s, of CaHPO42H2O.
0.32 g CaHPO 4 2H 2 O 1 mol CaHPO 4 2H 2 O
s
1.9 103 M
1 L soln
172.1 g CaHPO 4 2H 2 O
K sp [Ca 2 ][HPO 4 2 ] ( s )( s ) s 2 (1.9 103 ) 2 3.6 106
Thus the values are not quite consistent.
(b) The value of Ksp given in the problem is consistent with a smaller value of the molar
solubility. The reason is that not all of the solute ends up in solution as HPO42– ions.
HPO42– can act as either an acid or a base in water (see below), but it is base hydrolysis that
predominates.
H 3O (aq) PO43 (aq)
.
HPO 4 2 (aq) H 2 O(l)
K a 3 4.2 1013
OH (aq) H 2 PO 4 (aq)
HPO 4 2 (aq) H 2 O(l)
Kb
Kw
1.6 10 7
K2
69. (M) The solutions mutually dilute each other and, because the volumes are equal, the
concentrations are halved in the final solution: [Ca2+] = 0.00625 M, [SO42–] = 0.00760 M. We
cannot assume that either concentration remains constant during the precipitation. Instead, we
assume that precipitation proceeds until all of one reagent is used up. Equilibrium is reached
from that point.
927
Chapter 18: Solubility and Complex-Ion Equilibria
Equation: CaSO 4 (s)
Ca 2 (aq)
In soln
0.00625 M
Form ppt
0.00625 M
Not at equil
0M
Changes
xM
Equil:
xM
SO 4 2 (aq) K sp 9.1 10 6
0.00760 M
K sp [Ca 2 ][SO 4 2 ]
0.00625 M K sp ( x )(0.00135 x)
0.00135 M
K sp 0.00135 x
xM
x
(0.00135 x ) M
9.1 106
6.7 103 M
0.00135
{not a reasonable assumption!}
Solving the quadratic 0 = x2 + (1.35 × 10-2)x – 9.1 × 10-6 yields x = 2.4 × 10-3.
2.4×10-3 M final
% unprecipitated Ca =
×100% = 38% unprecipitated
0.00625 M initial
2+
70. (M) If equal volumes are mixed, each concentration is reduced to one-half of its initial value.
We assume that all of the limiting ion forms a precipitate, and then equilibrium is achieved by
the reaction proceeding in the forward direction to a small extent.
Ba 2 (aq)
BaCO3 (s)
Orig. soln:
0.00050 M
Form ppt:
0M
Equation:
Changes:
xM
Equil:
xM
CO32 (aq)
0.0010 M
0.0005 M
xM
(0.0005 x)M
K sp [Ba 2 ][CO32 ] 5.1109 (x)(0.0005 x) 0.0005 x
5.1 109
1 105 M The [Ba 2+ ] decreases from 5 × 10-4 M to 1 × 10-5 M.
0.0005
(x < 0.0005 M, thus the approximation was valid.)
x
% unprecipitated
1 10 5 M
100% 2%
5 10 4 M
Thus, 98% of the Ba2+ is precipitated as BaCO3(s).
71. (M) The pH of the buffer establishes [H3O+] = 10–pH = 10–3.00 = 1.0 × 10–3 M.
Now we combine the two equilibrium expressions, and solve the resulting expression for the
molar solubility of Pb(N3)2 in the buffer solution.
928
Chapter 18: Solubility and Complex-Ion Equilibria
2
Pb(N 3 ) 2 (s)
Pb (aq) 2 N 3 (aq)
K sp 2.5 10 9
2 HN 3 (aq) 2 H 2O(l)
2 H 3O (aq) 2 N 3 (aq)
1/K a 2
2
Pb(N 3 ) 2 (s) 2 H 3O (aq)
Pb (aq) 2 HN 3 (aq) 2 H 2 O (l)
Pb(N 3 ) 2 (s) 2 H 3O (aq)
1
2.8 109
5 2
(1.9 10 )
K K sp (1/ K a 2 ) 7.0
Pb 2 (aq) 2 HN 3 (aq) 2 H 2 O (l)
Buffer:
0.0010 M
0M
0M
Dissolving:
(buffer)
x M
2x M
Equilibrium:
0.0010 M
xM
2x M
K
[Pb 2 ][HN 3 ]2
[H 3O ]2
7.0
(x) (2 x) 2
(0.0010) 2
4x 3 7.0 (0.0010) 2
x
3
7.0(0.0010) 2
4
0.012 M
Thus, the molar solubility of Pb(N3)2 in a pH = 3.00 buffer is 0.012 M.
72.
(M) We first determine the equilibrium constant of the suggested reaction.
Mg 2 (aq) 2 OH (aq)
Mg(OH)2 (s)
K sp 1.8 1011
2 NH3 (aq) 2 H 2 O(l) 1/K b 2 1/(1.8 105 )2
2 NH 4 (aq) 2 OH (aq)
Mg 2 (aq) 2 NH3 (aq) 2 H 2 O(l)
Mg(OH)2 (s) 2 NH 4 (aq)
Initial:
1.00 M
0M
0M
Changes:
Equil:
K sp
K
x
Kb2
3
2 x M
(1.00 2 x) M
x M
xM
2x M
2x M
[Mg 2 ][NH 3 ]2
x(2 x) 2
1.8 1011
4 x3
0.056
(1.8 105 ) 2
[NH 4 ]2
(1.00 2 x) 2 1.002
0.056
0.24 M
4
Take this as a first approximation and cycle through again. 2x = 0.48
4x 3
K
0.056
(1.00 0.48) 2
x
3
0.056 (1.00 0.48) 2
0.16 M
4
Yet another cycle gives a somewhat more consistent value. 2x = 0.32
K
4x3
0.056
(1.00 0.32) 2
x
3
0.056 (1.00 0.38) 2
0.19 M
4
Another cycle gives more consistency. 2x = 0.38
929
Chapter 18: Solubility and Complex-Ion Equilibria
K
4x3
0.056
(1.00 0.38) 2
x
3
0.056 (1.00 0.38) 2
0.18 M [Mg 2 ]
4
The molar solubility of Mg(OH)2 in a 1.00 M NH4Cl solution is 0.18 M.
73. (D) First we determine the value of the equilibrium constant for the cited reaction.
Ca 2 (aq) CO32 (aq)
CaCO3 (s)
Ksp 2.8 109
H 2 O(l) HCO3 (aq)
H 3 O (aq) CO32 (aq)
1/K a 2 1/ 4.7 1011
Ca 2 (aq) HCO3 (aq) H 2 O(l)
CaCO3 (aq) H 3 O (aq)
K overall K sp 1/K a 2
K overall
2.8 109
60
4.7 1011
Ca 2 (aq) HCO3 (aq) H 2 O(l)
Equation: CaCO3 (s) H 3O (aq)
Initial:
Change:
10 pH
(buffer )
0M
x M
0M
x M
Equil:
10 pH
xM
xM
In the above set-up, we have assumed that the pH of the solution did not change because of the
dissolving of the CaCO3(s). That is, we have treated the rainwater as if it were a buffer solution.
(a) K
[Ca 2 ][HCO 3 ]
[H 3 O ]
60.
x2
3 10 6
x 60 3 10 6 1 10 2 M [Ca 2 ]
[Ca 2 ][HCO3 ]
x2
(b) K
60.
[H 3 O ]
6.3 105
x 60 6.3 105 6.1 102 M [Ca 2 ]
74. We use the Henderson-Hasselbalch equation to determine [H3O+] in this solution, and then use the
Ksp expression for MnS to determine [Mn2+] that can exist in this solution without precipitation
occurring.
930
Chapter 18: Solubility and Complex-Ion Equilibria
[C2 H 3 O 2 ]
0.500 M
pH pK a log
4.74 log
4.74 0.70 5.44
[HC2 H 3 O 2 ]
0.100 M
[H 3 O ] 105.44 3.6 106 M
Mn 2 (aq) H 2 S(aq) 2 H 2 O(l)
MnS(s) 2 H 3 O (aq)
K spa 3 107
Note that [H 2 S] [Mn 2 ] s, the molar solubility of MnS.
K spa
[Mn 2 ][H 2 S]
s2
7
3 10
[H 3 O ]2
(3.6 106 M) 2
mass MnS/L
s 0.02 M
0.02 mol Mn 2 1 mol MnS 87 g MnS
2 g MnS/L
1 L soln
1 mol Mn 2 1 mol MnS
75. (M)
(a) The precipitate is likely Ca3(PO4)2. Let us determine the %Ca of this compound (by mass)
%Ca
3 40.078 g Ca
100% 38.763% Ca
310.18 g Ca 3 (PO 4 ) 2
2
3 Ca 2 (aq) 2 HPO 4 (aq)
Ca 3 (PO 4 ) 2 (s) 2 H (aq)
(b) The bubbles are CO 2 (g) :
CO 2 (g) H 2 O(l)
“H 2 CO 3 (aq)”
Ca 2 (aq) H 2 CO3 (aq)
CaCO3 (s) 2 H (aq)
CaCO3 (s) + H 2 CO3 (aq)
Ca 2+ (aq) + 2 HCO3- (aq)
76.
precipitation
redissolving
(D)
(a) A solution of CO2(aq) has [CO32- ] = Ka[H2CO3] = 5.6 × 10-11. Since Ksp = 2.8 × 10-9 for
CaCO3, the [Ca2+] needed to form a precipitate from this solution can be computed.
2.8×10-9
[Ca ] =
=
= 50.M
[CO32- ] 5.6×10-11
2+
K sp
This is too high to reach by dissolving CaCl2 in solution. The reason why the technique works
is because the OH-(aq) produced by Ca(OH)2 neutralizes some of the HCO3-(aq) from the
ionization of CO2(aq), thereby increasing the [CO32-] above a value of 5.6 × 10-11.
(b) The equation for redissolving can be obtained by combining several equations.
931
Chapter 18: Solubility and Complex-Ion Equilibria
Ca 2+ (aq) + CO32- (aq)
CaCO3 (s)
K sp = 2.8×10-9
HCO3- (aq) + H 2O(l)
CO32- (aq) + H 3O + (aq)
1
1
=
K a 2 5.6×10-11
HCO3- (aq) +H 3O + (aq)
CO 2 (aq) + 2 H 2O(l)
K a1 = 4.2×10-7
Ca 2+ (aq) + 2 HCO3- (aq)
CaCO3 (s) + CO 2 (aq) + H 2 O(l)
K=
-9
2+
-7
K sp × K a1
Ka 2
- 2
[Ca ][HCO3 ]
(2.8×10 )( 4.2×10 )
= 2.1×10-5 =
-11
5.6×10
[CO 2 ]
2+
If CaCO3 is precipitated from 0.005 M Ca (aq) and then redissolved, [Ca2+] = 0.005 M and
[HCO32-] = 2×0.005 M = 0.010 M. We use these values in the above expression to compute
[CO2].
[Ca 2+ ][HCO3- ]2 (0.005)(0.010) 2
[CO 2 ]=
=
= 0.02 M
2.1×10-5
2.1×10-5
We repeat the calculation for saturated Ca(OH)2, in which [OH-] = 2 [Ca2+], after first
determining [Ca2+] in this solution.
K=
Ksp = [Ca2+][OH-]2 = 4 [Ca2+] = 5.5×10-6
[Ca2+]
3
5.5 106
= 0.011
4
[Ca 2+ ][HCO3- ]2 (0.011)(0.022) 2
=
= 0.25 M
M [CO 2 ]=
2.1×10-5
2.1×10-5
Thus, to redissolve the CaCO3 requires that [CO2] = 0.25 M if the solution initially is saturated
Ca(OH)2, but only 0.02 M CO2 if the solution initially is 0.005 M Ca(OH)2(aq).
A handbook lists the solubility of CO2 as 0.034 M. Clearly the CaCO3 produced cannot
redissolve if the solution was initially saturated with Ca(OH)2(aq).
77.
(M)
(a)
2
2
BaSO 4 (s)
Ba (aq) SO 4 (aq)
K sp 1.1 10-10
1
1
=
Ksp 5.1 10-9
Ba 2 (aq) CO 3 2- (aq)
BaCO 3 (s)
2
Sum BaSO 4 (s) +CO 3 2- (aq)
BaCO 3 (s) + SO 4 (aq)
Initial
Equil.
3M
(3-x )M
K overall
1.1 10-10
5.1 10-9
0.0216
0
x
(where x is the carbonate used up in the reaction)
K
[SO 4 2- ]
x
0.0216 and x 0.063
2[CO 3 ] 3 x
Since 0.063 M 0.050 M, the response is yes.
932
Chapter 18: Solubility and Complex-Ion Equilibria
2Ag + (aq) + 2 Cl- (aq)
2AgCl(s)
(b)
K sp = (1.8 10-10 ) 2
Ag 2 CO3 (s)
2 Ag + (aq) + CO32- (aq)
1/ K sp =
Ag 2 CO3 (s) + 2 Cl- (aq)
Sum 2AgCl(s) +CO32- (aq)
K overall =
1
8.5 10-12
(1.8 10-10 )2
=3.8 10-9
8.5 10-12
Initial
3M
0M
Equil.
(3-x )M
2x
(where x is the carbonate used up in the reaction)
K overall =
[Cl- ]2 (2x) 2
=
= 3.8 10-9 and x = 5.3 10-5 M
2[CO3 ] 3-x
Since 2x , (2(5.35 10-5 M)) , << 0.050 M, the response is no.
K sp 3.7 10-8
Mg 2+ (aq) 2 F- (aq)
MgF2 (s)
(c)
MgCO3 (s)
Mg 2+ (aq) CO32- (aq)
1/ K sp
MgCO3 (s) 2 F- (aq)
MgF2 (s) CO32- (aq)
sum
initial
equil.
3M
(3-x )M
(where x is the carbonate used up in the reaction)
K overall
K overall
1
3.5 10-8
3.7 10-8
1.1
3.5 10-8
0M
2x
[F ]2
(2x) 2
1.1 and x 0.769M
[CO32- ] 3 x
Since 2x , (2(0.769M)) 0.050 M, the response is yes.
78. (D) For ease of calculation, let us assume that 100 mL, that is, 0.100 L of solution, is to be
titrated. We also assume that [Ag+] = 0.10 M in the titrant. In order to precipitate 99.9% of the
Br– as AgBr, the following volume of titrant must be added.
volume titrant 0.999 100 mL
0.010 mmol Br 1 mmol Ag
1 mL sample
1 mmol Br
1 mL titrant
9.99 mL
0.10 mmol Ag
We compute [Ag+] when precipitation is complete.
[Ag ]
K sp
[Br ] f
5.0 10 13
5.0 10 8 M
0.001 0.01 M
Thus, during the course of the precipitation of AgBr, while 9.99 mL of titrant is added, [Ag+]
increases from zero to 5.0 × 10–8 M. [Ag+] increases from 5.0 × 10–8 M to 1.0 × 10–5 M from
the point where AgBr is completely precipitated to the point where Ag2CrO4 begins to
precipitate. Let us assume, for the purpose of making an initial estimate, that this increase in
[Ag+] occurs while the total volume of the solution is 110 mL (the original 100 mL plus 10 mL
933
Chapter 18: Solubility and Complex-Ion Equilibria
of titrant needed to precipitate all the AgBr.) The amount of Ag+ added during this increase is
the difference between the amount present just after AgBr is completely precipitated and the
amount present when Ag2CrO4 begins to precipitate.
1.0 10 5 mmol Ag
5.0 10 8 mmol Ag
110 mL
amount Ag added 110 mL
1 mL soln
1 mL soln
3
1.1 10 mmol Ag
1 mL titrant
volume added titrant 1.1 10 3 mmol Ag
1.1 10 2 mL titrant
0.10 mmol Ag
We see that the total volume of solution remains essentially constant. Note that [Ag+] has
increased by more than two powers of ten while the volume of solution increased a very small
degree; this is indeed a very rapid rise in [Ag+], as stated.
79. (D) The chemistry of aluminum is complex, however, we can make the following assumptions.
Consider the following reactions at various pH:
pH > 7
K K
K
= 1.43
overall
f sp
Al(OH)4-(aq)
Al(OH)3(s) + OH-
K = 1.31033
sp
3+
pH < 7 Al(OH)3(s)
Al (aq) + 3 OH
For the solubilities in basic solutions, consider the following equilibrium at pH = 13.00 and 11.00.
At pH = 13.00:
K K K
Initial :
–
Change: –
Equilibrium:
Kf = 1.43 =
= 1.43
ov
f sp
Al(OH)4-(aq)
Al(OH)3(s) + OH-
–
0.10
0
–x
+x
0.10 – x
x
x
x
(buffered at pH = 13.00) x = 0.143 M
0.10 x
0.10
At pH = 11.00:
Kf = 1.43 =
x
x
(buffered at pH = 11) x = 0.00143 M
0.0010 x
0.0010
Thus, in these two calculations we see that the formation of the aluminate ion [Al(OH)4] is
favored at high pH values. Note that the concentration of aluminum(III) increases by a factor of
one-hundred when the pH is increased from 11.00 to 13.00, and it will increase further still
above pH 13. Now consider the solubility equilibrium established at pH = 3.00, 4.00, and 5.00.
These correspond to [OH] = 1.0 1011 M, 1.0 1010 M, and 1.0 109 M, respectively.
K = 1.31033
sp
3+
Al(OH)3(s)
Al (aq) + 3 OH (aq)
At these pH values, [Al3+] in saturated aqueous solutions of Al(OH)3 are
[Al3+] = 1.3 1033 /(1.0 1011)3 = 1.3 M at pH = 3.00
934
Chapter 18: Solubility and Complex-Ion Equilibria
[Al3+] = 1.3 1033 /(1.0 1010)3 = 1.3 103 M at pH = 4.00
[Al3+] = 1.3 1033 /(1.0 109)3 = 1.3 106 M at pH = 5.00
These three calculations show that in strongly acidic solutions the concentration of
aluminum(III) can be very high and is certainly not limited by the precipitation of Al(OH)3(s).
However, this solubility decreases rapidly as the pH increases beyond pH 4.
At neutral pH 7, a number of equilibria would need to be considered simultaneously, including
the self-ionization of water itself. However, the results of just these five calculations do
indicate that concentration of aluminum(III) increases at both low and high pH values, with a
very low concentration at intermediate pH values around pH 7, thus aluminum(III)
concentration as a function of pH is a U-shaped (or V-shaped) curve.
80. (M) We combine the solubility product expression for AgCN(s) with the formation expression
for [Ag(NH3)2]+(aq).
Ag (aq) CN (aq)
AgCN(s)
Solubility:
Formation:
[Ag(NH 3 ) 2 ] (aq)
Ag (aq) 2NH 3 (aq)
K sp ?
K f 1.6 107
[Ag(NH3 )2 ] CN (aq) K overall K sp K f
Net reaction: AgCN(s) 2 NH3 (aq)
K overall
K sp
[[Ag(NH 3 ) 2 ] ][CN ] (8.8 106 ) 2
1.9 109 K sp 1.6 107
[NH 3 ]2
(0.200) 2
1.9 109
1.2 1016
1.6 107
Because of the extremely low solubility of AgCN in the solution, we assumed that [NH3] was
not altered by the formation of the complex ion.
81.
We combine the solubility product constant expression for CdCO3(s) with the formation
expression for [CdI4]2–(aq).
Solubility:
Cd 2 (aq) CO32 (aq)
CdCO3 (s)
K sp 5.2 1012
[CdI 4 ]2 (aq)
Formation: Cd 2 (aq) 4 I (aq)
Kf ?
[CdI4 ]2 (aq) CO32 (aq)
Net reaction: CdCO3 (s) 4 I (aq)
K overall K sp K f
K overall
Kf
[[CdI 4 ]2 ][CO32 ] (1.2 103 ) 2
1.4 106 5.2 1012 K f
4
4
[I ]
(1.00)
1.4 106
2.7 105
12
5.2 10
935
Chapter 18: Solubility and Complex-Ion Equilibria
Because of the low solubility of CdCO3 in this solution, we assumed that [I–] was not
appreciably lowered by the formation of the complex ion.
82. (M) We first determine [Pb2+] in this solution, which has [Cl-] = 0.10 M.
K sp 1.6 105 [Pb 2 ][Cl ]2
[Pb 2 ]
K sp
[Cl ]2
1.6 105
1.6 103 M
2
(0.10)
Then we use this value of [Pb2+] in the Kf expression to determine [[PbCl3]–].
Kf
[[PbCl 3 ] ]
[Pb 2 ][Cl ] 3
24
[[PbCl 3 ] ]
(1.6 10 3 )(0.10) 3
[[PbCl 3 ] ]
1.6 10 6
[[PbCl 3 ] ] 24 1.6 10 6 3.8 10 5 M
The solubility of PbCl2 in 0.10 M HCl is the following sum.
solubility = [Pb2+] + [[PbCl3]–] = 1.6 × 10–3 M + 3.8 × 10–5 M = 1.6 × 10–3 M
83. (M) Two of the three relationships needed to answer this question are the two solubility product
4.0 × 10–7 = [Pb2+][S2O32–]
expressions.
1.6 × 10–8 = [Pb2+][SO42–]
The third expression required is the electroneutrality equation, which states that the total
positive charge concentration must equal the total negative charge concentration:
[Pb2+] = [SO42–] + [S2O32–], provided [H3O+] = [OH-].
Or, put another way, there is one Pb2+ ion in solution for each SO42– ion and for each S2O32– ion.
We solve each of the first two expressions for the concentration of each anion, substitute these
expressions into the electroneutrality expression, and then solve for [Pb2+].
[SO 4
2
1.6 108
]
[Pb 2 ]
[S2 O3
2
4.0 107
]
[Pb 2 ]
[Pb 2 ]2 1.6 108 4.0 107 4.2 107
84.
1.6 108 4.0 107
[Pb ]
[Pb 2 ]
[Pb 2 ]
2
[Pb 2 ] 4.2 107 6.5 104 M
(D) The chemical equations are:
Pb 2 (aq) 2 Cl (aq) and
PbCl 2 (s)
Pb 2 (aq) 2 Br (aq)
PbBr2 (s)
The two solubility constant expressions, [Pb2+][Cl–]2 = 1.6 × 10–5, [Pb2+][Br–]2 = 4.0 × 10–5 and
the condition of electroneutrality, 2 [Pb2+] = [Cl–] + [Br–], must be solved simultaneously to
determine [Pb2+]. (Rearranging the electroneutrality relationship, one obtains:
[Pb 2 ] 12 [Cl ] 12 [Br ] .) First we solve each of the solubility constant expressions for the
concentration of the anion. Then we substitute these values into the electroneutrality expression
and solve the resulting equation for [Pb2+].
[Cl ]
1.6 10 5
[Pb 2 ]
[Br ]
4.0 10 5
[Pb 2 ]
2 [Pb 2 ]
1.6 10 5
4.0 10 5
[Pb 2 ]
[Pb 2 ]
2 [Pb 2 ]3 1.6 10 5 4.0 10 5 4.0 10 3 6.3 10 3 1.03 10 2
[Pb 2 ]3 5.2 10 3
[Pb 2 ]3 2.7 10 5
936
[Pb 2 ] 3.0 10 2 M
Chapter 18: Solubility and Complex-Ion Equilibria
85. (D) (a) First let us determine if there is sufficient Ag2SO4 to produce a saturated solution, in
which [Ag+] = 2 [SO42–].
2
2
2
K sp 1.4 10 5 [Ag ] 2 [SO 4 ] 4 [SO 4 ]3 [SO 4 ] 3
2
[SO 4 ]
1.4 10 5
0.015 M
4
2
2.50 g Ag 2 SO 4 1 mol Ag 2 SO 4
1 mol SO 4
0.0535 M
0.150 L
311.8 g Ag 2 SO 4 1 mol Ag 2 SO 4
Thus, there is more than enough Ag2SO4 present to form a saturated solution. Let us now see if
AgCl or BaSO4 will precipitate under these circumstances. [SO42–] = 0.015 M and [Ag+] = 0.030
M.
Q = [Ag+][Cl–] = 0.030 × 0.050 = 1.5 × 10–3 > 1.8 × 10–10 = Ksp, thus AgCl should precipitate.
Q = [Ba2+][SO42–] = 0.025 × 0.015 = 3.8 × 10–4 > 1.1 × 10–10 = Ksp, thus BaSO4 should precipitate.
Thus, the net ionic equation for the reaction that will occur is as follows.
BaSO 4 (s) 2 AgCl(s)
Ag 2 SO 4 (s) Ba 2 (aq) 2 Cl (aq)
(b)
Let us first determine if any Ag2SO4(s) remains or if it is all converted to BaSO4(s) and
AgCl(s). Thus, we have to solve a limiting reagent problem.
amount Ag 2 SO 4 2.50 g Ag 2 SO 4
amount BaCl 2 0.150 L
1 mol Ag 2 SO 4
8.02 10 3 mol Ag 2 SO 4
311.8 g Ag 2 SO 4
0.025 mol BaCl 2
3.75 10 3 mol aCl 2
1 L soln
Since the two reactants combine in a 1 mole to 1 mole stoichiometric ratio, BaCl2 is the limiting
reagent. Since there must be some Ag2SO4 present and because Ag2SO4 is so much more soluble
than either BaSO4 or AgCl, we assume that [Ag+] and [SO42–] are determined by the solubility of
Ag2SO4. They will have the same values as in a saturated solution of Ag2SO4.
[SO42–] = 0.015 M
[Ag+] = 0.030 M
We use these values and the appropriate Ksp values to determine [Ba2+] and [Cl–].
[Ba 2 ]
1.1 10 10
7.3 10 9 M
0.015
[Cl ]
1.8 10 10
6.0 10 9 M
0.030
Since BaCl2 is the limiting reagent, we can use its amount to determine the masses of
BaSO4 and AgCl.
mass BaSO 4 0.00375 mol BaCl 2
mass AgCl 0.00375 mol BaCl 2
1 mol BaSO 4 233.4 g BaSO 4
0.875 g BaSO 4
1 mol BaCl 2
1 mol BaSO 4
2 mol AgCl 143.3 g AgCl
1.07 g AgCl
1 mol BaCl 2 1 mol AgCl
937
Chapter 18: Solubility and Complex-Ion Equilibria
The mass of unreacted Ag2SO4 is determined from the initial amount and the amount that
reacts with BaCl2.
1 mol Ag 2 SO 4
mass Ag 2 SO4 0.00802 mol Ag 2 SO4 0.00375 mol BaCl2
1 mol BaCl2
mass Ag 2 SO4 1.33 g Ag 2 SO4 unreacted
311.8 g Ag 2 SO4
1 mol Ag 2 SO 4
Of course, there is some Ag2SO4 dissolved in solution. We compute its mass.
mass dissolved Ag 2SO 4 0.150 L
0.0150 mol Ag 2SO 4 311.8 g Ag 2SO 4
0.702 g dissolved
1 L soln
1 mol Ag 2SO 4
mass Ag 2SO 4 (s) 1.33 g 0.702 g 0.63 g Ag 2SO 4 (s)
86.
(M) To determine how much NaF must be added, determine what the solubility of BaF2 is
considering that the final concentration of F– must be 0.030 M.
Ksp = 1×10-6.
BaF2 (s)
Ba2+ (aq) +
0
+s
s
2F– (aq)
0
+2s
2s → (0.030 M)
K sp 1 106. s 0.030 s 9.0 104 ; Therefore, s 0.0011 M .
2
Since s is 0.0011 M, the F– contribution from BaF2 considering the common ion effect exerted
by NaF is 0.0022 M. Therefore, the total number of moles of NaF in a 1 L solution required is
therefore 0.0278 moles.
To determine how much BaF2 precipitates, we must first determine what the value of s is for a
saturated solution in absence of the common ion effect. That is,
2
K sp 1 106. s 2 s 4s 3 ; Therefore, s 0.0063 M .
Since in absence of the common ion effect from NaF the solubility of BaF2 is 0.0063 M, and
with NaF the solubility is 0.0011, the difference between the two (0.0063 – 0.0011 =
0.0052 M) has to be precipitated. The mass of the precipitate is therefore:
mass of BaF2
0.0052 mol BaF2 175.3 g BaF2
0.91 g
1 L solution
1 mol BaF2
938
Chapter 18: Solubility and Complex-Ion Equilibria
FEATURE PROBLEMS
87.
(M) Ca 2+ = SO 4
2
in the saturated solution. Let us first determine the amount of H 3O + in
2H 2O(l) + Na + aq
the 100.0 mL diluted effluent. H 3O + aq + NaOH aq
mmol H 3O + = 100.0 mL
8.25 mL base
0.0105 mmol NaOH 1 mmol H 3O +
10.00 mL sample
1 mL base
1 mmol NaOH
= 0.866 mmol H 3O + aq
Now we determine Ca 2+ in the original 25.00 mL sample, remembering that 2 H 3O + were
produced for each Ca 2+ .
Ca 2+ =
0.866 mmol H 3O + (aq)
1mmol Ca 2
2 mmol H 3O
25.00 mL
= 0.0173 M
Ksp = Ca 2+ SO4 2 = 0.0173 = 3.0 104 ; the K sp for CaSO4 is 9.1106 in Appendix D.
(D)
(a) We assume that there is little of each ion present in solution at equilibrium (that this is a
simple stoichiometric calculation). This is true because the K value for the titration
reaction is very large. Ktitration = 1/Ksp(AgCl) = 5.6 ×109. We stop the titration when just
AgCl s
enough silver ion has been added. Ag + aq + Cl aq
2
88.
V = 100.0 mL
29.5 mg Cl 1 mmol Cl 1 mmol Ag +
1mL
1000 mL
35.45 mg Cl 1 mmol Cl
0.01000 mmol AgNO3
= 8.32 mL
(b)
We first calculate the concentration of each ion as the consequence of dilution. Then we
determine the [Ag+] from the value of Ksp.
8.32 mL added
initial Ag + = 0.01000 M
= 7.68 104 M
108.3mL final volume
1mmol Cl
100.0 mL taken
35.45 mg Cl
= 7.68 104 M
1000 mL
108.3 mL final volume
29.5 mg Cl
initial Cl =
The slight excess of each ion will precipitate until the solubility constant is satisfied.
Ag + = Cl = Ksp 18
. 10 10 13
. 105 M
(c)
If we want Ag 2 CrO 4 to appear just when AgCl has completed precipitation,
Ag + = 1.3 105 M . We determine CrO 4 2 from the Ksp expression.
939
Chapter 18: Solubility and Complex-Ion Equilibria
2
Ksp = Ag + CrO42 = 1.11012 = 1.3105
(d)
2
CrO 2 ;
4
12
CrO 2 = 1.110
= 0.0065 M
4
2
5
1.310
If CrO 4 2 were greater than the answer just computed for part (c), Ag 2 CrO 4 would
appear before all Cl had precipitated, leading to a false early endpoint. We would
calculate a falsely low Cl for the original solution.
If CrO 4 2 were less than computed in part 3, Ag 2 CrO 4 would appear somewhat after
all Cl had precipitated, leading one to conclude there was more Cl in solution than
actually was the case.
(e)
89.
(D)
(a)
2
If it was Ag + that was being titrated, it would react immediately with the CrO 4 in the
sample, forming a red-orange precipitate. This precipitate would not likely dissolve, and
thus, very little if any AgCl would form. There would be no visual indication of the
endpoint.
We need to calculate the Mg 2+ in a solution that is saturated with Mg(OH)2.
K sp = 1.8 1011 = Mg 2+ OH = s 2s = 4 s 3
2
3
s=
2
1.8 1011
1.7 104 M = [Mg 2+ ]
4
(b)
Even though water has been added to the original solution, it remains saturated (it is in
equilibrium with the undissolved solid Mg OH 2 ). Mg 2+ = 1.7 104 M.
(c)
Although HCl(aq) reacts with OH , it will not react with Mg 2+ . The solution is simply
a more dilute solution of Mg2+.
100.0 mL initial volume
Mg 2+ = 1.7 104 M
= 2.8 105 M
100.0 + 500. mL final volume
(d)
In this instance, we have a dual dilution to a 275.0 mL total volume, followed by a
common-ion scenario.
1.7 104 mmol Mg 2+
0.065 mmol Mg 2+
25.00
mL
250.0
mL
1 mL
1 mL
initial Mg 2+ =
275.0 mL total volume
0.059 M
25.00 mL
initial OH =
1.7 104 mmol Mg 2+ 2 mmol OH
1 mL
1 mmol Mg 2+
3.1105 M
275.0 mL total volume
940
Chapter 18: Solubility and Complex-Ion Equilibria
Let’s see if precipitation occurs.
Qsp = Mg 2+ OH
2
b
gc
= 0.059 3.1 105
h
2
= 5.7 1011 1.8 1011 = Ksp
Thus, precipitation does occur, but very little precipitate forms. If OH goes down by
1.4 105 M (which means that Mg 2+ drops by 0.7 105 M ), then
OH = 1.7 105 M and
Mg 2+ = 0.059 M 0.7 105 M = 0.059 M , then Qsp Ksp and precipitation will
stop. Thus, Mg 2+ = 0.059 M .
(e)
Again we have a dual dilution, now to a 200.0 mL final volume, followed by a commonion scenario.
1.7 104 mmol Mg 2+
1 mL
4.3 105 M
200.0 mL total volume
50.00 mL
initial Mg 2+ =
150.0 mL initial volume
0.113 M
initial OH = 0.150 M
200.0 mL total volume
Now it is evident that precipitation will occur. Next we determine the Mg 2+ that can
exist in solution with 0.113 M OH . It is clear that Mg 2+ will drop dramatically to
satisfy the Ksp expression but the larger value of OH will scarcely be affected.
K sp = Mg 2+ OH = 1.8 1011 = Mg 2+ 0.0113 M
1.8 1011
Mg 2+ =
= 1.4 109 M
2
0.113
2
941
2
Chapter 18: Solubility and Complex-Ion Equilibria
SELF-ASSESSMENT EXERCISES
90.
(E)
(a) Ksp: The solubility product constant, which is the constant for the equilibrium established
between a solid solute and its ions in a saturated solution
(b) Kf: The formation constant of a complex ion, which is the equilibrium constant describing
the formation of a complex ion from a central ion (typically a metal cation) and the ligands
that attach to it
(c) Qsp: The ion product, which is the product of concentrations of the constituent ions of a
compound, and is used to determine if they meet the precipitation criteria or not.
(d) Complex ion: A polyatomic cation or anion composed of a central metal ion to which other
groups (molecules or ions) called ligands are coordinated/attached to.
91.
(E)
(a) Common-ion effect in solubility equilibrium: Where the solubility of a sparingly soluble
compound is suppressed by the presence of a quantity (often large) of one of the
compound’s ions from another source
(b) Fractional precipitation: A technique in which two or more ions in solution, each capable
of being precipitated by the same reagent, are separated by the proper use of that reagent:
One ion is precipitated, while the other(s) remains in solution.
(c) Ion-pair formation: Pair of ions held loosely together by electrostatic interactions in water
(d) Qualitative cation analysis: A qualitative method that is used to identify various the cations
present in a solution by stepwise addition of various anions in order to selectively
precipitate out various cations in steps.
92.
(E)
(a) Solubility and solubility product constant: Solubility is the number of moles of a
precipitate dissolved in a certain mass of water (or some other solvent.) The solubility
product constant is the constant for the equilibrium established between a solid solute and
its ions in a saturated solution.
(b) Common–ion effect and salt effect: The common ion effect is a consequence of Le
Châtelier’s principle, where the equilibrium concentration of all ions resulting from the
dissolution of a sparingly soluble precipitate is reduced because a large amount of one (or
more) of the ions from another source is present. The salt effect is the presence of a large
quantity of ions from a very soluble salt which slightly increases the solubility of a
sparingly soluble precipitate.
(c) Ion pair and ion product: An ion pair comprises two oppositely charged ions held together
by electrostatic forces. The ion product (Qsp) is the product of concentrations of the
constituent ions of a compound, and is used to determine if they meet the precipitation
criteria or not.
942
Chapter 18: Solubility and Complex-Ion Equilibria
93.
94.
(E) The answer is (d). See the reasoning below.
(a) Wrong, because the stoichiometry is wrong
(b) Wrong, because Ksp = [Pb2+]·[I-]2
(c) Wrong, because [Pb2+] = Ksp/[ I-]2
(d) Correct because of the [Pb2+]:2[I-] stoichiometry
(E) The answer is (a). BaSO 4 s Ba 2 SO 42 . If Na2SO4 is added, the common–ion effect
forces the equilibrium to the left and reduces [Ba2+].
95.
(E) The answer is (c). Choices (b) and (d) reduce solubility because of the common–ion effect.
In the case of choice (c), the diverse non-common–ion effect or the “salt effect” causes more
Ag2CrO4 to dissolve.
96.
(E) The answer is (b). The sulfate salt of Cu is soluble, whereas the Pb salt is insoluble.
97. (M) The answers are (c) and (d). Adding NH3 causes the solution to become basic (adding
OH–). Mg, Fe, Cu, and Al all have insoluble hydroxides. However, only Cu can form a
complex ion with NH3, which is soluble. In the case of (NH4)2SO4, it is slightly acidic and
dissolved readily in a base.
98.
(M) The answer is (a). CaCO3 is slightly basic, so it is more soluble in an acid. The only
option for an acid given is NH4Cl.
99.
(M) The answer is (c). Referring to Figure 18-7, it is seen that ammonia is added to an
aqueous H2S solution to precipitate more metal ions. Since ammonia is a base, increasing the
pH should cause more precipitation.
100. (M)
(a) H2C2O4 is a moderately strong acid, so it is more soluble in a basic solution.
(b) MgCO3 is slightly basic, so it is more soluble in an acidic solution.
(c) CdS is more soluble in acidic solutions, but the solubility is still so small that it is
essentially insoluble even in acidic solutions.
(d) KCl is a neutral salt, and therefore its solubility is independent of pH.
(e) NaNO3 is a neutral salt, and therefore its solubility is independent of pH.
(f) Ca(OH)2, a strong base, is more soluble in an acidic solution.
101. (E) The answer is NH3. NaOH(aq) precipitates both, and HCl(aq) precipitates neither.
Mg(OH)2 precipitates from an NH3(aq) solution but forms the soluble complex
Cu(NH3)4(OH)2.
102. (D) Al(OH)3 will precipitate. To demonstrate this, the pH of the acetate buffer needs to be
determined first, from which the OH– can be determined. The OH– concentration can be used
to calculate Qsp, which can then be compared to Ksp to see if any Al(OH)3 will precipitate. This
is shown below:
943
Chapter 18: Solubility and Complex-Ion Equilibria
Ac
0.35 M
pH pK a log
log 1.8 105 log
4.65
0.45 M
HAc
14 pH
4.467 1010 M
[OH ] 10
Al OH 3 Al3+ +3OH -
Qsp ( s )(3s )3
Qsp 0.275 4.467 1010 2.45 1029
3
Qsp K sp , therefore there will be precipitation.
103. (E) The answer is (b). Based on solubility rules, Cu3(PO4)2 is the only species that is sparingly
soluble in water.
104. (M) The answer is (d). The abbreviated work shown for each part calculates the molar
solubility (s) for all the salts. They all follow the basic outlined below:
M x A y xM + yA
K sp x s y s
x
y
and we solve for s.
(c) Mg 3 PO 4 2 3Mg 2 2PO34
(a) MgF2 Mg 2 2F
3.7 108 s (2s) 2 4s3
1 1025 3s 2s 108s5
s 2.1 103 M
s 3.9 106 M
3
2
(b) MgCO3 Mg 2 CO32
(d) Li3PO 4 3Li PO34
3.5 108 s s
3.2 109 3s s 27s 4
s 1.9 104 M
s 3.3 103 M
3
105. (E) The answer is (b). This is due to the “salt effect.” The more moles of salt there are
available, the greater the solubility. For (b), there are 0.300 moles of ions (3×0.100 M Na2S2O3).
106. (E) It will decrease the amount of precipitate. Since all salts of NO3– are highly soluble and
Ag+ and Hg2+ salts of I– are not, anything that forms a soluble complex ion with Ag+ and Hg2+
will reduce the amount of those ions available for precipitation with I– and therefore will
reduce the amount of precipitate.
944
Chapter 18: Solubility and Complex-Ion Equilibria
107. (M) No precipitate will form. To demonstrate this, one has to calculate [Ag+], and then, using
[I–], determine the Qsp and compare it to Ksp of AgI.
Since Kf is for the formation of the complex ion Ag(CN)2¯, its inverse is for the dissociation of
Ag(CN)2¯ to Ag+ and CN¯.
K dis 1 K f 1 5.6 1018 1.786 1019
The dissociation of Ag(CN)2¯ is as follows:
Ag CN 2 ¯(aq) Ag + +2CN -
x(1.05 x )
1.786 1019
(0.012 x )
We can simplify the calculations by noting that x is very small in relation to [Ag+] and [CN¯].
Therefore, x = 1.70×10-19 M.
K dis
Dissociation of AgI is as follows:
AgI Ag + + I-
Qsp 1.7 1019 2.0 3.4 1019
Since Qsp < Ksp, no precipitate will form.
108. (M) In both cases, the dissolution reaction is similar to reaction (18.5), for which K = Kf × Ksp.
This is an exceedingly small quantity for (a) but large for (b). CuCO3 is soluble in NH3(aq) and
CuS(s) is not.
109. (M) In this chapter, the concept of solubility of sparingly soluble precipitates is the overarching
concept. Deriving from this overall concept is the dissociation constant for the precipitate, Ksp
and molar solubility. Also emanating from the solubility concept are factors that affect
solubility: the common ion–effect, the salt effect, the effects of pH on solubility, and formation
of complex ions. Take a look at the subsection headings and problems for more refining of the
general and specific concepts.
945
CHAPTER 19
SPONTANEOUS CHANGE:
ENTROPY AND GIBBS ENERGY
PRACTICE EXAMPLES
1A
(E) In general, S 0 if ngas 0 . This is because gases are very dispersed compared to
liquids or solids; (gases possess large entropies). Recall that ngas is the difference
between the sum of the stoichiometric coefficients of the gaseous products and a similar
sum for the reactants.
(a)
ngas = 2 + 0 2 +1 = 1 . One mole of gas is consumed here. We predict S 0 .
(b)
ngas = 1+ 0 0 = +1. Since one mole of gas is produced, we predict S 0 .
1B (E) (a) The outcome is uncertain in the reaction between ZnS(s) and Ag 2O s . We have
used ngas to estimate the sign of entropy change. There is no gas involved in this
reaction and thus our prediction is uncertain.
(b)
2A
In the chlor-alkali process the entropy increases because two moles of gas have
formed where none were originally present ( ngas (1 1 0) (0 0) 2
(E) For a vaporization, Gvap
= 0 = Hvap
TS vap
. Thus, S vap
= Hvap
/ Tvap .
We substitute the given values. Svap
=
H vap
Tvap
=
20.2 kJ mol1
= 83.0 J mol1 K 1
29.79 + 273.15 K
2B
(E) For a phase change, Gtr = 0 = Htr TStr . Thus, Htr = TS tr . We substitute in the
given values. H tr = T Str = 95.5 + 273.2 K 1.09 J mol1 K 1 = 402 J/mol
3A
(M) The entropy change for the reaction is expressed in terms of the standard entropies of
the reagents.
S = 2S NH 3 g S N 2 g 3S H 2 g
= 2 192.5 J mol1K 1 191.6 J mol1 K 1 3 130.7 J mol1 K 1 = 198.7 J mol1 K 1
Thus to form one mole of NH 3 g , the standard entropy change is 99.4 J mol 1 K 1
946
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
3B
(M) The entropy change for the reaction is expressed in terms of the standard entropies of
the reagents.
S o = S o NO g + S o NO 2 g S o N 2 O 3 g
138.5 J mol1 K 1 = 210.8 J mol1 K 1 + 240.1 J mol1 K 1 S N 2 O3 g
= 450.9 J mol1 K 1 S N 2 O3 g
S N 2 O3 g = 450.9 J mol1 K 1 138.5 J mol1 K 1 = 312.4 J mol1 K
4A
(E) (a) Because ngas = 2 1 + 3 = 2 for the synthesis of ammonia, we would predict
S 0 for the reaction. We already know that H 0 . Thus, the reaction falls into
case 2, namely, a reaction that is spontaneous at low temperatures and nonspontaneous at high temperatures.
(b)
For the formation of ethylene ngas = 1 2 + 0 = 1 and thus S 0 . We are given
that H 0 and, thus, this reaction corresponds to case 4, namely, a reaction that is
non-spontaneous at all temperatures.
4B
(E) (a) Because ngas = +1 for the decomposition of calcium carbonate, we would predict
S 0 for the reaction, favoring the reaction at high temperatures. High
temperatures also favor this endothermic H o 0 reaction.
The “roasting” of ZnS(s) has ngas = 2 3 = 1 and, thus, S 0 . We are given that
(b)
H 0 ; thus, this reaction corresponds to case 2, namely, a reaction that is
spontaneous at low temperatures, and non-spontaneous at high ones.
5A
(E) The expression G = H TS is used with T = 298.15 K .
G o = H o T S o = 1648 kJ 298.15 K 549.3 J K 1 1 kJ / 1000 J
5B
= 1648 kJ +163.8 kJ = 1484 kJ
(M) We just need to substitute values from Appendix D into the supplied expression.
G = 2Gf NO 2 g 2Gf NO g Gf O 2 g
= 2 51.31 kJ mol1 2 86.55 kJ mol1 0.00 kJ mol1 = 70.48 kJ mol1
6A
(M) Pressures of gases and molarities of solutes in aqueous solution appear in
thermodynamic equilibrium constant expressions. Pure solids and liquids (including
solvents) do not appear.
(a)
K=
PSiCl
4
PCl2
Kp
(b)
K=
HOCl H + Cl
PCl
2
2
K = K p for (a) because all terms in the K expression are gas pressures.
947
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
6B
(M) We need the balanced chemical equation in order to write the equilibrium constant
expression. We start by translating names into formulas.
PbS s + HNO 3 aq Pb NO 3 2 aq + S s + NO g
The equation then is balanced with the ion-electron method.
oxidation : {PbS s Pb 2+ aq + S s + 2e
} 3
reduction :{NO3 aq + 4H + aq + 3e NO g + 2H 2O(l) } 2
net ionic : 3 PbS s + 2NO3 aq + 8H + aq 3Pb 2+ aq + 3S s + 2NO g + 4H 2 O(l)
In writing the thermodynamic equilibrium constant, recall that
neither pure solids (PbS(s) and S(s)) nor pure liquids H 2 O(l)
appear in the thermodynamic equilibrium constant expression. Note
also that we have written H + aq here for brevity even though we
understand that H 3O + aq is the acidic species in aqueous solution.
K
7A
2
[Pb 2 ]3 pNO
[NO3 ]2 [H ]8
(M) Since the reaction is taking place at 298.15 K, we can use standard free energies of
formation to calculate the standard free energy change for the reaction:
N 2 O 4 (g) 2 NO 2 (g)
G = 2Gf NO 2 g Gf N 2 O 4 g 2 51.31 kJ/mol 97.89 kJ/mol 4.73kJ
Gorxn = +4.73 kJ . Thus, the forward reaction is non-spontaneous as written at 298.15 K.
7B
(M) In order to answer this question we must calculate the reaction quotient and compare
it to the Kp value for the reaction:
N 2 O 4 (g) 2 NO 2 (g)
(0.5) 2
Qp
0.5
0.5
0.5 bar
0.5 bar
o
G rxn = +4.73 kJ = RTlnKp; 4.73 kJ/mol = (8.3145 10 –3 kJ/Kmol)(298.15 K)lnKp
Therefore, Kp = 0.148. Since Qp is greater than Kp, we can conclude that the reverse
reaction will proceed spontaneously, i.e. NO2 will spontaneously convert into N2O4.
8A
(D) We first determine the value of G and then set G = RT ln K to determine K .
G Gf Ag + aq + Gf [I (aq)] Gf AgI s
[(77.11 51.57) (66.19)] kJ/mol 91.73
948
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
91.73 kJ / mol
1000 J
G
ln K
37.00
1
1
RT
8.3145 J mol K 298.15 K 1 kJ
K e 37.00 8.5 1017
This is precisely equal to the value for the Ksp of AgI listed in Appendix D.
8B
(D) We begin by translating names into formulas.
MnO 2 s + HCl aq Mn 2 + aq + Cl2 aq Then we produce a balanced net ionic
equation with the ion-electron method.
oxidation : 2 Cl aq Cl2 g + 2e
reduction : MnO 2 s + 4H + aq + 2e Mn 2+ aq + 2H 2O(l)
net ionic : MnO 2 s + 4H + aq + 2Cl aq Mn 2+ aq + Cl2 g + 2H 2 O(l)
Next we determine the value of G for the reaction and then the value of K.
G = Gf Mn 2+ aq + Gf Cl2 g + 2Gf H 2 O l
Gf MnO 2 s 4Gf H + aq 2Gf Cl aq
= 228.1 kJ + 0.0 kJ + 2 237.1 kJ
465.1 kJ 4 0.0 kJ 2 131.2 kJ = +25.2 kJ
ln K
(25.2 103 J mol1 )
G
10.17
RT
8.3145 J mol-1 K 1 298.15 K
K e 10.2 4 105
Because the value of K is so much smaller than unity, we do not expect an appreciable
forward reaction.
9A
(M) We set equal the two expressions for G and solve for the absolute temperature.
G = H T S = RT ln K
T=
9B
H = T S RT ln K = T S R ln K
H o
114.1 103 J/mol
=
= 607 K
S o R ln K 146.4 8.3145 ln 150 J mol1 K 1
(D) We expect the value of the equilibrium constant to increase as the temperature
decreases since this is an exothermic reaction and exothermic reactions will have a larger
equilibrium constant (shift right to form more products), as the temperature decreases.
Thus, we expect K to be larger than 1000, which is its value at 4.3 102 K.
(a)
The value of the equilibrium constant at 25 C is obtained directly from the value of
G o , since that value is also for 25 C . Note:
G o = H o T S o = 77.1 kJ/mol 298.15 K 0.1213 kJ/mol K = 40.9 kJ/mol
ln K
(40.9 103 J mol1 )
G
16.5
RT
8.3145 J mol1 K 1 298.15 K
949
K e +16.5 1.5 107
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
(b)
First, we solve for G o at 75 C = 348 K
kJ 1000 J
J
G o = H o T S o = 77.1
348.15 K 121.3
mol 1 kJ
mol K
= 34.87 103 J/mol
Then we use this value to obtain the value of the equilibrium constant, as in part (a).
(34.87 103 J mol1 )
G
ln K
12.05
K e+12.05 1.7 105
-1
1
RT
8.3145 J mol K 348.15 K
As expected, K for this exothermic reaction decreases with increasing temperature.
10A (M) We use the value of Kp = 9.1 102 at 800 K and H o = 1.8 105 J / mol , for the
appropriate terms, in the van't Hoff equation.
5.8 102
1
1
1 9.66 8.3145
1.8 105 J/mol 1
ln
=
=
= 9.66;
2
1
1
9.1 10
8.3145 J mol K 800 K T K
T 800
1.8 105
1/T = 1.25 103 4.5 104 = 8.0 104
T = 1240 K 970 C
This temperature is an estimate because it is an extrapolated point beyond the range of the
data supplied.
10B (M) The temperature we are considering is 235 C = 508 K. We substitute the value of
Kp = 9.1 102 at 800 K and H o = 1.8 105 J/mol, for the appropriate terms, in the van't
Hoff equation.
Kp
Kp
1
1.8 105 J/mol 1
ln
=
= e +15.6 = 6 106
= +15.6 ;
2
2
1
1
9.1 10
8.3145 J mol K 800 K 508 K
9.110
K p = 6 106 9.1 102 = 5 109
INTEGRATIVE EXAMPLE
11A (D) The value of G can be calculated by finding the value of the equilibrium constant Kp
at 25 oC. The equilibrium constant for the reaction is simply given by K p p{N 2O5 (g)} .
o
The vapor pressure of N2O5(g) can be determined from the Clausius-Clapeyron eqution,
which is a specialized version of the van’t Hoff equation.
Stepwise approach:
We first determine the value of H sub .
ln
1
p2 H sub 1 1
H sub
1
760 mmHg
ln
-1
-1
p1
R T1 T2
100 mmHg 8.314Jmol K 7.5 273.15 32.4 273.15
2.028
5.81 10 4 J/mol
5
3.49 10
Using the same formula, we can now calculate the vapor pressure of N2O5 at 25 oC.
H sub
950
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
p3
p3
5.81 10 4 J/mol 1
1
1.46
ln
e1.46 4.31
-1 -1
100 mmHg 8.314Jmol K 280.7 298.2
100 mmHg
1 atm
0.567atm K p
p3 4.31 100mmHg
760 mmHg
G o RT ln K p (8.314 10 3 kJmol-1K -1 298.15K)ln(0.567) 1.42kJ/mol
Conversion pathway approach:
p
R ln 2
H sub 1 1
p
p1
H sub
ln 2
p1
R T1 T2
1 1
T T
1
2
760mmHg
2.028
100mmHg
Jmol-1 5.81 10 4 Jmol-1
1
1
3.49 10 5
7.5 273.15 32.4 273.15
8.314Jmol-1K -1 ln
H sub
p
H sub 1 1
p3 p1e
ln 3
p1
R T1 T2
p3 100mmHg e
H sub 1 1
R T1 T2
5.8110 4 Jmol-1 1
1 -1
K
8.314 JK -1 mol-1 280.7 298.2
431mmHg
1atm
0.567atm K p
760mmHg
G o RT ln K p (8.314 10 3 kJmol-1K -1 298.15K)ln(0.567) 1.42kJ/mol
11B (D) The standard entropy change for the reaction ( S o ) can be calculated from the known
values of H o and G o .
Stepwise approach:
H o G o 454.8kJmol-1 (323.1kJmol-1 )
o
o
o
o
441.7JK -1mol-1
G H T S S
T
298.15K
Plausible chemical reaction for the production of ethylene glycol can also be written as:
2C(s)+3H 2 (g)+O 2 (g)
CH 2OHCH 2 OH(l)
o
o
Since S o {S products
} {Sreac
tan ts } it follows that:
o
Srxn
S o (CH 2OHCH 2 OH(l)) [2 S o (C(s)) 3 S o (H 2 (g)) S o (O 2 (g))]
441.7JK -1mol-1 S o (CH 2 OHCH 2 OH(l)) [2 5.74JK -1mol-1 3 130.7JK -1mol-1 205.1JK -1mol-1 ]
S o (CH 2 OHCH 2 OH(l)) 441.7JK -1mol-1 608.68JK -1mol-1 167JK -1mol-1
951
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
Conversion pathway approach:
H o G o 454.8kJmol-1 (323.1kJmol-1 )
441.7JK -1mol-1
G o H o T S o S o
T
298.15K
441.7JK -1mol-1 S o (CH 2OHCH 2 OH(l)) [2 5.74JK -1mol-1 3 130.7JK -1mol-1 205.1JK -1mol-1 ]
S o (CH 2 OHCH 2 OH(l)) 441.7JK -1mol-1 608.68JK -1mol-1 167JK -1mol-1
EXERCISES
Spontaneous Change and Entropy
1.
2.
3.
4.
(E) (a) The freezing of ethanol involves a decrease in the entropy of the system. There is
a reduction in mobility and in the number of forms in which their energy can be
stored when they leave the solution and arrange themselves into a crystalline state.
(b)
The sublimation of dry ice involves converting a solid that has little mobility into a
highly dispersed vapor which has a number of ways in which energy can be stored
(rotational, translational). Thus, the entropy of the system increases substantially.
(c)
The burning of rocket fuel involves converting a liquid fuel into the highly dispersed
mixture of the gaseous combustion products. The entropy of the system increases
substantially.
(E) Although there is a substantial change in entropy involved in (a) changing H 2 O (1iq., 1
atm) to H 2 O (g, 1 atm), it is not as large as (c) converting the liquid to a gas at 10 mmHg.
The gas is more dispersed, (less ordered), at lower pressures. In (b), if we start with a solid
and convert it to a gas at the lower pressure, the entropy change should be even larger,
since a solid is more ordered (concentrated) than a liquid. Thus, in order of increasing S ,
the processes are: (a) (c) (b).
(E) The first law of thermodynamics states that energy is neither created nor destroyed (thus,
“The energy of the universe is constant”). A consequence of the second law of
thermodynamics is that entropy of the universe increases for all spontaneous, that is, naturally
occurring, processes (and therefore, “the entropy of the universe increases toward a
maximum”).
(E) When pollutants are produced they are usually dispersed throughout the environment.
These pollutants thus start in a relatively compact form and end up dispersed throughout a
large volume mixed with many other substances. The pollutants are highly dispersed, thus,
they have a high entropy. Returning them to their original compact form requires reducing
this entropy, which is a highly non-spontaneous process. If we have had enough foresight to
retain these pollutants in a reasonably compact form, such as disposing of them in a secure
landfill, rather than dispersing them in the atmosphere or in rivers and seas, the task of
permanently removing them from the environment, and perhaps even converting them to
useful forms, would be considerably easier.
952
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
5.
(E) (a) Increase in entropy because a gas has been created from a liquid, a condensed
phase.
(b)
Decrease in entropy as a condensed phase, a solid, is created from a solid and a gas.
(c)
For this reaction we cannot be certain of the entropy change. Even though the number
of moles of gas produced is the same as the number that reacted, we cannot conclude
that the entropy change is zero because not all gases have the same molar entropy.
(d)
6.
(E) (a) At 75 C , 1 mol H 2 O (g, 1 atm) has a greater entropy than 1 mol H 2 O (1iq., 1
atm) since a gas is much more dispersed than a liquid.
(b)
7.
8.
2H 2S g + 3O 2 g 2H 2 O g + 2SO 2 g Decrease in entropy since five moles of
gas with high entropy become only four moles of gas, with about the same quantity
of entropy per mole.
1 mol Fe
= 0.896 mol Fe has a higher entropy than 0.80 mol Fe, both (s)
55.8 g Fe
at 1 atm and 5 C , because entropy is an extensive property that depends on the
amount of substance present.
50.0 g Fe
(c)
1 mol Br2 (1iq., 1 atm, 8 C ) has a higher entropy than 1 mol Br2 (s, 1atm, 8 C )
because solids are more ordered (concentrated) substances than are liquids, and
furthermore, the liquid is at a higher temperature.
(d)
0.312 mol SO 2 (g, 0.110 atm, 32.5 C ) has a higher entropy than 0.284 mol O 2 (g,
15.0 atm, 22.3 C ) for at least three reasons. First, entropy is an extensive property
that depends on the amount of substance present (more moles of SO2 than O2). Second,
entropy increases with temperature (temperature of SO2 is greater than that for O2.
Third, entropy is greater at lower pressures (the O2 has a much higher pressure).
Furthermore, entropy generally is higher per mole for more complicated molecules.
(E) (a) Negative; A liquid (moderate entropy) combines with a solid to form another
solid.
(b)
Positive; One mole of high entropy gas forms where no gas was present before.
(c)
Positive; One mole of high entropy gas forms where no gas was present before.
(d)
Uncertain; The number of moles of gaseous products is the same as the number of
moles of gaseous reactants.
(e)
Negative; Two moles of gas (and a solid) combine to form just one mole of gas.
(M) The entropy of formation of a compound would be the difference between the absolute
entropy of one mole of the compound and the sum of the absolute entropies of the
appropriate amounts of the elements constituting the compound, with each species in its
most stable form.
953
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
Stepwise approach:
It seems as though CS2 1 would have the highest molar entropy of formation of the
compounds listed, since it is the only substance whose formation does not involve the
consumption of high entropy gaseous reactants. This prediction can be checked by
determining S of values from the data in Appendix D:
(a)
C graphite + 2H 2 g CH 4 g
Sfo CH 4 g = S o CH 4 g S o C graphite 2S o H 2 g
= 186.3 J mol1K 1 5.74 J mol1K 1 2 130.7 J mol1K 1
= 80.8 J mol1K 1
(b)
2C graphite + 3H 2 g + 12 O 2 g CH 3CH 2 OH l
1
Sfo CH3CH 2 OH l = S o CH3CH 2 OH l 2S o C graphite 3S o H 2 g S o O2 g
2
= 160.7 J mol1 K 1 2 5.74 J mol1 K 1 3 130.7 J mol1 K 1 12 205.1J mol1K 1
= 345.4 J mol1 K 1
(c)
C graphite + 2S rhombic CS2 l
Sf CS2 l = S o CS2 l S o C graphite 2S o S rhombic
= 151.3 J mol1 K 1 5.74 J mol1 K 1 2 31.80 J mol1 K 1
= 82.0 J mol1 K 1
Conversion pathway approach:
CS2 would have the highest molar entropy of formation of the compounds listed, because
it is the only substance whose formation does not involve the consumption of high
entropy gaseous reactants.
(a)
C graphite + 2H 2 g CH 4 g
(b)
Sfo CH 4 g = 186.3 J mol1K 1 5.74 J mol1K 1 2 130.7 J mol1K 1
80.8 J mol1K 1
2C graphite + 3H 2 g + 12 O 2 g CH 3CH 2 OH l
Sfo CH 3CH 2OH l = (160.7 2 5.74 3 130.7 12 205.1)J mol1 K 1
= 345.4 J mol1 K 1
(c)
C graphite + 2S rhombic CS2 l
Sf CS2 l = 151.3 J mol1 K 1 5.74 J mol1 K 1 2 31.80 J mol1 K 1
= 82.0 J mol1 K 1
954
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
Phase Transitions
9.
H vap
H f [H 2 O(g)] H f [H 2 O(l)] 241.8 kJ/mol (285.8 kJ/mol)
(M) (a)
44.0 kJ/mol
= S H 2 O g S o H 2 O l = 188.8 J mol1 K 1 69.91 J mol1K 1
= 118.9 J mol1K 1
.
There is an alternate, but incorrect, method of obtaining S vap
S
vap
S
vap
o
=
H vap
T
=
44.0 103 J/mol
= 148 J mol1 K 1
298.15 K
This method is invalid because the temperature in the denominator of the equation
must be the temperature at which the liquid-vapor transition is at equilibrium. Liquid
water and water vapor at 1 atm pressure (standard state, indicated by ) are in
equilibrium only at 100 C = 373 K.
(b)
The reason why Hvap
is different at 25 C from its value at 100 C has to do with
the heat required to bring the reactants and products down to 298 K from 373 K. The
specific heat of liquid water is higher than the heat capacity of steam. Thus, more
heat is given off by lowering the temperature of the liquid water from 100 C to
25 C than is given off by lowering the temperature of the same amount of steam.
Another way to think of this is that hydrogen bonding is more disrupted in water at
100 C than at 25 C (because the molecules are in rapid—thermal—motion), and
has
hence, there is not as much energy needed to convert liquid to vapor (thus H vap
a smaller value at 100 C . The reason why S vap
has a larger value at 25 C than at
100 C has to do with dispersion. A vapor at 1 atm pressure (the case at both
temperatures) has about the same entropy. On the other hand, liquid water is more
disordered (better able to disperse energy) at higher temperatures since more of the
hydrogen bonds are disrupted by thermal motion. (The hydrogen bonds are totally
disrupted in the two vapors).
10. (M) In this problem we are given standard enthalpies of the formation ( H of ) of liquid
and gas pentane at 298.15 K and asked to estimate the normal boiling point of
o
o
and furthermore comment on the significance of the sign of Gvap
.
pentane, Gvap
o
The general strategy in solving this problem is to first determine H vap
from the
known enthalpies of formation. Trouton’s rule can then be used to determine the
o
normal boiling point of pentane. Lastly, Gvap
,298 K can be calculated using
Gvap
= Hvap
TS vap
.
955
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
Stepwise approach:
o
Calculate H vap
from the known values of H of (part a):
C5H12(g)
C5H12(l)
H of -173.5 kJmol-1
-146.9 kJmol-1
o
H vap
146.9 (173.5)kJmol-1 26.6kJmol-1
Determine normal boiling point using Trouton’s rule (part a):
o
H vap
o
Svap
87Jmol-1K -1
Tnbp
Tnbp
o
H vap
o
Svap
26.6kJmol 1
306K
87kJK 1mol 1
1000
Tnbp 32.9 o C
Use Gvap
to calculate Gvap,298K
(part b):
= Hvap
TS vap
Gvap
= Hvap
TS vap
Gvap,298K
26.6kJmol-1 298.15K
87kJmol-1K -1
1000
Gvap,298K
0.66kJmol-1
(part c):
Comment on the value of Gvap,298K
The positive value of Gvap
indicates that normal boiling (having a vapor pressure
of 1.00 atm) for pentane should be non-spontaneous (will not occur) at 298. The
vapor pressure of pentane at 298 K should be less than 1.00 atm.
Conversion pathway approach:
C5H12(g)
C5H12(l)
H of -173.5 kJmol-1
-146.9 kJmol-1
o
H vap
146.9 (173.5)kJmol-1 26.6kJmol-1
S
o
vap
o
H vap
Tnbp
Gvap
= H vap
T Svap
11.
o
H vap
26.6kJmol-1
306K
o
87kJK -1mol-1
Svap
1000
87kJmol-1K -1
26.6kJmol-1 298.15K
0.66kJmol-1
1000
87Jmol K Tnbp
-1
-1
(M) Trouton's rule is obeyed most closely by liquids that do not have a high degree of
order within the liquid. In both HF and CH 3OH , hydrogen bonds create considerable order
within the liquid. In C6 H 5CH 3 , the only attractive forces are non-directional London
forces, which have no preferred orientation as hydrogen bonds do. Thus, of the three
choices, liquid C 6 H 5CH 3 would most closely follow Trouton’s rule.
956
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
12.
(E) H vap
H f [Br2 (g)] H f [Br2 (l)] 30.91 kJ/mol 0.00 kJ/mol = 30.91 kJ/mol
S
vap
=
H vap
Tvap
1
87 J mol K
1
or
Tvap =
H vap
S vap
30.91103 J/mol
= 3.5 102 K
1 1
87 J mol K
The accepted value of the boiling point of bromine is 58.8 C = 332 K = 3.32 102 K .
Thus, our estimate is in reasonable agreement with the measured value.
13.
(M) The liquid water-gaseous water equilibrium H 2 O (l, 0.50 atm) H 2 O (g, 0.50 atm)
can only be established at one temperature, namely the boiling point for water under 0.50
atm external pressure. We can estimate the boiling point for water under 0.50 atm external
pressure by using the Clausius-Clapeyron equation:
ln
o
P2 H vap 1 1
=
R T1 T2
P1
We know that at 373 K, the pressure of water vapor is 1.00 atm. Let's make P1 = 1.00
atm, P2 = 0.50 atm and T1 = 373 K. Thus, the boiling point under 0.50 atm pressure is T2.
To find T2 we simply insert the appropriate information into the Clausius-Clapeyron
equation and solve for T2:
ln
1
0.50 atm
40.7 kJ mol1
1
=
3
1
1
1.00 atm 8.3145 10 kJ K mol 373 K T2
1
1
-1.416 104 K =
373 K T2
Solving for T2 we find a temperature of 354 K or 81C. Consequently, to achieve an
equilibrium between gaseous and liquid water under 0.50 atm pressure, the temperature
must be set at 354 K.
14.
(M) Figure 12-19 (phase diagram for carbon dioxide) shows that at 60C and under 1
atm of external pressure, carbon dioxide exists as a gas. In other words, neither solid nor
liquid CO2 can exist at this temperature and pressure. Clearly, of the three phases,
gaseous CO2 must be the most stable and, hence, have the lowest free energy when T =
60 C and Pext = 1.00 atm.
Gibbs Energy and Spontaneous Change
15.
(E) Answer (b) is correct. Br—Br bonds are broken in this reaction, meaning that it is
endothermic, with H 0 . Since the number of moles of gas increases during the
reaction, S 0 . And, because G = H T S , this reaction is non-spontaneous
( G 0 ) at low temperatures where the H term predominates and spontaneous
( G 0 ) at high temperatures where the T S term predominates.
957
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
16.
(E) Answer (d) is correct. A reaction that proceeds only through electrolysis is a reaction
that is non-spontaneous. Such a reaction has G 0 .
17.
(E) (a)
H o 0 and S o 0 (since ngas 0 ) for this reaction. Thus, this reaction is case
2 of Table 19-1. It is spontaneous at low temperatures and non-spontaneous at high
temperatures.
(b)
We are unable to predict the sign of S o for this reaction, since ngas = 0 . Thus, no
strong prediction as to the temperature behavior of this reaction can be made. Since
Ho > 0, we can, however, conclude that the reaction will be non-spontaneous at low
temperatures.
(c)
18.
H o 0 and S o 0 (since ngas 0 ) for this reaction. This is case 3 of Table 19-1.
It is non-spontaneous at low temperatures, but spontaneous at high temperatures.
(E) (a) H o 0 and S o 0 (since ngas 0 ) for this reaction. This is case 4 of Table
19-1. It is non-spontaneous at all temperatures.
(b)
H o 0 and S o 0 (since ngas 0 ) for this reaction. This is case 1 of Table 19-1.
It is spontaneous at all temperatures.
(c)
H o 0 and S o 0 (since ngas 0 ) for this reaction. This is case 2 of Table 19-1.
It is spontaneous at low temperatures and non-spontaneous at high temperatures.
19.
(E) First of all, the process is clearly spontaneous, and therefore G 0 . In addition, the
gases are more dispersed when they are at a lower pressure and therefore S 0 . We also
conclude that H = 0 because the gases are ideal and thus there are no forces of attraction
or repulsion between them.
20.
(E) Because an ideal solution forms spontaneously, G 0 . Also, the molecules of solvent
and solute that are mixed together in the solution are in a more dispersed state than the
separated solvent and solute. Therefore, S 0 . However, in an ideal solution, the
attractive forces between solvent and solute molecules equal those forces between solvent
molecules and those between solute molecules. Thus, H = 0 . There is no net energy of
interaction.
21.
(M) (a) An exothermic reaction (one that gives off heat) may not occur spontaneously if,
at the same time, the system becomes more ordered (concentrated) that is, S o 0 .
This is particularly true at a high temperature, where the TS term dominates the
G expression. An example of such a process is freezing water (clearly exothermic
because the reverse process, melting ice, is endothermic), which is not spontaneous
at temperatures above 0 C .
(b)
A reaction in which S 0 need not be spontaneous if that process also is
endothermic. This is particularly true at low temperatures, where the H term
dominates the G expression. An example is the vaporization of water (clearly an
endothermic process, one that requires heat, and one that produces a gas, so S 0 ),
958
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
which is not spontaneous at low temperatures, that is, below100 C (assuming Pext =
1.00 atm).
22.
(M) In this problem we are asked to explain whether the reaction AB(g)A(g)+B(g) is
always going to be spontaneous at high rather than low temperatures. In order to answer
this question, we need to determine the signs of H , S and consequently G . Recall
that G H T S .
Stepwise approach:
Determine the sign of S:
We are generating two moles of gas from one mole. The randomness of the system
increases and S must be greater than zero.
Determine the sign of H:
In this reaction, we are breaking A-B bond. Bond breaking requires energy, so the reaction
must be endothermic. Therefore, H is also greater than zero.
Use G H T S to determine the sign of G :
G H T S . Since H is positive and S is positive there will be a temperature at
which T S will become greater than H . The reaction will be favored at high
temperatures and disfavored at low temperatures.
Conversion pathway approach:
S for the reaction is greater than zero because we are generating two moles of gas from
one mole. H for the reaction is also greater than zero because we are breaking A-B
(bond breaking requires energy). Because G H T S , there will be a temperature at
which T S will become greater than H . The reaction will be favored at high
temperatures and disfavored at low temperatures.
Standard Gibbs Energy Change
23. (M) H o = H f NH 4Cl s H f NH 3 g H f HCl g
314.4 kJ/mol (46.11 kJ/mol 92.31 kJ/mol) 176.0 kJ/mol
G o Gf NH 4Cl s Gf NH 3 g Gf HCl g
202.9 kJ/mol (16.48 kJ/mol 95.30 kJ/mol) 91.1 kJ/mol
G o H o T S o
H o G o 176.0 kJ/mol 91.1 kJ/mol 1000 J
S
285 J mol1
1 kJ
298 K
T
o
24.
(M) (a) G o = Gf C2 H 6 g Gf C2 H 2 g 2Gf H 2 g
= 32.82 kJ/mol 209.2 kJ/mol 2 0.00 kJ/mol = 242.0 kJ/mol
(b)
G o = 2Gf SO 2 g + Gf O 2 g 2Gf SO 3 g
= 2 300.2 kJ/mol + 0.00 kJ/mol 2 371.1 kJ/mol = +141.8 kJ/mol
959
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
(c)
G o = 3Gf Fe s + 4Gf H 2 O g Gf Fe3 O 4 s 4Gf H 2 g
= 3 0.00 kJ/mol + 4 228.6 kJ/mol 1015 kJ/mol 4 0.00 kJ/mol = 101 kJ/mol
(d)
G o = 2Gf Al3+ aq + 3Gf H 2 g 2Gf Al s 6Gf H + aq
= 2 485 kJ/mol + 3 0.00 kJ/mol 2 0.00kJ/mol 6 0.00 kJ/mol = 970. kJ/mol
25.
(M) (a) S o = 2 S o POCl3 1 2 S o PCl3 g S o O 2 g
= 2 222.4 J/K 2 311.7 J/K 205.1 J/K = 383.7 J/K
G = H o T S o = 620.2 103 J 298 K 383.7 J/K = 506 103 J = 506 kJ
o
(b)
26.
The reaction proceeds spontaneously in the forward direction when reactants and
products are in their standard states, because the value of G o is less than zero.
o
o
o
o
+
o
o
(M) (a) S = S Br2 l + 2 S HNO 2 aq 2 S H aq 2 S Br aq 2 S NO 2 g
= 152.2 J/K + 2 135.6 J/K 2 0 J/K 2 82.4 J/K 2 240.1 J/K = 221.6 J/K
G o = H o T S o = 61.6 103 J 298K 221.6 J/K = +4.4 103 J = +4.4 kJ
(b)
27.
The reaction does not proceed spontaneously in the forward direction when reactants and
products are in their standard states, because the value of G o is greater than zero.
(M) We combine the reactions in the same way as for any Hess's law calculations.
(a) N 2 O g N 2 g + 12 O 2 g
G o = 12 +208.4 kJ = 104.2 kJ
N 2 g + 2 O 2 g 2 NO 2 g
G o = +102.6 kJ
Net: N 2O g + 23 O 2 g 2NO 2 g
G o = 104.2 +102.6 = 1.6 kJ
This reaction reaches an equilibrium condition, with significant amounts of all
species being present. This conclusion is based on the relatively small absolute
value of G o .
(b)
2N 2 g + 6H 2 g 4NH 3 g
G o = 2 33.0 kJ = 66.0 kJ
4NH 3 g + 5O 2 g 4NO g + 6H 2 O l
4NO g 2N 2 g + 2O 2 g
G o = 1010.5 kJ
G o = 2 +173.1 kJ = 346.2 kJ
Net: 6H 2 g + 3O 2 g 6H 2 O l G o = 66.0 kJ 1010.5 kJ 346.2 kJ = 1422.7 kJ
This reaction is three times the desired reaction, which therefore has
G o = 1422.7 kJ 3 = 474.3 kJ.
The large negative G o value indicates that this reaction will go to completion at 25 C .
960
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
(c)
4NH 3 g + 5O 2 g 4NO g + 6H 2O l
G o = 1010.5 kJ
2N 2 g + O 2 g 2N 2 O g
G o = +208.4 kJ
4NO g 2N 2 g + 2O 2 g
G o = 2 +173.1 kJ = 346.2 kJ
4NH 3 g + 4O 2 g 2N 2 O g + 6H 2 O l G o = 1010.5 kJ 346.2 kJ + 208.4 kJ
= 1148.3kJ
28.
This reaction is twice the desired reaction, which, therefore, has G o = -574.2 kJ.
The very large negative value of the G o for this reaction indicates that it will go to
completion.
(M) We combine the reactions in the same way as for any Hess's law calculations.
(a)
COS g + 2CO 2 g SO 2 g + 3CO g
2CO g + 2H 2O g 2CO 2 g + 2H 2 g
G o = 246.4 kJ = +246.6 kJ
G o = 2 28.6 kJ = 57.2 kJ
COS g + 2H 2O g SO 2 g + CO g + 2H 2 g G o = +246.6 57.2 = +189.4 kJ
This reaction is spontaneous in the reverse direction, because of the large positive
value of G o
(b)
COS g + 2CO 2 g SO 2 g + 3CO g
3CO g + 3H 2O g 3CO 2 g + 3H 2 g
G o = 246.4 kJ = +246.6 kJ
G o = 328.6 kJ = 85.8 kJ
COS g + 3H 2 O g CO 2 g + SO 2 g + 3H 2 g G o = +246.6 85.8 = +160.8 kJ
This reaction is spontaneous in the reverse direction, because of the large positive value
of G o .
(c)
COS g + H 2 g CO g + H 2S g
G o = +1.4 kJ
CO g + H 2 O g CO 2 g + H 2 g
G o = 28.6 kJ = 28.6 kJ
COS g + H 2O g CO 2 g + H 2S g
G o = 1.4 kJ 28.6 kJ = 30.0 kJ
The negative value of the G o for this reaction indicates that it is spontaneous in the
forward direction.
29.
(D)
The combustion reaction is : C6 H 6 l + 152 O 2 g 6CO 2 g + 3H 2 O g or l
(a)
G o = 6Gf CO 2 g + 3Gf H 2 O l Gf C6 H 6 l 152 Gf O 2 g
(b)
G o = 6Gf CO 2 g + 3Gf H 2 O g Gf C6 H 6 l 152 Gf O 2 g
= 6 394.4 kJ + 3 237.1 kJ +124.5 kJ 152 0.00 kJ = 3202 kJ
= 6 394.4 kJ + 3 228.6 kJ +124.5 kJ 152 0.00 kJ = 3177 kJ
961
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
We could determine the difference between the two values of G o by noting the
difference between the two products: 3H 2 O l 3H 2O g and determining the
value of G o for this difference:
G o = 3Gf H 2 O g 3Gf H 2 O l = 3 228.6 237.1 kJ = 25.5 kJ
30.
(M) We wish to find the value of the H o for the given reaction: F2 g 2 F(g)
S o = 2S o F g S o F2 g = 2 158.8 J K 1 202.8 J K 1 = +114.8 J K 1
H o = G o + T S o = 123.9 103 J + 298 K 114.8 J/K = 158.1 kJ/mole of bonds
31.
The value in Table 10.3 is 159 kJ/mol, which is in quite good agreement with the value
found here.
(M) (a) Srxn = Sproducts - Sreactants
= [1 mol 301.2 J K1mol1 + 2 mol 188.8 J K1mol1] [2 mol 247.4 J K1mol1
+ 1 mol 238.5 J K1mol1] = 54.5 J K1
Srxn = = 0.0545 kJ K1
(b)
Hrxn= bonds broken in reactants (kJ/mol)) –bonds broken in products(kJ/ mol))
(c)
Grxn = Hrxn TSrxn = 616 kJ 298 K(0.0545 kJ K1) = 600 kJ
Since the Grxn is negative, the reaction is spontaneous, and hence feasible
= [4 mol (389 kJ mol1)N-H + 4 mol (222 kJ mol1)O-F]
[4 mol (301 kJ mol1)N-F + 4 mol (464 kJ mol1)O-H]
Hrxn= 616 kJ
(at 25 C ). Because both the entropy and enthalpy changes are negative, this
reaction will be more highly favored at low temperatures (i.e., the reaction is
enthalpy driven)
32.
(D) In this problem we are asked to find G at 298 K for the decomposition of
ammonium nitrate to yield dinitrogen oxide gas and liquid water. Furthermore, we are
asked to determine whether the decomposition will be favored at temperatures above or
below 298 K. In order to answer these questions, we first need the balanced chemical
equation for the process. From the data in Appendix D, we can determine Hrxn and
Srxn. Both quantities will be required to determine Grxn (Grxn=Hrxn-TSrxn).
Finally the magnitude of Grxn as a function of temperature can be judged depending on
the values of Hrxn and Srxn.
Stepwise approach:
First we need the balanced chemical equation for the process:
NH4NO3(s)
N2O(g) + 2H2O(l)
Now we can determine Hrxn by utilizing H of values provided in Appendix D:
H of
N2O(g) + 2H2O(l)
NH4NO3(s)
-1
-365.6 kJmol
82.05 kJmol-1 -285.6 kJmol-1
Hrxn=Hfproducts Hfreactants
Hrxn=[2 mol(285.8 kJ mol1) + 1 mol(82.05 kJ mol1)] [1 mol (365.6 kJ
mol1)]
962
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
Hrxn= 124.0 kJ
Similarly, Srxn can be calculated utilizing S values provided in Appendix D
NH4NO3(s)
N2O(g) + 2H2O(l)
-1 -1
o
S 15.1 Jmol K
219.9 Jmol-1K-1
69.91 Jmol-1K-1
Srxn=Sproducts Sreactants
Srxn=[2 mol 69.91 J K1mol1 + 1 mol 219.9 J K1mol1 ] [1 mol 151.1 J
K1mol1]
Srxn=208.6 J K1 = 0.2086 kJ K1
To find Grxn we can either utilize Gf values provided in Appendix D or Grxn=HrxnTSrxn:
Grxn=Hrxn-TSrxn=-124.0 kJ – 298.15 K0.2086 kJK-1
Grxn=-186.1 kJ
Magnitude of Grxn as a function of temperature can be judged depending on the values of
Hrxn and Srxn:
Since Hrxn is negative and Srxn is positive, the decomposition of ammonium nitrate is
spontaneous at all temperatures. However, as the temperature increases, the TS term gets
larger and as a result, the decomposition reaction shift towards producing more products.
Consequently, we can say that the reaction is more highly favored above 298 K (it will also
be faster at higher temperatures)
Conversion pathway approach:
From the balanced chemical equation for the process
N2O(g) + 2H2O(l)
NH4NO3(s)
we can determine Hrxn and Srxn by utilizing H of and S values provided in Appendix
D:
Hrxn=[2 mol(285.8 kJ mol1) + 1 mol(82.05 kJ mol1)] [1 mol (365.6 kJ mol1)]
Hrxn= 124.0 kJ
Srxn=[2 mol 69.91 J K1mol1 + 1 mol 219.9 J K1mol1 ] [1 mol 151.1 J
K1mol1]
Srxn=208.6 J K1 = 0.2086 kJ K1
Grxn=Hrxn-TSrxn=-124.0 kJ – 298.15 K0.2086 kJK-1
Grxn=-186.1 kJ
Since Hrxn is negative and Srxn is positive, the decomposition of ammonium nitrate is
spontaneous at all temperatures. However, as the temperature increases, the TS term gets
larger and as a result, the decomposition reaction shift towards producing more products.
The reaction is highly favored above 298 K (it will also be faster).
The Thermodynamic Equilibrium Constant
33.
(E) In all three cases, Keq = Kp because only gases, pure solids, and pure liquids are present
in the chemical equations. There are no factors for solids and liquids in Keq expressions,
and gases appear as partial pressures in atmospheres. That makes Keq the same as Kp for
these three reactions.
963
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
We now recall that K p = K c RT . Hence, in these three cases we have:
n
(a)
2SO 2 g + O 2 g 2SO3 g ; ngas = 2 2 +1 = 1; K = K p = K c RT
(b)
HI g 12 H 2 g + 12 I 2 g ;
(c)
NH 4 HCO3 s NH 3 g + CO 2 g + H 2O l ;
ngas = 2 0 = +2
34.
(M) (a)
(b)
(c)
K=
ngas = 1 12 + 12 = 0;
K = K p = K c RT
1
K = Kp = Kc
2
P{H 2 g }4
P{H 2 O g }4
Terms for both solids, Fe(s) and Fe 3O 4 s , are properly excluded from the
thermodynamic equilibrium constant expression. (Actually, each solid has an activity
of 1.00.) Thus, the equilibrium partial pressures of both H 2 g and H 2 O g do not
depend on the amounts of the two solids present, as long as some of each solid is
present. One way to understand this is that any chemical reaction occurs on the
surface of the solids, and thus is unaffected by the amount present.
We can produce H 2 g from H 2O g without regard to the proportions of Fe(s) and
Fe 3O 4 s with the qualification, that there must always be some Fe(s) present for the
production of H 2 g to continue.
35. (M) In this problem we are asked to determine the equilibrium constant and the change in
Gibbs free energy for the reaction between carbon monoxide and hydrogen to yield
methanol. The equilibrium concentrations of each reagent at 483K were provided. We
proceed by first determining the equilibrium constant. Gibbs free energy can be
calculated using G o RT ln K .
Stepwise approach:
First determine the equilibrium constant for the reaction at 483K:
CO(g)+2H 2 (g) CH3OH(g)
[CH 3OH ( g )]
0.00892
K
14.5
[CO( g )][ H 2 ( g )] 0.0911 0.0822 2
Now use G o RT ln K to calculate the change in Gibbs free energy at 483 K:
G o RT ln K
G o 8.314 483 ln(14.5)Jmol-1 1.1 10 4 Jmol-1
G o 11kJmol-1
Conversion pathway approach:
[CH 3OH ( g )]
0.00892
14.5
K
[CO( g )][ H 2 ( g )] 0.0911 0.0822 2
G o RT ln K 8.314 483 ln(14.5)Jmol-1 1.1 10 4 Jmol-1
G o 11kJmol-1
964
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
36. (M) Gibbs free energy for the reaction ( G o H o T S o ) can be calculated using H of
and S o values for CO(g), H2(g) and CH3OH(g) from Appendix D.
H o H of (CH 3OH(g)) [ H of (CO(g)) 2 H of (H 2 (g)]
H o 200.7kJmol-1 (110.5kJmol-1 0kJmol-1 ) 90.2kJmol-1
S o S o (CH 3OH(g)) [S o (CO(g)) 2S o (H 2 (g)]
S o 239.8JK 1mol-1 (197.7JK 1mol-1 2 130.7JK 1mol-1 ) 219.3JK -1mol-1
483K (219.3)kJK -1mol-1
o
-1
G 90.2kJmol
15.7kJmol-1
1000
Equilibrium constant for the reaction can be calculated using G o RT ln K
15.7 1000Jmol 1
G o
ln K
3.9 K e3.9 2.0 10 2
ln K
1
1
RT
8.314JK mol 483K
The values are different because in this case, the calculated K is the thermodynamic
equilibrium constant that represents the reactants and products in their standard states. In
Exercise 35, the reactants and products were not in their standard states.
Relationships Involving G, G o, Q and K
37. (M) G o = 2Gf NO g Gf N 2 O g 0.5 Gf O 2 g
= 2 86.55 kJ/mol 104.2 kJ/mol 0.5 0.00 kJ/mol = 68.9 kJ/mol
= RT ln K p = 8.3145 103 kJ mol1 K 1 298 K ln K p
ln K p =
38.
68.9 kJ/mol
= 27.8
8.3145 10 kJ mol1 K 1 298 K
3
K p = e 27.8 = 8 1013
o
(M) (a) G = 2Gf N 2 O5 g 2Gf N 2 O 4 g Gf O 2 g
= 2 115.1 kJ/mol 2 97.89 kJ/mol 0.00 kJ/mol = 34.4 kJ/mol
(b)
G o = RT ln K p
ln K p =
G o
34.4 103 J/mol
=
= 13.9
8.3145 J mol1 K 1 298 K
RT
K p = e 13.9 = 9 107
39.
(M) We first balance each chemical equation, then calculate the value of G o with data
from Appendix D, and finally calculate the value of Keq with the use of G o = RT ln K .
(a)
4HCl g + O 2 g 2H 2 O g + 2Cl2 s
G o = 2Gf H 2 O g + 2Gf Cl2 g 4Gf HCl g Gf O 2 g
kJ
kJ
kJ
kJ
kJ
= 2 228.6
= 76.0
4 95.30
+ 20
0
mol
mol
mol
mol
mol
o
3
G
+76.0 10 J/mol
K = e +30.7 = 2 1013
ln K =
=
= +30.7
1
1
RT
8.3145 J mol K 298 K
965
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
(b)
3Fe 2 O3 s + H 2 g 2Fe3O 4 s + H 2 O g
G o = 2Gf Fe3O 4 s + Gf H 2 O g 3Gf Fe 2 O3 s Gf H 2 g
= 2 1015 kJ/mol 228.6 kJ/mol 3 742.2 kJ/mol 0.00 kJ/mol
= 32 kJ/mol
ln K =
(c)
G o
32 103 J/mol
=
= 13;
RT
8.3145 J mol1 K 1 298 K
K = e +13 = 4 105
2Ag + aq + SO 4 2 aq Ag 2SO 4 s
G o = Gf Ag 2 SO 4 s 2Gf Ag + aq Gf SO 24 aq
= 618.4 kJ/mol 2 77.11kJ/mol 744.5 kJ/mol = 28.1kJ/mol
G o
28.1 103 J/mol
ln K =
=
= 11.3; K = e +11.3 = 8 104
1
1
RT
8.3145 J mol K 298 K
40.
(E) S o = S o {CO g }+ S o {H g } S o {CO g } S o {H O g }
2
2
2
= 213.7 J mol1 K 1 +130.7 J mol1 K 1 197.7 J mol1 K 1 188.8 J mol1K 1
= 42.1 J mol1 K 1
41.
(M) In this problem we need to determine in which direction the reaction
2SO 2 g + O 2 g 2SO3 g is spontaneous when the partial pressure of SO2, O2, and
SO3 are 1.010-4, 0.20 and 0.10 atm, respectively. We proceed by first determining the
standard free energy change for the reaction ( G o ) using tabulated data in Appendix D.
Change in Gibbs free energy for the reaction ( G ) is then calculated by employing the
equation G G o RT ln Q p ,where Qp is the reaction quotient. Based on the sign of
G , we can determine in which direction is the reaction spontaneous.
Stepwise approach:
First determine G o for the reaction using data in Appendix D:
2SO 2 g + O 2 g 2SO3 g
G o = 2Gfo SO3 g 2Gfo SO 2 g Gfo O 2 g
o
G 2 (371.1 kJ/mol) 2 (300.2 kJ/mol) 0.0 kJ/mol
G o 141.8kJ
Calculate G by employing the equation G G o RT ln Q p , where Qp is the reaction
quotient:
G G o RT ln Q p
Qp =
P{SO3 g }2
P{O 2 g }P{SO 2 g }2
966
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
0.10 atm
Qp =
0.20 atm 1.0 104
2
atm
2
5.0 106
G = 141.8 kJ + (8.3145 103 kJ/Kmol )(298 K)ln(5.0 106)
G = 141.8 kJ + 38.2 kJ = 104 kJ.
Examine the sign of G to decide in which direction is the reaction spontaneous:
Since G is negative, the reaction is spontaneous in the forward direction.
Conversion pathway approach:
G o = 2Gfo SO3 g 2Gfo SO 2 g Gfo O 2 g
o
G 2 (371.1 kJ/mol) 2 (300.2 kJ/mol) 0.0 kJ/mol=-141.8kJ
G G o RT ln Q p
Qp
0.10 atm
=
P{O g }P{SO g } 0.20 atm 1.0 10
2
P{SO3 g }2
2
2
2
4
atm
2
5.0 106
3
G = 141.8 kJ + (8.3145 10 kJ/Kmol )(298 K)ln(5.0 106)=-104 kJ.
Since G is negative, the reaction is spontaneous in the forward direction.
42.
(M) We begin by calculating the standard free energy change for the reaction:
H 2 g + Cl2 g 2HCl g
G = 2Gf HCl g Gf Cl2 g Gf H 2 g
2 (95.30 kJ/mol) 0.0 kJ/mol 0.0 kJ/mol 190.6 kJ
Now we can calculate G by employing the equation G G o RT ln Q p , where
0.5 atm
; Qp =
1
Qp =
P{H 2 g }P{Cl2 g }
0.5 atm 0.5 atm
P{HCl g }2
2
G = 190.6 kJ + (8.3145 103 kJ/Kmol )(298 K)ln(1)
G = 190.6 kJ + 0 kJ = 190.6 kJ.
Since G is negative, the reaction is spontaneous in the forward direction.
43.
(M) In order to determine the direction in which the reaction is spontaneous, we need to
calculate the non-standard free energy change for the reaction. To accomplish this, we will
employ the equation G G o RT ln Qc , where
1.0 103 M 1.0 105
[H O + aq ] [CH 3CO 2 aq ]
Qc = 3
; Qc =
(0.10 M)
[CH 3CO 2 H aq ]
2
G = 27.07 kJ + (8.3145 103 kJ/Kmol )(298 K)ln(1.0 105)
G = 27.07 kJ + (28.53 kJ) = 1.46 kJ.
Since G is negative, the reaction is spontaneous in the forward direction.
967
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
44.
(M) As was the case for exercise 39, we need to calculate the non-standard free energy
change for the reaction. Once again, we will employ the equation G G o RT ln Q , but
this time
+
1.0 103 M
[NH 4 aq ] [OH aq ]
1.0 105
Qc =
; Qc =
(0.10 M)
[NH 3 aq ]
2
G = 29.05 kJ + (8.3145 103 kJ/Kmol)(298 K)ln(1.0 105)
G = 29.05 kJ + (28.53 kJ) = 0.52 kJ.
Since G is positive, the reaction is spontaneous in the reverse direction.
45.
(E) The relationship So = (Go Ho)/T (Equation (b)) is incorrect. Rearranging this
equation to put Go on one side by itself gives Go = Ho + TSo. This equation is not
valid. The TSo term should be subtracted from the Ho term, not added to it.
46.
(E) The Go value is a powerful thermodynamic parameter because it can be used to
determine the equilibrium constant for the reaction at each and every chemically
reasonable temperature via the equation Go = RT ln K.
47.
(M) (a) To determine Kp we need the equilibrium partial pressures. In the ideal gas law,
each partial pressure is defined by P = nRT / V . Because R, T, and V are the same for
each gas, and because there are the same number of partial pressure factors in the
numerator as in the denominator of the Kp expression, we can use the ratio of amounts
to determine Kp .
Kp =
(b)
P CO(g) P H 2 O(g)
P CO2 (g) P H 2 (g)
=
n CO(g) n H 2 O(g)
n CO2 (g) n H 2 (g)
=
0.224 mol CO 0.224 mol H 2 O
= 0.659
0.276 mol CO2 0.276 mol H 2
G o1000K = RT ln K p = 8.3145 J mol1 K 1 1000. K ln 0.659
= 3.467 103 J/mol = 3.467 kJ/mol
(c)
0.0340 mol CO 0.0650 mol H 2 O
= 0.31 0.659 = K p
0.0750 mol CO 2 0.095 mol H 2
Since Qp is smaller than Kp , the reaction will proceed to the right, forming products,
Qp =
to attain equilibrium, i.e., G = 0.
48.
(M) (a) We know that K p = K c RT . For the reaction 2SO 2 g + O 2 g 2SO3 g ,
n
ngas = 2 2 +1 = 1 , and therefore a value of Kp can be obtained.
2.8 10 2
3.41 K
0.08206 L atm
1000 K
mol K
We recognize that K = K p since all of the substances involved in the reaction are
K p = K c RT
1
gases. We can now evaluate G o .
968
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
G o = RT ln K eq =
(b)
8.3145 J
1000 K ln 3.41 = 1.02 104 J/mol = 10.2 kJ/mol
mol K
We can evaluate Qc for this situation and compare the value with that of Kc = 2.8 102 to
determine the direction of the reaction to reach equilibrium.
2
0.72 mol SO3
2
2.50 L
[SO3 ]
45 2.8 102 K c
Qc
[SO 2 ]2 [O 2 ] 0.40 mol SO 2 0.18 mol O
2
2
2.50
L
2.50
L
Since Qc is smaller than Kc the reaction will shift right, producing sulfur trioxide
and consuming sulfur dioxide and molecular oxygen, until the two values are equal.
49.
(M) (a) K = K c
G o = RT ln K eq = 8.3145 103 kJ mol1 K 1 445 + 273 K ln 50.2 = 23.4 kJ
(b)
K = K p = K c RT
n g
= 1.7 1013 0.0821 298
1/2
= 8.4 1013
G o = RT ln K p = 8.3145 103 kJ mol1 K 1 298 K ln 8.4 1013
G o = +68.9 kJ/mol
(c)
K = K p = K c RT
n
= 4.61 103 0.08206 298 = 0.113
+1
G o = RT ln K p = 8.3145 10 3 kJ mol1 K 1 298K ln 0.113 = +5.40 kJ/mol
(d)
K = K c = 9.14 10
6
G o = RT ln K c = 8.3145 103 kJ mol1 K 1 298 K ln 9.14 106
G o = +28.7 kJ/mol
50.
(M) (a) The first equation involves the formation of one mole of Mg 2+ aq from
Mg OH 2 s and 2H + aq , while the second equation involves the formation of
only half-a-mole of Mg 2+ aq . We would expect a free energy change of half the
size if only half as much product is formed.
(b)
The value of K for the first reaction is the square of the value of K for the second
reaction. The equilibrium constant expressions are related in the same fashion.
2
1/2
Mg 2 + Mg 2 +
2
=
K1 =
= K 2
+
+ 2
H
H
(c)
The equilibrium solubilities will be the same regardless which expression is used.
The equilibrium conditions (solubilities in this instance) are the same no matter how
we choose to express them in an equilibrium constant expression.
969
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
51.
(E) G o = RT ln K p = 8.3145 103 kJ mol1K 1 298 K ln 6.5 1011 = 67.4 kJ/mol
CO g + Cl2 g COCl2 g
G o = 67.4 kJ/mol
C graphite + 12 O 2 g CO g
Gf = 137.2 kJ/mol
C graphite + 12 O 2 g + Cl2 g COCl2 g
Gf = 204.6 kJ/mol
Gf of COCl2 g given in Appendix D is 204.6 kJ/mol, thus the agreement is excellent.
52.
(M) In each case, we first determine the value of G o for the solubility reaction. From
that, we calculate the value of the equilibrium constant, K sp , for the solubility reaction.
(a)
AgBr s Ag + aq + Br aq
G o = Gf Ag + aq + Gf Br aq Gf AgBr s
= 77.11 kJ/mol 104.0 kJ/mol 96.90 kJ/mol = +70.0 kJ/mol
ln K =
(b)
70.0 103 J/mol
G o
=
= 28.2;
RT
8.3145 J mol1 K 1 298.15 K
K sp = e 28.2 = 6 1013
CaSO 4 s Ca 2+ aq + SO 4 2 aq
o
o
2
o
G o = Gf Ca 2+ aq + Gf SO 4 aq Gf CaSO 4 s
= 553.6 kJ/mol 744.5 kJ/mol 1332 kJ/mol = +34 kJ/mol
34 103 J/mol
G o
ln K =
=
= 14;
RT
8.3145 J mol1 K 1 298.15 K
(c)
K sp = e 14 = 8 107
Fe OH 3 s Fe3+ aq + 3OH aq
G o = Gf Fe 3+ aq + 3 Gf OH aq Gf Fe OH 3 s
= 4.7 kJ/mol + 3 157.2 kJ/mol 696.5 kJ/mol = +220.2 kJ/mol
ln K =
53.
220.2 103 J/mol
G o
=
= 88.83
RT
8.3145 J mol1 K 1 298.15 K
K sp = e 88.83 = 2.6 1039
(M)(a) We can determine the equilibrium partial pressure from the value of the equilibrium
constant.
G o
58.54 103 J/mol
G o = RT ln K p
ln K p =
=
= 23.63
RT
8.3145 J mol1K 1 298.15 K
K p = P{O 2 g }1/2 = e 23.63 = 5.5 1011
(b)
P{O 2 g } = 5.5 1011 = 3.0 1021 atm
2
Lavoisier did two things to increase the quantity of oxygen that he obtained. First, he ran
the reaction at a high temperature, which favors the products (i.e., the side with
molecular oxygen.) Second, the molecular oxygen was removed immediately after it was
formed, which causes the equilibrium to shift to the right continuously (the shift towards
products as result of the removal of the O2 is an example of Le Châtelier's principle).
970
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
54.
(D) (a) We determine the values of H o and S o from the data in Appendix D, and then
the value of G o at 25 C = 298 K.
H o = H f CH 3 OH g + H f H 2 O g H f CO 2 g 3 H f H 2 g
= 200.7 kJ/mol + 241.8 kJ/mol 393.5 kJ/mol 3 0.00 kJ/mol = 49.0 kJ/mol
S = S o CH 3OH g + S o H 2 O g S o CO 2 g 3 S o H 2 g
o
= (239.8 +188.8 213.7 3 130.7) J mol1K 1 = 177.2 J mol1K 1
G o = H o T S o = 49.0 kJ/mol 298 K 0.1772 kJ mol1K 1 = +3.81 kJ/mol
Because the value of G o is positive, this reaction does not proceed in the forward
direction at 25 C .
(b)
Because the value of H o is negative and that of S o is negative, the reaction is nonspontaneous at high temperatures, if reactants and products are in their standard states.
The reaction will proceed slightly in the forward direction, however, to produce an
equilibrium mixture with small quantities of CH 3OH g and H 2 O g . Also, because
the forward reaction is exothermic, this reaction is favored by lowering the
temperature. That is, the value of K increases with decreasing temperature.
(c)
o
G500K
= H o T S o = 49.0 kJ/mol 500.K 0.1772 kJ mol1K 1 = 39.6 kJ/mol
= 39.6 103 J/mol = RT ln K p
ln K p
(d)
G
39.6 103 J/mol
9.53;
RT
8.3145 J mol1 K 1 500. K
Reaction: CO 2 g +
Initial:
1.00 atm
Changes: x atm
Equil: (1.00 x ) atm
K p = 7.3 105 =
K p e 9.53 7.3 105
3H 2 g
CH 3OH g
1.00 atm
0 atm
3x atm
+x atm
x atm
1.00 3x atm
P CH 3OH P H 2 O
P CO 2 P H 2
0 atm
+x atm
x atm
x x
=
x2
1.00 x 1.00 3x
atm P CH 3OH Our assumption, that 3 x 1.00
3
x 7.3 105 8.5 103
+H 2O g
3
atm, is valid.
Go and K as Function of Temperature
55
(M)(a) S o = S o Na 2 CO3 s + S o H 2 O l + S o CO 2 g 2 S o NaHCO3 s
= 135.0
J
K mol
+ 69.91
J
K mol
+ 213.7
971
J
J
2 101.7
= 215.2
K mol
K mol
K mol
J
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
(b)
H o = H f Na 2 CO3 s + H f H 2 O l + H f CO 2 g 2H f NaHCO3 s
= 1131
(c)
kJ
mol
285.8
kJ
mol
393.5
kJ
kJ
2 950.8
= +91
mol
mol
mol
kJ
G o = H o T S o = 91 kJ/mol 298 K 215.2 103 kJ mol1K 1
= 91 kJ/mol 64.13 kJ/mol = 27 kJ/mol
(d)
G o = RT ln K
ln K =
27 103 J/mol
G o
=
= 10.9
RT
8.3145 J mol1 K 1 298 K
K = e 10.9 = 2 105
56.
(M) (a) S o = S o CH3 CH 2 OH g + S o H 2 O g S o CO g 2S o H 2 g S o CH3 OH g
J
J
J
J
J
+188.8
197.7
2 130.7
239.8
K mol
K mol
K mol
K mol
K mol
S o = 282.7
S o = 227.4
J
K mol
H o = H f CH CH
3
H o = 235.1
OH g + H f H 2 O g H f CO g 2H f H 2 g H f CH 3 OH g
2
kJ
kJ
kJ
kJ
kJ
241.8
110.5
2 0.00
200.7
mol
mol
mol
mol
mol
kJ
mol
kJ
= 165.7
mol
H o = 165.7
G o
kJ
kJ
kJ
+ 67.8
= 97.9
298K 227.4 103 KkJmol = 165.4 mol
mol
mol
(b)
H o 0 for this reaction. Thus it is favored at low temperatures. Also, because
ngas = +2 4 = 2 , which is less than zero, the reaction is favored at high pressures.
(c)
First we assume that neither S o nor H o varies significantly with temperature. Then
we compute a value for G o at 750 K. From this value of G o , we compute a value
for Kp .
G o = H o T S o = 165.7 kJ/mol 750.K 227.4 103 kJ mol1 K 1
= 165.7 kJ/mol + 170.6 kJ/mol = +4.9kJ/mol = RT ln K p
ln K p =
57.
4.9 103 J/mol
G o
=
= 0.79
RT
8.3145 J mol1 K 1 750.K
K p = e 0.79 = 0.5
(E) In this problem we are asked to determine the temperature for the reaction between
iron(III) oxide and carbon monoxide to yield iron and carbon dioxide given G o , H o ,
and S o . We proceed by rearranging G o =H o T S o in order to express the
temperature as a function of G o , H o , and S o .
972
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
Stepwise approach:
Rearrange G o =H o T S o in order to express T as a function of G o , H o , and S o :
G o =H o T S o
T S o =H o G o
H o G o
S o
Calculate T:
24.8 103 J 45.5 10 3 J
= 1.36 103 K
T=
15.2 J/K
Conversion pathway approach:
3
3
H o G o 24.8 10 J 45.5 10 J
o
o
o
= 1.36 103 K
G =H T S T
15.2 J/K
S o
T=
58.
(E) We use the van't Hoff equation with H o = 1.8 105 J/mol, T1 = 800 . K,
T2 = 100. C = 373 K, and K1 = 9.1 102 .
ln
K 2 H o 1 1
1
1.8 10 5 J/mol 1
=
=
= 31
1
1
K1
R T1 T2 8.3145 J mol K 800 K 373K
K2
K2
= e31 = 2.9 1013 =
K1
9.1 102
59.
K 2 = 2.9 1013 9.1102 = 3 1016
(M) We first determine the value of G o at 400 C , from the values of H o and S o ,
which are calculated from information listed in Appendix D.
H o = 2H f NH 3 g H f N 2 g 3H f H 2 g
= 2 46.11kJ/mol 0.00 kJ/mol 3 0.00 kJ/mol = 92.22 kJ/mol N 2
S o = 2 S o NH 3 g S o N 2 g 3S o H 2 g
= 2 192.5 J mol1 K 1 191.6 J mol1 K 1 3 130.7 J mol1 K 1 = 198.7 J mol1 K 1
G o = H o T S o = 92.22 kJ/mol 673 K 0.1987 kJ mol1 K 1
= +41.51 kJ/mol = RT ln K p
ln K p =
G o
41.51 103 J/mol
=
= 7.42;
8.3145 J mol1K 1 673K
RT
973
K p = e 7.42 = 6.0 104
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
60.
(M) (a) H o = H CO g + H H g H CO g H H O g
f
2
f 2
f
f 2
= 393.5 kJ/mol 0.00 kJ/mol 110.5 kJ/mol 241.8 kJ/mol = 41.2 kJ/mol
S o = S o CO 2 g + S o H 2 g S o CO g S o H 2 O g
= 213.7 J mol1 K 1 +130.7 J mol1 K 1 197.7 J mol1 K 1 188.8 J mol1 K 1
= 42.1 J mol1 K 1
G o H o T S o 41.2kJ/mol 298.15K (42.1 10 3 )kJ/molK
G o 41.2kJ/mol 12.6kJ/mol 28.6kJ/mol
(b)
G o = H o T S o = 41.2 kJ/mol 875 K 42.1 10 3 kJ mol1 K 1
= 41.2 kJ/mol + 36.8 kJ/mol = 4.4 kJ/mol = RT ln K p
ln K p =
61.
G o
4.4 103 J/mol
=
= +0.60
8.3145 J mol1K 1 875K
RT
K p = e +0.60 = 1.8
(M) We assume that both H o and S o are constant with temperature.
H o = 2H f SO 3 g 2H f SO 2 g H f O 2 g
= 2 395.7 kJ/mol 2 296.8 kJ/mol 0.00 kJ/mol = 197.8 kJ/mol
S o = 2S o SO3 g 2 S o SO 2 g S o O 2 g
= 2 256.8 J mol1 K 1 2 248.2 J mol1 K 1 205.1 J mol1 K 1
= 187.9 J mol1 K 1
G o = H o T S o = RT ln K
T=
H o = T S o RT ln K
T=
H o
S o R ln K
197.8 103 J/mol
650 K
187.9 J mol1 K 1 8.3145 J mol1 K 1 ln 1.0 106
This value compares very favorably with the value of T = 6.37 102 that was obtained in
Example 19-10.
62.
(E) We use the van't Hoff equation to determine the value of H o (448 C = 721 K and
350 C = 623 K) .
K1 H o 1 1
H o 1
1 H o
50.0
ln
=
= ln
= 0.291 =
2.2 10 4
=
K2
R T2 T1
R 623 721
R
66.9
H o
0.291
=
= 1.3 103 K;
R
2.2 104 K -1
H o = 1.3 103 K 8.3145 J K -1 mol-1 = 11 103 J mol-1 = 11 kJ mol-1
974
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
63.
(M) (a) ln
K 2 H o 1 1
57.2 103 J/mol 1
1
=
=
= 2.11
1
1
K1
R T1 T2 8.3145 J mol K 298 K 273 K
K2
= e 2.11 = 0.121
K1
(b)
K 2 = 0.121 0.113 = 0.014 at 273 K
K 2 H o 1 1
1
57.2 10 3 J/mol 1
0.113
ln
=
=
= 2.180
= ln
1
1
K1
R T1 T2 8.3145 J mol K T1 298 K
1.00
1
1 2.180 8.3145 1
K = 3.17 10 4 K 1
3
T 298K =
57.2 10
1
1
1
=
3.17 104 = 3.36 103 3.17 104 = 3.04 103 K 1 ;
T1 298
64.
T1 = 329 K
(D) First we calculate G o at 298 K to obtain a value for Keq at that temperature.
G o = 2Gf NO 2 g 2Gf NO g Gf O 2 g
= 2 51.31 kJ/mol 2 86.55 kJ/mol 0.00 kJ/mol = 70.48 kJ/mol
70.48 103 J/mol K
G
28.43
K e 28.43 2.2 1012
8.3145 J
RT
298.15 K
mol K
Now we calculate H o for the reaction, which then will be inserted into the van't Hoff equation.
H o = 2H f NO 2 g 2H f NO g H f O 2 g
ln K
= 2 33.18 kJ/mol 2 90.25 kJ/mol 0.00 kJ/mol = 114.14 kJ/mol
K 2 H o 1 1 114.14 103 J/mol 1
1
ln
=
=
= 9.26
1
1
K1
R T1 T2 8.3145 J mol K 298 K 373 K
K2
= e 9.26 = 9.5 105 ; K 2 = 9.5 105 2.2 1012 = 2.1108
K1
Another way to find K at 100 ºC is to compute H o 114.14 kJ/mol from Hf values
o
and S o 146.5 J mol1 K 1 from S o values. Then determine Go(59.5 kJ/mol), and
find Kp with the expression G o = RT ln Kp . Not surprisingly, we obtain the same
result, K p = 2.2 108 .
65.
(M) First, the van't Hoff equation is used to obtain a value of H o . 200 C = 473K and
260 C = 533K .
1
H o 1 1
H o
K
2.15 1011
1
ln 2 =
=
ln
=
6.156
=
8
1
1
K1
R T1 T2
4.56 10
8.3145 J mol K 533K 473 K
6.156
6.156 = 2.9 105 H o
H o =
= 2.1 105 J / mol = 2.1 102 kJ / mol
2.9 105
975
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
Another route to H o is the combination of standard enthalpies of formation.
CO g + 3H 2 g CH 4 g + H 2O g
H o = H f CH 4 g + H f H 2 O g H f CO g 3H f H 2 g
= 74.81 kJ/mol 241.8 kJ/mol 110.5 3 0.00 kJ/mol = 206.1 kJ/mol
Within the precision of the data supplied, the results are in good agreement.
(D) (a)
t, C
T,K
1/ T , K 1
30.
303
3.30 103
50.
323
3.10 103
3.90 104
7.849
70.
343
2.92 103
6.27 103
5.072
100.
373
2.68 103
2.31 101
1.465
Kp
lnKp
1.66 105
11.006
Plot of ln(Kp) versus 1/T
0.0026
0.0029
1/T(K-1)
0.0032
0
-2
ln Kp
66.
-4
-6
-8
-10
y = -15402.12x + 39.83
The -12
slope of this graph is H o / R = 1.54 104 K
H o = 8.3145 J mol1K 1 1.54 104 K = 128 103 J/mol = 128 kJ/mol
(b)
When the total pressure is 2.00 atm, and both gases have been produced from NaHCO 3 s ,
P{H 2 O g } = P{CO 2 g } = 1.00 atm
K p = P{H 2 O g }P{CO 2 g } = 1.00 1.00 = 1.00
Thus, ln K p = ln 1.00 = 0.000 . The corresponds to 1/T = 2.59 103 K 1 ;
T = 386 K .
We can compute the same temperature from the van't Hoff equation.
976
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
ln
K 2 H o 1 1
128 10 3 J/mol 1
1.66 10 5
1
=
=
=
ln
= 11.006
K1
R T1 T2 8.3145 J mol1 K 1 T1 303 K
1.00
1
1 11.006 8.3145 1
K = 7.15 10 4 K 1
3
T 303 K =
128 10
1
1
1
7.15 104 = 3.30 103 7.15 104 = 2.59 103 K 1 ; T1 = 386 K
=
T1 303
This temperature agrees well with the result obtained from the graph.
Coupled Reactions
67.
(E) (a) We compute G o for the given reaction in the following manner
H o = H f TiCl4 l + H f O 2 g H f TiO 2 s 2H f Cl2 g
= 804.2 kJ/mol + 0.00 kJ/mol 944.7 kJ/mol 2 0.00 kJ/mol
= +140.5 kJ/mol
S o = S o TiCl4 l + S o O 2 g S o TiO 2 s 2 S o Cl2 g
= 252.3 J mol 1 K 1 + 205.1 J mol1 K 1 50.33 J mol 1 K 1 2 223.1 J mol 1 K 1
= 39.1 J mol1 K 1
G o = H o T S o = +140.5 kJ/mol 298 K 39.1103 kJ mol1 K 1
= +140.5 kJ/mol +11.6 kJ/mol = +152.1 kJ/mol
Thus the reaction is non-spontaneous at 25 C . (we also could have used values of Gf
to calculate G o ).
(b)
For the cited reaction, G o = 2 Gf CO 2 g 2 Gf CO g Gf O 2 g
G o = 2 394.4 kJ/mol 2 137.2 kJ/mol 0.00 kJ/mol = 514.4 kJ/mol
Then we couple the two reactions.
TiO 2 s + 2Cl2 g
TiCl4 l + O 2 g
G o = +152.1 kJ/mol
2CO g + O 2 g
2CO 2 g
G o = 514.4 kJ/mol
_________________________________________________________________
TiO 2 s + 2Cl2 g + 2CO g TiCl4 l + 2CO 2 g ; G o = 362.3 kJ/mol
The coupled reaction has G o 0 , and, therefore, is spontaneous.
977
o
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
68.
(E) If G o 0 for the sum of coupled reactions, the reduction of the oxide with carbon is
spontaneous.
(a)
NiO s Ni s + 12 O 2 g
G o = +115 kJ
C s + 12 O 2 g CO g
G o = 250 kJ
Net : NiO s + C s Ni s + CO g
G o = +115 kJ 250 kJ = 135 kJ
Therefore the coupled reaction is spontaneous
(b)
MnO s Mn s + 12 O 2 g
G o = +280 kJ
C s + 12 O 2 g CO g
G o = 250 kJ
Net : MnO s +C s
Mn s + CO g
G o = +280 kJ 250 kJ = +30 kJ
Therefore the coupled reaction is non-spontaneous
(c)
TiO 2 s
Ti s + O 2 g
G o = +630 kJ
2C s + O 2 g
2CO g
G o = 2 250 kJ = 500 kJ
Net : TiO 2 s + 2C s
Ti s + 2CO g
G o = +630 kJ 500 kJ = +130 kJ
Therefore the coupled reaction is non-spontaneous
69.
(E) In this problem we need to determine if the phosphorylation of arginine with ATP is a
spontaneous reaction. We proceed by coupling the two given reactions in order to
calculate Gto for the overall reaction. The sign of Gto can then be used to determine
whether the reaction is spontaneous or not.
Stepwise approach:
First determine Gto for the coupled reaction:
ADP+P
ATP+H 2O
G to 31.5kJmol-1
arginine+P
phosphorarginine+H 2O G to 33.2kJmol-1
___________________________________________
ATP+arginine
phosphorarginine+ADP
G o (31.5 33.2)kJmol-1 1.7kJmol-1
Examine the sign of Gto :
Gto 0 . Therefore, the reaction is not spontaneous.
Conversion pathway approach:
Gto for the coupled reaction is:
ATP+arginine
phosphorarginine+ADP
G o (31.5 33.2)kJmol-1 1.7kJmol-1
Since Gto 0 ,the reaction is not spontaneous.
978
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
70.
(E) By coupling the two reactions, we obtain:
Glu - +NH +4
Gln+H 2 O G to 14.8kJmol-1
ATP+H 2O
ADP+P G to 31.5kJmol-1
_________________________________________
Glu - +NH +4 +ATP
Gln+ADP+P G o (14.8 31.5)kJmol-1 16.7kJmol-1
Therefore, the reaction is spontaneous.
INTEGRATIVE AND ADVANCED EXERCISES
71. (M) (a) The normal boiling point of mercury is that temperature at which the mercury
vapor pressure is 1.00 atm, or where the equilibrium constant for the vaporization
equilibrium reaction has a numerical value of 1.00. This is also the temperature where
G 0 , since G RT ln K eq and ln (1.00) 0 .
Hg(l) Hg(g)
H H f [Hg(g)] H f [Hg(l)] 61.32 kJ/mol 0.00 kJ/mol 61.32 kJ/mol
S S [Hg(g)] S [Hg(l)] 175.0 J mol 1 K 1 76.02 J mol 1 K 1 99.0 J mol 1 K 1
0 H T S 61.32 10 3 J/mol T 99.0 J mol 1 K 1
61.32 10 3 J/mol
T
619 K
99.0 J mol 1 K 1
(b) The vapor pressure in atmospheres is the value of the equilibrium constant, which is
related to the value of the free energy change for formation of Hg vapor.
Gf [Hg(g)] 31.82 kJ/mol RT ln K eq
31.82 103 J/mol
K e 12.84 2.65 106 atm
12.84
8.3145 J mol1 K 1 298.15 K
Therefore, the vapor pressure of Hg at 25ºC is 2.65×10-6 atm.
ln K
72. (M) (a)
(b) FALSE;
TRUE;
It is the change in free energy for a process in which
reactants and products are all in their standard states (regardless of
whatever states might be mentioned in the statement of the problem).
When liquid and gaseous water are each at 1 atm at 100 °C (the normal
boiling point), they are in equilibrium, so that G = G° = 0 is only true
when the difference of the standard free energies of products and reactants
is zero. A reaction with G = 0 would be at equilibrium when products
and reactants were all present under standard state conditions and the
pressure of H2O(g) = 2.0 atm is not the standard pressure for H2O(g).
G 0. The system is not at equilibrium.
979
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
(c) FALSE;
G° can have only one value at any given temperature, and that is the
value corresponding to all reactants and products in their standard states,
so at the normal boiling point G° = 0 [as was also the case in answering
part (a)]. Water will not vaporize spontaneously under standard conditions
to produce water vapor with a pressure of 2 atmospheres.
G 0. The process of transforming water to vapor at 2.0 atm pressure
at 100°C is not a spontaneous process; the condensation (reverse) process
is spontaneous. (i.e. for the system to reach equilibrium, some H2O(l)
must form)
73. (D) G 12 Gf [Br2 (g)] 12 Gf [Cl 2 (g)] Gf [BrCl(g)]
12 (3.11 kJ/mol) 12 (0.00 kJ/mol) (0.98 kJ/mol) 2.54 kJ/mol RT ln K p
(d) TRUE;
ln K p
2.54 10 3 J/mol
G
1.02
RT
8.3145 J mol 1 K 1 298.15 K
K p e 1.02 0.361
For ease of solving the problem, we double the reaction, which squares the value of the
equilibrium constant. K eq (0.357) 2 0.130
Reaction: 2 BrCl(g) Br2 (g) Cl2 (g)
Initial:
1.00 mol
0 mol
0 mol
x mol
x mol
Changes: 2 x mol
Equil:
(1.00 2 x)mol x mol
x mol
P{Br2 (g)} P{Cl 2 (g)} [n{Br2 (g)}RT / V ][n{Cl 2 (g)}RT / V ] n{Br2 (g)} n{Cl 2 (g)}
Kp
[n{BrCl(g)}RT / V ] 2
P{BrCl(g)}2
n{BrCl(g)}2
x2
x
(0.361) 2
0.361
x 0.361 0.722 x
2
1.00 2 x
(1.00 2 x)
0.361
x
0.210 mol Br2 0.210 mol Cl 2 1.00 2 x 0.580 mol BrCl
1.722
74. (M) First we determine the value of Kp for the dissociation reaction. If I2(g) is 50% dissociated,
then for every mole of undissociated I2(g), one mole of I2(g) has dissociated, producing two
moles of I(g). Thus, the partial pressure of I(g) is twice the partial pressure of I2(g)
( I 2 (g) 2 I(g) ).
Ptotal 1.00 atm PI 2 (g) PI(g) PI 2 (g) 2 PI 2 (g) 3 PI 2 (g)
PI 2 (g ) 0.333 atm
Kp
PI(g)
2
PI 2 (g )
(0.667) 2
1.34
0.333
ln K p 0.293
H 2 H f [I(g)] H f [I 2 (g)] 2 106.8 kJ/mol 62.44 kJ/mol 151.2 kJ/mol
S 2 S [I(g)] S [I 2 (g)] 2 180.8 J mol 1 K 1 260.7 J mol 1 K 1 100.9 J mol 1 K 1
Now we equate two expressions for G and solve for T.
980
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
G H T S RT ln K p 151.2 10 3 100.9T 8.3145 T 0.293
151.2 10 3
1535 K 1.5 10 3 K
98.5
75. (M) (a) The oxide with the most positive (least negative) value of Gf is the one that
most readily decomposes to the free metal and O2(g), since the decomposition is the
reverse of the formation reaction. Thus the oxide that decomposes most readily is
Ag2O(s).
151.2 10 3 100.9 T 2.44 T 98.4 T
T
(b) The decomposition reaction is 2 Ag 2 O(s)
4 Ag(s) O 2 (g) For this reaction
K p PO 2 (g) . Thus, we need to find the temperature where K p 1.00 .Since
G RT ln K p
and ln (1.00) 0 , we wish to know the temperature where G 0 .
Note also that the decomposition is the reverse of the formation reaction. Thus, the
following values are valid for the decomposition reaction at 298 K.
H 31.05 kJ/mol
G 11.20 kJ/mol
We use these values to determine the value of S for the reaction.
H G
G H T S T S H G S
T
3
3
31.05 10 J/mol 11.20 10 J/mol
S
66.6 J mol 1 K 1
298 K
Now we determine the value of T where G 0 .
H G 31.05 10 3 J/mol 0.0 J/mol
T
466 K 193 C
S
66.6 J mol 1 K 1
76. (M) At 127 °C = 400 K, the two phases are in equilibrium, meaning that
G rxn 0 H rxn TS rxn [H f (yellow) H f (red)] T [ S (yellow) S (red)]
[102.9 (105.4)] 10 3 J 400 K [ S ( yellow) 180 J mol 1 K 1 ]
2.5 10 3 J/mol 400 K S ( yellow) 7.20 10 4 J/mol
(7.20 10 4 2.5 10 3 ) J/mol
S ( yellow)
186 J mol 1 K 1
400 K
Then we compute the value of the “entropy of formation” of the yellow form at 298 K.
S f S [HgI 2 ] S [Hg(l)] S [I 2 (s)] [186 76.02 116.1] J mol1 K 1 6 J mol1 K 1
Now we can determine the value of the free energy of formation for the yellow form.
kJ
J
1kJ
kJ
Gf H f T S f 102.9
[298 K ( 6
)
] 101.1
mol
K mol 1000 J
mol
77. (M) First we need a value for the equilibrium constant. 1% conversion means that 0.99 mol
N2(g) are present at equilibrium for every 1.00 mole present initially.
K Kp
PNO(g) 2
PN2 (g) PO2 (g)
[n{NO(g)}RT/V]2
n{NO(g)}2
[n{N 2 (g)}RT/V][n{O 2 (g)}RT/V] n{N 2 (g)} n{O 2 (g)}
981
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
Reaction:
Initial:
Changes(1% rxn):
Equil:
N 2 (g)
O 2 (g)
1.00 mol 1.00 mol
0.010 mol 0.010 mol
0.99 mol
0.99 mol
2 NO(g)
0 mol
0.020 mol
0.020 mol
K
(0.020) 2
4.1 104
(0.99)(0.99)
The cited reaction is twice the formation reaction of NO(g), and thus
H = 2H°f [NO(g)] = 2 90.25 kJ/mol = 180.50 kJ/mol
S° = 2S°[NO(g)] - S°[N 2 (g)] - S°[O 2 (g)]
=2 (210.7 J mol-1 K)-191.5 J mol-1 K -1 -205.0 J mol-1 K -1 = 24.9 J mol-1 K -1
G° = -RTlnK = -8.31447 JK -1mol-1 (T)ln(4.110-4 ) = 64.85(T)
G° = H°-TS° = 64.85(T) = 180.5 kJ/mol - ( T)24.9 J mol-1 K -1
180.5 10 J/mol = 64.85(T) + ( T)24.9 J mol-1 K -1 = 89.75(T)
T = 2.01 103 K
2+
78. (E) Sr(IO3)2(s)
Sr (aq) + 2 IO3 (aq)
G = (2 mol×(-128.0 kJ/mol) + (1 mol ×-500.5 kJ/mol)) (1 mol × -855.1 kJ/mol) = -0.4 kJ
G = -RTlnK = -8.31447 JK-1mol-1(298.15 K)ln K = -0.4 kJ = -400 J
ln K = 0.16 and K = 1.175 = [Sr2+][IO3-]2
Let x = solubility of Sr(IO3)2
[Sr2+][IO3-]2 = 1.175 = x(2x)2 = 4x3 x = 0.665 M for a saturated solution of Sr(IO3)2.
79. (M) PH2O = 75 torr or 0.0987 atm.
K = (PH2O)2 = (0.0987)2 = 9.74 × 10-3
G = (2 mol×(-228.6 kJ/mol) + (1 mol)×-918.1 kJ/mol) – (1 mol)×-1400.0 kJ/mol = 309.0 kJ
H = (2 mol×(-241.8 kJ/mol) + (1 mol)×-1085.8.1 kJ/mol) – (1 mol)×-1684.3 kJ/mol = 114.9 kJ
S = (2 mol×(188.J/K mol) + (1 mol×146.J/K mol) + (1 mol)×221.3.J/K mol = 302.3 J/K mol
G = -RT ln K eq = -8.3145 J mol-1 K -1 T ln (9.74 10-3 ) = 38.5(T)
G° = 38.5(T) = H°-TS° = 114,900 J mol-1 - (T) 302.3J K -1 mol-1
114,900 = 340.8 K -1 (T) Hence: T = 337 K = 64 o C
131 103 J/mol
G
80. (D) (a)
G RT ln K
52.8
ln K
RT
8.3145 J mol1 K 1 298.15 K
K e52.8 1.2 1023 atm
760 mmHg
8.9 1021 mmHg
1atm
Since the system cannot produce a vacuum lower than 10–9 mmHg, this partial pressure
of CO2(g) won't be detected in the system.
982
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
(b) Since we have the value of G for the decomposition reaction at a specified
temperature (298.15 K), and we need H and S for this same reaction to
determine P{CO 2 (g)} as a function of temperature, obtaining either H or S will
enable us to determine the other.
(c) H H f [CaO(s)] H f [CO 2 (g)] H f [CaCO 3 (s)]
635.1 kJ/mol 393.5 kJ/mol (1207 kJ/mol) 178 kJ/mol
G H T S
S
T S H G
H G
T
178 kJ/mol 131 kJ/mol 1000 J
1.6 102 J mol1 K 1
298. K
1 kJ
K 1.0 109 mmHg
1 atm
1.3 1012
760 mmHg
H T S RT ln K
G H T S RT ln K eq
T
S
178 103 J/mol
H
4.6 102 K
S R ln K 1.6 102 J mol1 K 1 8.3145 J mol1 K 1 ln(1.3 1012 )
81. (D) H H f [PCl 3 (g)] H f [Cl 2 (g)] H f [PCl 5 (g)]
287.0 kJ/mol 0.00 kJ/mol (374.9 kJ/mol) 87.9 kJ/mol
S S [PCl3 (g)] S [Cl2 (g)] S [PCl5 (g)]
311.8 J mol-1 K -1 + 223.1 J mol-1 K -1 364.6 J mol-1 K -1 = +170.3 J mol-1 K -1
G H T S 87.9 10 3 J/mol 500 K 170.3 J mol 1 K 1
Gº 2.8 103 J/mol RT ln K p
ln K p
2.8 10 J/mol
G
0.67
RT
8.3145 J mol1 K 1 500 K
Pi [PCl 5 ]
Reaction:
Initial:
Changes:
Equil:
K p e0.67 0.51
nRT 0.100 mol 0.08206 L atm mol 1 K 1 500 K
2.74 atm
V
1.50 L
PCl5 (g)
2.74 atm
x atm
(2.74 x) atm
PCl3 (g)
Cl2 (g)
0 atm
x atm
x atm
0 atm
x atm
x atm
983
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
Kp
P[PCl3 ] P[Cl2 ]
xx
0.51
P[PCl5 ]
2.74 x
x 2 0.51(2.74 x) 1.4 0.51 x x 2 0.51 x 1.4 0
b b 2 4ac 0.51 0.26 5.6
0.96 atm, 1.47 atm
2a
2
Ptotal PPCl5 PPCl3 PCl 2 (2.74 x) x x 2.74 x 2.74 0.96 3.70 atm
x
82. (M) The value of H determined in Exercise 64 is H 128 kJ/mol . We use any one of
the values of K p K eq to determine a value of G . At 30 °C = 303 K,
G RT ln K eq (8.3145 J mol 1 K 1 )(303 K ) ln(1.66 10 5 ) 2.77 10 4 J/mol
Now we determine S . G H T S
H G 128 10 3 J/mol 2.77 10 4 J/mol
S
331 J mol 1 K 1
T
303 K
By using the appropriate S° values in Appendix D, we calculate S 334 J mol 1 K 1 .
83. (M) In this problem we are asked to estimate the temperature at which the vapor pressure of
cyclohexane is 100 mmHg. We begin by using Trouton’s rule to determine the value of
H vap for cyclohexane. The temperature at which the vapor pressure is 100.00 mmHg can
then be determined using Clausius–Clapeyron equation.
Stepwise approach:
Use Trouton’s rule to find the value of H vap :
H vap Tnbp S vap 353.9 K 87 J mol 1 K 1 31 10 3 J/mol
Next, use Clausius–Clapeyron equation to find the required temperature:
100 mmHg
P H vap 1 1
ln 2
ln
760 mmHg
P1
R T1 T2
31 10 3 J/mol
1
1
2.028
1
1
8.3145 J mol K 353.9 K T
1 2.028 8.3145
1
5.4 10 4
31 10 3
353.9 T
1
1
2.826 10 3
3.37 10 3 K 1
T
T
T 297 K 24 C
984
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
Conversion pathway approach:
P H vap 1 1 Tnbp Svap 1 1
ln 2
T T
P1
R T1 T2
R
1
2
1 1
R
P2
1
1
R
P2
T T T S ln P T T T S ln P
1
2
nbp
vap
1
2
1
nbp
vap
1
1
1
8.314JK -1mol-1
100mmHg
ln
3.37 10 3
-1
-1
T2 353.9K 353.9K 87JK mol
760mmHg
T2 297K=24 oC
84.
92. (M)(a)
1
2Ag(s) O 2 (g) Ag 2 O
2
G of G of (Ag 2 O(s))-{2G of ( Ag(s))
1
G of (O 2 )}
2
1
G of -11.2 kJ-{2(0) (0)} 11.2 kJ Ag 2 O is thermodynamically stable at 25oC
2
o
o
(b) Assuming H , S are constant from 25-200oC (not so, but a reasonable assumption !)
1 o
1
S (O 2 )} 121.3 (2(42.6) (205.1)) 66.5 J/K
2
2
(473 K)(-66.5 J/K)
G o 31.0 kJ
H o T S o 0.45 kJ
1000 J/kJ
thermodynamically unstable at 200oC
So So (Ag 2 O)-{2 So (Ag(s))
85. (M) G = 0 since the system is at equilibrium. As well, G = 0 because this process is
under standard conditions. Since G = H - TS = 0. = H = TS =273.15 K×21.99 J
Since we are dealing with 2 moles of ice melting, the
K-1 mol-1 = 6.007 kJ mol-1.
values of H and S are doubled. Hence, H = 12.01 kJ and S = 43.98 J K-1.
Note: The densities are not necessary for the calculations required for this question.
86. (D) First we determine the value of Kp that corresponds to 15% dissociation. We represent
the initial pressure of phosgene as x atm.
Reaction: COCl2 (g) CO(g)
Initial :
Changes:
Equil:
x atm
0.15 x atm
0.85 x atm
Cl2 (g)
0 atm
0 atm
0.15 x atm 0.15 x atm
0.15 x atm
0.15 x atm
985
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
Ptotal 0.85 x atm 0.15 x atm 0.15 x atm 1.15 x atm 1.00 atm
Kp
PCO PCl2
PCOCl 2
x
1.00
0.870 atm
1.15
(0.15 0.870) 2
0.0230
0.85 0.870
Next we find the value of H for the decomposition reaction.
ln
K 1 H 1 1
6.7 10 9
H 1
1 H
ln
15.71
(1.18 10 3 )
2
K2
R T2 T1
R 668 373.0
R
4.44 10
H
15.71
1.33 10 4 ,
3
R
1.18 10
H 1.33 10 4 8.3145 111 10 3 J/mol 111 kJ/mol
And finally we find the temperature at which K = 0.0230.
K1 H 1 1
0.0230
111 103 J/mol 1
1
ln
0.658
1
1
K2
R T2 T1
0.0444 8.3145 J mol K 668 K T
1
1 0.658 8.3145
1
1
4.93 10 5 1.497 103
1.546 10 3
3
T
T
668 T
111 10
T 647 K 374 C
ln
87. (D) First we write the solubility reaction for AgBr. Then we calculate values of H and
S for the reaction: AgBr(s) Ag (aq) Br (aq) K eq K sp [Ag ][Br ] s 2
H H f [Ag (aq)] H f [Br (aq)] H f [AgBr(s)]
105.6 kJ/mol 121.6 kJ/mol (100.4 kJ/mol) 84.4 kJ/mol
S S [Ag (aq)] S [Br (aq)] S [AgBr(s)]
72.68 J mol 1 K 1 82.4 J mol 1 K 1 107.1 J mol 1 K 1 48.0 J mol 1 K 1
These values are then used to determine the value of G for the solubility reaction, and the
standard free energy change, in turn, is used to obtain the value of K.
G H T S 84.4 103 J mol1 (100 273) K 48.0 J mol1 K 1 66.5 103 J/mol
66.5 103
G
ln K
21.4
K K sp e 21.4 5.0 1010 s 2
8.3145 J
RT
373K
mol K
And now we compute the solubility of AgBr in mg/L.
187.77 g AgBr 1000 mg
s 5.0 10 10
4.2 mg AgBr/L
1 mol AgBr
1g
986
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
88. (M) S298.15 = S274.68 + Sfusion + Sheating
298.15
12,660 J mol-1
J
J
S298.15 = 67.15 J K- mol-1 +
+
97.78
+ 0.0586
T 274.68
274.68
274.68 K
mol K
mol K 2
S298.15 = 67.15 J K-1 mol-1 + 46.09 J K-1 mol-1 + 8.07 J K-1 mol-1 = 121.3 J K-1 mol-1
89. (M) S = Ssolid + Sfusion + Sheating + Svaporization + Spressure change
J
134.0
dT
-1
298.15
9866 J mol
33,850 J mol-1
-1
-1
mol
K
+
+
S = 128.82 J K mol +
278.68
298.15 K
T
278.68 K
95.13 torr
+ 8.3145 J K-1 mol-1 × ln
760 torr
S = 128.82 J K-1 mol-1 + 35.40 J K-1 mol-1 + 9.05 J K-1 mol-1 + 113.5 J K-1 mol-1 + (-17.28 J K-1 mol-1 )
S = 269.53 J K-1 mol-1
90. (D) Start by labeling the particles A, B and C. Now arrange the particles among the states.
One possibility includes A, B, and C occupying one energy state (=0,1,2 or 3). This counts as
one microstate. Another possibility is two of the particles occupying one energy state with the
remaining one being in a different state. This possibility includes a total of three microstates.
The final set of combinations is one with each particle being in different energy state. This
combination includes a total of six microstates. Therefore, the total number of microstates in the
system is 10. See pictorial representation on the following page illustrating the three different
cases.
987
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
All particles are in a single
energy level ( i.e. =0)
=3
=2
=1
=0
1 microstate
A
B
C
Two particles are in one energy level (i.e. =0) and the remaining
particle is in a different level (i.e. =2)
=3
=3
C
=2
=2
=1
=0
=3
B
=1
B
A
=0
1 microstate
A
=2
=1
C
A
=0
1 microstate
3 microstates
C
B
1 microstate
All particles are in different energy levels (i.e. =0, 1 and 2)
=3
=2
=1
=0
=3
C
B
A
=2
=1
=0
1 microstate
=3
=2
=1
=0
=3
C
A
B
C
A
1 microstate
=1
=0
1 microstate
=3
B
=2
=2
=1
=0
B
A
C
1 microstate
6 microstates
=3
A
C
B
=2
=1
=0
1 microstate
A
B
C
1 microstate
91. (M) (a) In the solid as the temperature increases, so do the translational, rotational, and
vibrational degrees of freedom. In the liquid, most of the vibrational degrees of freedom are
saturated and only translational and rotational degrees of freedom can increase. In the gas phase,
all degrees of freedom are saturated. (b) The increase in translation and rotation on going from
solid to liquid is much less than on going from liquid to gas. This is where most of the change in
entropy is derived.
988
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
92. (D) Because KNO3 is a strong electrolyte, its solution reaction will be:
KNO3 ( s ) H 2O K (aq ) NO3 (aq )
This reaction can be considered to be at equilibrium when the solid is in contact with a
saturated solution, i.e. the conditions when crystallization begins. The solubility, s, of the
salt, in moles per liter, can be calculated from the amount of salt weighted out and the
volume of the solution. The equilibrium constant K for the reaction will be:
K=[K + (aq)][NO-3 (aq)]=(s)(s)=s 2
In the case of 25.0 mL solution at 340 K, the equilibrium constant K is:
m
20.2g
n 0.200mol
n(KNO3 )
0.200mol s=
8.0molL-1
1
M 101.103gmol
V
0.0250L
K s 2 8 2 64
The equilibrium constant K can be used to calculate G for the reaction using G=-RTlnK:
G 8.314JK -1mol-1 340K ln 64 12kJmol-1
The values for K and G at all other temperatures are summarized in the table below.
Volume (mL)
25.0
29.2
33.4
37.6
41.8
46.0
51.0
T/(K)
340
329
320
313
310
306
303
1/T (K-1)
0.00294
0.00304
0.00312
0.00320
0.00322
0.00327
0.00330
s (molL-1)
8.0
6.9
6.0
5.3
4.8
4.3
3.9
K
64
48
36
28
23
18.5
15
lnK
4.2
3.9
3.6
3.3
3.1
2.9
2.7
G (kJmol-1)
-12
-11
-9.6
-8.6
-8.0
-7.4
-6.8
The plot of lnK v.s. 1/T provides H (slope=-H/R) and S (y-intercept=S/R) for the
reaction:
4.5
y = 16.468 - 4145.7x R= 0.99156
ln K
4
3.5
3
2.5
0.0029
0.003
0.0031
0.0032
0.0033
0.0034
-1
1/T (K )
H slope R 4145.7 8.314JK mol 34.5kJmol1
-1
-1
S y int ercept R 16.468 8.314JK -1mol-1 136.9JK -1mol-1
989
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
H for the crystallization process is -35.4 kJmol-1. It is negative as expected because
crystallization is an exothermic process. Furthermore, the positive value for S shows that
crystallization is a process that decreases the entropy of a system.
FEATURE PROBLEMS
93.
(M) (a) The first method involves combining the values of Gfo . The second uses
G o = H o T S o
G o = Gf H 2 O g Gf H 2 O l
= 228.572 kJ/mol 237.129 kJ/mol = +8.557 kJ/mol
H o = H f H 2 O g H f H 2 O l
= 241.818 kJ/mol 285.830 kJ/mol = +44.012 kJ/mol
S = S o H 2 O g S o H 2 O l
o
= 188.825 J mol1 K 1 69.91 J mol1 K 1 = +118.92 J mol1 K 1
G o = H o T S o
= 44.012 kJ/mol 298.15 K 118.92 103 kJ mol1 K 1 = +8.556 kJ/mol
(b) We use the average value: G o = +8.558 103 J/mol = RT ln K
8558 J/mol
ln K =
= 3.452; K = e 3.452 = 0.0317 bar
8.3145 J mol1 K 1 298.15 K
(c)
P{H 2 O} = 0.0317 bar
(d)
ln K =
1atm
760 mmHg
= 23.8 mmHg
1.01325 bar
1atm
8590 J/mol
= 3.465 ;
8.3145 J mol1 K 1 298.15 K
K = e 3.465 = 0.03127 atm ;
760 mmHg
P{H 2 O} = 0.0313atm
= 23.8 mmHg
1atm
94.
(D) (a) When we combine two reactions and obtain the overall value of G o , we subtract
the value on the plot of the reaction that becomes a reduction from the value on the
plot of the reaction that is an oxidation. Thus, to reduce ZnO with elemental Mg, we
subtract the values on the line labeled “ 2 Zn + O 2 2 ZnO ” from those on the line
labeled “ 2 Mg + O 2 2 MgO ”. The result for the overall G o will always be
negative because every point on the “zinc” line is above the corresponding point on
the “magnesium” line
(b)
In contrast, the “carbon” line is only below the “zinc” line at temperatures above about
1000 C . Thus, only at these elevated temperatures can ZnO be reduced by carbon.
990
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
The decomposition of zinc oxide to its elements is the reverse of the plotted reaction,
the value of G o for the decomposition becomes negative, and the reaction becomes
spontaneous, where the value of G o for the plotted reaction becomes positive. This
occurs above about 1850 C .
(d)
The “carbon” line has a negative slope, indicating that carbon monoxide becomes
more stable as temperature rises. The point where CO(g) would become less stable
than 2C(s) and O2(g) looks to be below 1000 C (by extrapolating the line to lower
temperatures). Based on this plot, it is not possible to decompose CO(g) to C(s) and
O 2 g in a spontaneous reaction.
(e)
0
Standard Free Energy Change
(kJ)
(c)
500
1000
1500
2000
0
-100
1
-200
Reaction 1
2 CO(g) + O2(g) 2 CO2(g)
-300
2
-400
Reaction 2
C(s) + O2(g) CO2(g)
-500
3
-600
-700
Reaction 3
2 C(s) + O2(g) 2 CO(g)
Temperature (°C)
All three lines are straight-line plots of G vs. T following the equation G = H TS.
The general equation for a straight line is given below with the slightly modified
Gibbs Free-Energy equation as a reference: G = STH (here H assumed
constant)
y = mx + b
(m = S = slope of the line)
Thus, the slope of each line multiplied by minus one is equal to the S for the oxide
formation reaction. It is hardly surprising, therefore, that the slopes for these lines
differ so markedly because these three reactions have quite different S values (S
for Reaction 1 = -173 J K1, S for Reaction 2 = 2.86 J K1, S for Reaction 3
= 178.8 J K1)
(f)
Since other metal oxides apparently have positive slopes similar to Mg and Zn, we
can conclude that in general, the stability of metal oxides decreases as the
temperature increases. Put another way, the decomposition of metal oxides to their
elements becomes more spontaneous as the temperature is increased. By contrast,
the two reactions involving elemental carbon, namely Reaction 2 and Reaction 3,
have negative slopes, indicating that the formation of CO2(g) and CO(g) from
graphite becomes more favorable as the temperature rises. This means that the G
for the reduction of metal oxides by carbon becomes more and more negative with
991
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
increasing temperature. Moreover, there must exist a threshold temperature for each
metal oxide above which the reaction with carbon will occur spontaneously. Carbon
would appear to be an excellent reducing agent, therefore, because it will reduce
virtually any metal oxide to its corresponding metal as long as the temperature
chosen for the reaction is higher than the threshold temperature (the threshold
temperature is commonly referred to as the transition temperature).
Consider for instance the reaction of MgO(s) with graphite to give CO2(g) and Mg metal:
2 MgO(s) + C(s) 2 Mg(s) + CO2(g) Srxn = 219.4 J/K and rxn = 809.9 kJ
o
H rxn
809.9 kJ
Ttransition =
=
= 3691 K = Tthreshold
o
0.2194 kJ K 1
Srxn
Consequently, above 3691 K, carbon will spontaneously reduce MgO to Mg metal.
With a 36% efficiency and a condenser temperature (Tl) of 41 °C = 314 K,
Th 314
T T
efficiency h 1 100% 36%
0.36 ;
Th
Th
Th = (0.36 T h) + 314 K; 0.64 Th = 314 K; Th = 4.9 102 K
95. (D) (a)
(b)
The overall efficiency of the power plant is affected by factors other than the
thermodynamic efficiency. For example, a portion of the heat of combustion of the
fuel is lost to parts of the surroundings other than the steam boiler; there are
frictional losses of energy in moving parts in the engine; and so on. To compensate
for these losses, the thermodynamic efficiency must be greater than 36%. To obtain
this higher thermodynamic efficiency, Th must be greater than 4.9 102 K.
(c)
The steam pressure we are seeking is the vapor pressure of water at 4.9 102 K. We
also know that the vapor pressure of water at 373 K (100 °C) is 1 atm. The enthalpy of
vaporization of water at 298 K is H° = Hf°[H2O(g) Hf°[H2O(l)] = 241.8 kJ/mol
(285.8 kJ/mol) = 44.0 kJ/mol. Although the enthalpy of vaporization is somewhat
temperature dependent, we will assume that this value holds from 298 K to 4.9 102
K, and make appropriate substitutions into the Clausius-Clapeyron equation.
P2
1
44.0 kJ mol1
1
3
3
3
ln
=
= 5.29 10 2.68 10 2.04 10
3
1
1 atm 8.3145 10 kJ mol 373 K 490 K
P
P
ln 2 = 3.39 ; 2 = 29.7 ; P2 30 atm
1 atm
1 atm
The answer cannot be given with greater certainty because of the weakness of the
assumption of a constant H°vapn.
(d)
It is not possible to devise a heat engine with 100% efficiency or greater. For 100%
efficiency, Tl = 0 K, which is unattainable. To have an efficiency greater than 100 %
would require a negative Tl, which is also unattainable.
992
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
96. (D) (a) Under biological standard conditions:
ADP3– + HPO42– + H+ ATP4–+ H2O Go' = 32.4 kJ/mol
If all of the energy of combustion of 1 mole of glucose is employed in the
conversion of ADP to ATP, then the maximum number of moles ATP produced is
2870 kJ mol1
88.6 moles ATP
32.4 kJ mol1
In an actual cell the number of ATP moles produced is 38, so that the efficiency is:
number of ATP's actually produced
38
Efficiency =
0.43
Maximum number of ATP's that can be produced 88.6
Thus, the cell’s efficiency is about 43%.
Maximum number =
(b)
(c)
The previously calculated efficiency is based upon the biological standard state. We
now calculate the Gibbs energies involved under the actual conditions in the cell. To
do this we require the relationship between G and G' for the two coupled
reactions. For the combustion of glucose we have
6
aCO
aH6 O
o
2
2
G G RT ln
6
aglu aO
2
For the conversion of ADP to ATP we have
aATP aH O
o
2
G G RT ln
aADP aP H + / 107
i
Using the concentrations and pressures provided we can calculate the Gibbs energy
for the combustion of glucose under biological conditions. First, we need to replace
the activities by the appropriate effective concentrations. That is,
6
p / po
aH6 O
2
CO
2
G G o RT ln
6
o
o
glu / glu p / p O
2
using aH2O 1 for a dilute solution we obtain
0.050 bar /1 bar 6 16
G G RT ln
glu /1 0.132 bar /1 bar 6
The concentration of glucose is given in mg/mL and this has to be converted to
molarity as follows:
1.0 mg
g
1000 mL
1
[glu] =
0.00555 mol L1 ,
1
mL 1000 mg
L
180.16 g mol
where the molar mass of glucose is 180.16 g mol1.
Assuming a temperature of 37 oC for a biological system we have, for one mole of
glucose:
993
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
6
0.050 16
G 2870 103 J 8.314 JK 1 310K ln
0.00555 /1 0.132 6
3
2.954 10
G 2870 103 J 8.314 JK 1 310K ln
0.00555
G 2870 103 J 8.314 JK 1 310 K ln 0.5323
G 2870 103 J 8.314 JK 1 310K 0.6305
G 2870 103 J 1.625 103 J
G 2872 103 J
In a similar manner we calculate the Gibbs free energy change for the conversion of
ADP to ATP:
ATP / 1 1
G G o RT ln
7
ADP / 1 Pi / 1 ( H / 10 )
0.0001
G 32.4 103 J 8.314 JK 1 310 K ln
7
7
0.0001 0.0001 (10 /10 )
G 32.4 103 J 8.314 JK 1 310 K ln 104
G 32.4 103 J 8.314 JK 1 310 K 9.2103) 32.4 103 J 23.738 103 J 56.2 103 J
(d) The efficiency under biological conditions is
Efficiency =
number of ATP's actually produced
38
0.744
Maximum number of ATP's that can be produced 2872/56.2
Thus, the cell’s efficiency is about 74%.
The theoretical efficiency of the diesel engine is:
Th T1
1923 873
100%
100% 55%
Th
1923
Thus, the diesel’s actual efficiency is 0.78 55 % = 43 %.
The cell’s efficiency is 74% whereas that of the diesel engine is only 43 %. Why is
there such a large discrepancy? The diesel engine supplies heat to the surroundings,
which is at a lower temperature than the engine. This dissipation of energy raises the
temperature of the surroundings and the entropy of the surroundings. A cell operates
under isothermal conditions and the energy not utilized goes only towards changing
the entropy of the cell’s surroundings. The cell is more efficient since it does not
heat the surroundings.
Efficiency
994
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
97.
(E) (a) In this case CO can exist in two states, therefore, W=2. There are N of these
states in the crystal, and so we have
S k ln 2 N 1.381 10 23 JK -1 6.022 10 23 mol-1 ln 2 5.8JK -1mol-1
(b) For water, W=3/2, which leads to
3
S k ln( )N 1.381 10 23 JK -1 6.022 10 23 mol-1 ln1.5 3.4JK -1mol-1
2
SELF-ASSESSMENT EXERCISES
98.
(E) (a) Suniv or total entropy contains contributions from the entropy change of the
system ( Ssys ) and the surroundings ( Ssurr ). According to the second law of
thermodynamics, Suniv is always greater then zero.
(b) G of or standard free energy of formation is the free energy change for a reaction in
which a substances in its standard state is formed from its elements in their reference
forms in their standard states.
(c) For a hypothetical chemical reaction aA+bB cC+dD , the equilibrium constant K is
[C]c [D]d
defined as K
.
[A]a [B]b
99.
(E) (a) Absolute molar entropy is the entropy at zero-point energy (lowest possible
energy state) and it is equal to zero.
(b) Coupled reactions are spontaneous reactions (G<0) that are obtained by pairing
reactions with positive G with the reactions with negative G.
(c) Trouton’s rule states that for many liquids at their normal boiling points, the standard
molar entropy of vaporization has a value of approximately 87 Jmol-1K-1.
(d) Equilibrium constant K for a certain chemical reaction can be evaluated using either
G of or H of in conjunction with S o (which are often tabulated). The relationship used
to calculate K is G o RT ln K .
100.
(E) (a) A spontaneous process is a process that occurs in a system left to itself; once
started, no external action form outsize the system is necessary to make the process
continue. A nonspontaneous process is a process that will not occur unless some external
action is continuously applied.
(b) Second law of thermodynamics states that the entropy of universe is always greater
than zero or in other words that all spontaneous processes produce an increase in the
entropy of the universe. The third law of thermodynamics states that the entropy of
perfect pure crystal at 0K is zero.
(c) G is the Gibbs free energy defined as G=H-TS. G0 is the standard free energy
change.
101.
(E) Second law of thermodynamics states that all spontaneous processes produce an
increase in the entropy of the universe. In other words, Suniv Ssys Ssurr 0 .
Therefore, the correct answer is (d).
995
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
102.
(E) The Gibbs free energy is a function of H, S and temperature T. It cannot be used
to determine how much heat is absorbed from the surroundings or how much work the
system does on the surroundings. Furthermore, it also cannot be used to determine the
proportion of the heat evolved in an exothermic reaction that can be converted to various
forms of work. Since Gibbs free energy is related to the equilibrium constant of a
chemical reaction ( G RT ln K ) its value can be used to access the net direction in
which the reaction occurs to reach equilibrium. Therefore, the correct answer is (c).
103.
(M) In order to answer this question, we must first determine whether the entropy
change for the given reaction is positive or negative. The reaction produces three moles
of gas from two moles, therefore there is an increase in randomness of the system, i.e.
entropy change for the reaction is positive. Gibbs free energy is a function of enthalpy,
entropy and temperature ( G H T S ). Since H 0 and S 0 , this reaction
will be spontaneous at any temperature. The correct answer is (a).
(M) Recall that G o RT ln K . If G o 0 , then it follows that G o RT ln K 0 .
Solving for K yields: ln K 0 K e0 1 . Therefore, the correct answer is (b).
104.
105.
(E) In this reaction, the number of moles of reactants equals the number of moles of
products. Therefore, K is equal to Kp and Kc. The correct answers are (a) and (d).
106.
(M) (a) The two lines will intersect at the normal melting point of I2(s) which is 113.6
C. (b) G o for this process must be equal to zero because solid and liquid are at
equilibrium and also in their standard states.
o
107.
(M) (a) No reaction is expected because of the decrease in entropy and the expectation
that the reaction is endothermic. As a check with data from Appendix D, G o =326.4
kJmol-1 for the reaction as written-a very large value. (b) Based on the increase in
entropy, the forward reaction should occur, at least to some extent. For this reaction
G o =75.21 kJmol-1. (c) S is probably small, and H is probably also small (one Cl-Cl
bond and one Br-Br bonds are broken and two Br-Cl bonds are formed). G o should be
small and the forward reaction should occur to a significant extent. For this reaction
G o =-5.07 kJmol-1.
108.
(M) (a) Entropy change must be accessed for the system and its surroundings ( Suniv ),
not just for the system alone. (b) Equilibrium constant can be calculated from
G o ( G o RT ln K ), and K permits equilibrium calculations for nonstandard
conditions.
109.
(D)
H
H [(C5 H 10 (g)] H [(C5 H 10 (l)] 77.2kJ/mol (105.9kJ/mol) 28.7kJ/mol . Now
o
vap
(a)
o
f
First we need to determine
o
H vap
which is simply equal to:
o
f
we use Trouton’s rule to calculate the boiling point of cyclopentane:
996
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
S
o
vap
o
H vap
Tbp
87Jmol K Tbp
-1
o
H vap
-1
87Jmol-1K -1
28.7 1000Jmol-1
330K
87Jmol-1K -1
Tbp 330K 273.15K 57 o C
o
o
o
(b) If we assume that H vap
and Svap
are independent of T we can calculate Gvap
:
87
kJK -1mol-1 2.8kJmol-1
1000
o
(c) Because Gvap,298
K >0, the vapor pressure is less than 1 atm at 298 K, consistent with
-1
o
o
o
Gvap,298
K H vap T Svap 28.7kJmol 298.15K
Tbp=57 oC.
110.
(M) (a) We can use the data from Appendix D to determine the change in enthalpy and
entropy for the reaction:
H o H of (N 2O(g)) 2H of (H 2O(l)) H of (NH 4 NO3 (s))
H o 82.05kJmol-1 2 (285.8kJmol-1 ) (365.6kJmol-1 ) 124kJmol-1
S o S o (N 2O(g)) 2S o (H 2O(l)) S o (NH 4 NO3 (s))
S o 219.9JK -1mol-1 2 69.91JK -1mol-1 151.1JK -1mol-1 208.6JK -1mol-1
(b) From the values of H o and S o determined in part (a) we can calculate G o at
298K:
G o H o T S o
208.6kJmol-1K -1
186.1kJmol-1
1000
can also be calculated directly using G of values tabulated in
G o 124kJmol-1 298K
Alternatively, G o
Appendix D.
(c) The equilibrium constant for the reaction is calculated using G o RT ln K :
G o RT ln K 186.1 1000Jmol-1 8.314JK -1mol-1 298K ln K
186100Jmol-1 2477.6 ln K ln K 75.1 K e 75.1 4.1 10 32
(d) The reaction has H o 0 and S o 0 . Because G o H o T S o , the reaction
will be spontaneous at all temperatures.
111.
(M) Recall from exercise 104 that G o 0 when K=1. Therefore, we are looking for
the diagram with smallest change in Gibbs free energy between the products and the
reactants. The correct answer is diagram (a). Notice that diagrams (b) and (c) represent
chemical reactions with small and large values of equilibrium constants, respectively.
112.
(M) Carbon dioxide is a gas at room temperature. The melting point of carbon dioxide is
expected to be very low. At room temperature and normal atmospheric pressure this
process is spontaneous. The entropy of the universe if positive.
997
CHAPTER 20
ELECTROCHEMISTRY
PRACTICE EXAMPLES
Cathode, reduction:
{Ag + aq + e Ag s } 3
Net reaction
Sc s + 3Ag aq Sc
___________________________________
1B
+
(E) Oxidation of Al(s) at the anode:
3+
aq + 3Ag s
Al s Al
3+
Reduction of Ag aq at the cathode: Ag aq + e
+
+
aq + 3e
Ag s
cathode
(E) The conventions state that the anode material is written first, and the cathode material is written
last.
eAnode, oxidation:
Sc s Sc 3+ aq + 3e
anode
1A
salt bridge
Ag
Al
NO3
-
Al3+
K
+
Ag+
________________
Overall reaction in cell: Al s + 3Ag + aq Al 3+ aq + 3Ag s
2A
2B
Diagram: Al(s)|Al3+ (aq)||Ag + (aq)|Ag(s)
(E) Anode, oxidation: Sn s Sn 2 + aq + 2e
Cathode, reduction:
{ Ag aq 1e Ag( s )} 2
___________________________________________
Overall reaction in cell: Sn( s ) 2 Ag (aq ) Sn 2 (aq ) 2 Ag( s )
(E) Anode, oxidation: {In s In 3+ aq + 3e } 2
Cathode, reduction:
{Cd 2 aq 2e Cd(s)} 3
___________________________________________
Overall reaction in cell: 2In(s) 3Cd 2 (aq) 2In 3 (aq) 3Cd(s)
3A
(M) We obtain the two balanced half-equations and the half-cell potentials from Table 20-1.
Oxidation: {Fe 2+ aq Fe3+ aq + e } 2
Reduction: Cl2 g + 2e 2 Cl aq
E o = +1.358V
Net: 2Fe 2+ aq + Cl2 g 2Fe3+ aq + 2Cl aq ;
3B
E o = 0.771V
o
Ecell
= +1.358V 0.771V = +0.587 V
(M) Since we need to refer to Table 20-1, in any event, it is perhaps a bit easier to locate the
two balanced half-equations in the table. There is only one half-equation involving both
Fe 2+ aq and Fe 3+ aq ions. It is reversed and written as an oxidation below. The half-
equation involving MnO 4 aq is also written below. [Actually, we need to know that in
acidic solution Mn 2+ aq is the principal reduction product of MnO 4 aq .]
Oxidation: {Fe 2+ aq Fe3+ aq + e } 5
Reduction: MnO 4 aq + 8 H + aq + 5e Mn 2+ aq + 4 H 2 O(l)
E o = 0.771V
E o = +1.51V
Net: MnO 4 aq + 5 Fe 2+ aq + 8 H + aq Mn 2+ aq + 5 Fe3+ aq + 4H 2 O(l)
998
Chapter 20: Electrochemistry
o
= +1.51 V 0.771 V = +0.74 V
Ecell
4A
(M) We write down the oxidation half-equation following the method of Chapter 5, and
obtain the reduction half-equation from Table 20-1, along with the reduction half-cell
potential.
Oxidation: {H 2 C2 O 4 (aq)
2 CO 2 (aq) 2 H + (aq) 2 e } 3
E {CO 2 /H 2 C2 O 4 }
E o = +1.33V
aq +14 H + aq + 6 e 2 Cr 3+ aq + 7 H 2O(l)
2
Net: Cr2 O7 aq + 8 H + aq + 3H 2 C2 O 4 aq 2 Cr 3+ aq + 7 H 2 O(l) + 6 CO 2 g
Reduction: Cr2 O7
2
o
Ecell
= +1.81V = +1.33V E o {CO2 /H 2 C2 O4 };
4B
E o{CO2 /H 2 C2 O4 } = 1.33V 1.81V = 0.48V
(M) The 2nd half-reaction must have O 2 g as reactant and H 2 O(l) as product.
Oxidation: {Cr 2+ (aq)
Cr 3+ (aq) e } 4
E {Cr 3 /Cr 2 }
Reduction: O 2 g + 4H + aq + 4e 2H 2 O(l)
E o = +1.229V
Net: O 2 g + 4H aq 4 Cr (aq) 2H 2 O(l) + 4Cr
+
2+
3
2
Ecell 1.653V 1.229V E {Cr / Cr } ;
o
5A
o
3+
aq
______________________________________
E {Cr /Cr 2 } 1.229V 1.653V 0.424V
o
3
(M) First we write down the two half-equations, obtain the half-cell potential for each, and
o
. From that value, we determine G o
then calculate Ecell
Oxidation: {Al s Al3+ aq +3e } 2
E o = +1.676V
Reduction: {Br2 l +2e 2 Br aq } 3
E o = +1.065V
____________________________
o
Net: 2 Al s + 3Br2 l 2 Al3+ aq + 6 Br aq Ecell
= 1.676 V +1.065 V = 2.741V
o
= 6 mol e
G o = nFEcell
5B
96, 485 C
2.741V = 1.587 106 J = 1587 kJ
1mol e
(M) First we write down the two half-equations, one of which is the reduction equation from
the previous example. The other is the oxidation that occurs in the standard hydrogen
electrode.
Oxidation: 2H 2 g 4H + aq + 4e
Reduction: O 2 g + 4 H + aq + 4e 2 H 2 O(l)
Net:
2 H 2 g + O2 g 2 H 2O l
n = 4 in this reaction.
This net reaction is simply twice the formation reaction for H 2 O(l) and, therefore,
o
G o = 2Gf H 2 O l = 2 237.1 kJ = 474.2 103 J = nFEcell
474.2 103 J
G o
o
Ecell =
=
1.229 V E o , as we might expect.
96,485C
nF
4 mol e
mol e
o
6A
(M) Cu(s) will displace metal ions of a metal less active than copper. Silver ion is one
example.
999
Chapter 20: Electrochemistry
Oxidation: Cu s Cu 2+ aq + 2e
6B
E o = 0.340V (from Table 20.1)
Reduction: {Ag + aq + e Ag s } 2
E o = +0.800V
Net: 2Ag + aq + Cu s Cu 2+ aq + 2Ag s
o
Ecell
= 0.340 V + 0.800 V = +0.460 V
(M) We determine the value for the hypothetical reaction's cell potential.
E o = +2.713V
Oxidation: {Na s Na + aq + e } 2
Reduction: Mg 2+ aq +2e Mg s
E o = 2.356 V
Net: 2 Na s +Mg 2+ aq 2 Na + aq + Mg s
o
Ecell
= 2.713V 2.356 V = +0.357 V
The method is not feasible because another reaction occurs that has a more positive cell
potential, i.e., Na(s) reacts with water to form H 2 g and NaOH(aq):
Oxidation: {Na s Na + aq + e } 2
E o = +2.713V
Reduction: 2H 2O+2e- H 2 (g)+2OH - (aq)
E o =-0.828V
o
Ecell
= 2.713 V 0.828 V = +1.885 V .
7A
(M) The oxidation is that of SO42 to S2O82, the reduction is that of O2 to H2O.
Reduction:
aq S2O82 aq + 2e } 2
O 2 g + 4 H + aq + 4e 2H 2O(l)
Net:
O 2 g + 4 H aq + 2 SO 4
Oxidation: {2 SO 4
2
+
E o = 2.01V
E o = +1.229 V
________________
2
aq S2O8 aq + 2H 2O(l)
2
Ecell = -0.78 V
Because the standard cell potential is negative, we conclude that this cell reaction is
nonspontaneous under standard conditions. This would not be a feasible method of producing
peroxodisulfate ion.
7B
(M) (1)
The oxidation is that of Sn 2+ aq to Sn 4 + aq ; the reduction is that of O2 to
H2O.
E o = 0.154 V
Oxidation: {Sn 2 + aq Sn 4 + aq + 2e } 2
Reduction: O 2 g + 4 H + aq + 4e 2H 2O(l)
E o = +1.229 V
___________
Net:
o
= +1.075 V
O 2 g + 4H + aq + 2Sn 2+ aq 2Sn 4+ aq + 2H 2 O(l) Ecell
Since the standard cell potential is positive, this cell reaction is spontaneous under
standard conditions.
(2)
The oxidation is that of Sn(s) to Sn 2+ aq ; the reduction is still that of O2 to H2O.
Oxidation: {Sn s Sn 2 + aq + 2e }
2
Reduction: O 2 g + 4 H + aq + 4e 2 H 2O(l)
E o = +0.137 V
E o = +1.229 V
_______________
Net:
O 2 g + 4 H + aq + 2Sn s 2Sn 2+ aq + 2 H 2O(l)
o
Ecell
= 0.137 V +1.229 V = +1.366 V
1000
Chapter 20: Electrochemistry
The standard cell potential for this reaction is more positive than that for situation (1).
Thus, reaction (2) should occur preferentially. Also, if Sn 4 + aq is formed, it should
react with Sn(s) to form Sn 2+ aq .
Oxidation: Sn s Sn 2 + aq + 2e
E o = +0.137 V
Reduction: Sn 4+ aq + 2 e Sn 2+ aq
Net: Sn
8A
4+
E o = +0.154 V
___________________________________
aq + Sn s 2 Sn aq
2+
E
o
cell
= +0.137 V + 0.154 V = +0.291V
(M) For the reaction 2 Al s + 3Cu 2+ aq 2 Al3+ aq + 3Cu s we know n = 6 and
o
= +2.013 V . We calculate the value of Keq .
Ecell
o
nEcell
0.0257
6 (+2.013)
ln K eq ; ln K eq =
=
= 470; K eq = e 470 = 10204
n
0.0257
0.0257
The huge size of the equilibrium constant indicates that this reaction indeed will go to
essentially 100% to completion.
o
Ecell
=
8B
o
(M) We first determine the value of Ecell
from the half-cell potentials.
– E o = +0.137 V
Oxidation: Sn s Sn 2+ aq +2 e
Reduction: Pb 2+ aq +2 e Pb s
E o = 0.125 V
Net: Pb 2+ aq + Sn s Pb s + Sn 2+ aq
o
Ecell
= +0.137 V 0.125 V = +0.012 V
_________________________________
o
nEcell
0.0257
2 (+0.012)
ln K eq
ln K eq =
=
=0.93 K eq = e0.93 = 2.5
n
0.0257
0.0257
The equilibrium constant's small size (0.001 < K <1000) indicates that this reaction will not
go to completion.
o
Ecell
=
9A
(M) We first need to determine the standard cell voltage and the cell reaction.
E o = +1.676 V
Oxidation: {Al s Al3+ aq + 3 e } 2
Reduction: {Sn 4+ aq + 2 e Sn 2+ aq } 3
Net: 2 Al s + 3 Sn
4+
E o = +0.154 V
aq 2 Al aq + 3 Sn aq
3+
2+
_________________________
o
Ecell
= +1.676 V + 0.154 V = +1.830 V
Note that n = 6 . We now set up and substitute into the Nernst equation.
Al3+ Sn 2+
0.36 M 0.54 M
0.0592
0.0592
=E
log
= +1.830
log
3
n
6
Sn 4 +
0.086 M 3
= +1.830 V 0.0149 V = +1.815 V
2
Ecell
9B
3
2
3
o
cell
(M) We first need to determine the standard cell voltage and the cell reaction.
E o = 1.358 V
Oxidation: 2 Cl 1.0 M Cl2 1 atm + 2e
Reduction: PbO 2 s + 4 H + aq + 2 e Pb 2+ aq + 2 H 2 O(l)
E o = +1.455 V
__________________________________________
1001
Chapter 20: Electrochemistry
Net: PbO 2 s + 4 H + 0.10 M + 2 Cl 1.0 M Cl 2 1 atm + Pb 2+ 0.050 M + 2 H 2O(l)
o
= 1.358 V +1.455 V = +0.097 V
Note that n = 2 . Substitute values into the Nernst
Ecell
equation.
P{Cl2 } Pb 2+
1.0 atm 0.050 M
0.0592
0.0592
o
Ecell =Ecell
log
= +0.097
log
4
2
4
2
n
2
0.10 M 1.0 M
H + Cl
= +0.097 V 0.080 V = +0.017 V
10A (M) The cell reaction is 2 Fe3+ 0.35 M + Cu s 2 Fe 2+ 0.25 M + Cu 2+ 0.15 M with
o
n = 2 and Ecell
= 0.337 V + 0.771 V = 0.434 V
concentrations into the Nernst equation.
Next, substitute this voltage and the
2
Ecell
2
Fe 2+ Cu 2+
0.25 0.15
0.0592
0.0592
o
= Ecell
log
= 0.434
log
= 0.434 0.033
2
2
n
2
Fe3+
0.35
Ecell = +0.467 V Thus the reaction is spontaneous under standard conditions as written.
10B (M) The reaction is not spontaneous under standard conditions in either direction when
Ecell = 0.000 V . We use the standard cell potential from Example 20-10.
2
Ecell
2
Ag +
0.0592
o
= Ecell
log 2+ ;
n
Hg
Ag +
0.0592
0.000 V = 0.054 V
log 2+
2
Hg
2
2
Ag +
0.054 2
= 1.82;
log 2+ =
0.0592
Hg
Ag +
= 101.82 = 0.0150
2+
Hg
o
11A (M) In this concentration cell Ecell
= 0.000 V because the same reaction occurs at anode and
cathode, only the concentrations of the ions differ. Ag + = 0.100 M in the cathode
compartment. The anode compartment contains a saturated solution of AgCl(aq).
Ksp = 1.8 1010 = Ag + Cl = s 2 ;
s 1.8 1010 1.3 105 M
Now we apply the Nernst equation. The cell reaction is
Ag + 0.100 M Ag + 1.3 10 5 M
Ecell = 0.000
0.0592
1.3 105 M
log
= +0.23 V
1
0.100 M
11B (D) Because the electrodes in this cell are identical, the standard electrode potentials are
o
numerically equal and subtracting one from the other leads to the value Ecell
= 0.000 V.
However, because the ion concentrations differ, there is a potential difference between the
1002
Chapter 20: Electrochemistry
two half cells (non-zero nonstandard voltage for the cell). Pb 2+ = 0.100 M in the cathode
compartment, while the anode compartment contains a saturated solution of PbI 2 .
We use the Nernst equation (with n = 2 ) to determine Pb 2+ in the saturated solution.
Ecell = +0.0567 V = 0.000
0.0592
xM
log
;
2
0.100 M
log
xM
2 0.0567
=
= 1.92
0.0592
0.100 M
xM
= 101.92 = 0.012; Pb 2+
= x M = 0.012 0.100 M = 0.0012 M;
anode
0.100 M
I = 2 0.0012 M 0.0024 M
K sp = Pb 2 + I = 0.0012 0.0024 = 6.9 10 9 compared with 7.1 109 in Appendix
D
2
2
12A (M) From Table 20-1 we choose one oxidations and one reductions reaction so as to get the
least negative cell voltage. This will be the most likely pair of ½ -reactions to occur.
E o = 0.535 V
Oxidation: 2I aq I 2 s + 2 e
2 H 2 O(l) O 2 g + 4 H + aq + 4 e
Reduction: K + aq + e K s
E o = 1.229 V
E o = 2.924 V
2 H 2 O(l) + 2 e H 2 g + 2 OH aq
E o = 0.828 V
The least negative standard cell potential 0.535 V 0.828 V = 1.363V occurs when I2 s is
produced by oxidation at the anode, and H 2 g is produced by reduction at the cathode.
12B (M) We obtain from Table 20-1 all the possible oxidations and reductions and choose one of each
to get the least negative cell voltage. That pair is the most likely pair of half-reactions to occur.
Oxidation: 2 H 2 O(l) O2 g + 4 H + aq + 4e
E o = 1.229V
Ag s Ag + aq + e
E o = 0.800V
Reduction: Ag + aq + e Ag s
E o = +0.800V
[We cannot further oxidize NO3 aq or Ag + aq .]
2 H 2 O(l) + 2e H 2 g + 2 OH aq
E o = 0.828V
Thus, we expect to form silver metal at the cathode and Ag + aq at the anode.
13A (M) The half-cell equation is Cu 2+ aq + 2e Cu s , indicating that two moles of
electrons are required for each mole of copper deposited. Current is measured in amperes, or
coulombs per second. We convert the mass of copper to coulombs of electrons needed for the
reduction and the time in hours to seconds.
1003
Chapter 20: Electrochemistry
1 molCu
2 mole 96,485 C
3.735 104 C
63.55g Cu 1 molCu 1mole
1.89 amperes
60 min 60 s
1.98 104 s
5.50 h
1h
1min
12.3 g Cu
Current
13B (D) We first determine the moles of O 2 g produced with the ideal gas equation.
1 atm
738 mm Hg
2.62 L
760 mm Hg
moles O 2 (g)
0.104 mol O 2
0.08206 L atm
26.2 273.2 K
mol K
Then we determine the time needed to produce this amount of O2.
4 mol e 96,485 C
1s
1h
elapsed time = 0.104 mol O 2
= 5.23 h
1 mol O 2 1mol e
2.13 C 3600 s
INTEGRATIVE EXAMPLE
14A (D) In this problem we are asked to determine E o for the reduction of CO2(g) to C3H8(g) in
an acidic solution. We proceed by first determining G o for the reaction using tabulated
o
for the reaction can be determined using
values for G of in Appendix D. Next, Ecell
o
G o zFEcell
. Given reaction can be separated into reduction and oxidation. Since we are
in acidic medium, the reduction half-cell potential can be found in Table 20.1. Lastly, the
oxidation half-cell potential can be calculated using
o
Ecell
E o (reduction half-cell) E o (oxidation half-cell) .
Stepwise approach
First determine G o for the reaction using tabulated values for G of in Appendix D:
G
o
f
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
-23.3 kJ/mol
0 kJ/mol
-394.4 kJ/mol -237.1 kJ/mol
G o 3 G of (CO2 (g)) 4 G of (H 2O(l)) [G of (C3 H 8 (g)) 5 G of (O2 (g))]
G o 3 (394.4) 4 (237.1) [23.3 5 0]kJ/mol
G o 2108kJ/mol
o
o
In order to calculate Ecell
for the reaction using G o zFEcell
, z must be first determined.
We proceed by separating the given reaction into oxidation and reduction:
Reduction: 5{O2(g)+4H+(aq)+4e- 2H2O(l)} Eo=+1.229 V
Oxidation: C3H8(g) + 6H2O(l) 3CO2(g) + 20H+ + 20e- Eo=xV
_____________________________________________________
Overall: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Eo=+1.229 V-xV
o
o
can now be calculated using G o zFEcell
:
Since z=20, Ecell
1004
Chapter 20: Electrochemistry
o
G o zFEcell
2108 1000J/mol 20mol e-
96485C
o
Ecell
1mol e
2108 1000
V 1.092V
20 96485
Finally, E o (reduction half-cell) can be calculated using
o
Ecell
o
Ecell
E o (reduction half-cell) E o (oxidation half-cell) :
1.092 V = 1.229 V – E o (oxidation half-cell) V
E o (oxidation half-cell) =1.229 V – 1.092 V = 0.137 V
Therefore, Eo for the reduction of CO2(g) to C3H8(g) in an acidic medium is 0.137 V.
Conversion pathway approach:
Reduction: 5{O2(g)+4H+(aq)+4e- 2H2O(l)} Eo=+1.229 V
Oxidation: C3H8(g) + 6H2O(l) 3CO2(g) + 20H+ + 20e- Eo=x V
_____________________________________________________
Overall: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Eo=+1.229 V-x V
-23.3 kJ/mol
0 kJ/mol
-394.4 kJ/mol -237.1 kJ/mol
G of
G o 3 G of (CO2 (g)) 4 G of (H 2O(l)) [G of (C3 H 8 (g)) 5 G of (O2 (g))]
G o 3 (394.4) 4 (237.1) [23.3 5 0]kJ/mol
G o 2108kJ/mol
o
G o zFEcell
2108 1000J/mol 20mol e-
96485C
o
Ecell
1mol e
2108 1000
V 1.092V
20 96485
1.092 V = 1.229 V – E o (oxidation half-cell) V
o
Ecell
E o (oxidation half-cell) =1.229 V – 1.092 V = 0.137 V
14B (D) This is a multi component problem dealing with a flow battery in which oxidation
occurs at an aluminum anode and reduction at a carbon-air cathode. Al3+ produced at the
anode is complexed with OH- anions from NaOH(aq) to form [Al(OH)4]-.
Stepwise approach:
Part (a): The flow battery consists of aluminum anode where oxidation occurs and the formed
Al3+ cations are complexes with OH- anions to form [Al(OH)4]-. The plausible half-reaction
for the oxidation is:
Oxidation: Al(s) + 4OH-(aq) [Al(OH)4]- + 3eThe cathode, on the other hand consists of carbon and air. The plausible half-reaction for the
reduction involves the conversion of O2 and water to form OH- anions (basic medium):
Reduction: O2(g) + 2H2O(l) + 4e- 4OH-(aq)
Combining the oxidation and reduction half-reactions we obtain overall reaction for the
process:
Oxidation: {Al(s) + 4OH-(aq) [Al(OH)4]- + 3e-}4
1005
Chapter 20: Electrochemistry
Reduction: {O2(g) + 2H2O(l) + 4e- 4OH-(aq)} 3
____________________________________________
Overall: 4Al(s) + 4OH-(aq) +3O2(g) + 6H2O(l) 4[Al(OH)4]-(aq)
o
as well as Eo for
Part(b): In order to find Eo for the reduction, use the known value for Ecell
the reduction half-reaction from Table 20.1:
o
Ecell
E o (reduction half-cell) E o (oxidation half-cell)
o
Ecell
0.401V E o (oxidation half-cell) 2.73V
E o (oxidation half-cell) 0.401V 2.73V 2.329V
o
o
Part (c): From the given value for Ecell
(+2.73V) first calculate G o using G o zFEcell
(notice that z=12 from part (a) above):
96485C
o
G o zFEcell
12mol e -
2.73V
1mol e G o 3161kJ/mol
Given the overall reaction (part (a)) and G of for OH-(aq) anions and H2O(l), we can
calculate the Gibbs energy of formation of the aluminate ion, [Al(OH)4]-:
Overall reaction:
4Al(s) + 4OH-(aq) + 3O2(g) + 6H2O(l) 4[Al(OH)4]-(aq)
o
-157 kJ/mol 0 kJ/mol -237.2 kJ/mol
x
G f 0 kJ/mol
G o 4 x [4 0 4 (157) 3 0 6 (237.2)kJ/mol 3161kJ/mol
4 x 3161 2051.2 5212.2kJ/mol
x 1303kJ/mol
Therefore, G of ([Al(OH )4 ] ) 1303kJ/mol
Part(d): First calculate the number of moles of electrons:
1mol enumber of mol e current(C / s) time(s)
96485C
60 min 60s
C 1mol e
number of mol e 4.00h
10.0
1h
1min
s 96485C
number of mol e 1.49mol e
Now, use the oxidation half-reaction to determine the mass of Al(s) consumed:
1mol Al 26.98g Al
mass(Al) 1.49mol e -
13.4g
3mol e1mol Al
Conversion pathway approach:
Part (a):
Oxidation: {Al(s) + 4OH-(aq) [Al(OH)4]- + 3e-}4
Reduction: {O2(g) + 2H2O(l) + 4e- 4OH-(aq)} 3
____________________________________________
Overall: 4Al(s) + 4OH-(aq) +3O2(g) + 6H2O(l) 4[Al(OH)4]-(aq)
Part (b):
1006
Chapter 20: Electrochemistry
o
E o (reduction half-cell) E o (oxidation half-cell)
Ecell
o
0.401V E o (oxidation half-cell) 2.73V
Ecell
E o (oxidation half-cell) 0.401V 2.73V 2.329V
Part (c):
96485C
o
12mol e -
2.73V
G o zFEcell
1mol e G o 3161kJ/mol
Overall reaction:
4Al(s) + 4OH-(aq) + 3O2(g) + 6H2O(l) 4[Al(OH)4]-(aq)
G of 0 kJ/mol
-157 kJ/mol 0 kJ/mol -237.2 kJ/mol
x
G o 4 x [4 0 4 (157) 3 0 6 (237.2)kJ/mol 3161kJ/mol
4 x 3161 2051.2 5212.2kJ/mol
x G of ([Al(OH )4 ] ) 1303kJ/mol
Part (d):
1mol enumber of mol e current(C / s) time(s)
96485C
60 min 60s
C 1mol enumber of mol e 4.00h
10.0
1h
1min
s 96485C
number of mol e 1.49mol e
1mol Al 26.98g Al
mass(Al) 1.49mol e-
13.4g
3mol e1mol Al
EXERCISES
Standard Electrode Potential
1.
(E) (a)
If the metal dissolves in HNO 3 , it has a reduction potential that is smaller than
E o NO3 aq /NO g = 0.956 V . If it also does not dissolve in HCl, it has a
reduction potential that is larger than E o H + aq /H 2 g = 0.000 V . If it displaces
Ag + aq from solution, then it has a reduction potential that is smaller than
E o Ag + aq /Ag s = 0.800 V . But if it does not displace Cu 2 + aq from solution,
then its reduction potential is larger than
E o Cu 2+ aq /Cu s = 0.340 V
Thus, 0.340 V E o 0.800 V
(b)
If the metal dissolves in HCl, it has a reduction potential that is smaller than
E o H + aq /H 2 g = 0.000 V . If it does not displace Zn 2+ aq from solution, its
reduction potential is larger than E o Zn 2+ aq /Zn s = 0.763 V . If it also does not
displace Fe 2+ aq from solution, its reduction potential is larger than
1007
Chapter 20: Electrochemistry
E o Fe2+ aq /Fe s = 0.440 V .
0.440 V E o 0.000 V
2.
(E) We would place a strip of solid indium metal into each of the metal ion solutions and see
if the dissolved metal plates out on the indium strip. Similarly, strips of all the other metals
would be immersed in a solution of In3+ to see if indium metal plates out. Eventually, we
will find one metal whose ions are displaced by indium and another metal that displaces
indium from solution, which are adjacent to each other in Table 20-1. The standard
electrode potential for the In/In3+(aq) pair will lie between the standard reduction potentials
for these two metals. This technique will work only if indium metal does not react with
water, that is, if the standard reduction potential of In 3+ aq / In s is greater than about
1.8 V . The inaccuracy inherent in this technique is due to overpotentials, which can be as
much as 0.200 V. Its imprecision is limited by the closeness of the reduction potentials for
the two bracketing metals
3.
(M) We separate the given equation into its two half-equations. One of them is the reduction
of nitrate ion in acidic solution, whose standard half-cell potential we retrieve from Table 20-1
and use to solve the problem.
Oxidation: {Pt(s) 4 Cl (aq)
[PtCl4 ]2 (aq) 2 e } 3 ;
E o {[PtCl4 ]2 (aq)/Pt(s)}
Reduction: {NO3 aq +4 H + aq +3e NO g +2 H 2O(l)} 2 ; E o = +0.956 V
Net: 3Pt s + 2 NO 3 aq + 8 H + aq +12 Cl aq 3PtCl 4
2
o
cell
E
aq + 2 NO g + 6H 2O l
2
0.201V 0.956 V E {[PtCl 4 ] (aq)/Pt(s)}
o
E {[PtCl4 ]2 (aq)/Pt(s)} 0.956 V 0.201V 0.755 V
o
4.
(M) In this problem, we are dealing with the electrochemical reaction involving the oxidation
o
3.20V , we
of Na(in Hg) to Na+(aq) and the reduction of Cl2(s) to Cl-(aq). Given that Ecell
+
o
are asked to find E for the reduction of Na to Na(in Hg). We proceed by separating the
given equation into its two half-equations. One of them is the reduction of Cl 2 (g) to Cl (aq)
whose standard half-cell potential we obtain from Table 20-1 and use to solve the problem.
Stepwise approach:
Separate the given equation into two half-equations:
E {Na (aq) / Na(in Hg)}
Oxidation:{Na(in Hg) Na (aq) e } 2
Reduction: Cl2 (g) 2e 2Cl (aq)
E 1.358 V
Net: 2Na(in Hg) Cl2 (g) 2 Na (aq) 2 Cl (aq)
Ecell
3.20 V
o
E o (reduction half-cell) E o (oxidation half-cell) to solve for
Use Ecell
E o (oxidation half-cell) :
E o 3.20 V=+1.385 V E o{Na (aq) / Na(in Hg)}
cell
E {Na (aq) / Na(in Hg)} 1.358 V 3.20 V 1.84 V
Conversion pathway approach:
Oxidation:{Na(in Hg) Na (aq) e } 2
E {Na (aq) / Na(in Hg)}
1008
Chapter 20: Electrochemistry
Reduction: Cl2 (g) 2e 2Cl (aq)
E 1.358 V
Net: 2Na(in Hg) Cl2 (g) 2 Na (aq) 2 Cl (aq)
Ecell
3.20 V
3.20V 1.358 E o {Na (aq) / Na(in Hg)}
E o {Na (aq) / Na(in Hg)} 1.358V 3.20V 1.84V
5.
(M) We divide the net cell equation into two half-equations.
Oxidation: {Al s + 4 OH aq [Al(OH) 4 ] aq + 3 e } 4 ;
E o {[Al(OH) 4 ] (aq)/Al(s)}
Reduction: {O 2 g + 2 H 2O(l) + 4 e 4 OH aq } 3 ;
E o = +0.401V
Net: 4 Al s + 3O 2 g + 6 H 2 O(l) + 4 OH aq 4[Al(OH) 4 ] aq
o
Ecell = 2.71V
o
Ecell
=2.71V=+0.401V E o {[Al(OH) 4 ] (aq)/Al(s)}
E o {[Al(OH) 4 ] (aq)/Al(s)} = 0.401V 2.71V = 2.31V
6.
(M) We divide the net cell equation into two half-equations.
Oxidation: CH 4 g + 2H 2O(l) CO 2 g + 8 H + aq + 8e
E o CO 2 g /CH 4 g
Reduction: {O 2 g + 4H + aq + 4 e 2H 2 O(l)} 2
E o = +1.229 V
Net: CH 4 g + 2 O 2 g CO 2 g + 2 H 2 O(l)
o
Ecell
=1.06 V
o
Ecell
1.06 V 1.229 V E o CO2 g /CH 4 g
E o CO 2 g /CH 4 g 1.229 V 1.06 V 0.17 V
7.
(M) (a) We need standard reduction potentials for the given half-reactions from Table 10.1:
Eo=+0.800 V
Ag+(aq)+e- Ag(s)
Zn2+(aq)+2e- Zn(s) Eo=-0.763 V
Cu2+(aq)+2e- Cu(s) Eo=+0.340 V
Al3+(aq)+3e- Al(s) Eo=-1.676 V
Therefore, the largest positive cell potential will be obtained for the reaction involving the
oxidation of Al(s) to Al3+(aq) and the reduction of Ag+(aq) to Ag(s):
o
1.676V 0.800V 2.476V
Al(s) + 3Ag+(aq) 3Ag(s)+Al3+(aq) Ecell
Ag is the anode and Al is the cathode.
(b) Reverse to the above, the cell with the smallest positive cell potential will be obtained for
the reaction involving the oxidation of Zn(s) to Zn2+(aq) (anode) and the reduction of Cu+2(aq)
to Cu(s) (cathode):
o
Zn(s) + Cu2+(aq) Cu(s)+Zn2+(aq) Ecell
0.763V 0.340V 1.103V
8.
(M) (a) The largest positive cell potential will be obtained for the reaction:
Zn(s) + 4NH3(aq) + 2VO2+(aq) + 4H+(aq) [Zn(NH3)4]2+(aq)+2V3+(aq)+2H2O(l)
o
Ecell
E o (reduction half-cell) E o (oxidation half-cell) 0.340V (1.015V ) 1.355V
Zn is the anode and VO2+ is the cathode.
1009
Chapter 20: Electrochemistry
(b) Reverse to the above, the cell with the smallest positive cell potential will be obtained for the
reaction involving the oxidation of Ti2+(aq) to Ti3+(aq) (anode) and the reduction of Sn+2(aq) to
Sn(s) (cathode):
2Ti2+(aq) + Sn2+(aq) 2Ti3+(aq)+Sn(aq)
o
Ecell
E o (reduction half-cell) E o (oxidation half-cell) 0.14V (0.37V ) 0.23V
Predicting Oxidation-Reduction Reactions
9.
(E) (a) Ni2+, (b) Cd.
10.
(E) (a) potassium, (b) barium.
11.
(M) (a)
Oxidation: Sn s Sn 2+ aq + 2 e
Reduction: Pb 2+ aq +2 e Pb s
Net: Sn s + Pb
(b)
2+
E o = 0.125 V
aq Sn aq + Pb s
2+
Oxidation: 2I aq I 2 s + 2e
________________________________
o
cell
E
= +0.012 V
Spontaneous
E o = 0.535 V
Reduction: Cu 2+ aq +2e Cu s
Net: 2I aq + Cu 2+ aq Cu s + I 2 s
(c)
E o = +0.137 V
E o = +0.340 V
________________________
o
Ecell = 0.195 V Nonspontaneous
Oxidation: {2H 2O(l) O 2 g + 4H + aq + 4 e } 3
E o = 1.229 V
Reduction: {NO3 (aq) + 4H + aq +3e NO g +2 H 2O(l)} 4
E o = +0.956 V
_______________________
Net: 4 NO3 aq + 4 H + aq 3O 2 g + 4 NO g + 2 H 2 O(l)
_
o
Ecell
= 0.273 V
Nonspontaneous
(d)
Oxidation: Cl aq + 2 OH aq OCl aq + H 2 O(l) + 2e E o = 0.890 V
Reduction: O3 g + H 2 O(l) + 2 e O 2 g + 2OH aq
E o = +1.246 V
Net: Cl aq + O3 g OCl aq + O 2 g (basic solution)
o
Ecell
= +0.356 V
Spontaneous
12.
13.
(M) It is more difficult to oxidize Hg(l) to Hg 22 + 0.797 V than it is to reduce H+ to H 2
(0.000 V); Hg(l) will not dissolve in 1 M HCl. The standard reduction of nitrate ion to NO(g)
in acidic solution is strongly spontaneous in acidic media ( +0.956 V ). This can help overcome
the reluctance of Hg to be oxidized. Hg(l) will react with and dissolve in the HNO 3 aq .
(M) (a)
Oxidation: Mg s Mg 2+ aq + 2 e
Reduction: Pb 2+ aq + 2 e Pb s
Net: Mg s + Pb
2+
aq Mg aq
2+
E o = +2.356 V
E o = 0.125 V
+ Pb s
1010
_____________________________________
o
Ecell
= +2.231V
Chapter 20: Electrochemistry
This reaction occurs to a significant extent.
(b)
Oxidation: Sn s Sn 2+ aq + 2 e
E o = +0.137 V
Reduction: 2 H + aq H 2 g
Net: Sn s + 2 H
+
E o = 0.000 V
aq Sn aq
2+
+ H2 g
________________________________________
o
cell
E
= +0.137 V
This reaction will occur to a significant extent.
(c)
Oxidation: Sn 2+ aq Sn 4+ aq + 2 e
Reduction: SO 4
(d)
2
E o = 0.154 V
aq + 4H + aq +2 e SO2 g +2H 2O(l)
E o = + 0.17 V
________________________
Net: Sn 2+ aq + SO4 2 aq + 4 H + aq Sn 4+ aq + SO2 g + 2H 2 O(l)
This reaction will occur, but not to a large extent.
o
Ecell
= + 0.02V
Oxidation: {H 2O 2 aq O 2 g + 2 H + aq + 2e } 5
E o = 0.695 V
Reduction: {MnO 4 aq +8H + aq +5e - Mn 2+ aq +4H 2 O(l)} 2
-
E o = +1.51V
o
Net: 5H 2 O2 aq +2MnO4- aq +6H + aq 5O 2 g +2Mn 2+ aq +8H 2 O(l) Ecell
= +0.82V
This reaction will occur to a significant extent.
(e)
Oxidation: 2 Br aq Br2 aq + 2 e
E o = 1.065 V
Reduction: I 2 s + 2e 2 I aq
E o = +0.535 V
Net: 2 Br aq + I 2 s Br2 aq + 2 I aq
o
Ecell
= 0.530 V
__________________________________
This reaction will not occur to a significant extent.
14.
(M) In this problem we are asked to determine whether the electrochemical reaction between
Co(s) and Ni2+(aq) to yield Co2+(aq) + Ni(s) will proceed to completion based on the known
o
Ecell
value. This question can be answered by simply determining the equilibrium constant.
Stepwise approach
o
First comment on the value of Ecell
:
o
The relatively small positive value of Ecell
for the reaction indicates that the reaction will
proceed in the forward direction, but will stop short of completion. A much larger positive
o
would be necessary before we would conclude that the reaction goes to
value of Ecell
completion.
0.0257
o
ln K eq :
Calculate the equilibrium constant for the reaction using Ecell
n
o
n Ecell
0.0257
2 0.02
o
Ecell
ln K eq ln K eq
2
0.0257
n
0.0257
K eq e 2 7
Comment on the value of Keq:
1011
Chapter 20: Electrochemistry
Keq is small. A value of 1000 or more is needed before we can describe the reaction as one
that goes to completion.
Conversion pathway approach:
0.0257
o
ln K eq
Ecell
n
o
n Ecell
ln K eq
0.0257
o
n Ecell
20.02
K eq e 0.0257 e 0.0257 7
Keq is too small. The reaction does not go to completion.
o
15. (M) If Ecell
is positive, the reaction will occur. For the reduction of Cr2O72 to Cr 3+ aq :
Cr2 O7
2
aq +14 H + aq + 6 e 2 Cr 3+ aq + 7 H 2O(l)
E o = +1.33V
If the oxidation has E o smaller (more negative) than 1.33V , the oxidation will not occur.
(a)
Sn 2+ aq Sn 4+ aq +2 e
E o = 0.154 V
Hence, Sn 2+ aq can be oxidized to Sn 4 + aq by Cr2O 27 aq .
(b)
I 2 s + 6 H 2O(l) 2 IO3 aq +12 H + aq +10e
E o = 1.20 V
I2 s can be oxidized to IO3 aq by Cr2O 27 aq .
(c)
Mn 2+ aq + 4 H 2 O(l) MnO 4 aq +8 H + aq +5e
E o = 1.51V
Mn 2+ aq cannot be oxidized to MnO 4 aq by Cr2O 27 aq .
16.
(M) In order to reduce Eu3+ to Eu2+, a stronger reducing agent than Eu2+ is required. From
the list given, Al(s) and H2C2O4(aq) are stronger reducing agents. This is determined by
looking at the reduction potentials (-1.676 V for Al3+/Al(s) and -0.49 V for CO2,
H+/H2C2O4(aq)), are more negative than -0.43 V). Co(s), H2O2 and Ag(s) are not strong
enough reducing agents for this process. A quick look at their reduction potentials shows
that they all have more positive reduction potentials than that for Eu3+ to Eu2+ (-0.277 V for
Co2+/Co(s), +0.695 V for O2, H+/H2O2(aq) and +0.800 V for Ag+/Ag(s).
17.
(M) (a)
Oxidation: {Ag s Ag + aq + e } 3
E o = 0.800 V
Reduction: NO3 aq + 4 H + aq + 3 e NO g + 2 H 2 O(l)
______________
E o = +0.956 V
_____________
o
Net: 3Ag s +NO3 aq +4 H + aq 3Ag + aq +NO g +2 H 2 O(l) Ecell
= +0.156 V
Ag(s) reacts with HNO 3 aq to form a solution of AgNO 3 aq .
(b)
Oxidation: Zn s Zn 2+ aq + 2 e
E o = +0.763V
Reduction: 2 H + aq +2 e H 2 g
E o = 0.000 V
Net: Zn s + 2 H + aq Zn 2+ aq + H 2 g
o
Ecell
= +0.763V
Zn(s) reacts with HI(aq) to form a solution of ZnI 2 aq .
1012
Chapter 20: Electrochemistry
(c)
Oxidation: Au s Au 3+ aq + 3e
E o = 1.52 V
Reduction: NO3 aq + 4 H + aq + 3e NO g + 2H 2 O(l)
E o = +0.956 V
o
Net: Au s + NO3 aq + 4 H + aq Au 3+ aq + NO g + 2 H 2 O(l) ; Ecell
= 0.56 V
Au(s) does not react with 1.00 M HNO 3 aq .
18.
(M) In each case, we determine whether Ecell
is greater than zero; if so, the reaction will
occur.
(a) Oxidation: Fe s Fe 2+ aq + 2 e
E o 0.440 V
Reduction: Zn 2+ aq + 2e Zn s
E o = 0.763V
Net: Fe s + Zn 2+ aq Fe 2+ aq + Zn s
o
Ecell
0.323V
The reaction is not spontaneous under standard conditions as written
(b)
Oxidation: {2Cl aq Cl2 g + 2 e } 5
E o = 1.358 V
Reduction: {MnO 4 aq +8 H + aq +5e Mn 2+ aq +4 H 2 O(l)}×2 ; E o = 1.51V
Net: 10Cl aq + 2 MnO 4 aq +16 H + aq 5Cl2 g + 2 Mn 2+ aq + 8 H 2O(l)
o
Ecell
= +0.15 V . The reaction is spontaneous under standard conditions as written.
(c)
Oxidation: {Ag s Ag + aq + e } 2
E o = 0.800 V
Reduction: 2 H + aq + 2 e H 2 g
E o = +0.000 V
Net: 2 Ag s + 2 H + aq 2 Ag + aq + H 2 g
o
Ecell
= 0.800 V
The reaction is not spontaneous under standard conditions as written.
(d)
Oxidation: {2 Cl aq Cl2 g + 2 e } 2
E o = 1.358 V
Reduction: O 2 g + 4 H + aq + 4 e 2 H 2 O(l)
E o = +1.229 V
_______________________________________
Net: 4 Cl aq + 4 H + aq + O 2 g 2 Cl2 g + 2 H 2 O(l)
o
Ecell
= 0.129 V
The reaction is not spontaneous under standard conditions as written.
Galvanic Cells
19.
(M) (a)
Oxidation: {Al s Al3+ aq + 3e } 2
Reduction: {Sn 2+ aq + 2e Sn s } 3
Net:
(b)
2 Al s + 3 Sn 2+ aq 2 Al3+ aq + 3 Sn s
Oxidation: Fe 2+ aq Fe3+ aq + e
Net:
(c)
Fe
3+
Oxidation: {Cr(s) Cr 2+ (aq)+2e- } 3
1013
____________________________
o
Ecell
= +1.539 V
E o = +0.800 V
aq + Ag aq Fe aq +Ag s
+
E o = 0.137 V
E o = 0.771 V
Reduction: Ag + aq + e Ag s
2+
E o = +1.676 V
____________________
o
cell
E
= +0.029 V
E o = 0.90V
Chapter 20: Electrochemistry
Reduction: {Au 3+ (aq)+3e- Au(s)} 2
E o = 1.52V
____________________
Net:
(d)
3Cr(s)+2Au 3+ (aq) 3Cr 2+ (aq)+2Au(s)
Oxidation: 2H 2 O(l) O 2 (g)+4H + (aq)+4e-
o
Ecell
= 2.42V
E o = 1.229V
Reduction: O 2 (g)+2H 2O(l)+4e- 4OH - (aq)
E o = 0.401V
____________________
Net:
20.
H 2 O(l) H (aq)+OH (aq)
+
-
o
cell
E
= 0.828V
(M) In this problem we are asked to write the half-reactions, balanced chemical equation
o
for a series of electrochemical cells.
and determine Ecell
(a) Stepwise approach
First write the oxidation and reduction half-reactions and find Eo values from
Appendix D:
Oxidation: Cu(s) Cu 2+ (aq)+2e E o = 0.340 V
E o = +0.520 V
Reduction: Cu + (aq)+e- Cu(s)
In order to obtain balanced net equation, the reduction half-reaction needs to be
multiplied by 2:
2Cu + (aq)+2e- 2Cu(s)
Add the two half-reactions to obtain the net reaction:
Cu(s) Cu 2+ (aq)+2e2Cu + (aq)+2e- 2Cu(s)
____________________
Net:
2Cu (aq) Cu (aq)+Cu(s)
o
Determine Ecell
:
+
2+
o
Ecell
=-0.340V+0.520V= +0.18 V
Conversion pathway approach:
Oxidation: Cu(s) Cu 2+ (aq)+2e-
E o = 0.340 V
Reduction: {Cu + (aq)+e- Cu(s)} 2
E o = +0.520 V
___________
Net:
2Cu (aq) Cu (aq)+Cu(s)
Follow the same methodology for parts (b), (c), and (d).
Oxidation: Ag(s)+I - (aq) AgI(s)+e-
E o = 0.152 V
Reduction: AgCl(s)+e- Ag(s)+Cl- (aq)
E o = 0.2223 V
+
(b)
2+
E
o
cell
0.18V
____________________
Net:
(c)
AgCl(s)+I (aq) AgI(s)+Cl (aq)
-
-
o
cell
E
= +0.3743 V
Oxidation: {Ce3+ (aq) Ce 4+ (aq)+e- } 2
E o = 1.76 V
Reduction: I 2 (s)+2e- 2I - (aq)
E o = +0.535 V
____________________
Net:
2Ce (aq)+I 2 (s) 2Ce (aq)+2I (aq)
3+
4+
1014
-
o
Ecell
= -1.225 V
Chapter 20: Electrochemistry
(d)
Oxidation: {U(s) U 3+ (aq)+3e- } 2
E o = 1.66V
Reduction: {V 2+ (aq)+2e- V(s)} 3
E o = 1.13V
____________________
Net:
21.
2U(s)+3V (aq) 2U (aq)+3V(s)
2+
3+
o
Ecell
= +0.53 V
is greater than zero; if so, the reaction will
(M) In each case, we determine whether Ecell
occur.
(a) Oxidation: H 2 (g) 2H + 2 e aq
E o 0.000 V
Reduction: F2 (g) + 2 e 2F aq
E o = 2.866 V
o
Net: H 2 (g) + F2 (g) 2 H + (aq) + 2 F- (aq) Ecell
2.866 V
The reaction is spontaneous under standard conditions as written
(b)
Oxidation: Cu(s) Cu 2+ (aq) + 2 e
E o = 0.340 V
Reduction: Ba 2+ (aq) + 2 e Ba(s) ;
E o = 2.92 V
o
Net: Cu(s) + Ba 2+ (aq) Cu 2+ (aq) + Ba(s)
Ecell
= 3.26 V .
The reaction is not spontaneous under standard conditions as written.
(c)
(d)
Oxidation: Fe2+ (aq) Fe3+ (aq) + e 2
E o = 0.771 V
Reduction: Fe 2+ (aq) + 2e Fe(s)
E o = 0.440 V
Net: 3 Fe2+ (aq) Fe(s) + 2 Fe3+ (aq)
The reaction is not spontaneous as written.
o
Ecell
= 1.211 V
Oxidation: 2 Hg(l) + 2 Cl- Hg 2 Cl2 (s) + 2 e
E o = (0.2676) V
Reduction: 2 HgCl2 (aq) + 2 e Hg 2 Cl2 (s) + 2 Cl- (aq) E o = +0.63V
o
Net: 2 Hg(l) + 2 HgCl2 (aq) 2 Hg 2 Cl2 (s)
Ecell
= 0.36 V
(divide by 2 to get Hg(l) + HgCl 2 (aq) Hg 2 Cl2 (s) )
The reaction is spontaneous under standard conditions as written.
1015
Chapter 20: Electrochemistry
(M) (a)
e-
c
a
th
o
d
e
a
n
o
d
e
salt bridge
Cu
Pt
Fe3+
Cu2+
Fe2+
Anode, oxidation: Cu s Cu 2+ aq + 2 e
E o = 0.340 V
Cathode, reduction: {Fe3+ aq + e Fe 2+ aq } 2 ; E o = +0.771V
Net: Cu s + 2 Fe3+ aq Cu 2+ aq + 2 Fe 2+ aq
o
Ecell
= +0.431V
(b)
e-
c
a
th
o
d
e
a
n
o
d
e
salt bridge
Pb
Al
Al3+
Pb2+
Anode, oxidation: {Al s Al3+ aq + 3e } 2
E o = +1.676 V
Cathode, reduction: {Pb 2+ aq + 2 e Pb s } 3 ;
Net: 2Al s + 3Pb 2+ aq 2Al3+ aq + 3Pb s
E o = 0.125 V
o
Ecell
= +1.551 V
(c)
e-
c
a
th
o
d
e
salt bridge
a
n
o
d
e
22.
Pt
Pt
H+
H2O
Cl-
Cl2(g)
O2(g)
Anode, oxidation: 2 H 2 O(l) O 2 g + 4 H + aq + 4 e
E o = 1.229 V
Cathode, reduction: {Cl2 g + 2 e 2 Cl aq } 2 E o = +1.358 V
1016
Chapter 20: Electrochemistry
o
Net: 2 H 2 O(l) + 2 Cl2 g O 2 g + 4 H + aq + 4 Cl aq Ecell
= +0.129 V
(d)
e-
c
a
th
o
d
e
a
n
o
d
e
salt bridge
Pt
Zn
Zn+2
HNO3
NO(g)
Anode, oxidation: {Zn s Zn 2+ aq + 2 e } 3
E o = +0.763V
Cathode, reduction: {NO3 aq +4 H + aq +3e NO g +2 H 2 O(l)} 2 ; E o = +0.956 V
o
Net: 3Zn s + 2NO3 aq + 8H + aq 3Zn 2+ aq + 2NO g + 4H 2 O(l) Ecell
= +1.719 V
23.
(M) In each case, we determine whether Ecell
is greater than zero; if so, the reaction will
occur.
(a) Oxidation: Ag(s) Ag + (aq)+e E o 0.800 V
Reduction: Fe3+ (aq)+e - Fe 2+ (aq)
E o = 0.771 V
Net: Ag(s)+Fe3+ (aq) Ag + (aq)+Fe 2+ (aq)
o
Ecell
0.029 V
The reaction is not spontaneous under standard conditions as written.
(b)
Oxidation: Sn(s) Sn 2+ (aq)+2eReduction: Sn 4+ (aq)+2e- Sn +2 (aq)
E o 0.137 V
E o = 0.154 V
Net: Sn(s)+Sn 4+ (aq) 2Sn 2+ (aq)
(c)
o
Ecell
0.291V
The reaction is spontaneous under standard conditions as written.
Oxidation: 2Br - (aq) Br2 (l)+2e E o 1.065V
Reduction: 2Hg 2+ (aq)+2e- Hg +2 (aq)
E o = 0.630 V
o
Ecell
0.435V
The reaction is not spontaneous under standard conditions as written.
Oxidation: {NO-3 (aq)+2H + (aq)+e - NO 2 (g)+H 2O(l)} 2 E o 2.326V
Net: 2Br - (aq)+2Hg 2+ (aq) Br2 (l)+Hg +2
(d)
Reduction: Zn(s) Zn 2+ (aq)+2e-
E o = 1.563V
o
Net: 2NO-3 (aq)+4H + (aq)+Zn(s) 2NO 2 (g)+2H 2O(l)+Zn 2 (aq) Ecell
0.435V
The reaction is not spontaneous under standard conditions as written.
1017
Chapter 20: Electrochemistry
(M) (a)
Fe s Fe 2 + aq Cl aq Cl2 g Pt s
Oxidation: Fe s Fe 2+ aq + 2 e
E o = +1.358 V
Net: Fe s + Cl2 g Fe 2+ aq + 2Cl aq
o
Ecell
= +1.798 V
Zn s Zn 2+ aq Ag aq Ag s
Oxidation: Zn s Zn 2+ aq + 2 e
Reduction: {Ag + aq + e Ag s } 2
E o = +0.800 V
Net: Zn s + 2 Ag + aq Zn 2+ aq + 2 Ag s
o
Ecell
= +1.563V
Pt s Cu + aq ,Cu 2 + aq Cu + aq Cu s
Oxidation: Cu + aq Cu 2+ aq + e
Reduction: Cu + aq + e Cu s
E o = +0.520 V
Net: 2 Cu + aq Cu 2+ aq + Cu s
o
Ecell
= +0.361V
Mg s Mg 2 + aq Br aq Br2 l Pt s
Oxidation: Mg s Mg 2+ aq + 2 e
E o = +2.356 V
Reduction: Br2 l + 2 e 2 Br aq
E o = +1.065 V
Net: Mg s + Br2 l Mg 2+ aq + 2 Br aq
o
Ecell
= +3.421V
e-
34(a)
24(a)
anode
salt bridge
Pt
Fe
e-
34(b)
24(b)
cathode
Cl2(g)
anode
salt bridge
Ag
Zn
Cl-
Zn2+
Fe2+
esalt bridge
Pt
Cu+ Cu2+
Cu
Mg
Cu+
salt bridge
Pt
Mg2+
1018
Ag+
e-
34(d)
24(d)
anode
34(c)
24(c)
anode
(d)
E o = 0.159 V
cathode
(c)
E o = +0.763V
cathode
(b)
E o = +0.440 V
Reduction: Cl2 g + 2 e 2Cl aq
cathode
24.
Br2 Br-
Chapter 20: Electrochemistry
o
G o , Ecell
, and K
25.
(M) (a)
Oxidation: {Al s Al3+ aq +3e } 2
E o = +1.676 V
Reduction: {Cu 2+ aq +2 e Cu s } 3
E o = +0.337 V
Net: 2 Al s + 3Cu 2+ aq 2 Al3+ aq + 3Cu s
o
Ecell
= +2.013V
______________________
o
G o = nFEcell
= 6 mol e 96, 485 C/mol e 2.013V
G o = 1.165 106 J = 1.165 103 kJ
(b)
Oxidation: {2 I aq I 2 s +2 e } 2
E o = 0.535 V
Reduction: O 2 g + 4 H + aq + 4 e 2 H 2 O(l)
E o = +1.229 V
_________________________________________
o
Net: 4 I aq + O 2 g + 4 H + aq 2 I 2 s + 2 H 2 O(l) Ecell
= +0.694 V
o
G o = nFEcell
= 4 mol e 96, 485 C/mol e 0.694 V = 2.68 105 J = 268 kJ
(c)
Oxidation: {Ag s Ag + aq + e } 6
Reduction: Cr2 O7
2
E o = 0.800 V
aq +14 H + aq +6 e 2 Cr 3+ aq +7 H 2O(l)
Net: 6 Ag s + Cr2 O7
2
E o = +1.33V
aq +14 H + aq 6 Ag + aq + 2 Cr 3+ aq + 7 H 2O(l)
o
Ecell
= 0.800 V +1.33V = +0.53V
o
G o = nFEcell
= 6mol e 96,485C/mol e 0.53V = 3.1105 J = 3.1102 kJ
26.
(M) In this problem we need to write the equilibrium constant expression for a set of redox
o
from
reactions and determine the value of K at 25 oC. We proceed by calculating Ecell
standard electrode reduction potentials (Table 20.1). Then we use the expression
o
G o = nFEcell
= RTln K to calculate K.
(a)
Stepwise approach:
o
Determine Ecell
from standard electrode reduction potentials (Table 20.1):
Oxidation: Ni(s) Ni2+ (aq)+2e-
E o 0.257 V
Reduction: {V 3+ (aq)+e- V 2+ (aq)} 2
E o 0.255 V
_________________________
Net: Ni(s)+2V (aq) Ni (aq)+2V (aq)
3+
2+
2+
o
Use the expression G o = nFEcell
= RTln K to calculate K:
1019
o
Ecell
=0.003V
Chapter 20: Electrochemistry
o
G o = nFEcell
= RTln K;
C
0.003V 578.9J
mol
G o RT ln K 578.9J 8.314JK 1mol 1 298.15K ln K
G o 2mole 96485
ln K 0.233 K e0.233 1.26
2
Ni2+ V 2+
K 1.26
2
3+
V
Conversion pathway approach:
o
from standard electrode reduction potentials (Table 20.1):
Determine Ecell
Oxidation: Ni(s) Ni2+ (aq)+2e-
E o 0.257 V
E o 0.255 V
Reduction: {V 3+ (aq)+e- V 2+ (aq)} 2
_________________________
Net: Ni(s)+2V 3+ (aq) Ni2+ (aq)+2V 2+ (aq)
o
Ecell
=0.003V
o
nFEcell
n 96485Cmol 1
Eo
G = nFE = RTln K ln K eq
RT
8.314JK 1mol 1 298.15K cell
n
2 mol e 0.003V
o
ln K eq
Ecell
0.233
0.0257
0.0257
o
o
cell
Ni2+ V 2+
0.233
K eq e
1.26
2
3+
V
2
Similar methodology can be used for parts (b) and (c)
(b) Oxidation: 2Cl aq Cl2 g + 2 e
E o = 1.358 V
Reduction: MnO 2 s + 4H + aq + 2e Mn 2+ aq + 2H 2O(l)
E o = +1.23V
o
Net: 2 Cl aq + MnO 2 s + 4 H + aq Mn 2+ aq + Cl 2 g + 2 H 2O(l) Ecell
= 0.13V
(c)
Mn 2+ P{Cl2 g }
2 mol e 0.13 V
10.1
5
ln K eq =
= 10.1 ; K eq = e
= 4 10 =
2
4
0.0257
Cl H +
Oxidation: 4 OH aq O 2 g + 2H 2 O(l) + 4 e
E o = 0.401 V
Reduction: {OCl (aq)+H 2 O(l)+2e Cl (aq)+2 OH } 2
E o 0.890 V
Net: 2OCI (aq)
2CI (aq) O 2 (g)
o
Ecell
0.489 V
4 mol e (0.489 V)
76.1
ln K eq
0.0257
27.
Cl P O 2 (g)
1 10
2
OCl
2
K eq e
76.1
33
o
(M) First calculate Ecell
from standard electrode reduction potentials (Table 20.1). Then use
o
= RTln K to determine G o and K.
G o = nFEcell
1020
Chapter 20: Electrochemistry
(a)
Oxidation: {Ce 3 (aq) Ce 4 (aq) e } 5
E o = 1.61 V
Reduction: MnO-4 (aq)+8H + (aq)+5e- Mn 2+ (aq)+4H 2O(l)
E o = +1.51V
o
Net: MnO-4 (aq)+8H + (aq)+5Ce 3+ (aq) Mn 2+ (aq)+4H 2 O(l)+5Ce 4+ (aq) Ecell
0.100V
(b)
o
= RTln K;
G o = nFEcell
G o 5 96485
C
(0.100V ) 48.24kJmol-1
mol
(c)
G o RT ln K 48.24 1000Jmol-1 8.314JK -1mol-1 298.15K ln K
ln K 19.46 K e19.46 3.5 109
(d) Since K is very small the reaction will not go to completion.
28.
o
(M) First calculate Ecell
from standard electrode reduction potentials (Table 20.1). Then use
o
= RTln K to determine G o and K.
G o = nFEcell
(a)
(b)
(c)
(d)
29.
Oxidation: Pb 2+ (aq) Pb 4+ (aq)+2e Reduction: Sn 4+ (aq)+2e - Sn 2+ (aq)
Eo 0.180V
Eo 0.154V
Net: Pb 2+ (aq)+Sn 4+ (aq) Pb 4+ (aq)+Sn 2+ (aq)
o
Ecell
0.026V
o
G o = nFEcell
= RTln K G o 2 96485
C
(0.026) 5017Jmol-1
mol
G o RT ln K 5017Jmol-1 8.314JK -1mol-1 298.15K ln K
ln K 2.02 K e2.02 0.132
The value of K is small and the reaction does not go to completion.
(M) (a)
o
o
is positive
A negative value of Ecell
( 0.0050 V ) indicates that G o = nFEcell
which in turn indicates that Keq is less than one K eq 1.00 ; G o = RTlnKeq .
Keq =
Cu 2+
2
Sn 2+
Cu +
2
Sn 4+
Thus, when all concentrations are the same, the ion product, Q, equals 1.00. From the
negative standard cell potential, it is clear that Keq must be (slightly) less than one.
Therefore, all the concentrations cannot be 0.500 M at the same time.
(b)
In order to establish equilibrium, that is, to have the ion product become less than 1.00,
and equal the equilibrium constant, the concentrations of the products must decrease
and those of the reactants must increase. A net reaction to the left (towards the
reactants) will occur.
1021
Chapter 20: Electrochemistry
30.
(D) (a)
First we must calculate the value of the equilibrium constant from the standard
cell potential.
o
nEcell
0.0257
2 mol e ( 0.017) V
o
Ecell =
ln K eq ; ln K eq =
=
= 1.32
n
0.0257
0.0257
K eq = e 1.32 = 0.266
To determine if the described solution is possible, we compare
Keq with Q. Now K eq =
BrO-3 Ce 4+
2
. Thus, when
2
2
H + BrO-4 Ce3+
4+
3+
BrO 4 = Ce 0.675 M , BrO 3 = Ce 0.600 M and pH=1 ( H + =0.1M
0.600 0.675 2
= 112.5 0.266 = K eq . Therefore, the
the ion product, Q =
0.12 0.675 0.600 2
described situation can occur
(b)
31.
In order to establish equilibrium, that is, to have the ion product (112.5) become equal
to 0.266, the equilibrium constant, the concentrations of the reactants must increase and
those of the products must decrease. Thus, a net reaction to the left (formation of
reactants) will occur.
(M) Cell reaction: Zn s + Ag 2O s ZnO s + 2Ag s . We assume that the cell operates at
298 K.
G o = Gfo ZnO s + 2Gfo Ag s Gfo Zn s Gfo Ag 2O s
= 318.3 kJ/mol + 2 0.00 kJ/mol 0.00 kJ/mol 11.20 kJ/mol
o
= 307.1 kJ/mol = nFEcell
o
=
Ecell
32.
G o
307.1 103 J/mol
=
= 1.591V
2 mol e /mol rxn 96, 485C/mol e
nF
(M) From equation (20.15) we know n = 12 and the overall cell reaction. First we must
compute value of G o .
96485 C
o
G o = n FEcell
2.71 V = 3.14 106 J = 3.14 103 kJ
= 12 mol e
1 mol e
Then we will use this value, the balanced equation and values of Gf to calculate
o
Gfo Al OH 4 .
4 Al s + 3O 2 g + 6 H 2O(l) + 4 OH aq 4 Al OH 4 aq
G o = 4Gfo Al OH 4 4Gfo [Al s ] 3Gfo [O 2 g ] 6Gfo [H 2 O l ] 4Gfo [OH aq ]
1022
Chapter 20: Electrochemistry
3.14 103 kJ = 4Gfo Al OH 4 4 0.00 kJ 3 0.00 kJ 6 237.1 kJ 4 157.2
= 4Gfo Al OH 4 + 2051.4 kJ
Gfo Al OH 4 = 3.14 103 kJ 2051.4 kJ 4 = 1.30 103 kJ/mol
33.
(D) From the data provided we can construct the following Latimer diagram.
IrO2
(IV)
0.223 V
Ir3+ 1.156 V
(III)
Ir (Acidic conditions)
(0)
Latimer diagrams are used to calculate the standard potentials of non-adjacent half-cell
couples. Our objective in this question is to calculate the voltage differential between IrO2
and iridium metal (Ir), which are separated in the diagram by Ir3+. The process basically
involves adding two half-reactions to obtain a third half-reaction. The potentials for the two
half-reactions cannot, however, simply be added to get the target half-cell voltage because
the electrons are not cancelled in the process of adding the two half-reactions. Instead, to
find E1/2 cell for the target half-reaction, we must use free energy changes, which are additive.
To begin, we will balance the relevant half-reactions in acidic solution:
4 H+(aq) + IrO2(s) + e Ir3+(aq) + 2 H2O(l)
Ir3+(aq)+ 3 e Ir(s)
E1/2red(a) = 0.223 V
E1/2red(b) = 1.156V
4 H+(aq) + IrO2(s) + 4e 2 H2O(l) + Ir(s)
E1/2red(c) = ?
E1/2red(c) E1/2red(a) + E1/2red(b) but G(a) + G(b) = G(c) and G = nFE
4F(E1/2red(c)) = 1F(E1/2red(a)) + 3F(E1/2red(b)
4F(E1/2red(c)) = 1F(0.223) + 3F(1.156
1F (0.223) 3F (1.156) 1(0.223) 3(1.156)
E1/2red(c) =
=
= 0.923 V
4 F
4
In other words, E(c) is the weighted average of E(a) and E(b)
34.
(D) This question will be answered in a manner similar to that used to solve 31.
Let's get underway by writing down the appropriate Latimer diagram:
H2MoO4 0.646 V
(VI)
?V
MoO2
(IV)
Mo3+
(III)
(Acidic conditions)
0.428 V
This time we want to calculate the standard voltage change for the 1 e reduction of MoO2 to
Mo3+. Once again, we must balance the half-cell reactions in acidic solution:
H2MoO4(aq) + 2 e + 2 H+(aq) MoO2(s) + 2 H2O(l)
MoO2(s) + 4 H+(aq) + 1 e Mo3+(aq) + 2 H2O(l)
E1/2red(a) = 0.646 V
E1/2red(b) = ? V
H2MoO4(aq) + 3 e + 6 H+(aq) Mo3+(aq) + 4 H2O(l)
E1/2red(c) = 0.428 V
So, 3F(E1/2red(c)) = 2F(E1/2red(a)) + 1F(E1/2red(b)
3F(0.428 V) = 2F(0.646) + 1F(E1/2red(b)
1FE1/2red(b) = 3F(0.428 V) + 2F(0.646)
1023
Chapter 20: Electrochemistry
E1/2red(c) =
3F (0.428 V) 2F (0.646)
= 1.284 V 1.292 V = 0.008 V
1F
Concentration Dependence of Ecell —the Nernst Equation
35.
(M) In this problem we are asked to determine the concentration of [Ag+] ions in
electrochemical cell that is not under standard conditions. We proceed by first determining
o
Ecell
. Using the Nerst equation and the known value of E, we can then calculate the
concentration of [Ag+].
Stepwise approach:
o
First, determine Ecell
:
Oxidation: Zn s Zn 2+ aq + 2 e
E o = +0.763V
Reduction: {Ag + aq + e Ag s } 2
E o = +0.800 V
Net: Zn s + 2 Ag + aq Zn 2+ aq + 2Ag s
o
Ecell
= +1.563V
Use the Nerst equation and the known value of E to solve for [Ag]+:
Zn 2+
1.00
0.0592
0.0592
o
log
= +1.563V
log 2 = +1.250 V
E =Ecell
2
2
n
x
Ag +
1.00 M 2 (1.250 1.563)
log
10.6 ; x 2.5 1011 5 106 M
2
x
0.0592
Therefore, [Ag+] = 5 × 10-6 M
Conversion pathway approach:
E o = +0.763V
Oxidation: Zn s Zn 2+ aq + 2 e
Reduction: {Ag + aq + e Ag s } 2
E o = +0.800 V
Net: Zn s + 2 Ag + aq Zn 2+ aq + 2Ag s
o
cell
E=E
Zn 2+
Zn 2+
0.0592
n (E E o )
log
log
cell
2
2
n
0.0592
Ag +
Ag +
Zn 2+
2
Ag
+
10
Ag +
36.
o
Ecell
= +1.563V
n
o
( E Ecell
)
0.0592
Zn 2+
Ag
1.00
2
(1.2501.563)
0.0592
+
10
n
o
( E Ecell
)
0.0592
5 106
10
(M) In each case, we employ the equation Ecell = 0.0592 pH .
(a)
Ecell = 0.0592 pH = 0.0592 5.25 = 0.311 V
(b)
pH = log 0.0103 = 1.987
Ecell = 0.0592 pH = 0.0592 1.987 = 0.118 V
1024
Chapter 20: Electrochemistry
(c)
H + C2 H 3O 2
x2
x2
5
Ka =
= 1.8 10 =
0.158 x 0.158
HC 2 H 3O 2
x 0.158 1.8 105 1.7 103 M
pH log (1.7 103 ) 2.77
Ecell = 0.0592 pH = 0.0592 2.77 = 0.164 V
37.
o
(M) We first calculate Ecell
for each reaction and then use the Nernst equation to calculate
Ecell .
(a) Oxidation: {Al s Al3+ 0.18 M + 3 e } 2
E o = +1.676 V
Reduction: {Fe 2+ 0.85 M + 2 e Fe s } 3
E o = 0.440 V
Net: 2 Al s + 3Fe 2+ 0.85 M 2 Al3+ 0.18 M + 3Fe s
o
Ecell
= +1.236 V
2
Al3+
0.18
0.0592
0.0592
log
= 1.236 V
log
1.249 V
3
3
6
n
0.85
Fe2+
2
o
Ecell = Ecell
(b)
Oxidation: {Ag s Ag + 0.34 M + e } 2
E o = 0.800 V
Reduction: Cl2 0.55 atm + 2 e 2 Cl 0.098 M
E o = +1.358 V
o
Net: Cl2 0.55atm + 2 Ag s 2 Cl 0.098 M + 2 Ag + 0.34 M ; Ecell
= +0.558 V
2
2
Cl Ag +
0.0592
0.34 0.098
log
= +0.558
log
= +0.638 V
2
o
cell
Ecell = E
38.
=
0.0592
n
2
P{Cl 2 g }
0.55
0.40 M + 2 e
Reduction: {Cr 3+ 0.35 M +1e Cr 2+ 0.25 M } 2
Net: 2 Cr 3+ 0.35 M + Mn s 2 Cr 2+ 0.25 M + Mn 2+ 0.40 M
(M) (a)
Oxidation: Mn s Mn
2
2+
E o = +1.18 V
E o = 0.424 V
o
Ecell
= +0.76 V
2
Cr 2+ Mn 2+
0.25 0.40
0.0592
0.0592
log
= +0.76 V
log
= +0.78 V
2
2
n
2
Cr 3+
0.35
2
o
Ecell = Ecell
(b)
Oxidation: {Mg s Mg 2+ 0.016M + 2e } 3
E o = +2.356 V
Reduction:{ Al OH 0.25M +3 e 4OH 0.042 M +Al s } 2 ; E o = 2.310V
4
Net: 3 Mg s +2[Al OH ] 0.25M 3Mg
4
o
Ecell
= +0.046 V
1025
___________
2
0.016 M +8OH 0.042M +2Al s ;
Chapter 20: Electrochemistry
3
8
Mg 2 + OH
0.016 0.042
0.0592
0.0592
o
= Ecell
log
= +0.046
log
2
6
6
0.25 2
Al OH
4
= 0.046 V + 0.150 V = 0.196 V
3
Ecell
39.
8
(M) All these observations can be understood in terms of the procedure we use to balance
half-equations: the ion—electron method.
(a)
The reactions for which E depends on pH are those that contain either H + aq or
OH aq in the balanced half-equation. These reactions involve oxoacids and
oxoanions whose central atom changes oxidation state.
(b)
(c)
H + aq will inevitably be on the left side of the reduction of an oxoanion because
reduction is accompanied by not only a decrease in oxidation state, but also by the loss
2
of oxygen atoms, as in ClO3 CIO 2 , SO 4 SO 2 , and NO3 NO . These
oxygen atoms appear on the right-hand side as H 2 O molecules. The hydrogens that are
added to the right-hand side with the water molecules are then balanced with H + aq
on the left-hand side.
If a half-reaction with H + aq ions present is transferred to basic solution, it may be
re-balanced by adding to each side OH aq ions equal in number to the H + aq
originally present. This results in H 2 O(l) on the side that had H + aq ions (the left
side in this case) and OH aq ions on the other side (the right side.)
40.
(M) Oxidation: 2 Cl aq Cl2 g + 2 e
Reduction: PbO 2 s + 4 H + aq + 2 e Pb 2+ aq + 2 H 2 O(l)
E o = 1.358 V
E o = +1.455 V
o
Net: PbO 2 s + 4 H + aq + 2 Cl aq Pb 2+ aq + 2 H 2 O(l) + Cl2 g ; Ecell
= +0.097 V
We derive an expression for Ecell that depends on just the changing H + .
o
cell
Ecell = E
0.0592
P{Cl 2 }[Pb 2 ]
(1.00 atm)(1.00 M)
log
= +0.097 0.0296 log
4
2
2
[H ] [Cl ]
[H ]4 (1.00) 2
0.097 4 0.0296 log[H ] 0.097 0.118log[H + ] 0.097 0.118 pH
(a)
Ecell = +0.097 + 0.118 log 6.0 = +0.189 V
Forward reaction is spontaneous under standard conditions
(b)
Ecell = +0.097 + 0.118 log 1.2 = +0.106 V
Forward reaction is spontaneous under standard conditions
1026
Chapter 20: Electrochemistry
(c)
Ecell = +0.097 0.118 4.25 = 0.405 V
Forward reaction is nonspontaneous under standard conditions
The reaction is spontaneous in strongly acidic solutions (very low pH), but is
nonspontaneous under standard conditions in basic, neutral, and weakly acidic solutions.
41.
(M) Oxidation: Zn s Zn 2+ aq 2 e
E o = +0.763V
Reduction: Cu 2+ aq + 2 e Cu s
E o = +0.337 V
Net: Zn s + Cu 2+ aq Cu s + Zn 2+ aq
o
Ecell
= +1.100 V
(a)
We set E = 0.000V , Zn 2+ = 1.00M , and solve for Cu 2+ in the Nernst equation.
o
cell
Ecell = E
log
(b)
Zn 2+
0.0592
log
;
2
Cu 2+
0.000 = 1.100 0.0296 log
1.0 M
0.000 1.100
=
= 37.2;
2+
0.0296
Cu
1.0 M
Cu 2+
Cu 2+ = 1037.2 = 6 1038 M
If we work the problem the other way, by assuming initial concentrations of
Cu 2+
= 6 1038 M and
= 1.0 M and Zn 2+
= 0.0 M , we obtain Cu 2+
initial
initial
final
Zn 2+
= 1.0 M . Thus, we would conclude that this reaction goes to completion.
final
42.
(M) Oxidation:
Sn s Sn 2+ aq + 2 e
Reduction: Pb 2+ aq + 2 e Pb s
Net:
Sn s + Pb 2 + aq Sn 2+ aq + Pb s
E o = +0.137 V
E o = 0.125 V
o
Ecell
= +0.012 V
Now we wish to find out if Pb 2+ aq will be completely displaced, that is, will Pb 2+ reach
0.0010 M, if Sn 2+ is fixed at 1.00 M? We use the Nernst equation to determine if the cell
voltage still is positive under these conditions.
Sn 2+
0.0592
0.0592
1.00
o
Ecell = Ecell
log
= +0.012
log
= +0.012 0.089 = 0.077 V
2+
2
2
0.0010
Pb
The negative cell potential tells us that this reaction will not go to completion under the conditions
stated. The reaction stops being spontaneous when Ecell = 0 . We can work this the another way as
well: assume that Pb 2 + = 1.0 x M and calculate Sn 2+ = x M at equilibrium, that is,
Sn 2+
0.0592
0.0592
x
o
where Ecell = 0 .
Ecell = 0.00 = Ecell
log
= +0.012
log
2+
2
2
1.0 x
Pb
1027
Chapter 20: Electrochemistry
2 0.012
2.6
x
=
= 0.41 x = 10 0.41 1.0 x = 2.6 2.6 x x =
= 0.72 M
0.0592
3.6
1.0 x
We would expect the final Sn 2+ to equal 1.0 M (or at least 0.999 M) if the reaction went to
log
completion. Instead it equals 0.72 M and consequently, the reaction fails to go to completion.
43.
o
(M) (a)
The two half-equations and the cell equation are given below. Ecell
= 0.000 V
+
Oxidation: H 2 g 2 H 0.65 M KOH + 2 e
Reduction:
2 H + 1.0 M + 2 e H 2 g
2 H + 1.0 M 2 H + 0.65 M KOH
Net:
Kw
1.00 1014 M 2
H + =
=
= 1.5 1014 M
base
0.65 M
OH
2
H + base
1.5 10 14
0.0592
0.0592
Ecell = E
log
= 0.000
log
2
2
2
1.0 2
H + acid
For the reduction of H 2 O(l) to H 2 g in basic solution,
o
cell
(b)
2
= +0.818 V
2 H 2 O(l) + 2 e 2 H 2 g + 2 OH aq , E o = 0.828 V . This reduction is the
44.
reverse of the reaction that occurs in the anode of the cell described, with one small
difference: in the standard half-cell, [OH] = 1.00 M, while in the anode half-cell in
the case at hand, [OH] = 0.65 M. Or, viewed in another way, in 1.00 M KOH, [H+]
is smaller still than in 0.65 M KOH. The forward reaction (dilution of H + ) should be
even more spontaneous, (i.e. a more positive voltage will be created), with 1.00 M
o
KOH than with 0.65 M KOH. We expect that Ecell
(1.000 M NaOH) should be a
little larger than Ecell (0.65 M NaOH), which, is in fact, the case.
(M) (a)
Because NH 3 aq is a weaker base than KOH(aq), [OH] will be smaller than in
the previous problem. Therefore the [H+] will be higher. Its logarithm will be less
negative, and the cell voltage will be less positive. Or, viewed as in Exercise 41(b), the
difference in [H+] between 1.0 M H+ and 0.65 M KOH is greater than the difference in
[H+] between 1.0 M H+ and 0.65 M NH3. The forward reaction is “less spontaneous”
and Ecell is less positive.
(b)
Reaction:
NH 3 aq + H 2 O(l)
0.65M
x M
0.65 x M
NH +4 aq +
0M
+x M
xM
Initial:
Changes:
Equil:
NH 4 + OH
xx
x2
= 1.8 105 =
Kb =
NH3
0.65 x 0.65
x = OH = 0.65 1.8 105 3.4 103 M;
1028
OH aq
0 M
+x M
xM
[H 3O + ]
1.00 1014
2.9 1012 M
3.4 103
Chapter 20: Electrochemistry
Ecell = E
45.
o
cell
[ H ]2base
(2.9 1012 ) 2
0.0592
0.0592
log 2 0.000
log
0.683 V
[ H ]acid
(10
. )2
2
2
(M) First we need to find Ag + in a saturated solution of Ag 2 CrO 4 .
1.1 10 12
6.5 10 5 M
4
o
The cell diagrammed is a concentration cell, for which Ecell = 0.000 V , n = 1,
3
12
K sp = Ag + CrO 2
4
= 2 s s = 4 s = 1.1 10 s
2
2
Ag +
= 2 s = 1.3 104 M
anode
3
Cell reaction: Ag s + Ag + 0.125 M Ag s + Ag + 1.3 10 4 M
o
Ecell = Ecell
46.
4
0.0592
1.3 10 M
log
= 0.000 + 0.177 V = 0.177 V
1
0.125 M
(M) First we need to determine Ag + in the saturated solution of Ag 3 PO 4 .
o
The cell diagrammed is a concentration cell, for which Ecell
= 0.000V, n = 1 .
Cell reaction: Ag s + Ag + 0.140 M Ag s + Ag + x M
0.0592
xM
xM
0.180
o
log
; log
=
= 3.04
Ecell = 0.180V = Ecell
1
0.140 M
0.140 M 0.0592
x M = 0.140 M 103.04 = 0.140 M 9.110 4 = 1.3 104 M = Ag +
K sp = Ag + PO34 = 3s s = 1.3 10
3
47.
(D) (a)
3
1.3 10
4 3
4
Oxidation: Sn s Sn 2+ 0.075 M + 2 e
Reduction: Pb 2 + 0.600 M + 2 e Pb s
anode
3 = 9.5 1017
E o = +0.137 V
E o = 0.125 V
_____
o
Net: Sn s + Pb 2+ 0.600 M Pb s + Sn 2+ 0.075 M ; Ecell
= +0.012 V
o
cell
Ecell = E
Sn 2+
0.0592
0.075
log
= 0.012 0.0296 log
= 0.012 + 0.027 = 0.039 V
2+
2
0.600
Pb
(b)
As the reaction proceeds, [Sn2+] increases while [Pb2+] decreases. These changes cause
the driving force behind the reaction to steadily decrease with the passage of time.
This decline in driving force is manifested as a decrease in Ecell with time.
(c)
When Pb 2+ = 0.500 M = 0.600 M 0.100 M , Sn 2+ = 0.075 M + 0.100 M , because
the stoichiometry of the reaction is 1:1 for Sn 2+ and Pb 2+ .
Sn 2+
0.0592
0.175
o
Ecell = Ecell
log
= 0.012 0.0296 log
= 0.012 + 0.013 = 0.025 V
2+
2
0.500
Pb
1029
Chapter 20: Electrochemistry
(d)
Reaction:
Initial:
Changes:
Final:
Ecell = E
o
cell
Sn(s) +
Pb 2+ aq
0.600 M
x M
0.600 x M
Pb s
+
Sn 2 + aq
0.075 M
+x M
0.075 + x M
Sn 2+
0.0592
0.075 + x
log
= 0.020 = 0.012 0.0296 log
2+
2
0.600 x
Pb
0.075 + x Ecell 0.012 0.020 0.012
0.075 + x
=
=
= 0.27;
= 100.27 = 0.54
0.0296
0.0296
0.600 x
0.600 x
0.324 0.075
0.075 + x = 0.54 0.600 x = 0.324 0.54 x; x =
= 0.162 M
1.54
Sn 2+ = 0.075 + 0.162 = 0.237 M
log
(e)
48.
Here we use the expression developed in part (d).
0.075 + x Ecell 0.012 0.000 0.012
log
=
=
= +0.41
0.600 x
0.0296
0.0296
0.075 + x
= 10+0.41 = 2.6; 0.075 + x = 2.6 0.600 x = 1.6 2.6 x
0.600 x
1.6 0.075
x=
= 0.42 M
3.6
2+
2+
Sn = 0.075 + 0.42 = 0.50 M;
Pb = 0.600 0.42 = 0.18 M
(D) (a)
Oxidation: Ag s Ag + 0.015 M + e
Reduction: Fe3+ 0.055 M + e Fe 2+ 0.045 M
E o = 0.800 V
E o = +0.771V
o
Net: Ag s + Fe3+ 0.055 M Ag + 0.015 M + Fe 2+ 0.045 M Ecell
= 0.029 V
o
cell
Ecell = E
Ag + Fe 2+
0.0592
0.015 0.045
log
= 0.029 0.0592 log
3+
1
0.055
Fe
= 0.029 V + 0.113 V = +0.084 V
(b)
As the reaction proceeds, Ag + and Fe 2+ will increase, while Fe 3+ decrease. These
changes cause the driving force behind the reaction to steadily decrease with the passage
of time. This decline in driving force is manifested as a decrease in Ecell with time.
(c)
When Ag + = 0.020 M = 0.015 M + 0.005 M , Fe 2+ = 0.045 M + 0.005 M = 0.050 M and
Fe 3+ = 0.055 M 0.005 M = 0.500 M , because, by the stoichiometry of the reaction, a mole
of Fe 2+ is produced and a mole of Fe 3+ is consumed for every mole of Ag + produced.
1030
Chapter 20: Electrochemistry
Ecell = E
o
cell
Ag + Fe 2+
0.0592
0.020 0.050
log
= 0.029 0.0592log
3+
1
0.050
Fe
= 0.029 V + 0.101 V = +0.072 V
(d)
Reaction:
Initial:
Changes:
Final:
Ag s +
Fe3+ 0.055 M
0.055 M
x M
0.055 x M
Ag + 0.015 M +
0.015 M
+x M
0.015 + x M
Fe 2 + 0.045 M
0.045 M
+x M
0.045 + x M
Ag + Fe 2 +
0.0592
0.015 + x 0.045 + x
log
= 0.029 0.0592 log
3+
1
Fe
0.055 x
0.015 + x 0.045 + x = Ecell + 0.029 = 0.010 + 0.029 = 0.66
log
0.0592
0.0592
0.055 x
0.015 + x 0.045 + x = 10 0.66 = 0.22
0.055 x
o
Ecell = Ecell
0.00068 + 0.060 x + x 2 = 0.22 0.055 x = 0.012 0.22 x
x 2 + 0.28 x 0.011 = 0
b b 2 4ac 0.28 (0.28) 2 + 4 0.011
x=
=
= 0.035 M
2a
2
Ag + = 0.015 M + 0.035 M = 0.050 M
Fe 2+ = 0.045 M + 0.035 M = 0.080 M
Fe 3+ = 0.055 M 0.035 M = 0.020 M
(e)
We use the expression that was developed in part (d).
log
0.015 + x 0.045 + x = Ecell + 0.029 = 0.000 + 0.029 = 0.49
0.0592
0.0592
0.055 x
0.015 + x 0.045 + x = 10 0.49 = 0.32
0.055 x
0.00068 + 0.060 x + x 2 = 0.32 0.055 x = 0.018 0.32 x
x 2 + 0.38 x 0.017 = 0
b b 2 4ac 0.38 (0.38) 2 + 4 0.017
x=
=
= 0.040 M
2a
2
Ag + = 0.015 M + 0.040 M = 0.055 M
Fe 2+ = 0.045 M + 0.040 M = 0.085 M
Fe 3+ = 0.055 M 0.040 M = 0.015 M
1031
Chapter 20: Electrochemistry
49.
(M) First we will need to come up with a balanced equation for the overall redox reaction.
Clearly, the reaction must involve the oxidation of Cl(aq) and the reduction of Cr2O72(aq):
14 H+(aq) + Cr2O72(aq) + 6 e 2 Cr3+(aq) + 7 H2O(l)
{Cl(aq) 1/2 Cl2(g) + 1 e} 6
E1/2red = 1.33 V
E1/2ox = 1.358 V
_____________________________________________
14 H+(aq) + Cr2O72(aq) + 6 Cl(aq) 2 Cr3+(aq) + 7 H2O(l) + 3 Cl2(g) Ecell = 0.03 V
A negative cell potential means, the oxidation of Cl(aq) to Cl2(g) by Cr2O72(aq) at standard
conditions will not occur spontaneously. We could obtain some Cl2(g) from this reaction by
driving it to the product side with an external voltage. In other words, the reverse reaction is
the spontaneous reaction at standard conditions and if we want to produce some Cl2(g) from
the system, we must push the non-spontaneous reaction in its forward direction with an
external voltage, (i.e., a DC power source). Since Ecell is only slightly negative, we could
also drive the reaction by removing products as they are formed and replenishing reactants as
they are consumed.
50.
50.
(D) We proceed by first deriving a balanced equation for the reaction occurring in the cell:
Oxidation: Fe(s) Fe 2+ (aq)+2e-
Reduction: {Fe 3+ (aq)+e- Fe 2+ (aq)} 2
Net:
Fe(s)+2Fe 3+ (aq) 3Fe 2+ (aq)
(a) G o and the equilibrium constant Keq can be calculated using
o
G o = nFEcell
RT ln K eq :
o
= 2 96485Cmol-1 1.21V 233.5kJmol-1
G o = nFEcell
G o RT ln K 8.314JK -1mol-1 298.15K ln K 233.5 1000Jmol-1
ln K 94.2 K e94.2 8.1 1040
(b) Before calculating voltage using the Nernst equation, we need to re-write the net reaction
to take into account concentration gradient for Fe2+(aq):
Oxidation: Fe(s) Fe 2+ (aq,1.0 10 3M)+2e-
Reduction: {Fe 3+ (aq,1.0 10 3M)+e - Fe 2+ (aq,0.10M)} 2
Net:
Fe(s)+2Fe 3+ (aq,1.0 10-3 ) Fe 2+ (aq,1.0 10 -3M)+2Fe 2+ (aq,0.10M)
Therefore,
1.0 10 3 (0.10)2
Q
10
(1.0 10 3 )2
Now, we can apply the Nerst equation to calculate the voltage:
0.0592
0.0592
o
Ecell Ecell
log Q =1.21 V
log10 1.18 V
2
n
(c) From parts (a) and (b) we can conclude that the reaction between Fe(s) and Fe3+(aq) is
spontaneous. The reverse reaction (i.e. disproportionation of Fe2+(aq)) must therefore be
nonspontaneous.
1032
Chapter 20: Electrochemistry
Batteries and Fuel Cells
51.
(M) Stepwise approach:
(a) The cell diagram begins with the anode and ends with the cathode.
Cell diagram: Cr s |Cr 2 + aq ,Cr 3+ aq ||Fe 2 + aq , Fe 3+ aq |Fe s
(b)
Oxidation: Cr 2+ aq Cr 3+ aq + e
E o = +0.424 V
Reduction: Fe3+ aq + e Fe 2+ aq
E o = +0.771V
Net: Cr 2+ aq + Fe3+ aq Cr 3+ aq + Fe 2+ aq
o
Ecell
= +1.195 V
Conversion pathway approach:
Cr s |Cr 2 + aq ,Cr 3+ aq ||Fe 2 + aq , Fe 3+ aq |Fe s
E o = +0.424 V
E o = +0.771V
o
Ecell
= 0.424V 0.771V 1.195V
52.
(M) (a)
Oxidation: Zn s Zn 2+ aq + 2 e
E o = +0.763V
Reduction: 2MnO 2 s + H 2 O(l) + 2 e Mn 2 O3 s + 2OH aq
Acid-base: {NH 4 aq + OH aq NH 3 g + H 2 O l }
+
2
Complex: Zn 2+ aq + 2NH 3 aq + 2Cl aq Zn NH 3 2 Cl2 s
___________________________________________________
Net: Zn s + 2 MnO 2 s + 2 NH 4 + aq + 2 Cl aq Mn 2 O 3 s + H 2 O l + [Zn NH 3 2 ]Cl 2 s
(b)
o
G o = nFEcell
= 2 mol e 96485 C/mol e 1.55 V = 2.99 105 J/mol
This is the standard free energy change for the entire reaction, which is composed of
the four reactions in part (a). We can determine the values of G o for the acid-base
and complex formation reactions by employing the appropriate data from Appendix D
and pK f = 4.81 (the negative log of the Kf for [Zn(NH3)2]2+).
Gao b = RTlnK b = 8.3145 J mol1 K 1 298.15 K ln 1.8 105 = 5.417 104 J/mol
2
2
o
Gcmplx
= RTlnK f = 8.3145 J mol 1 K 298.15 K ln 104.81 = 2.746 104 J/mol
o
o
o
Then Gtotal
= Gredox
+ Gao b + Gcmplx
o
o
o
Gredox
= Gtotal
Gao b Gcmplx
= 2.99 105 J/mol 5.417 104 + 2.746 104 J/mol = 3.26 105 J/mol
Thus, the voltage of the redox reactions alone is
3.26 105 J
o
E =
= 1.69 V 1.69V = +0.763V + E o MnO 2 /Mn 2 O3
2 mol e 96485 C / mol e
E o MnO 2 /Mn 2O3 = 1.69V 0.763V = +0.93V
The electrode potentials were calculated by using equilibrium constants from Appendix D.
These calculations do not take into account the cell’s own internal resistance to the flow of
electrons, which makes the actual voltage developed by the electrodes less than the
theoretical values derived from equilibrium constants. Also because the solid species
1033
Chapter 20: Electrochemistry
(other than Zn) do not appear as compact rods, but rather are dispersed in a paste, and
since very little water is present in the cell, the activities for the various species involved
in the electrochemical reactions will deviate markedly from unity. As a result, the
equilibrium constants for the reactions taking place in the cell will be substantially
different from those provided in Appendix D, which apply only to dilute solutions and
reactions involving solid reactants and products that posses small surface areas. The
actual electrode voltages, therefore, will end up being different from those calculated here.
53.
(M) (a)
Cell reaction: 2H 2 g + O 2 g 2H 2 O l
Grxn = 2Gf H 2 O l = 2 237.1 kJ/mol = 474.2 kJ/mol
G o
474.2 103 J/mol
o
=
= 1.229 V
Ecell =
4 mol e 96485 C/mol e
nF
o
(b)
o
Anode, oxidation: {Zn s Zn 2+ aq + 2e } 2
Cathode, reduction: O 2 g + 4 H + aq + 4e 2H 2 O(l)
E o = +0.763V
E o = +1.229 V
o
Net: 2 Zn s + O 2 g + 4 H + aq 2 Zn 2+ aq + 2 H 2 O(l) Ecell
= +1.992 V
(c)
54.
Anode, oxidation: Mg s Mg 2+ aq + 2e
E o = +2.356 V
Cathode, reduction: I 2 s + 2e 2 I aq
E o = +0.535 V
Net: Mg s + I 2 s Mg 2+ aq + 2 I aq
o
Ecell
= +2.891V
2+
Zn(s)
Zn (aq) + 2 e
Precipitation: Zn2+(aq) + 2 OH(aq)
Zn(OH)2(s)
Reduction:
2 MnO2(s) + H2)(l) + 2 e
Mn2O3(s) + 2 OH (aq)
(M) (a)
Oxidation:
__
Net : Zn(s) + 2 MnO2(s) + H2O(l) + 2 OH(aq)
Mn2O3(s) + Zn(OH)2(s)
(b)
In 50, we determined that the standard voltage for the reduction reaction is +0.93 V (n
= 2 e). To convert this voltage to an equilibrium constant (at 25 C) use:
nE o
2(0.93)
K red 1031.42 3 1031
31.4 ;
log K red
0.0592 0.0592
2+
and for Zn(s)
Zn (aq) + 2 e (E = 0.763 V and n = 2 e )
nE o
2(0.763)
25.8 ;
log K ox
K ox 1025.78 6 1025
0.0592 0.0592
Gtotal = Gprecipitation + Goxidation + Greduction
1
Gtotal = RT ln
+ (RT lnKox) + (RT lnKred)
K sp, Zn(OH)2
1034
Chapter 20: Electrochemistry
55.
Gtotal = RT (ln
1
K sp, Zn(OH)2
+ lnKox + lnKred)
1
kJ
(298 K) (ln
+ ln(6.0 1025) + ln(2.6 1031))
17
1.2 10
K mol
Gtotal = 423 kJ = nFEtotal
423103 J
Hence, Etotal = Ecell =
= 2.19 V
(2 mol)(96485 C mol1 )
Gtotal = 0.0083145
(M) Aluminum-Air Battery: 2 Al(s) + 3/2 O2(g) Al2O3(s)
Zinc-Air Battery:
Zn(s) + ½ O2(g) ZnO(s)
Iron-Air Battery
Fe(s) + ½ O2(g) FeO(s)
Calculate the quantity of charge transferred when 1.00 g of metal is consumed in each cell.
Aluminum-Air Cell:
1 mol Al(s)
3 mol e
96, 485C
1.00 g Al(s)
1.07 104 C
26.98 g Al(s) 1 mol Al(s) 1 mol e
Zinc-Air Cell:
1.00 g Zn(s)
1 mol Zn(s)
2 mol e
96, 485 C
2.95 103 C
65.39 g Zn(s) 1 mol Zn(s) 1 mol e
Iron-Air Cell:
1.00 g Fe(s)
1 mol Fe(s)
2 mol e
96, 485 C
3.46 103 C
55.847 g Fe(s) 1 mol Fe(s) 1 mol e
As expected, aluminum has the greatest quantity of charge transferred per unit mass
(1.00 g) of metal oxidized. This is because aluminum has the smallest molar mass and forms
the most highly charged cation (3+ for aluminum vs 2+ for Zn and Fe).
56.
(M) (a)
A voltaic cell with a voltage of 0.1000 V would be possible by using two halfcells whose standard reduction potentials differ by approximately 0.10 V, such as the
following pair.
Oxidation: 2 Cr 3+ aq + 7 H 2O(l) Cr2O7 2 aq +14 H + aq + 6 e E o = 1.33 V
Reduction: {PbO2 s + 4 H + aq + 2e Pb2+ aq + 2 H 2O(l)} 3
E o = +1.455 V
______________________________
o
Net: 2Cr aq +3PbO 2 s +H 2 O(l) Cr2 O 7 aq +3Pb aq +2H aq Ecell
=0.125 V
The voltage can be adjusted to 0.1000 V by a suitable alteration of the concentrations.
Pb 2+ or H + could be increased or Cr 3+ could be decreased, or any combination
2-
3+
2+
+
of the three of these.
(b)
To produce a cell with a voltage of 2.500 V requires that one start with two half-cells
whose reduction potentials differ by about that much. An interesting pair follows.
Oxidation: Al s Al3+ aq + 3e
1035
E o = +1.676 V
Chapter 20: Electrochemistry
Reduction: {Ag + aq + e Ag s } 3
Net:
Al s + 3Ag aq Al
+
3+
aq + 3Ag s
E o = +0.800 V
___________________________
o
cell
E
= +2.476 V
Again, the desired voltage can be obtained by adjusting the concentrations. In this case
increasing Ag + and/or decreasing Al 3+ would do the trick.
Since no known pair of half-cells has a potential difference larger than about 6 volts,
we conclude that producing a single cell with a potential of 10.00 V is currently
impossible. It is possible, however, to join several cells together into a battery that
delivers a voltage of 10.00 V. For instance, four of the cells from part (b) would deliver
~10.0 V at the instant of hook-up.
(M) Oxidation (anode):
Li(s) Li+ (aq)+e E o 3.040V
Reduction (cathode): MnO 2 (s)+2H 2 O(l)+e- Mn(OH)3 (s)+OH - (aq) E o 0.20V
(c)
57.
________________________________________________________________________________________________________
Net: MnO 2 (s)+2H 2 O(l)+Li(s) Mn(OH)3 (s)+OH - (aq)+Li (aq )
Cell diagram:
Li( s ), Li (aq ) KOH ( satd ) MnO2 ( s ), Mn(OH )3 ( s )
58.
(M) (a) Oxidation (anode): Zn(s) Zn +2 (aq)+2eReduction (cathode): Br2 (l)+2e- 2Br - (aq)
o
Ecell
2.84V
E o 0.763V
E o 1.065V
________________________________________________________________________________________________________
Net:
Zn(s)+Br2 (l) Zn 2+ (aq)+2Br - (aq)
(b) Oxidation (anode): {Li(s) Li+ (aq)+e- } 2
Reduction (cathode): F2 (g)+2e- 2F - (aq)
o
Ecell
1.828V
o
E 3.040V
E o 2.866V
________________________________________________________________________________________________________
Net:
2Li(s)+F2 (g) 2Li+ (aq)+2F - (aq)
o
Ecell
4.868V
Electrochemical Mechanism of Corrosion
59.
60.
(M) (a)
Because copper is a less active metal than is iron (i.e. a weaker reducing agent),
this situation would be similar to that of an iron or steel can plated with a tin coating
that has been scratched. Oxidation of iron metal to Fe2+(aq) should be enhanced in the
body of the nail (blue precipitate), and hydroxide ion should be produced in the vicinity
of the copper wire (pink color), which serves as the cathode.
(b)
Because a scratch tears the iron and exposes “fresh” metal, it is more susceptible to
corrosion. We expect blue precipitate in the vicinity of the scratch.
(c)
Zinc should protect the iron nail from corrosion. There should be almost no blue
precipitate; the zinc corrodes instead. The pink color of OH should continue to form.
(M) The oxidation process involved at the anode reaction, is the formation of Fe2+(aq). This
occurs far below the water line. The reduction process involved at the cathode, is the formation
of OH aq from O2(g). It is logical that this reaction would occur at or near the water line
close to the atmosphere (which contains O2). This reduction reaction requires O2(g) from the
atmosphere and H2O(l) from the water. The oxidation reaction, on the other hand simply
1036
Chapter 20: Electrochemistry
requires iron from the pipe together with an aqueous solution into which the Fe 2 + aq can
disperse and not build up to such a high concentration that corrosion is inhibited.
Anode, oxidation:
Fe s Fe 2 + aq + 2e
Cathode, reduction:
O 2 g + 2H 2 O(l) + 4e 4OH aq
61.
(M)During the process of corrosion, the metal that corrodes loses electrons. Thus, the metal
in these instances behaves as an anode and, hence, can be viewed as bearing a latent negative
polarity. One way in which we could retard oxidation of the metal would be to convert it
into a cathode. Once transformed into a cathode, the metal would develop a positive charge
and no longer release electrons (or oxidize). This change in polarity can be accomplished by
hooking up the metal to an inert electrode in the ground and then applying a voltage across
the two metals in such a way that the inert electrode becomes the anode and the metal that
needs protecting becomes the cathode. This way, any oxidation that occurs will take place at
the negatively charged inert electrode rather than the positively charged metal electrode.
62.
(M) As soon as the iron and copper came into direct contact, an electrochemical cell was
created, in which the more powerfully reducing metal (Fe) was oxidized. In this way, the
iron behaved as a sacrificial anode, protecting the copper from corrosion. The two halfreactions and the net cell reaction are shown below:
Anode (oxidation)
Cathode (reduction)
Net:
Fe(s) Fe2+(aq) + 2 e
Cu2+(aq) + 2 e Cu(s)
Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s)
E = 0.440 V
E = 0.337 V
Ecell = 0.777 V
Note that because of the presence of iron and its electrical contact with the copper, any
copper that does corrode is reduced back to the metal.
Electrolysis Reactions
63.
(M) Here we start by calculating the total amount of charge passed and the number of moles
of electrons transferred.
60 s 2.15 C 1 mol e
mol e = 75 min
= 0.10 mol e
1 min
1s
96485 C
1 mol Zn 2+
1 mol Zn 65.39 g Zn
= 3.3 g Zn
2 mol e
1 mol Zn 2+
1 mol Zn
(a)
mass Zn = 0.10 mol e
(b)
1 mol Al3+ 1 mol Al 26.98 g Al
mass Al = 0.10 mol e
= 0.90 g Al
3 mol e 1 mol Al3+ 1 mol Al
(c)
mass Ag = 0.10 mol e
1 mol Ag + 1 mol Ag 107.9 g Ag
= 11 g Ag
1 mol e 1 mol Ag + 1 mol Ag
1037
Chapter 20: Electrochemistry
(d)
64.
mass Ni = 0.10 mol e
1 mol Ni 2+ 1 mol Ni 58.69 g Ni
= 2.9 g Ni
2 mol e 1 mol Ni 2+ 1 mol Ni
(M) We proceed by first writing down the net electrochemical reaction. The number of
moles of hydrogen produced in the reaction can be calculated from the reaction
stoichiometry. Lastly, the volume of hydrogen can be determined using ideal gas law.
Stepwise approach:
The two half reactions follow: Cu 2 + aq + 2 e Cu s and 2 H + aq + 2 e H 2 g
Thus, two moles of electrons are needed to produce each mole of Cu(s) and two moles of
electrons are needed to produce each mole of H 2 g . With this information, we can
compute the moles of H 2 g that will be produced.
1 mol Cu 2 mol e 1mol H 2 g
= 0.0516 mol H 2 g
63.55g Cu 1 mol Cu
2 mol e
Then we use the ideal gas equation to find the volume of H 2 g .
0.08206 L atm
0.0516 mol H 2
(273.2 28.2) K
mol K
Volume of H 2 (g)
1.27 L
1 atm
763 mmHg
760 mmHg
This answer assumes the H 2 g is not collected over water, and that the H2(g) formed is the
only gas present in the container (i.e. no water vapor present)
Conversion pathway approach:
Cu 2 + aq + 2 e Cu s and 2 H + aq + 2 e H 2 g
mol H 2 g = 3.28 g Cu
1 mol Cu 2 mol e 1mol H 2 g
= 0.0516 mol H 2 g
63.55g Cu 1 mol Cu
2 mol e
0.08206 L atm
(273.2 28.2)K
0.0516 molH 2
molK
V (H 2 (g))
1.27 L
1 atm
763 mmHg
760 mmHg
mol H 2 g = 3.28 g Cu
65.
(M) Here we must determine the standard cell voltage of each chemical reaction. Those
chemical reactions that have a negative voltage are the ones that require electrolysis.
(a)
Oxidation: 2 H 2 O(l) 4H + aq + O 2 g + 4 e
E o = 1.229 V
Reduction: {2 H + aq + 2 e H 2 g } 2
E o = 0.000 V
Net: 2 H 2 O(l) 2 H 2 g + O 2 g
o
Ecell
= 1.229 V
This reaction requires electrolysis, with an applied voltage greater than +1.229V .
(b)
Oxidation: Zn s Zn 2+ aq + 2 e
E o = +0.763 V
Reduction: Fe 2+ aq + 2 e Fe s
E o = 0.440 V
1038
Chapter 20: Electrochemistry
Net: Zn s + Fe2+ aq Fe s + Zn 2+ aq
o
Ecell
= +0.323V
This is a spontaneous reaction under standard conditions.
(c)
Oxidation: {Fe 2 + aq Fe 3+ aq + e } 2
E o = 0.771 V
Reduction: I 2 s + 2 e 2I aq
E o = +0.535 V
o
Net: 2 Fe 2 + aq + I2 s 2 Fe 3+ aq + 2 I aq
Ecell
= 0.236 V
This reaction requires electrolysis, with an applied voltage greater than +0.236V .
(d)
Oxidation: Cu s Cu 2+ aq + 2 e
E o = 0.337 V
Reduction: {Sn 4+ aq + 2 e Sn 2+ aq } 2
E o = +0.154 V
o
Net:
= 0.183V
Cu s + Sn 4 + aq Cu 2 + aq + Sn 2+ aq Ecell
This reaction requires electrolysis, with an applied voltage greater than +0.183V .
66.
(M) (a)
Because oxidation occurs at the anode, we know that the product cannot be H 2
(H2 is produced from the reduction of H 2 O ), SO 2 , (which is a reduction product of
2
2
SO 4 ), or SO 3 (which is produced from SO 4 without a change of oxidation state;
it is the dehydration product of H 2SO 4 ). It is, in fact O2 that forms at the anode.
The oxidation of water at the anode produces the O2(g).
(b)
Reduction should occur at the cathode. The possible species that can be reduced are
H 2 O to H 2 g , K + aq to K s , and SO 24 aq to perhaps SO 2 g . Because
potassium is a highly active metal, it will not be produced in aqueous solution. In order
for SO 24 aq to be reduced, it would have to migrate to the negatively charged
cathode, which is not very probable since like charges repel each other. Thus, H 2 g
is produced at the cathode.
(c)
At the anode:
2 H 2 O(l) 4H + aq + O 2 g + 4 e
At the cathode:
{2H + aq + 2 e H 2 g } 2
Net cell reaction: 2 H 2 O(l) 2 H 2 g + O 2 g
E o = 1.229 V
E o = 0.000 V
o
Ecell
= 1.229 V
A voltage greater than 1.229 V is required. Because of the high overpotential required
for the formation of gases, we expect that a higher voltage will be necessary.
67.
(M) (a)
(b)
68.
The two gases that are produced are H 2 g and O 2 g .
At the anode: 2 H 2 O(l) 4 H + aq + O 2 g + 4 e
E o = 1.229 V
At the cathode: {2H + aq + 2 e H 2 g } 2
E o = 0.000 V
Net cell reaction: 2 H 2 O(l) 2 H 2 g + O 2 g
o
Ecell
= 1.229 V
(M) The electrolysis of Na 2 SO 4 aq produces molecular oxygen at the anode.
E o O 2 g /H 2 O = 1.229 V . The other possible product is S2O 28 aq . It is however,
1039
Chapter 20: Electrochemistry
unlikely to form because it has a considerably less favorable half-cell potential.
2
2
E o S2 O8 aq /SO 4 aq = 2.01 V .
H 2 g is formed at the cathode.
3600 s 2.83 C 1 mol e 1 mol O 2
mol O 2 = 3.75 h
= 0.0990 mol O 2
1h
1s
96485 C 4 mol e
The vapor pressure of water at 25 C , from Table 12-2, is 23.8 mmHg.
nRT 0.0990 mol 0.08206 L atm mol1 K 1 298 K
2.56 L O 2 (g)
V
1 atm
P
(742 23.8) mmHg
760 mmHg
69.
Zn 2+ aq + 2e Zn s
60 s 1.87 C 1 mol e 1 mol Zn 65.39 g Zn
mass of Zn = 42.5 min
= 1.62 g Zn
1 min
1s
96, 485 C 2 mol e 1 mol Zn
(M) (a)
(b)
2I aq I2 s + 2e
time needed = 2.79 g I2
70.
(M) (a)
1mol I2 2 mol e 96, 485 C 1 s 1 min
= 20.2 min
253.8g I2 1mol I2 1 mol e 1.75C 60 s
Cu 2+ aq + 2e Cu s
mmol Cu 2+ consumed = 282 s
2.68 C 1 mol e 1 mol Cu 2+ 1000 mmol
1s
96, 485 C 2 mol e
1mol
= 3.92 mmol Cu 2+
decrease in Cu 2+ =
3.92 mmol Cu 2+
= 0.00922 M
425 mL
final Cu 2+ = 0.366 M 0.00922 M = 0.357 M
(b)
mmol Ag + consumed = 255 mL 0.196 M 0.175 M = 5.36 mmol Ag +
time needed = 5.36 mmol Ag +
1 mol Ag +
1 mol e
96485C
1s
1000 mmol Ag + 1 mol Ag + 1 mol e 1.84 C
= 281 s 2.8 102 s
71.
(M) (a)
(b)
charge = 1.206 g Ag
current =
1 mol Ag
1 mol e 96,485 C
= 1079 C
107.87 g Ag 1 mol Ag 1 mol e
1079 C
= 0.7642 A
1412 s
1040
Chapter 20: Electrochemistry
72.
(D) (a)
(b)
Anode, oxidation: 2H 2 O(l) 4H + aq + 4e + O 2 g
Cathode, reduction: {Ag + aq + e Ag s } 4
E o = +0.800 V
Net: 2H 2 O(l) + 4Ag + aq 4H + aq + O 2 g + 4Ag s
o
Ecell
= 0.429 V
charge = 25.8639 25.0782 g Ag
current =
(c)
E o = 1.229V
1 mol Ag
1 mol e 96, 485 C
= 702.8C
107.87 g Ag 1 mol Ag 1 mol e
702.8 C
1h
= 0.0976 A
2.00 h 3600 s
The gas is molecular oxygen.
1 mol e 1mol O 2
L atm
702.8
C
0.08206
(23+273) K
96485 C 4 mol e
mol K
nRT
V
1 atm
P
755 mmHg
760 mmHg
1000 mL
0.0445 L O 2
44.5 mL of O 2
1L
73.
(D) (a) The electrochemical reaction is:
Anode (oxidation):
{Ag(s) Ag + (aq)+e- (aq)} 2
Cathode (reduction):
Cu 2+ (aq)+2e- Cu(s)
E o 0.800V
Eo 0.340V
___________________________________________________________________________________________________
Net:
2Ag(s)+Cu 2+ (aq) Cu(s)+2Ag + (aq)
Therefore, copper should plate out first.
charge
1h
(b) current =
= 0.75A charge 6750C
2.50hmin 3600 s
o
Ecell
0.46V
1 mole- 1mol Cu 63.546 g Cu
=2.22 g Cu
96485 C 2 mol e1 mol Cu
(c) The total mass of the metal is 3.50 g out of which 2.22g is copper. Therefore, the mass
of silver in the sample is 3.50g-2.22g=1.28g or (1.28/3.50)x100=37%.
mass=6750 C
74.
First, calculate the number of moles of electrons involved in the electrolysis:
60 sec 1mole e1.20C / s 32.0 min
0.0239mol e1min 96485C
From the known mass of platinum, determine the number of moles:
1mol Pt
0.0109molPt
2.12gPt
195.078gPt
Determine the number of electrons transferred:
0.0239 mol e 2.19
0.0109 mol Pt
(D)
1041
Chapter 20: Electrochemistry
Therefore, 2.19 is shared between Pt2+ and Ptx+. Since we know the mole ratio between
Pt2+ and Ptx+, we can calculate x:
2.19=0.90(+2) + 0.10 (x).
2.19 1.80 0.10 x x 4
(a) The oxidation state of the contaminant is +4.
INTEGRATIVE AND ADVANCED EXERCISES
75. (M) Oxidation:
Reduction:
V 3 H 2 O
VO 2 2 H e
Ag e
Ag(s)
E o 0.800 V
V 3 H 2 O Ag
VO 2 2 H Ag(s)
Net:
0.439 V E o a 0.800 V
VO 2 2 H e
V 3 H 2 O
Reduction:
Eocell 0.439 V
E a 0.800 V 0.439 V 0.361 V
V 2
V 3 e
Oxidation:
Net:
E a
V 2 VO 2 2 H
2 V 3 H 2 O
E ob
E o 0.361 V
Eocell 0.616 V
0.616 V E o b 0.361 V E o b 0.361 V 0.616 V 0.255 V
Thus, for the cited reaction:
V 3 e
V 2
E o = 0.255 V
76. (M)The cell reaction for discharging a lead storage battery is equation (20.24).
PbSO 4 (s) 2 H 2 O(l)
Pb(s) PbO 2 (s) 2 H (aq) 2 HSO 4 (aq)
The half-reactions with which this equation was derived indicates that two moles of electrons
are transferred for every two moles of sulfate ion consumed. We first compute the amount of
H2SO4 initially present and then the amount of H2SO4 consumed.
5.00 mol H 2 SO 4
initial amount H 2 SO 4 1.50 L
7.50 mol H 2 SO 4
1 L soln
2
1 mol H 2 SO 4
3600 s 2.50 C 1 mol e 2 mol SO 4
amount H 2 SO 4 consumed 6.0 h
2
96485 C
1h
1s
2 mol e
1 mol SO 4
0.56 mol H 2 SO 4
final [H 2 SO 4 ]
7.50 mol 0.56 mol
4.63 M
1.50 L
77. (M) The cell reaction is 2 Cl (aq) 2 H 2 O(l)
2 OH (aq) H 2 (g) Cl 2 (g)
We first determine the charge transferred per 1000 kg Cl2.
1 mol Cl2
1000 g
2 mol e
96, 485 C
charge 1000 kg Cl2
2.72 109 C
1 kg
70.90 g Cl2 1 mol
Cl2 1 mol e
1042
Chapter 20: Electrochemistry
1J
1 kJ
9.38 10 6 kJ
1 V C 1000 J
1
W
s
1h
1 kWh
(b) energy 9.38 10 9 J
2.61 10 3 kWh
1J
3600 s 1000 W h
(a) energy 3.45 V 2.72 10 9 C
78.
(D) We determine the equilibrium constant for the reaction.
Oxidation : Fe(s)
Fe (aq) 2e _
E 0.440 V
Reduction : {Cr 3 (aq) e _
Cr 2 (aq)} 2
3+
Net:
Fe(s) + 2 Cr (aq)
G nFE
o
cell
o
E 0.424 V
Fe (aq) + 2 Cr (aq)
2+
RT ln K ln K
2+
o
nFEcell
RT
2 mol e 96485 Coul/mol e _ 0.016 V
ln K
1.2
8.3145 J mol -1 K -1 298 K
_
Reaction: Fe(s) 2 Cr 3 (aq) Fe 2 (aq)
Initial :
1.00 M
0M
Changes :
–2x M
x M
x M
Equil:
(1.00 – 2x)M
K
E°cell = +0.016 V
K e1..2 3.3
2Cr 2 (aq)
0 M
2x M
2x M
[Fe 2 ][Cr 2 ]2
x (2 x) 2
3.3
[Cr 3 ]2
(1.00 2 x) 2
Let us simply substitute values of x into this cubic equation to find a solution. Notice that x
cannot be larger than 0.50, (values > 0.5 will result in a negative value for the denominator.
x 0.40
0.40 (0.80) 2
K
6.4 3.3
(1.00 0.80) 2
x 0.35
0.35 (0.70) 2
K
1.9 3.3
(1.00 0.70) 2
x 0.37
K
0.37 (0.74) 2
3.0 3.3
(1.00 0.74) 2
x 0.38
K
0.38 (0.76) 2
3.8 3.3
(1.00 0.76) 2
Thus, we conclude that x = 0.37 M = [Fe2+].
79. (D) First we calculate the standard cell potential.
Oxidation: Fe 2 (aq)
Fe3 (aq) e
Reduction: Ag (aq) e
Ag(s)
E o 0.771 V
E o 0.800 V
o
Net:
Fe 2 (aq) Ag (aq)
Fe3 (aq) Ag(s)
Ecell
0.029 V
Next, we use the given concentrations to calculate the cell voltage with the Nernst equation.
Ecell Eocell
0.0592
1
log
[Fe3 ]
2
[Fe ][Ag ]
0.029 0.0592 log
0.0050
0.0050 2.0
0.029 0.018 0.047 V
The reaction will continue to move in the forward direction until concentrations of reactants
decrease and those of products increase a bit more. At equilibrium, Ecell = 0, and we have the
following.
1043
Chapter 20: Electrochemistry
o
cell
E
0.0592
[Fe3 ]
0.029
log
1
[Fe 2 ][Ag ]
Reaction: Fe2 (aq) Ag (aq)
Initial:
0.0050 M
2.0 M
x M
x M
Changes:
Equil:
(0.0050 x) M (2.0 - x) M
[Fe3 ]
0.029
0.49
log
2
[Fe ][Ag ] 0.0592
Fe3 (aq)
0.0050 M
x M
(0.0050 x ) M
Ag(s)
0.0050 x
[Fe3 ]
0.0050 x
100.49 3.1
2
[Fe ][Ag ]
(0.0050 x)(2.0 x) 2.0(0.0050 x)
0.026
6.2(0.0050 x) 0.0050 x 0.031 6.2 x 7.2 x 0.026 x
0.0036 M
7.2
Note that the assumption that x 2.0 is valid. [Fe2+] = 0.0050 M – 0.0036 M = 0.0014 M
80. (D) We first note that we are dealing with a concentration cell, one in which the standard
oxidation reaction is the reverse of the standard reduction reaction, and consequently its
standard cell potential is precisely zero volts. For this cell, the Nernst equation permits us to
determine the ratio of the two silver ion concentrations.
0.0592
[Ag (satd AgI)]
0.0860 V
log
Ecell 0.000 V
1
[Ag (satd AgCl, x M Cl )]
log
0.0860
[Ag (satd AgI)]
1.45
[Ag (satd AgCl, x M Cl )] 0.0592
[Ag (satd AgI)]
101.45 0.035
[Ag (satd AgCl, x M Cl )]
We can determine the numerator concentration from the solubility product expression for AgI(s)
K sp [Ag ][I ] 8.5 1017 s 2
s 8.5 1017 9.2 109 M
This permits the determination of the concentration in the denominator.
9.2 10 9
[Ag (satd AgCl, x M Cl )]
2.6 10 7 M
0.035
We now can determine the value of x. Note: Cl– arises from two sources, one being the dissolved
AgCl.
K sp [Ag ][Cl ] 1.8 10 10 (2.6 10 7 )(2.6 10 7 x) 6.8 10 14 2.6 10 7 x
1.8 10 10 6.8 10 14
x
6.9 10 4 M [Cl ]
7
2.6 10
81. (M) The Faraday constant can be evaluated by measuring the electric charge needed to
produce a certain quantity of chemical change. For instance, let’s imagine that an electric
circuit contains the half-reaction Cu 2 (aq) 2 e Cu(s) . The electrode on which the
solid copper plates out is weighed before and after the passage of electric current. The mass
gain is the mass of copper reduced, which then is converted into the moles of copper
reduced. The number of moles of electrons involved in the reduction then is computed from
the stoichiometry for the reduction half-reaction. In addition, the amperage is measured
during the reduction, and the time is recorded. For simplicity, we assume the amperage is
constant. Then the total charge (in coulombs) equals the current (in amperes, that is,
1044
Chapter 20: Electrochemistry
coulombs per second) multiplied by the time (in seconds). The ratio of the total charge (in
coulombs) required by the reduction divided by the amount (in moles) of electrons is the
Faraday constant. To determine the Avogadro constant, one uses the charge of the electron,
1.602 × 10–19 C and the Faraday constant in the following calculation.
NA
96,485 C
1 electron
electrons
6.023 10 23
19
1 mol electrons 1.602 10 C
mole
82. (M) In this problem we are asked to determine G of for N2H4(aq) using the electrochemical
data for hydrazine fuel cell. We first determine the value of G o for the cell reaction, a
reaction in which n = 4. G of can then be determined using data in Appendix D.
Stewise approach:
Calculate G o for the cell reaction (n=4):
96,485 C
o
4 mol e-
1.559 V 6.017 10 5 J 601.7 kJ
G o n FEcell
1 mol e
Using the data in Appendix D, determine G of for hydrazine (N2H4):
-601.7 kJ Gf o[N 2 (g)] 2 Gf o[H 2 O(l)] Gf o[N 2 H 4 (aq)] Gf o[O 2 (g)]
0.00 kJ 2 ( 237.2 kJ) Gf o[N 2 H 4 (aq)] 0.00 kJ
Gf o[N 2 H 4 (aq)] 2 ( 237.2 kJ) 601.7 kJ 127.3 kJ
Conversion pathway approach:
96,485 C
o
4 mol e-
1.559 V 6.017 10 5 J 601.7 kJ
G o n FEcell
1 mol e
o
-601.7 kJ Gf [N 2 (g)] 2 Gf o[H 2 O(l)] Gf o[N 2 H 4 (aq)] Gf o[O 2 (g)]
0.00 kJ 2 ( 237.2 kJ) Gf o[N 2 H 4 (aq)] 0.00 kJ
Gf o[N 2 H 4 (aq)] 2 ( 237.2 kJ) 601.7 kJ 127.3 kJ
83. (M) In general, we shall assume that both ions are present initially at concentrations of 1.00 M.
Then we shall compute the concentration of the more easily reduced ion when its reduction
potential has reached the point at which the other cation starts being reduced by electrolysis. In
performing this calculation we use the Nernst equation, but modified for use with a halfreaction. We find that, in general, the greater the difference in E° values for two reduction
half-reactions, the more effective the separation.
(a) In this case, no calculation is necessary. If the electrolysis takes place in aqueous solution,
H2(g) rather than K(s) will be produced at the cathode. Cu2+ can be separated from K+ by
electrolysis.
(b) Cu 2 (aq) 2 e Cu(s) E o 0.340 V
Ag (aq) e Ag(s) E o 0.800 V
Ag+ will be reduced first. Now we ask what [Ag+] will be when E = +0.337 V.
0.0592
1
1
0.800 0.340
0.337 V 0.800 V
log
log
7.77
1
0.0592
[Ag ]
[Ag ]
1045
Chapter 20: Electrochemistry
[Ag+] = 1.7 × 10–8 M. Separation of the two cations is essentially complete.
(c) Pb 2 (aq) 2 e Pb(s) E o 0.125 V
Sn 2 (aq) 2 e Sn(s)
E o 0.137 V
Pb2+ will be reduced first. We now ask what [Pb2+] will be when E° = –0.137 V.
0.0592
1
1
2(0.137 0.125)
0.137 V 0.125 V
log
log
0.41
2
2
2
0.0592
[Pb ]
[Pb ]
[Pb2+] = 10–0.41 = 0.39 M Separation of Pb2+ from Sn2+ is not complete.
84. (D) The efficiency value for a fuel cell will be greater than 1.00 for any exothermic reaction
( H o 0 ) that has G that is more negative than its H o value. Since G o H o T S o , this
means that the value of S o must be positive. Moreover, for this to be the case, ngas is usually
greater than zero. Let us consider the situation that might lead to this type of reaction. The
combustion of carbon-hydrogen-oxygen compounds leads to the formation of H2O(l) and
CO2(g). Since most of the oxygen in these compounds comes from O2(g) (some is present in
the C-H-O compound), there is a balance in the number of moles of gas produced—CO2(g)—
and those consumed—O2(g)—which is offset in a negative direction by the H2O(l) produced.
Thus, the combustion of these types of compounds will only have a positive value of ngas if
the number of oxygens in the formula of the compound is more than twice the number of
hydrogens. By comparison, the decomposition of NOCl(g), an oxychloride of nitrogen, does
produce more moles of gas than it consumes. Let us investigate this decomposition reaction.
NOCl(g)
12 N 2 (g) 12 O 2 (g) 12 Cl 2 (g)
H o 12 H of [ N 2 (g)] 12 H of [O 2 (g)] 12 H of [Cl 2 (g)] H of [ NOCl(g)]
0.500 (0.00 kJ/mol 0.00 kJ/mol 0.00 kJ/mol) 51.71 kJ/mol 51.71 kJ/mol
G o 12 G of [ N 2 (g)] 12 G of [O 2 (g)] 12 G of [Cl 2 (g)] G of [ NOCl(g)]
0.500 (0.00 kJ/mol 0.00 kJ/mol 0.00 kJ/mol) 66.08 kJ/mol 66.08 kJ/mol
G o 66.08 kJ/mol
1.278
H o 51.71 kJ/mol
Yet another simple reaction that meets the requirement that G o H o is the combustion of
graphite: C(graphite) O 2 (g)
CO 2 (g) We see from Appendix D that
o
G f [CO 2 (g)] 394.4 kJ/mol is more negative than H fo [CO 2 (g)] 393.5 kJ/mol . (This
reaction is accompanied by an increase in entropy; S =213.7-5.74-205.1 =2.86 J/K, = 1.002.)
G o H o is true of the reaction in which CO(g) is formed from the elements. From
Appendix D, H fo {CO(g)} 110.5 kJ/mol , and Gfo {CO(g)} 137.2 kJ/mol , producing
= (–137.2/–110.5) = 1.242.
Note that any reaction that has > 1.00 will be spontaneous under standard conditions at all
temperatures. (There, of course, is another category, namely, an endothermic reaction that has
S o 0 . This type of reaction is nonspontaneous under standard conditions at all temperatures. As
such it consumes energy while it is running, which is clearly not a desirable result for a fuel cell.)
1046
Chapter 20: Electrochemistry
85. (M) We first write the two half-equations and then calculate a value of G o from
thermochemical data. This value then is used to determine the standard cell potential.
Oxidation: CH 3 CH 2 OH(g) 3 H 2 O(l)
2 CO 2 ( g ) 12 H (aq) 12 e _
Reduction: {O 2 (g ) 4 H (aq) 4e _
2 H 2 O(l)} 3
Overall:
CH 3 CH 2 OH(g) 3 O 2 (g )
3 H 2 O(l) 2 CO 2 ( g )
Thus, n 12
(a) G o 2Gfo [CO 2 (g)] 3G of [ H 2 O(l)] Gfo [CH 3 CH 2 OH(g)] 3Gfo [O 2 (g)]
2( 394.4 kJ/mol) 3( 237.1 kJ/mol) (168.5 kJ/mol) 3(0.00 kJ/mol) 1331.6 kJ/mol
1331.6 103 J/mol
G
Ecell
1.1501 V
n F
12 mol e 96, 485 C/mol e
(b)
E cell
E [O 2 (g)/H 2 O] E [CO 2 (g ) /CH 3 CH 2 OH(g )] 1.1501 V
1.229 V E [CO 2 (g ) / CH 3 CH 2 OH(g)]
E [CO 2 (g ) / CH 3 CH 2 OH(g)] 1.229 1.1501 0.079 V
86. (M) First we determine the change in the amount of H+ in each compartment.
Oxidation: 2 H 2 O(l) O 2 ( g ) 4 H (aq) 4 e
Reduction: 2 H (aq) 2 e H 2 (g)
60 s 1.25 C 1 mol e 1 mol H
0.165 mol H
1 min
1s
96, 485 C 1 mol e
amount H 212 min
Before electrolysis, there is 0.500 mol H2PO4– and 0.500 mol HPO42– in each compartment. The
electrolysis adds 0.165 mol H+ to the anode compartment, which has the effect of transforming
0.165 mol HPO42– into 0.165 mol H2PO4–, giving a total of 0.335 mol HPO42– (0.500 mol – 0.165
mol) and 0.665 mol H2PO4– (0.500 mol + 0.165 mol). We can use the Henderson-Hasselbalch
equation to determine the pH of the solution in the anode compartment.
2
pH pK a2 log
2
[HPO 4 ]
0.335 mol HPO 4 / 0.500 L
7.20 log
6.90
[H 2 PO 4 ]
0.665 mol H 2 PO 4 / 0.500 L
Again we use the Henderson-Hasselbalch equation in the cathode compartment. After electrolysis
there is 0.665 mol HPO42– and 0.335 mol H2PO4–. Again we use the Henderson-Hasselbalch
equation.
2
pH pK a2 log
[HPO 4 ]
7.20 log
[H 2 PO 4 ]
0.665 mol HPO 4
2
/ 0.500 L
7.50
0.335 mol H 2 PO 4 / 0.500 L
87. (M) We first determine the change in the amount of M2+ ion in each compartment.
3600 s 0.500 C 1mol e- 1mol M 2+
=0.0933 mol M 2+
M 2+ =10.00h
1h
1s
96485 C
2mol e This change in amount is the increase in the amount of Cu2+ and the decrease in the amount of
Zn2+. We can also calculate the change in each of the concentrations.
1047
Chapter 20: Electrochemistry
0.0933 mol Cu 2
0.0933 mol Zn 2
2
[Cu ]
0.933 M [Zn ]
0.933 M
0.1000 L
0.1000 L
Then the concentrations of the two ions are determined.
[Cu 2 ] 1.000 M 0.933 M 1.933 M [Zn 2 ] 1.000 M 0.933 M 0.067 M
Now we run the cell as a voltaic cell, first determining the value of Ecell°.
Oxidation : Zn(s)
Zn 2 (aq) 2 e E 0.763 V
2
Reduction : Cu 2 (aq) 2 e Cu(s)
E 0.340 V
2
Net: Zn(s) Cu (aq)
Cu(s)
Ecell
1.103 V
Then we use the Nernst equation to determine the voltage of this cell.
0.067 M
0.0592
[Zn 2 ]
0.0592
1.103
1.103 0.043 1.146 V
Ecell E cell
log
log
2
2
2
1.933 M
[Cu ]
88. (M)(a)
Anode: Zn(s)
Zn 2 (aq) 2e
E 0.763 V
Cathode: {AgCl(s) + e- (aq)
Ag(s) + Cl- (1 M)} × 2
E°=+0.2223 V
2+
Net: Zn(s) + 2AgCl(s)
Zn (aq) + 2Ag(s) + 2Cl (1 M) E°cell = +0.985 V
(b)
The major reason why this electrode is easier to use than the standard hydrogen
electrode is that it does not involve a gas. Thus there are not the practical difficulties
involved in handling gases. Another reason is that it yields a higher value of E cell ,
thus, this is a more spontaneous system.
(c)
Oxidation:
Reduction:
Ag(s)
Ag (aq) e
AgCl(s) e
Ag(s) Cl (aq)
E o 0.800 V
E 0.2223 V
o
Net:
AgCl(s)
Ag (aq) Cl (aq)
Ecell
0.578 V
The net reaction is the solubility reaction, for which the equilibrium constant is Ksp.
G nFE RT ln K sp
ln K sp
nFE
RT
96485 C
1 J
(0.578 V)
1 V.C
1 mol e
22.5
1
1
8.3145 J mol K 298.15 K
1 mol e
K sp e 22.5 1.7 1010
This value is in good agreement with the value of 1.8 × 10-10 given in Table 18-1.
89.
(D) (a)
Ag(s) Ag+(aq) + e-E = -0.800 V
AgCl(s) + e Ag(s) + Cl (aq)
E = 0.2223 V
AgCl(s)
Ag+(aq) + Cl-(aq) Ecell = -0.5777 V
[1.00][1.00 10-3 ]
0.0592
[Ag + ][Cl- ]
0.0592
=
-0.5777
log
V
log
Ecell = Ecell
1
1
1
1
(b)
10.00 mL of 0.0100 M CrO42- + 100.0 mL of 1.00 × 10-3 M Ag+ (Vtotal = 110.0 mL)
1048
= -0.400 V
Chapter 20: Electrochemistry
Concentration of CrO42-after dilution: 0.0100 M×10.00 mL /110.00 mL = 0.000909 M
Concentration of Ag+ after dilution: 0.00100 M×100.0 mL /110.00 mL = 0.000909 M
K sp 1.11012
2Ag+(aq) +
CrO42-(aq)
Ag2CrO4(s)
Initial
0.000909 M
0.000909 M
Change(100% rxn)
-0.000909 M
-0.000455 M
New initial
0M
0.000455 M
Change
+2x
+x
Equilibrium
2x
0.000455 M+x 0.000455 M
-12
2
1.1 × 10 = (2x) (0.000454)
x = 0.0000246 M Note: 5.4% of 0.000455 M (assumption may be considered valid)
(Answer would be x = 0.0000253 using method of successive approx.)
[Ag+] = 2x = 0.0000492 M (0.0000506 M using method of successive approx.)
0.0592
0.0592
log [ Ag ][Cl- ] 0.5777
log [1.00M ][4.92 104 M ]
1
1
0.323 V (-0.306 V for method of successive approximations)
Ecell E cell
(c)
10.00 mL 0.0100 M NH3 + 10.00 mL 0.0100 M CrO42- + 100.0 mL 1.00×10-3 M Ag+
(Vtotal = 120.0 mL)
Concentration of NH3 after dilution: 10.0 M×10.00 mL /120.00 mL = 0.833 M
Concentration of CrO42-after dilution: 0.0100 M×10.00 mL /110.00 mL = 0.000833 M
Concentration of Ag+ after dilution: 0.00100 M×100.0 mL /110.00 mL = 0.000833 M
In order to determine the equilibrium concentration of free Ag+(aq), we first consider
complexation of Ag+(aq) by NH3(aq) and then check to see if precipitation occurs.
Initial
Change(100% rxn)
New initial
Change
Equilibrium
Equilibrium (x 0)
Ag+(aq)
+
0.000833 M
-0.000833 M
0M
+x
x
x
K f 1.610
2NH3(aq)
0.833 M
-0.00167 M
0.831 M
+2x
(0.831+2x) M
0.831 M
7
Ag(NH3)2+(aq)
0M
+0.000833 M
0.000833 M
-x
(0.000833 –x) M
0.000833 M
x = 7.54×10-11 M = [Ag+]
1.6 × 107 = 0.000833 /x(0.831)2
Note: The assumption is valid
Now we look to see if a precipitate forms: Qsp = (7.54×10-11)2(0.000833) = 4.7 ×10-24
Since Qsp < Ksp (1.1 × 10-12), no precipitate forms and [Ag+] = 7.54 × 10-11 M
0.0592
0.0592
log [Ag + ][Cl- ]= -0.5777 V log [1.00M][7.54 × 10-11 M]
1
1
= 0.0215 V
E cell = E°cell E cell
90.
(M) We assume that the Pb2+(aq) is “ignored” by the silver electrode, that is, the silver
electrode detects only silver ion in solution.
1049
Chapter 20: Electrochemistry
Oxidation : H 2 (g)
2 H (aq) 2 e
E 0.000 V
Reduction : {Ag (aq) e
Ag(s)} 2
H 2 (g) 2 Ag
Net :
Ecell Ecell
log
E 0.800 V
2 H (aq) 2 Ag(s) E cell 0.800 V
0.0592
2
log
[H ]2
[Ag ]2
2(0.800 0.503)
1.00 2
10.0
2
0.0592
[Ag ]
0.503 V 0.800 V
0.0592
2
log
1.002
[Ag ]2
1.00 2
10 10.0 1.0 1010
[Ag ] 2
[Ag + ]2 =1.0 10-10 M 2 [Ag + ]=1.0 10-5
mass Ag 0.500 L
1.0 10 5 mol Ag 1 mol Ag 107.87 g Ag
5.4 10 4 g Ag
1 L soln
1 mol Ag
1 mol Ag
5.4 10 4 g Ag
100 % 0.051% Ag (by mass)
1.050 g sample
91. (M) 250.0 mL of 0.1000 M CuSO4 = 0.02500 moles Cu2+ initially.
% Ag
moles of Cu2+ plated out =
1 mol e- 1 mol Cu 2+
3.512C
1368 s
0.02490 mol Cu 2+
96,485 C
s
2 mole
moles of Cu2+ in solution = 0.02500 mol – 0.02490 mol = 0.00010 mol Cu2+
[Cu2+] = 0.00010 mol Cu2+/0.250 L = 4.0 × 10-4 M
Cu2+(aq) +
0.00040 M
-0.00040M
0M
+x
x
Initial
Change(100% rxn)
New initial
Change
Equilibrium
K f 1.110
4NH3(aq)
0.10 M
maintained
0.10 M
maintained
0.10 M
13
Cu(NH3)42+(aq)
0M
+0.00040 M
0.00040 M
-x
(0.00040 –x) M 0.00040 M
[Cu(NH 3 ) 4 2+ ]
0.00040
1.1 1013
Kf =
=
[Cu 2+ ] = 3.6 1013 M
2+
4
2+
4
[Cu ][NH3 ] [Cu ](0.10)
Hence, the assumption is valid. The concentration of Cu(NH3)42+(aq) = 0.00040 M which is
40 times greater than the 1 × 10-5 detection limit. Thus, the blue color should appear.
92.
(M) First we determine the molar solubility of AgBr in 1 M NH3 .
Sum
AgBr(s) 2NH 3 (aq) Ag(NH 3 ) 2 (aq) Br (aq)
Initial
1.00 M
Equil.
K
1.00-2s
0 M
s
[Ag(NH 3 ) 2 ][Br ]
s2
8.0 106
[NH 3 (aq)]
(1 2 s ) 2
So [Ag ]
K sp
[Br ]
5.0 1013
1.8 1010 M
3
2.81 10
1050
K K sp K f 8.0 106
0M
s
{s AgBr molar solubility}
s 2.81 10 3 M (also [Br ])
Chapter 20: Electrochemistry
AgBr(s) Ag Br
K sp 5.0 1013
Ag 2NH 3 Ag(NH 3 ) 2
K f 1.6 107
(sum) AgBr(s) 2NH 3 (aq) Ag(NH 3 ) 2 Br K K sp K f 8.0 106
Now, let's construct the cell, guessing that the standard hydrogen electrode is the anode.
Oxidation: H 2 (g)
2 H 2 e
E o 0.000 V
Reduction: {Ag (aq) e
Ag(s) } 2
E o 0.800 V
Net:
2 Ag (aq) H 2 (g)
2 Ag(s) 2 H (aq)
From the Nernst equation:
o
Ecell
0.800V
0.0592
[ H ]2
0.0592
12
o 0.0592
0.800 V
E E log 10 Q E log
log10
n
n
2
[ Ag ]2
(1.78 1010 ) 2
and E 0.223 V. Since the voltage is positive, our guess is correct and the standard
o
hydrogen electrode is the anode (oxidizing electrode).
93. (M) (a)
Anode:
2H2O(l) 4 e- + 4 H+(aq) + O2(g)
Cathode:
2 H2O(l) + 2 e- 2 OH-(aq) + H2(g)
Overall:
2 H2O(l) + 4 H2O(l) 4 H+(aq) + 4 OH-(aq) + 2 H2(g) + O2(g)
2 H2O(l) 2 H2(g) + O2(g)
(b)
21.5 mA = 0.0215 A or 0.0215 C s-1 for 683 s
mol H 2SO 4 =
0.0215 C
s
683s
1 mol e96485C
1 mol H +
1 mol e
-
1 mol H 2SO 4
2 mol H +
= 7.61 10-5 mol H 2SO 4
7.61 10-5 mol H 2SO 4 in 10.00 mL. Hence [H 2SO 4 ] = 7.61 10-5 mol / 0.01000 L = 7.61 10-3 M
94.
(D) First we need to find the total surface area
Outer circumference = 2r = 2(2.50 cm) = 15.7 cm
Surface area = circumference × thickness = 15.7 cm × 0.50 cm = 7.85 cm2
Inner circumference = 2r = 2(1.00 cm) = 6.28 cm
Surface area = circumference × thickness = 6.28 cm × 0.50 cm = 3.14 cm2
Area of small circle = r2 = (1.00 cm)2 = 3.14 cm2
Area of large circle = r2 = (2.50 cm)2 = 19.6 cm2
Total area = 7.85 cm2 + 3.14 cm2 + 2×(19.6 cm2) – 2×(3.14 cm2) = 43.91 cm2
Volume of metal needed = surface area × thickness of plating = 43.91 cm2 0.0050 cm = 0.22 cm3
8.90 g
1 mol Ni
2 mol e- 96485 C
Charge required = 0.22 cm 3
6437.5 C
58.693 g Ni 1 mol Ni 1 mol ecm 3
Time = charge/time = 6437.5 C / 1.50 C/s = 4291.7 s or 71.5 min
1051
Chapter 20: Electrochemistry
95. (M) The overall reaction for the electrolytic cell is:
o
Ecell
1.103V
Cu(s)+Zn 2+ (aq) Cu 2+ (aq)+Zn(s)
2+
Next, we calculate the number of moles of Zn (aq) plated out and number of moles of
Cu2+(aq) formed:
0.500C
60 min 60s
1 mol e- 1 mol Cu 2+
10 h
0.0935mol Cu 2+
n(Cu 2+ ) =
96,485 C
s
1h
1min
2 mol e
2+
2+
n(Zn ) n(Cu )=0.0935mol
Initially, solution contained 1.00molL-1 0.100L=0.100 mol Zn2+(aq). Therefore, at the end
of electrolysis we are left with:
6.5 10-3mol
n(Zn 2+ ) LEFT =(0.100-0.0935)mol=6.5 10-3mol [Zn 2+ ]=
=6.5 10-2 M
0.1L
0.0935mol
n(Cu 2+ ) FORMED =0.0935mol [Cu 2+ ]=
=0.936M
0.1L
The new potential after the cell was switched to a voltaic one can be calculated using Nernst
equation:
0.0592
[Zn 2 ]
0.0592
6.5 103 M
log
1.103
log
1.103 0.064 1.167 V
Ecell Ecell
2
2
0.935 M
[Cu 2 ]
96.
(M) (a)
The metal has to have a reduction potential more negative than -0.691 V, so that
its oxidation can reverse the tarnishing reaction’s -0.691 V reduction potential.
Aluminum is a good candidate, because it is inexpensive, readily available, will not react
with water and has an Eo of -1.676 V. Zinc is also a possibility with an Eo of -0.763 V,
but we don't choose it because there may be an overpotential associated with the
tarnishing reaction.
(b)
Oxidation:
{Al(s)
Al3 (aq) 3 e } 2
Reduction:
{Ag 2 S(s) 2 e-
Ag(s) S2 (aq) } 3
Net:
2 Al(s) 3 Ag 2 S(s)
6Ag(s) 2 Al3 (aq) 3 S2 (aq)
(c)
The dissolved NaHCO3(s) serves as an electrolyte. It would also enhance the electrical
contact between the two objects.
(d)
There are several chemicals involved: Al, H2O, and NaHCO3. Although the aluminum
plate will be consumed very slowly because silver tarnish is very thin, it will,
nonetheless, eventually erode away. We should be able to detect loss of mass after
many uses.
Overall:
C3H8(g) + 5 O2(g)
3 CO2(g) + 4 H2O(l)
Anode:
6 H2O(l) + C3H8(g) 3 CO2(g) + 20 H+(g) + 20 eCathode: 20 e- + 20 H+(g) + 5 O2(g) 10 H2O(l)
97. (M) (a)
Grxn = 3(-394.4 kJ/mol) + 4(-237.1 kJ/mol) – 1(-23.3 kJ/mol) = -2108.3 kJ/mol
1052
Chapter 20: Electrochemistry
Grxn = -2108.3 kJ/mol = -2,108,300 J/mol
Grxn = -nFEcell = -20 mol e- × (96485 C/mol e-) × Ecell
Ecell = +1.0926 V
20 e- + 20 H+(g) + 5 O2(g)
Hence
Ecell = Ecathode Eanode
10 H2O(l) Ecathode = +1.229 V
+1.0926 V = +1.229 V Eanode
Eanode = +0.136 V (reduction potential for 3 CO2(g) + 20 H+(g) + 20 e- 6 H2O(l) + C3H8(g))
(b) Use thermodynamic tables for 3 CO2(g) + 20 H+(g) + 20 e-
6 H2O(l) + C3H8(g)
Grxn = 6(-237.1 kJ/mol) + 1(-23.3 kJ/mol) [(-394.4 kJ/mol) + 20( 0kJ/mol)]= -262.7 kJ/mol
Gred = -262.7 kJ/mol = -262,700 J/mol = -nFEred = -20 mol e-×(96,485 C/mol e-)×Ered
Ered = 0.136 V (Same value, as found in (a))
o
98. (D) (a) Equation 20.15 ( G o zFEcell
) gives the relationship between the standard Gibbs
energy of a reaction and the standard cell potential. Gibbs free energy also varies with
temperature ( G o H o T S o ). If we assume that H o and S o do not vary
significantly over a small temperature range, we can derive an equation for the temperature
o
:
variation of Ecell
H o T S o
G H T S zFE E
zF
Considering two different temperatures one can write:
H o T1S o
H o T2 S o
o
o
Ecell
(T1 )
and Ecell
(T2 )
zF
zF
o
o
o
H T1S
H T2 S o
o
o
Ecell (T1 ) Ecell (T2 )
zF
zF
o
o
T S T2 S
S o
o
o
Ecell
(T1 ) Ecell
(T2 ) 1
T2 T1
zF
zF
(b) Using this equation, we can now calculate the cell potential of a Daniel cell at 50 oC:
10.4JK 1mol 1
o
o
Ecell
(25 o C) Ecell
(50 o C)
50 25 K 0.00135
2 96485Cmol 1
o
Ecell
(50 o C) 1.103V 0.00135 1.104V
o
o
o
o
cell
o
cell
99. (D) Recall that under non-standard conditions G G o RT ln K .
G o H o T S o and G zFEcell one obtains:
zFEcell H o T S o RT lnQ
For two different temperatures (T1 and T2) we can write:
1053
Substituting
Chapter 20: Electrochemistry
zFEcell (T1 ) H o T1S o RT1 lnQ
zFEcell (T2 ) H o T2 S o RT2 lnQ
H o T1S o RT1 lnQ H o T2 S o RT2 lnQ
Ecell (T1 ) Ecell (T2 )
zF
o
T S RT1 lnQ T2 S o RT2 lnQ
Ecell (T1 ) Ecell (T2 ) 1
zF
o
T S RT1 lnQ T2 S o RT2 lnQ
Ecell (T1 ) Ecell (T2 ) 1
zF
zF
o
S R lnQ
S o R lnQ
Ecell (T1 ) Ecell (T2 ) T1
T2
zF
zF
S o R lnQ
Ecell (T1 ) Ecell (T2 ) (T1 T2 )
zF
o
o
The value of Q at 25 oC can be calculated from Ecell
and Ecell . First calculate Ecell
:
2+
Oxidation:
Cu(s) Cu (aq)+2e
Eo 0.340V
3+
2+
Reduction:
2Fe (aq)+2e 2Fe (aq)
Eo 0.771V
_____________________________________________________________________________________________________________
o
Cu(s)+2Fe 3 Cu 2+ (aq)+2Fe 2
Ecell
0.431V
0.0592
0.370 0.431
logQ 0.296 logQ 0.431 0.370 0.061
2
logQ 0.206 Q 10 0.206 1.61
Now, use the above derived equation and solve for S o :
S o 8.314 ln1.61
0.394 0.370 (50 25)
2 96485
Overall:
S o 3.96
) S o 3.96 185.3 S o 189.2JK -1
192970
o
o
we can calculate G o , K (at 25 oC) and H o :
Since G H o T S o zFEcell
o
G o zFEcell
2 96485 0.431 83.2kJ
0.024 25 (
G o RT ln K 83.2 1000 8.314 298.15 ln K
ln K 33.56 K e33.56 3.77 1014
189.2
-83.2kJ=H o 298.15
kJ H o 83.2 56.4 26.8kJ
1000
o
Since we have H and S o we can calculate the value of the equilibrium constant at 50
o
C:
189.2
G o H o T S o 26.8kJ (273.15 50)K
kJK -1 87.9kJ
1000
-1
-1
87.9 1000J 8.314JK mol (273.15 50)K ln K
ln K 32.7 K e 32.7 1.59 1014
1054
Chapter 20: Electrochemistry
Choose the values for the concentrations of Fe2+, Cu2+ and Fe3+ that will give the value of
the above calculated Q. For example:
2
Fe 2+ Cu 2+
Q=
=1.61
2
Fe 3+
0.12 1.61
1.61
0.12
Determine the equilibrium concentrations at 50 oC. Notice that since Q1. S o , H o and U o cannot be used alone to determine whether a
particular electrochemical reaction will have a positive or negative value.
106. (M) The half-reactions for the first cell are:
Anode (oxidation):
X(s) X + (aq)+e-
E Xo / X
Cathode (reduction): 2H + (aq)+2e - H 2 (g) E o 0V
Since the electrons are flowing from metal X to the standard hydrogen electrode,
E Xo / X 0V .
The half-reactions for the second cell are:
Anode (oxidation):
X(s) X + (aq)+eCathode (reduction):
Y2+ +2e - Y(s)
E Xo / X
EYo2 /Y
Since the electrons are flowing from metal X to metal Y, E Xo / X + EYo2 /Y >0.
From the first cell we know that E Xo / X 0V . Therefore, E Xo / X > EYo2 /Y .
107. (M) The standard reduction potential of the Fe2+(aq)/Fe(s) couple can be determined from:
Fe 2+ (aq) Fe 3+ (aq)+e- Eo 0.771V
Fe 3+ (aq)+3e- Fe(s) E o 0.04V
Overall:
Fe 2+ (aq)+2e- Fe(s)
We proceed similarly to the solution for 100:
1060
Chapter 20: Electrochemistry
Eo
n E
o
i
i
n
0.771 (1) 0.04 3
0.445V
31
i
SELF-ASSESSMENT EXERCISES
108. (E) (a) A standard electrode potential E o measures the tendency for a reduction process to
occur at an electrode.
(b) F is the Faraday constant, or the electric charge per mole of electrons (96485 C/mol).
(c) The anode is the electrode at which oxidation occurs.
(d) The cathode is the electrode at which reduction occurs.
109. (E) (a) A salt bridge is a devise used to connect the oxidation and reduction half-cells of a
galvanic (voltaic) cell.
(b) The standard hydrogen electrode (abbreviated SHE), also called normal hydrogen
electrode (NHE), is a redox electrode which forms the basis of the thermodynamic scale of
oxidation-reduction potentials. By definition electrode potential for SHE is 0.
(c) Cathodic protection is a technique commonly used to control the corrosion of a metal
surface by making it work as a cathode of an electrochemical cell. This is achieved by
placing in contact with the metal to be protected another more easily corroded metal to act as
the anode of the electrochemical cell.
(d) A fuel cell is an electrochemical cell that produces electricity from fuels. The essential
process in a fuel cell is fuel+oxygen oxidation products.
110. (E) (a) An overall cell reaction is a combination of oxidation and reduction halfreactions.
(b) In a galvanic (voltaic) cell, chemical change is used to produce electricity. In an
electrolytic cell, electricity is used to produce a nonspontaneous rection.
(c) In a primary cell, the cell reaction is not reversible. In a secondary cell, the cell
reaction can be reversed by passing electricity through the cell (charging).
o
(d) Ecell
refers to the standard cell potential (the ionic species are present in aqueous
solution at unit activity (approximately 1M), and gases are at 1 bar pressure (approximately 1
atm).
111. (M) (a) False. The cathode is the positive electrode in a voltaic cell and negative in
electrolytic cell.
(b) False. The function of the salt bridge is to permit the migration of the ions not
electrons.
(c) True. The anode is the negative electrode in a voltaic cell.
(d) True.
(e) True. Reduction always occurs at the cathode of an electrochemical cell. Because of
the removal of electrons by the reduction half-reaction, the cathode of a voltaic cell is
positive. Because of the electrons forced onto it, the cathode of an electrolytic cell is
negative. For both types, the cathode is the electrode at which electrons enter the cell.
1061
Chapter 20: Electrochemistry
(f) False. Reversing the direction of the electron flow changes the voltaic cell into an
electrolytic cell.
(g) True. The cell reaction is an oxidation-reduction reaction.
112. (M) The correct answer is (b), Hg 2 (aq ) is more readily reduced than H (aq ) .
113. (M) Under non-standard conditions, apply the Nernst equation to calculate Ecell :
0.0592
o
Ecell Ecell
log Q
z
0.0592
0.10
Ecell 0.66
log
0.63V
2
0.01
The correct answer is (d).
114. (E) (c) The displacement of Ni(s) from the solution will proceed to a considerable extent,
but the reaction will not go to completion.
115. (E) The gas evolved at the anode when K2SO4(aq) is electrolyzed between Pt electrodes is
most likely oxygen.
116. (M) The electrochemical reaction in the cell is:
Anode (oxidation):
{Al(s) Al3+ (aq)+3e- } 2
Cathode (reduction): {H 2 (g)+2e- 2H + (aq)} 3
Overall:
2Al(s)+3H 2 (g) 2Al 3+ (aq)+6H + (aq)
1mol Al
3 mol H 2
4.5g Al
0.250mol H 2
26.98g Al 2 mol Al
22.4L H 2
0.250 mol H 2
=5.6L H 2
1mol H 2
117. (E)
The correct answer is (a) G .
118. (M) Anode (oxidation): {Zn(s) Zn 2+ (aq)+2e- } 3
Cathode (reduction): {NO-3 (aq)+4H + (aq)+3e - NO(g)+2H 2 O(l)} 2
E o 0.763V
E o 0.956V
_________________________________________________________________________________________________________
Overall:
o
3Zn(s)+2NO-3 (aq)+8H + (aq) 3Zn 2+ (aq)+2NO(g)+4H 2O(l) Ecell
1.719V
Cell diagram:
Zn(s) Zn 2+ (1M) H + (1M),NO-3 (1M) NO(g,1atm) Pt(s)
119. (M) Apply the Nernst equation:
1062
Chapter 20: Electrochemistry
0.0592
log Q
z
0.0592
0.108 0
log x 2 log x 2 3.65
2
2
3.65
x 10
x 0.0150 M
pH log(0.0150) 1.82
o
Ecell Ecell
o
, we can calculate K for the given reaction:
120. (M) (a) Since we are given Ecell
RT
o
Ecell
ln K
nF
8.314 JK 1mol 1 298 K
0.0050V
ln K ln K 0.389
2 96485Cmol 1
K e0.389 0.68
Since for the given conditions Q=1, the system is not at equilibrium.
(b) Because Q>K, a net reaction occurs to the left.
o
121. (M) (a) Fe(s)+Cu 2+ (1M) Fe 2+ (1M)+Cu(s) , Ecell
0.780V , electron flow from B to A
2+
+
4+
o
(b) Sn (1M)+2Ag (1M) Sn (aq)+2Ag(s) , Ecell 0.646V , electron flow from A to B.
o
(c) Zn(s)+Fe 2+ (0.0010M) Zn 2+ (0.10M)+Fe(s) , Ecell
0.264V ,electron flow from A to
B.
122. (M) (a)
(b)
(c)
(d)
Cl2(g) at anode and Cu(s) at cathode.
O2(g) at anode and H2(g) and OH-(aq) at cathode.
Cl2(g) at anode and Ba(l) at cathode.
O2(g) at anode and H2(g) and OH-(aq) at cathode.
1063
CHAPTER 24
COMPLEX IONS AND COORDINATION COMPOUNDS
PRACTICE EXAMPLES
1A
(E) There are two different kinds of ligands in this complex ion, I and CN. Both are
monodentate ligands, that is, they each form only one bond to the central atom. Since there
are five ligands in total for the complex ion, the coordination number is 5: C.N. = 5 . Each
CN ligand has a charge of 1, as does the I ligand. Thus, the O.S. must be such that:
O.S. + 4 +1 1 = 3 = O.S. 5 . Therefore, O.S. = +2 .
1B
(E) The ligands are all CN . Fe3+ is the central metal ion. The complex ion is
Fe CN 6 .
3
2A
(E) There are six “Cl” ligands (chloride), each with a charge of 1. The platinum metal
2
center has an oxidation state of +4 . Thus, the complex ion is PtCl6 , and we need two
K + to balance charge: K 2 PtCl6 .
2B
(E) The “SCN” ligand is the thiocyanato group, with a charge of 1bonding to the
central metal ion through the sulfur atom. The “NH3” ligand is ammonia, a neutral ligand.
There are five (penta) ammine ligands bonded to the metal. The oxidation state of the
cobalt atom is +3 . The complex ion is not negatively charged, so its name does not end
with “-ate”. The name of the compound is pentaamminethiocyanato-S-cobalt(III) chloride.
3A
(M) The oxalato ligand must occupy two cis- positions. Either the two NH3 or the two Cl
ligands can be coplanar with the oxalate ligand, leaving the other two ligands axial. The
other isomer has one NH3 and one Cl ligand coplanar with the oxalate ligand. The
structures are sketched below.
ox
Cl
3B
Cl
Co NH3
NH3
ox
H3N
NH3
Co Cl
Cl
ox
H3N
Cl
Co NH3
Cl
(M) We start out with the two pyridines, C5 H 5 N , located cis to each other. With this
assignment imposed, we can have the two Clligands trans and the two CO ligands cis, the
two CO ligands trans and the two Cl ligands cis, or both the Clligands and the two CO
ligands cis. If we now place the two pyridines trans, we can have either both other pairs
trans, or both other pairs cis. There are five geometric isomers. They follow, in the order
described.
1163
Chapter 24: Complex Ions and Coordination Compounds
4A
(M) The F ligand is a weak field ligand. MnF6
2
is an octahedral complex. Mn 4+ has
2
three 3d electrons. The ligand field splitting diagram for MnF6 is sketched below. There
are three unpaired electrons.
eg
E
t2g
4B
(M) Co2+ has seven 3d electrons. Cl is a weak field ligand. H2O is a moderate field
ligand. There are three unpaired electrons in each case. The number of unpaired electrons
has no dependence on geometry for either metal ion complex.
CoCl
Co H 2O 6
2
4
5A
E
2+
E
(M) CN is a strong field ligand. Co 2+ has seven 3d electrons. In the absence of a crystal
field, all five d orbitals have the same energy. Three of the seven d electrons in this case
will be unpaired. We need an orbital splitting diagram in which there are three orbitals of
the same energy at higher energy. This is the case with a tetrahedral orbital diagram.
t2 g
eg
In absence of a
crystal field
eg
t2 g
Tetrahedral geometry
Octahedralgeometry
Thus, Co CN 4 must be tetrahedral (3 unpaired electrons) and not octahedral (1
unpaired electron) because the magnetic behavior of a tetrahedral arrangement would
agree with the experimental observations (3 unpaired electrons).
2
5B
(M) NH3 is a strong field ligand. Cu 2+ has nine 3d electrons. There is only one way to
arrange nine electrons in five d-orbitals and that is to have four fully occupied orbitals (two
electrons in each orbital), and one half-filled orbital. Thus, the complex ion must be
paramagnetic to the extent of one unpaired electron, regardless of the geometry the ligands
adopt around the central metal ion.
6A
(M) We are certain that Co H 2O 6
2
is octahedral with a moderate field ligand.
Tetrahedral CoCl 4 has a weak field ligand. The relative values of ligand field splitting
2
for the same ligand are t = 0.44 o . Thus, Co H 2O 6
1164
2+
absorbs light of higher energy,
Chapter 24: Complex Ions and Coordination Compounds
blue or green light, leaving a light pink as the complementary color we observe. CoCl 4
absorbs lower energy red light, leaving blue light to pass through and be seen.
2
CO
NC5H5
Cl Mo Cl
OC
NC5H5
6B
Cl
Cl
NC5H5
OC Mo CO
Cl
NC5H5
NC5H5
CO
Cl Mo Cl
OC
NC5H5
NC5H5
Cl Mo CO
OC
NC5H5
NC5H5
CO
Cl Mo CO
Cl
NC5H5
(M) In order, the two complex ions are Fe H 2O 6 and Fe CN 6 . We know that
CN is a strong field ligand; it should give rise to a large value of o and absorb light of
4
2+
the shorter wavelength. We would expect the cyano complex to absorb blue or violet light
and thus K 4 Fe CN 6 3 H 2 O should appear yellow. The compound
Fe H 2O 6 NO 3 2 , contains the weak field ligand H2O and thus should be green. (The
weak field would result in the absorption of light of long wavelength (namely, red light),
which would leave green as the color we observe.)
INTEGRATIVE EXAMPLE
A
(M) (a) There is no reaction with AgNO3 or en, so the compound must be transchlorobis(ethylenediamine)nitrito-N-cobalt(III) nitrite
+
H2
N
Cl
H2
N
Co
H2N
NO2
NO2-
N
H2
(b) If the compound reacts with AgNO3, but not with en, it must be transbis(ethylenediamine)dinitrito-N-cobalt(III) chloride.
+
H
N
NO2
H
N
Co
HN
NO2
Cl-
N
H
(c) If it reacts with AgNO3 and en and is optically active, it must be cisbis(ethylenediamine)dinitrito-N-cobalt(III) chloride.
+
H2
N
NO2
H2
N
Co
H2N
NO2
1165
N
H2
Cl-
Chapter 24: Complex Ions and Coordination Compounds
B
(D) We first need to compute the empirical formula of the complex compound:
0.236
1 mol Pt
=0.236 mol Pt
=1 mol Pt
46.2 g Pt
0.236
195.1 g Pt
0.946
1 mol Cl
=0.946 mol Cl
=4 mol Pt
33.6 g Cl
0.236
35.5 g Cl
1.18
1 mol N
=1.18 mol N
=5 mol N
16.6 g N
0.236
14.01 g N
3.6
1 mol H
=3.6 mol H
=15 mol H
3.6 g Pt
0.236
1gH
The nitrogen ligand is NH3, apparently, so the empirical formula is Pt(NH3)5Cl4.
T
0.74 o C
0.4m . The effective
The effective molality of the solution is m
K fp 1.86 o C / m
molality is 4 times the stated molality, so we have 4 particles produced per mole of Pt
complex, and therefore 3 ionizable chloride ions. We can write this in the following way:
[Pt(NH3)5Cl][Cl]3. Only one form of the cation (with charge 3+) shown below will exist.
+3
NH3
H3N
NH3
Pt
H3N
NH3
Cl
EXERCISES
Nomenclature
1.
(E) (a)
(b)
(c)
2.
CrCl4 NH 3 2
Fe CN 6
3
diamminetetrachlorochromate(III) ion
hexacyanoferrate(III) ion
Cr en 3 Ni CN 4 . tris(ethylenediamine)chromium(III) tetracyanonickelate(II)
3
2
ion
Co NH 3 6
The coordination number of Co is 6; there are six
monodentate NH 3 ligands attached to Co. Since the NH 3 ligand is neutral, the
oxidation state of cobalt is +2 , the same as the charge for the complex ion;
hexaamminecobalt(II) ion.
(E) (a)
2+
1166
Chapter 24: Complex Ions and Coordination Compounds
(b)
(c)
(d)
(e)
(f)
3.
4.
AlF6 The coordination number of Al is 6; F is monodentate. Each F has a 1
charge; thus the oxidation state of Al is +3 ; hexafluoroaluminate(III) ion.
3
2
Cu CN 4 The coordination number of Cu is 4; CN is monodentate. CN has a
1 charge; thus the oxidation state of Cu is +2 ; tetracyanocuprate(II) ion
CrBr2 NH 3 4
The coordination number of Cr is 6; NH 3 and Br are
monodentate. NH 3 has no charge; Br has a 1 charge. The oxidation state of
chromium is +3 ; tetraamminedibromochromium(III) ion
+
4
Co ox 3 The coordination number of Co is 6; oxalate is bidentate. C2O 24 ox
has a 2 charge; thus the oxidation state of cobalt is +2 ; trioxalatocobaltate(II) ion.
Ag S2 O 3 2 The coordination number of Ag is 2; S2 O 32 is monodentate. S2 O 32
has a 2 charge; thus the oxidation state of silver is +1 ; dithiosulfatoargentate(I)
ion. ( Although +1 is by far the most common oxidation state for silver in its
compounds, stable silver(III) complexes are known. Thus, strictly speaking, silver is
not a non-variable metal, and hence when naming silver compounds, the oxidation
state(s) for the silver atom(s) should be specified).
3
(M) (a) Co OH H 2 O 4 NH 3
2+
amminetetraaquahydroxocobalt(III) ion
(b)
Co ONO 3 NH 3 3
triamminetrinitrito-O-cobalt(III)
(c)
Pt H 2O 4 PtCl6
tetraaquaplatinum(II) hexachloroplatinate(IV)
(d)
Fe ox 2 H 2O 2
diaquadioxalatoferrate(III) ion
(e)
Ag 2 HgI 4
silver(I) tetraiodomercurate(II)
(M) (a) K 3 Fe CN 6
potassium hexacyanoferrate(III)
(b)
Cu en 2
(c)
Al OH H 2O 5 Cl2
pentaaquahydroxoaluminum(III) chloride
(d)
CrCl en 2 NH 3 SO 4
amminechlorobis(ethylenediammine)chromium(III) sulfate
(e)
Fe en 3 4 Fe CN 6 3
tris(ethylenediamine)iron(III) hexacyanoferrate(II)
2+
bis(ethylenediamine)copper(II) ion
1167
Chapter 24: Complex Ions and Coordination Compounds
Bonding and Structure in Complex Ions
5.
(E) The Lewis structures are grouped together at the end.
(a)
H2O has 1×2 + 6 = 8 valence electrons, or 4 pairs.
(b)
CH 3 NH 2 has 4 + 3 1+ 5 + 2 1 = 14 valence electrons, or 7 pairs.
(c)
ONO has 2 6 + 5 +1 = 18 valence electrons, or 9 pairs. The structure has a 1
formal charge on the oxygen that is singly bonded to N.
(d)
SCN has 6 + 4 + 5 +1 = 16 valence electrons, or 8 pairs. This structure,
appropriately, gives a 1 formal charge to N.
H H
H
N C H
O N O
S C N
O
H
H H
(a)
6.
(b)
(c)
(M) The Lewis structures are grouped together at the end.
(a)
OH- has 6 +1 + 1 = 8 valence electrons, or 4 pairs.
(b)
SO42- has 6 + 6 4 + 2 = 32 valence electrons, or 16 pairs.
(c)
C2O42- has 2 4 + 6 4 + 2 = 34 valence electrons, or 17 pairs.
(d)
SCN has 6 + 4 + 5 +1 = 16 valence electrons, or 8 pairs. This structure,
appropriately, gives a 1 formal charge to N.
O
O
O
O H
2+
S
(a)
O
O O
O
O
O
7.
(d)
N C
S
O
monodentate
or bidentate
(b)
(c)
(d)
is square planar by analogy with PtCl2 NH 3 2 in Figure
24-5. The other two complex ions are octahedral.
(M) We assume that PtCl 4
2-
(a)
Cl
Cl
Pt
Cl
Cl
2
(b)
H2O
H2O
2+
NH3
NH3
Co NH3
OH2
1168
(c)
Cl
H2O
H2O
OH2
Cr OH2
OH2
2+
Chapter 24: Complex Ions and Coordination Compounds
8.
(M) The structures for Pt ox 2
2
and Cr ox 3
3
are drawn below. The structure of
Fe EDTA is the same as the generic structure for M EDTA drawn in Figure 24-23,
with M n+ = Fe 2+ .
O
(a)
(b)
3O
O
O
O
2
2
C
C
O
C
2-
C
C
Pt
O
O
O
O
C C
O
O Cr
O
C O
C O
O C
O
O
O C
(c)
9.
See Figure 24-23.
(M) (a)
NH3
NH3
H3N
Cr
H3N
+
(b)
3-
O
O
C C
O
OSO 3
O
O Co O
C O
C O
O C
O
O
O C
NH3
pentaamminesulfateochromium(III)
trioxalatocobaltate(III)
(c)
fac-triamminedichloronitro-O-cobalt(III)
NH3
NH3
Cl
Co NH3
ONO
mer-triammine-cis-dichloronitro-O-cobalt(III)
NH3
NH3
Cl Co ONO
Cl
Cl
NH3
mer-triammine-trans-dichloronitro-O-cobalt(III)
NH3
NH3
Cl Co Cl
ONO
10.
NH3
(M) (a) pentaamminenitroto-N-cobalt(III) ion
copper(II)
O2N
H3N
NH3
NH3
Co NH3
(b) ethylenediaminedithiocyanato-S-
2+
H2N
NCS
NH3
Cu
CH2
CH2
NH2
SCN
1169
Chapter 24: Complex Ions and Coordination Compounds
(c)
hexaaquanickel(II) ion
2+
H2O
OH2
H2O Ni OH2
H2O
H2O
Isomerism
11.
(E) (a) cis-trans isomerism cannot occur with tetrahedral structures because all of the
ligands are separated by the same angular distance from each other. One ligand
cannot be on the other side of the central atom from another.
(b)
(c)
12.
Square planar structures can show cis-trans isomerism. Examples are drawn following,
with the cis-isomer drawn on the left, and the trans-isomer drawn on the right.
A
A
B
B
M
M
A
B
cis-isomer
B
A
trans-isomer
Linear structures do not display cis-trans isomerism; there is only one way to bond
the two ligands to the central atom.
(M) (a) CrOH NH 3 5
2+
NH3
NH3
HO Cr NH3
H3N
NH3
2+
(b)
(c)
has one isomer.
CrCl2 H 2 O NH 3 3 has three isomers.
+
+
+
Cl
Cl
Cl
NH3
Cl
NH3
H
O
Cr
NH3
2
H2O Cr NH3
H2O Cr Cl
H3N
H3N
H3N
Cl
NH3
NH3
+
CrCl2 en 2 has two geometric isomers, cis- and trans-.
+
+
Cl
C
N
Cl
C
Cl
N Cr N
N
Cr
N
C N
C
N
Cl
C
C
N
C
C
+
1170
Chapter 24: Complex Ions and Coordination Compounds
(d)
(e)
CrCl 4 en has only one isomer since the ethylenediamine (en) ligand cannot
bond trans to the central metal atom.
Cl
Cl
Cl Cr N
Cl
C
N
C
Cr en 3
3+
N
has only one geometric isomer; it has two optical isomers.
3+
C
C
N
N
N Cr
C N
N
C
13.
C
C
(M) (a) There are three different square planar isomers, with D, C, and B, respectively,
trans to the A ligand. They are drawn below.
B
A
(b)
14.
D
D
C
A
C
Tetrahedral ZnABCD
are drawn above.
D
2+
B
A
B
C
A
B
Zn
Zn
Pt
Pt
Pt
C
B
A
D
C
D
does display optical isomerism. The two optical isomers
(M) There are a total of four coordination isomers. They are listed below. We assume that
the oxidation state of each central metal ion is +3 .
Co en 3 Cr ox 3
tris(ethylenediamine)cobalt(III) trioxalatochromate(III)
Co ox en 2 Cr ox 2 en bis(ethylenediamine)oxalatocobalt(III)
(ethylenediamine)dioxalatochromate(III)
Cr ox en 2 Co ox 2 en bis(ethylenediamine)oxalatochromium(III)
(ethylenediamine)dioxalatocobaltate(III)
Cr en 3 Co ox 3
tris(ethylenediamine)chromium(III) trioxalatocobaltate(III)
1171
Chapter 24: Complex Ions and Coordination Compounds
15.
(M) The cis-dichlorobis(ethylenediamine)cobalt(III) ion is optically active. The two optical
isomers are drawn below. The trans-isomer is not optically active: the ion and its mirror
image are superimposable.
Cl
Cl
C
Cl
N Co N N
C
N
C C
C
C
N C
cis-isomers
16.
C N
Cl
Co N
C
N
C
N
Cl
Co
N
N
Cl
C
N C
trans-isomer
(M) Complex ions (a) and (b) are identical; complex ions (a) and (d) are geometric
isomers; complex ions (b) and (d) are geometric isomers; complex ion (c) is distinctly
different from the other three complex ions (it has a different chemical formula).
Crystal Field Theory
17.
(E) In crystal field theory, the five d orbitals of a central transition metal ion are split into
two (or more) groups of different energies. The energy spacing between these groups often
corresponds to the energy of a photon of visible light. Thus, the transition-metal complex
will absorb light with energy corresponding to this spacing. If white light is incident on the
complex ion, the light remaining after absorption will be missing some of its components.
Thus, light of certain wavelengths (corresponding to the energies absorbed) will no longer
be present in the formerly white light. The resulting light is colored. For example, if blue
light is absorbed from white light, the remaining light will be yellow in color.
18.
(E) The difference in color is due to the difference in the value of , which is the ligand field
splitting energy. When the value of is large, short wavelength light, which has a blue color,
is absorbed, and the substance or its solution appears yellow. On the other hand, when the
value of is small, light of long wavelength, which has a red or yellow color, is absorbed,
and the substance or its solution appears blue. The cyano ligand is a strong field ligand,
producing a large value of , and thus yellow complexes. On the other hand, the aqua ligands
are weak field ligands, which produce a small value of , and hence blue complexes.
19. (M) We begin with the 7 electron d-orbital diagram for Co2+ [Ar]
The strong field and weak field diagrams for octahedral complexes follow, along with the
number of unpaired electrons in each case.
Strong Field
Weak Field
eg
eg
(1 unpaired
electron)
(3 unpaired
electrons)
t2g
1172
t2g
Chapter 24: Complex Ions and Coordination Compounds
20.
(M) We begin with the 8 electron d-orbital diagram for Ni 2+ Ar
The strong field and weak field diagrams for octahedral complexes follow, along with the
number of unpaired electrons in each case.
Strong Field
Weak Field
eg
eg
t2g
(2 unpaired
electrons)
(2 unpaired
electrons)
t2g
The number of unpaired electrons is the same in both cases.
21.
(M) (a) Both of the central atoms have the same oxidation state, +3. We give the electron
configuration of the central atom to the left, then the completed crystal field diagram
in the center, and finally the number of unpaired electrons. The chloro ligand is a
weak field ligand in the spectrochemical series.
Mo 3+
Kr 4d 3
weak field
Kr
t2g
3 unpaired electrons; paramagnetic
4d
eg
The ethylenediamine ligand is a strong field ligand in the spectrochemical series.
Co 3+
Ar 3d 6
eg
strong field
Ar
t2g
no unpaired electrons; diamagnetic
(b)
2
In CoCl 4 the oxidation state of cobalt is 2+. Chloro is a weak field ligand. The
electron configuration of Co 2+ is Ar 3d 7 or Ar
The tetrahedral ligand field diagram is
shown on the right.
weak field
t2g
eg
3 unpaired electrons
22.
(M) (a) In Cu py 4 the oxidation state of copper is +2. Pyridine is a strong field
ligand. The electron configuration of Cu 2+ is Ar 3d 9 or Ar
2+
There is no possible way that an odd number of electrons can be paired up, without
at least one electron being unpaired. Cu py 4
1173
2+
is paramagnetic.
Chapter 24: Complex Ions and Coordination Compounds
(b)
3
In Mn CN 6 the oxidation state of manganese is +3. Cyano is a strong field
ligand. The electron configuration of Mn 3+ is Ar 3d 4 or Ar
The ligand field diagram follows, on the left-hand side. In FeCl4 the oxidation
state of iron is +3. Chloro is a weak field ligand. The electron configuration of Fe 3+
is Ar 3d 5 or [Ar]
The ligand field diagram follows, below.
t2g
eg
strong field
weak field
eg
t2g
2 unpaired electrons
5 unpaired electrons
23. (M) The electron configuration of Ni 2+
3
than in Mn CN 6 .
is Ar 3d 8 or Ar
There are more unpaired electrons in FeCl 4
Ammonia is a strong field ligand. The ligand field diagrams follow, octahedral at left,
tetrahedral in the center and square planar at right.
Octahedral
Tetrahedral
eg
Square Planar
t2g
t2g
dx2-y2
dxy
dz2
dxz, dyz
eg
Since the octahedral and tetrahedral configurations have the same number of unpaired
electrons (that is, 2 unpaired electrons), we cannot use magnetic behavior to determine
whether the ammine complex of nickel(II) is octahedral or tetrahedral. But we can
determine if the complex is square planar, since the square planar complex is diamagnetic
with zero unpaired electrons.
24.
(M) The difference is due to the fact that Fe CN 6
4
is a strong field complex ion, while
Fe H 2O 6 is a weak field complex ion. The electron configurations for an iron atom
and an iron(II) ion, and the ligand field diagrams for the two complex ions follow.
2+
Fe CN 6
iron atom
iron(II) ion
Ar
3d
[Ar] 3d
4
eg
4s
t2g
4s
1174
[Fe H 2 O 6 ] 2+
eg
t2g
Chapter 24: Complex Ions and Coordination Compounds
Complex-Ion Equilibria
Zn NH3
Zn OH 2 s + 4 NH3 aq
4
25. (E) (a)
(b)
2+
aq + 2 OH aq
Cu OH s
Cu 2+ aq + 2 OH aq
2
The blue color is most likely due to the presence of some unreacted [Cu(H2O)4]2+ (pale blue)
Cu NH3
Cu OH 2 s + 4 NH3 aq
4
Cu NH 3 4
26.
2+
2+
aq,
dark blue + 2 OH aq
Cu H 2 O aq + 4 NH 4 + aq
aq + 4 H3O+ aq
4
2+
CuCl4 2 aq, yellow
(M) (a) CuCl2 s + 2 Cl aq
2 CuCl4
2
aq + 4H 2O(l)
CuCl4 aq, yellow Cu H 2O 4 aq, pale blue + 4 Cl aq
2
2+
or
2 Cu H 2 O 2 Cl2 aq, green + 4 Cl aq
(b)
Either
Cu H 2 O aq + 4 Cl aq
CuCl4 aq + 4 H 2O(l)
4
Or
Cu H 2O
Cu H 2 O 2 Cl2 aq + 2H 2 O(l)
4
2
2+
2+
aq + 2 Cl aq
The blue solution is that of [Cr(H2O)6]2+. This is quickly oxidized to Cr 3+ by O 2 g
from the atmosphere. The green color is due to CrCl2 H 2 O 4 .
3+
4 [Cr(H 2 O)6 ]2 (aq, blue)+ 4 H + (aq) +8 Cl (aq) + O 2 (g)
4 [CrCl2 (H 2O) 4 ] (aq, green)+10 H 2 O(l)
Over a period of time, we might expect volatile HCl(g) to escape, leading to the
formation of complex ions with more H 2 O and less Cl .
HCl g
H + aq + Cl aq
CrCl 2 H 2O 4 aq, green + H 2 O(l) CrCl H 2O 5
+
1175
2+
aq, blue-green + Cl aq
Chapter 24: Complex Ions and Coordination Compounds
CrCl H 2O 5
2+
Cr H 2 O aq, blue + Cl aq
aq, blue-green + H 2O(l)
6
3+
Actually, to ensure that these final two reactions actually do occur in a timely
fashion, it would be helpful to dilute the solution with water after the chromium
metal has dissolved.
27.
28.
(M) Co en 3 should have the largest overall Kf value. We expect a complex ion with
polydentate ligands to have a larger value for its formation constant than complexes that
contain only monodentate ligands. This phenomenon is known as the chelate effect. Once
one end of a polydentate ligand becomes attached to the central metal, the attachment of
the remaining electron pairs is relatively easy because they already are close to the central
metal (and do not have to migrate in from a distant point in the solution).
3+
(M) (a) Zn NH 3 4
2+
4 = K1 K2 K3 K4 = 3.9 102 2.1 102 1.0 102 50.= 4.1 108 .
(b)
Ni H 2O 2 NH 3 4
2+
4 = K1 K2 K3 K4 = 6.3 102 1.7 102 54 15 = 8.7 107
29.
(M) First: Fe H 2 O 6
3+
Fe H 2 O en aq + 2H 2O(l)
aq + en aq
4
3+
K1 = 104.34
Second: Fe H 2 O 4 en
Third:
3+
Fe H 2 O 2 en 2
aq +
3+
en aq Fe H 2O 2 en 2
aq +
en aq Fe en 3
3+
3+
aq + 2H 2O(l)
aq + 2H 2O(l)
K2 = 103.31
K3 = 102.05
Fe H 2 O 6 aq + 3 en aq Fe en 3 aq + 6H 2 O(l)
Net:
K f = K1 K2 K3
log Kf = 4.34 + 3.31+ 2.05 = 9.70
Kf = 109.70 = 5.0 109 = 3
3+
30.
3+
(E) Since the overall formation constant is the product of the individual stepwise formation
constants, the logarithm of the overall formation constant is the sum of the logarithms of
the stepwise formation constants.
log Kf = log K1 + log K2 + log K3 + log K4 = 2.80 +1.60 + 0.49 + 0.73 = 5.62
Kf = 105.62 = 4.2 105
1176
Chapter 24: Complex Ions and Coordination Compounds
31.
(M) (a) Aluminum(III) forms a stable (and soluble) hydroxo complex but not a stable
ammine complex.
Al H 2 O OH aq + H 2 O(l)
Al H 2 O 3 OH 3 s + OH aq
2
4
(b)
Although zinc(II) forms a soluble stable ammine complex ion, its formation constant
is not sufficiently large to dissolve highly insoluble ZnS. However, it is sufficiently
large to dissolve the moderately insoluble ZnCO 3 . Said another way, ZnS does not
produce sufficient Zn 2+ to permit the complex ion to form.
Zn NH 3
ZnCO3 s + 4 NH 3 aq
4
(c)
2+
aq + CO32 aq
Chloride ion forms a stable complex ion with silver(I) ion, that dissolves the AgCl(s)
that formed when Cl is low.
AgCl s Ag + aq + Cl aq
and
Ag + aq + 2 Cl aq AgCl2 aq
AgCl2 aq
Overall: AgCl s + Cl aq
32.
(M) (a) Because of the large value of the formation constant for the complex ion,
Co NH 3 6 aq , the concentration of free Co 3+ aq is too small to enable it to
oxidize water to O2 g . Since there is not a complex ion present, except, of course,
3+
Co H 2 O 6 aq , when CoCl 3 is dissolved in water, the Co 3+ is sufficiently
high for the oxidation-reduction reaction to be spontaneous.
3+
4 Co3+ aq + 2 H 2 O(l) 4 Co 2+ aq + 4 H + aq + O 2 g
(b)
Although AgI(s) is often described as insoluble, there is actually a small
concentration of Ag + aq present because of the solubility equilibrium:
2 AgI(s) Ag(aq) +I-(aq)
These silver ions react with thiosulfate ion to form the stable dithiosulfatoargentate(I)
complex ion:
Ag S2 O3 aq
Ag + aq + 2S2 O32 aq
2
1177
Chapter 24: Complex Ions and Coordination Compounds
Acid-Base Properties
33.
(E) Al H 2 O 6
Al H 2O 6
3+
3+
aq is capable of releasing H+:
AlOH H 2 O aq + H 3O + aq
aq + H 2O(l)
5
2+
The value of its ionization constant ( pKa = 5.01) approximates that of acetic acid.
34.
(M) (a) CrOH H 2 O 5
(b)
CrOH H 2 O 5
2+
2+
aq + OH aq Cr OH 2 H 2O 4 aq +
+
aq +
H 3O + aq Cr H 2 O 6
3+
aq +
H 2 O(l)
H 2 O(l)
Applications
35.
Ksp =5.01013
+
AgBr s
Ag aq + Br aq
(M) (a) Solubility:
K f =1.710
Ag + aq + 2S2 O32 aq
Ag S2 O3 2
13
Cplx. Ion Formation:
Ag S2 O3
Net: AgBr s + 2S2 O32 aq
2
3
aq
+ Br - (aq)
3
aq
K overall = K sp K f
K overall = 5.0 1013 1.7 1013 = 8.5
With a reasonably high S2 O 3
(b)
2
, this reaction will go essentially to completion.
NH 3 aq cannot be used in the fixing of photographic film because of the relatively
small value of Kf for Ag NH 3 2 aq , Kf = 1.6 107 . This would produce a
value of K = 8.0 106 in the expression above, which is nowhere large enough to
drive the reaction to completion.
+
36.
aq Co NH aq e
Reduction: H O aq 2e 2OH aq
(M) Oxidation: Co NH 3 6
2
Net:
2+
3+
3 6
2 E 0.10V.
E 0.88V
2
H 2 O 2 aq + 2 Co NH 3 6
2+
aq 2 Co NH3 6 aq + 2OH (aq)
3+
E cell = +0.88 V+ 0.10 V = +0.78 V.
The positive value of the standard cell potential indicates that this is a spontaneous reaction.
37.
(M) To make the cis isomer, we must use ligands that show a strong tendency for
directing incoming ligands to positions that are trans to themselves. I− has a stronger
tendency than does Cl− or NH3 for directing incoming ligands to the trans positions, and
so it is beneficial to convert K2[PtCl4] to K2[PtI4] before replacing ligands around Pt with
NH3 molecules.
1178
Chapter 24: Complex Ions and Coordination Compounds
(M) Transplatin is more reactive and is involved in more side reactions before it reaches
its target inside of cancer cells. Thus, although it is more reactive, it is less effective at
killing cancer cells.
cis-Pt(NH3)2Cl2
trans-Pt(NH3)2Cl2
38.
Cl
NH3
Pt
Cl
NH3
Pt
H3N
Cl
Cl
NH3
INTEGRATIVE AND ADVANCED EXERCISES
(M)
39.
(a)
(b)
(c)
(d)
(e)
cupric tetraammine ion
dichlorotetraamminecobaltic chloride
platinic(IV) hexachloride ion
disodium copper tetrachloride
dipotassium antimony(III) pentachloride
[Cu(NH3)4]2+
[CoCl2(NH3)4]Cl
[PtCl6]2–
Na2[CuCl4]
K2[SbCl5]
tetraamminecopper(II) ion
tetraamminedichlorocobalt(III) chloride
hexachloroplatinate(IV) ion
sodium tetrachlorocuprate(II)
potassium pentachloroantimonate(III)
40. (E)[Pt(NH3)4][PtCl4] tetraammineplatinum(II) tetrachloroplatinate(II)
41. (M) The four possible isomers for [CoCl2(en)(NH3)2]+ are sketched below:
NH3
Cl
Cl
NH2
Co
Cl
H3N
Cl
NH2
Cl
Co
H 2N
H 3N
NH3
Cl
NH2
H3N
Co
H 2N
H3N
H2N
Cl
NH2
Co
Cl
NH3
H 2N
NH3
42. (M)The color of the green solid, as well as that of the green solution, is produced by the complex
ion [CrCl2(H2O)4]+. Over time in solution, the chloro ligands are replaced by aqua ligands,
producing violet [Cr(H2O)6]3+(aq). When the water is evaporated, the original complex is reformed as the concentration of chloro ligand, [Cl–], gets higher and the chloro ligands replace the
aqua ligands.
3
H2 O
Cr
H2O
Cl
OH2
Cl
OH2
OH2
+2H20
H2O
Cr
H2 O
OH2
OH2
Cl
OH2
2Cl
+2H20
H2O
Cr
H2O
OH2
OH2
Cl
43. (M)The chloro ligand, being lower in the spectrochemical series than the ethylenediamine ligand,
is less strongly bonded to the central atom than is the ethylenediamine ligand. Therefore, of the
two types of ligands, we expect the chloro ligand to be replaced more readily. In the cis isomer,
the two chloro ligands are 90° from each other. This is the angular spacing that can be readily
spanned by the oxalato ligand, thus we expect reaction with the cis isomer to occur rapidly. On
the other hand, in the trans isomer, the two chloro ligands are located 180° from each other. After
1179
Chapter 24: Complex Ions and Coordination Compounds
the chloro ligands are removed, at least one end of one ethylenediamine ligand would have to be
relocated to allow the oxalato ligand to bond as a bidentate ligand. Consequently, replacement of
the two chloro ligands by the oxalato ligand should be much slower for the trans isomer than for
the cis isomer.
44. (M) Cl2 + 2 e- 2 Cl[Pt(NH3)4]2+ [Pt(NH3)4]4+ + 2 eCl2 + [Pt(NH3)4]2+ [Pt(Cl2NH3)4]2+
Cl
H 3N
Pt
NH3
H3N
H 3N
+ Cl-Cl
NH3
Pt
NH3
H3N
NH3
Cl
45. (M) The successive acid ionizations of a complex ion such as [Fe(OH)6]3+ are more nearly
equal in magnitude than those of an uncharged polyprotic acid such as H3PO4 principally
because the complex ion has a positive charge. The second proton is leaving a species which
has one fewer positive charge but which is nonetheless positively charged. Since positive
charges repel each other, successive ionizations should not show a great decrease in the
magnitude of their ionization constants. In the case of polyprotic acids, on the other hand,
successive protons are leaving a species whose negative charge is increasingly greater with
each step. Since unlike charges attract each other, it becomes increasingly difficult to
remove successive protons.
46. (M)
-1
NH3
Pt
Cl
+1
NH3
Cl
K+
Cl
Pt
Pt
Cl
NH3 H3N
(c)
+2
NH3
Pt
H3N
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
-2
Cl
Cl
2Cl-
NH3
Cl-
NH3
(b)
(a)
H3N
NH3
Cl
C
o
n
d
u
ct
iv
it
y
Cl
Cl
Number of
Cl ligands
2K+
Pt
Cl
(e)
(d)
47. (M) (a)
Fe(H 2 O)63+ (aq)
Initial 0.100 M
Equil. (0.100 x) M
K 9.0 10
H 2 O(l)
-4
Fe(H 2 O)5 OH 2 (aq)
0M
x
{where x is the molar quantity of Fe(H 2O)63+ (aq) hydrolyzed}
[[Fe(H 2 O)5 OH]2+ ] [H 3 O + ]
x2
9.0 10-4
K
=
3+
0.100-x
[[Fe(H 2 O)6 ] ]
Solving, we find x 9.5 103 M, which is the [H 3 O + ] so:
pH log(9.5 103 ) 2.02
1180
H3 O
0M
x
Chapter 24: Complex Ions and Coordination Compounds
(b)
[Fe(H 2 O)6 ]3+ (aq) H 2 O(l)
initial 0.100 M
equil. (0.100 x) M
K 9.0 10
-4
[Fe(H 2 O)5OH]2 (aq)
H 3O
0M
xM
0.100 M
(0.100 x) M
{where x is [[Fe(H 2 O)6 ]3+ ] reacting}
K
[[Fe(H 2 O)5OH]2 ] [H 3O + ] x (0.100 x)
9.0 10-4
3+
[[Fe(H 2 O)6 ] ]
(0.100-x )
Solving, we find x 9.0 10-4 M, which is the [[Fe(H 2 O)5 OH]2 ]
(c)
We simply substitute [[Fe(H 2 O)5 OH]2 ] = 1.0 10-6 M into the K a expression
with [Fe(H 2 O)6 ]3+ 0.100 M and determine the concentration of H 3 O +
[H 3 O + ]
K a [[Fe(H 2 O)6 ]3+ ] 9.0 10-4 (0.100 M)
90. M
[[Fe(H 2 O)5 OH]2 ]
[1.0 10-6 M]
To maintain the concentration at this level requires an impossibly high concentration of H 3 O +
48. (D) Let us first determine the concentration of the uncomplexed (free) Pb2+(aq). Because of
the large value of the formation constant, we assume that most of the lead(II) ion is present as
the EDTA complex ion ([Pb(EDTA)]2-). Once equilibrium is established, we can see if there
is sufficient lead(II) ion present in solution to precipitate with the sulfide ion. Note: the
concentration of EDTA remains constant at 0.20 M)
210
18
Reaction:
Pb2+(aq) + EDTA4-(aq)
Initial:
0M
0.20 M
0.010 M
Change:
+x M
constant
-x M
Equilibrium:
x M 0.20 M
Kf =
[Pb(EDTA)]2-(aq)
(0.010-x) M 0.010 M
[Pb(EDTA)2- ]
0.010
= 2 1018 =
2+
40.20(x)
[Pb ][EDTA ]
x = 2.5 10-20 M = [Pb2+ ]
Now determine the [S2- ] required to precipitate PbS from this solution. K sp = [Pb2 ][S 2 ] 8 1028
[S2- ]=
8 10-28
= 2.5 10-8 M Recall (i) that a saturated H 2S solution is ~0.10 M H 2S
-20
(5 10 )
and (ii) That the [S2- ] = K a2 = 1 10-14
Qsp = [Pb2 ][S 2 ] 2.5 10-20 1 10-14 2.5 10-34
Qsp << K sp
K sp 8 1028
Hence, PbS(s) will not precipitate from this solution.
49. (M) The formation of [Cu(NH3)4]2+(aq) from [Cu(H2O)4]2+(aq) has K1 = 1.9 × 104, K2 = 3.9 ×
103, K3 = 1.0 × 103, and K4 = 1.5 × 102. Since these equilibrium constants are all considerably
larger than 1.00, one expects that the reactions they represent, yielding ultimately the ion
[Cu(NH3)4]2+(aq), will go essentially to completion. However, if the concentration of NH3
were limited to less than the stoichiometric amount, that is, to less than 4 mol NH3 per mol of
1181
Chapter 24: Complex Ions and Coordination Compounds
Cu2+(aq), one would expect that the ammine-aqua complex ions would be present in
significantly higher concentrations than the [Cu(NH3)4]2+(aq) ions.
50. (D) For [Ca(EDTA)]2–, Kf = 4 × 1010 and for [Mg(EDTA)]2–, Kf = 4 × 108. In Table 18-1, the
least soluble calcium compound is CaCO3, Ksp = 2.8 × 10–9, and the least soluble magnesium
compound is Mg3(PO4)2, Ksp = 1 × 10–25. We can determine the equilibrium constants for
adding carbonate ion to [Ca(EDTA)]2–(aq) and for adding phosphate ion to [Mg(EDTA)]2–(aq)
as follows:
Ca 2 (aq) EDTA 4 (aq)
[Ca(EDTA)]2 (aq)
CaCO3 (s)
Precipitation: Ca 2 (aq) CO32 (aq)
Instability:
K ppt 1/2.8 10-9
CaCO3 (s) + EDTA 4 (aq) K K i K ppt
[Ca(EDTA)]2 + CO32 (aq)
Net:
1
(4 10 )( 2.8 10-9 )
K
K i 1/4 1010
10
9 10-3
The small value of the equilibrium constant indicates that this reaction does not proceed
very far toward products.
2
4
{[Mg(EDTA)]2 (aq)
Mg (aq) EDTA (aq)} 3
Precipitation: 3 Mg2 (aq) 2 PO43 (aq)
Mg3 (PO4 )2 (s)
Instability:
Net :
K
Ki3 1/(4 108 )3
Kppt 1/1 1025
4
3 [Mg(EDTA)] 2-+2 PO43- (aq)
Mg3 (PO4 )2 (s) +3 EDTA (aq)
K Ki3 Kppt
1
0.16
(4 10 ) (1 1025 )
8 3
This is not such a small value, but again we do not expect the formation of much product,
particularly if the [EDTA4– ] is kept high. We can approximate what the concentration of
precipitating anion must be in each case, assuming that the concentration of complex ion is
0.10 M and that of [EDTA4– ] also is 0.10 M.
Reaction:
K
CaCO3 (s) EDTA 4 (aq)
[Ca(EDTA)]2 CO32 (aq)
[EDTA 4 ]
0.10 M
9 10-3
2
2
[[Ca(EDTA)] ][CO3 ]
0.10 M [CO32 ]
[CO32 ] 1 102 M
This is an impossibly high [CO32–].
Reaction:
Mg 3 (PO 4 ) 2 (s) 3 EDTA 4 (aq)
3[Mg(EDTA)]2 2 PO 43 (aq)
[EDTA 4 ]3
(0.10 M)3
K
0.16
[[Mg(EDTA)]2 ]3[PO 43 ]2
(0.10 M)3 [PO 43 ]2
[PO 43 ] 2.5 M
Although this [PO43–] is not impossibly high, it is unlikely that it will occur without
the deliberate addition of phosphate ion to the water. Alternatively, we can substitute
[[M(EDTA)]2– ] = 0.10 M and [EDTA4– ] = 0.10 M into the formation constant
[[M(EDTA)]2 ]
expression: K f
[M 2 ][EDTA 4 ]
1182
Chapter 24: Complex Ions and Coordination Compounds
This substitution gives [M 2 ]
1
, hence, [Ca2+] = 2.5 × 10–11 M and [Mg2+] = 2.5 × 10–9 M,
Kf
which are concentrations that do not normally lead to the formation of precipitates unless the
concentrations of anions are substantial. Specifically, the required anion concentrations are [CO32–] =
1.1 × 102 M for CaCO3, and [PO43–] = 2.5 M, just as computed above.
51. (M) If a 99% conversion to the chloro complex is achieved, which is the percent conversion
necessary to produce a yellow color, [[CuCl4]2–] = 0.99 × 0.10 M = 0.099 M, and
[[Cu(H2O)4]2+] = 0.01 × 0.10 M = 0.0010 M. We substitute these values into the formation
constant expression and solve for [Cl–].
[[CuCl 4 ] 2 ]
0.099 M
4.2 10 5
Kf
4
2
[[Cu(H 2 O) 4 ] ] [Cl ]
0.0010 M [Cl ] 4
4
[Cl ]
0.099
0.0010 4.2 10 5
0.12 M
This is the final concentration of free chloride ion in the solution. If we wish to account for all
chloride ion that has been added, we must include the chloride ion present in the complex ion.
total [Cl–] = 0.12 M free Cl– + (4 × 0.099 M) bound Cl– = 0.52 M
52. (M)(a)
Oxidation:
2H 2 O(l)
O 2 (g) 4 H (aq) 4 e3
2
Reduction: {Co ((aq) e
Co (aq)}
Net:
-
E -1.229 V
E 1.82 V
4 Co3 (aq) 2 H2 O(l)
4 Co2 (aq) 4 H (aq) O2 (g) Ecell 0.59 V
[Co(NH 3 )6 ]3 (aq)
(b) Reaction: Co3 (aq) 6 NH 3 (aq)
Initial: 1.0 M
Changes: xM
Equil: (1.0 x)M
0.10 M
constant
0.10 M
0M
xM
xM
[[Co(NH 3 )6 ]3 ]
x
Kf
4.5 1033
3
6
(1.0 x)(0.10)6
[Co ][NH 3 ]
Thus [[Co(NH3)6]3+] = 1.0 M because K is so large, and
1
[Co 3 ]
2.2 10 -28 M
4.5 10 27
1183
x
4.5 1027
1.0 x
Chapter 24: Complex Ions and Coordination Compounds
(c) The equilibrium and equilibrium constant for the reaction of NH3 with water follows.
NH 4 (aq) OH (aq)
NH 3 (aq) H 2 O(l)
[NH 4 ][OH - ] [OH ]2
K b 1.8 10
[NH 3 ]
0.10
5
[OH ]
0.10(1.8 105 ) 0.0013 M
In determining the [OH– ], we have noted that [NH4+] = [OH– ] by stoichiometry, and also that
Kw
1.0 10 14
7.7 10 12 M
[NH3] = 0.10 M, as we assumed above. [H 3 O ]
0.0013
[OH ]
We use the Nernst equation to determine the potential of reaction (24.12) at the conditions
described.
E E cell
[Co 2 ]4 [H ]4 P[O 2 ]
0.0592
log
4
[Co3 ]4
0.0592
(1 104 ) 4 (7.7 1012 ) 4 0.2
E 0.59
log
0.59 V 0.732 V 0.142 V
4
(2.2 1028 ) 4
The negative cell potential indicates that the reaction indeed does not occur.
53. (M) We use the Nernst equation to determine the value of [Cu2+]. The cell reaction follows.
Cu(s) 2 H (aq)
Cu 2 (aq) H 2 (g)
E E
E cell 0.337 V
2
[Cu 2 ]
[Cu ]
0.0592
0.0592
log
0.08
V
0
.
337
log
2
2
[H ] 2
(1.00) 2
log [Cu 2 ]
2 (0.337 0.08)
14.1
0.0592
[Cu 2 ] 8 10 15 M
Now we determine the value of Kf.
Kf
[[Cu(NH 3 ) 4 ] 2 ]
2
[Cu ] [NH 3 ]
4
1.00
1 1014
4
15
8 10 (1.00)
This compares favorably with the value of Kf = 1.1 × 1013 given in Table 18-2, especially
considering the imprecision with which the data are known. (Ecell = 0.08 V is known to but
one significant figure.)
1184
Chapter 24: Complex Ions and Coordination Compounds
54. (M) We first determine [Ag+] in the cyanide solution.
[[Ag(CN) 2 ] ]
0.10
5.6 1018
2
[Ag ][CN ]
[Ag ] (0.10.) 2
The cell reaction is as follows. It has
0.10
[Ag ]
1.8 10 18
5.6 1018 (0.10) 2
E cell 0.000 V ; the same reaction occurs at both anode and cathode and thus the Nernstian
voltage is influenced only by the Ag+ concentration.
Kf
Ag (0.10 M)
Ag (0.10 M [Ag(CN) 2 ] , 0.10 M KCN)
E E
[Ag ]CN
0.0592
1.8 10 18
log
0.000
0.0592
log
0.99 V
1
0.10
[Ag ]
55. (M)
+
+2
OH2
H3N
Co
H3N
OH2
NH3
Cl-
NH3
Cl
H3N
Co
H3N
NH3
2Cl-
NH3
OH2
A 0.10 mol/L solution of these compounds would result in ion concentration of
0.20 mol/L for [Co Cl (H2O(NH3)4)]Cl•H2O or 0.30 mol/L for [Co(H2O)2 (NH3)4]Cl2.
Observed freezing point depression = -0.56 C.
T = -Kf×m×i = -0.56 deg = -1.86 mol kg-1 deg(0.10 mol/L)×i
Hence, i = 3.0 mol/kg or 3.0 mol/L. This suggests that the compound is [Co(H2O)2 (NH3)4]Cl2.
56. (M) Sc(H2O)63+ and Zn(H2O)42+ have outer energy shells with 18 electrons, and thus they do not
have any electronic transitions in the energy range corresponding to visible light. Fe(H2O)63+ has 17
electrons (not 18) in its outer shell; thus, it does have electronic transitions in the energy range
corresponding to visible light.
57. (M) The Cr3+ ion would have 3 unpaired electrons, each residing in a 3d orbital and would be sp3d2
hybridized. The hybrid orbitals would be hybrids of 4s, 4p, and 3d (or 4d) orbitals. Each Cr–NH3
coordinate covalent bond is a σ bond formed when a lone pair in an sp3 orbital on N is directed
toward an empty sp3d2 orbital on Cr3+. The number of unpaired electrons predicted by valence bond
theory would be the same as the number of unpaired electrons predicted by crystal field theory.
58. (M) Since these reactions involve Ni2+ binding to six identical nitrogen donor atoms, we can
assume the H for each reaction is approximately the same. A large formation constant indicates
a more negative free energy. This indicates that the entropy term for these reactions is different.
The large increase in entropy for each step, can be explained by the chelate effect. The same
number of water molecules are displaced by fewer ligands, which is directly related to entropy
change.
1185
Chapter 24: Complex Ions and Coordination Compounds
59. (M)
This
compound
has
a
nonsuperimposable
mirror
image
(therefore it is optically active). These
are enantiomers.
One enantiomer
rotates plane polarized light clockwise,
while the other rotates plane polarized
light counter-clockwise. Polarimetry
can be used to determine which isomer
rotates plane polarized light in a
particular direction.
Co(acac)3
CH3
H3C
H3C
CH3
O
O
H3C
CH3
O
O
O
Co
O
Co
O
O
O
O
O
O
H3C
CH3
CH3
CH3
H3C
H3C
This
compound
has
a
superimposable mirror image
(therefore it is optically inactive).
These are the same compound.
This compound will not rotate
plane polarized light (net rotation
= 0).
trans-[Co(acac)2(H2O)2]Cl2
H3C
O
H3C
CH3
OH2
O
O
O
O
O
Co
Co
O
CH3
OH2
O
OH2
OH2
H3C
CH3
H3C
CH3
This compound has a nonsuperimposable mirror image
(therefore it is optically active).
These are enantiomers.
One
enantiomer
rotates
plane
polarized light clockwise, while
the other rotates plane polarized
light counter-clockwise.
By
using a polarimeter, we can
determine which isomer rotates
plane polarized light in a
particular direction.
cis-[Co(acac)2(H2O)2]Cl2
CH3
H3C
H3C
O
O
Co
O
CH3
O
OH2
H2O
OH
HO
O
O
Co
O
O
H3 C
CH3
OH2
H2O
3+
2+
60. (D)[Co(H2O)6] (aq) + e [Co(H2O)6] (aq)
log K = nE/0.0592 = (1)(1.82)/0.0592 = 30.74
[Co(en)3]3+(aq) + 6 H2O(l)
[Co(H2O)6]2+(aq) + 3 en
[Co(H2O)6]3+(aq) + e[Co(en)3]3+(aq) + e
K = 1030.74
[Co(H2O)6]3+(aq) + 3 en
[Co(en)3]2+(aq) + 6 H2O(l)
[Co(H2O)6]2+(aq)
[Co(en)3]2+(aq)
1186
K = 1/1047.30
K = 1012.18
K = 1030.74
Koverall = ?
Chapter 24: Complex Ions and Coordination Compounds
Koverall = 1/1047.30×1012.18×1030.74 = 10-4.38 = 0.0000417
E = (0.0592/n)log K = (0.0592/1)log(0.0000417) = -0.26 V
4 Co3+(aq) + 4 e- 4 Co2+(aq)
Ecathode = 1.82 V
+
2 H2O(l) 4 H (aq) + O2(g) + 4 e Eanode = 1.23 V
4 Co3+(aq) +2 H2O(l) 4 H+(aq) + O2(g) + 4 Co2+(aq)
Ecell = Ecathode– Eanode = 1.82V – 1.23 V = +0.59 V (Spontaneous)
4 [Co(en)3]3+(aq) + 4 e- 4 [Co(en)3]2+(aq)
Ecathode = -0.26 V
Eanode = 1.23 V
2 H2O(l) 4 H+(aq) + O2(g) + 4 e3+
+
4 [Co(en)3] (aq) +2 H2O(l) 4 H (aq) + O2(g) + 4 [Co(en)3]2+(aq)
Ecell = Ecathode– E = -0.26 V – 1.23 V = -1.49 V (non-spontaneous)
61. (M) There are two sets of isomers for Co(gly)3. Each set comprises two enantiomers (nonsuperimposable mirror images). One rotates plane polarized light clockwise (+), the other counterclockwise (). Experiments are needed to determine which enantiomer is (+) and which is ().
O
H2 N
O
O
O
NH2
O
O
O
Co
O
O
Co
NH2
H2N
H2N
Co
H2N
O
O
O
O
OO
NH2
O
O
NH2
H2N
NH2
H2N
O
NH2
O
O
O
Co
O
O
O
(+)-fac-tris(glycinato)cobalt(III)
(+)-mer-tris(glycinato)cobalt(III)
()-fac-tris(glycinato)cobalt(III)
()-mer-tris(glycinato)cobalt(III)
[Co(gly)2Cl(NH3)] has 10 possible isomer, four pairs of enantiomers and two achiral isomers.
O
O
NH3
O
O
O
Co
H2N
O
O
O
Cl
NH2
Co
Cl
H3N
NH2
H2N
NH2
H2N
Cl
Co
NH3
O
NH3
O
O
H2
N
Co
H2N
O
O
O
O
Cl
1187
O
Chapter 24: Complex Ions and Coordination Compounds
H2N
O
O
Cl
O
Co
O
O
NH2
H3N
H3N
O
O
H2N
NH3
O
O
O
NH3
Co
NH2
O
NH2
Co
Cl
Cl
Co
O
O
H2 N
O
O
NH2
O
NH2
O
Co
NH3
H2N
O
Cl
Cl
Co
O
H3N
H2N
NH2
O
Cl
H2 N
O
O
O
O
62. (M) The coordination compound is face-centered cubic, K+ occupies tetrahedral holes, while
PtCl62- occupies octahedral holes.
63.
(M)
# NH3
0
1
2
3
Formula:
K2[PtCl6]
K[PtCl5(NH3)]
PtCl4(NH3)2
[PtCl3(NH3)3]Cl
Total # ions
3
2
0
2
(per formula
unit)
# NH3
4
5
6
Formula:
[PtCl2(NH3)4]Cl2
[PtCl(NH3)5]Cl3
[Pt(NH3)6]Cl4
Total # ions
3
4
5
(per formula
unit)
1188
Chapter 24: Complex Ions and Coordination Compounds
FEATURE PROBLEMS
A trigonal prismatic structure predicts three geometric isomers for [CoCl2(NH3)4]+,
64. (M)(a)
which is one more than the actual number of geometric isomers found for this complex ion.
All three geometric isomers arising from a trigonal prism are shown below.
Cl
Cl
Cl
Cl
Cl
Cl
(i)
(ii)
(iii)
The fact that the trigonal prismatic structure does not afford the correct number of isomers is a
clear indication that the ion actually adopts some other structural form (i.e., the theoretical model
is contradicted by the experimental result). We know now of course, that this ion has an
octahedral structure and as a result, it can exist only in cis and trans configurations.
(b) All attempts to produce optical isomers of [Co(en)3]3+ based upon a trigonal prismatic structure are
shown below. The ethylenediamine ligand appears as an arc in diagrams below:
(i)
(ii)
(iii)
Only structure (iii), which has an ethylenediamine ligand connecting the diagonal corners of a
face can give rise to optical isomers. Structure (iii) is highly unlikely, however, because the
ethylenediamine ligand is simply too short to effectively span the diagonal distance across the
face of the prism. Thus, barring any unusual stretching of the ethylenediamine ligand, a trigonal
prismatic structure cannot account for the optical isomerism that was observed for [Co(en)3]3+.
65. (D) Assuming that each hydroxide ligand bears its normal 1– charge, and that each ammonia
ligand is neutral, the total contribution of negative charge from the ligands is 6 –. Since the
net charge on the complex ions is 6+, the average oxidation state for each Co atom must be
+3 (i.e., each Co in the complex can be viewed as a Co3+ 3d 6 ion surrounded by six
ligands.) The five 3d orbitals on each Co are split by the octahedrally arranged ligands into
three lower energy orbitals, called t2g orbitals, and two higher energy orbitals, called eg
orbitals. We are told in the question that the complex is low spin. This is simply another
way of saying that all six 3d electrons on each Co are paired up in the t2g set as a result of the
eg and t2g orbitals being separated by a relatively large energy gap (see below). Hence, there
should be no unpaired electrons in the hexacation (i.e., the cation is expected to be
diamagnetic).
1189
Chapter 24: Complex Ions and Coordination Compounds
eg
Low Spin Co3+
(Octahedral Environment):
Relatively Large Crystal Field Splitting
t2g
The Lewis structures for the two optical isomers (enantiomers) are depicted below:
NH3
NH3
Co
H3N
H3N
H
HO
Co
O
O
Co
H3N
H3N
H3N
H
H
O H
O H
H3N
NH3
H3N
Co
NH3
OH H
H3N
NH3
H O Co O
Co
NH
O
3
H O
H NH3
H3N
O
H
Co
H3N
NH3
NH3
NH3
NH3
O
Co
NH3
H3N
NH3
Non-superimposable mirror images of one another
(D) The data used to construct a plot of hydration energy as a function of metal ion atomic
number is collected in the table below. The graph of hydration energy (kJ/mol) versus
metal ion atomic number is located beneath the table.
Metal ion Atomic number
Ca2+
20
Sc2+
21
Ti2+
22
2+
V
23
2+
Cr
24
2+
Mn
25
Fe2+
26
Co2+
27
2+
Ni
28
2+
Cu
29
2+
Zn
30
Hydration energy
2468 kJ/mol
2673 kJ/mol
2750 kJ/mol
2814 kJ/mol
2799 kJ/mol
2743 kJ/mol
2843 kJ/mol
2904 kJ/mol
2986 kJ/mol
2989 kJ/mol
2936 kJ/mol
2+
Hydration Enthalpy of M ions (Z = 20-30)
(a)
20
Hydration Energy (kJ/mol)
66.
22
24
26
28
-2600
-2800
-3000
Atomic Number (Z)
1190
30
Chapter 24: Complex Ions and Coordination Compounds
(b)
When a metal ion is placed in an octahedral field of ligands, the five dorbitals are split into eg and t2g subsets, as shown in the diagram below:
dz 2 dx2 - y 2
eg
3
/5o
d-orbitals
(symmerical field)
2
/5o
dxy dyz dxz
t2g
(octahedral field)
Since water is a weak field ligand, the magnitude of the splitting is relatively small. As a
consequence, high-spin configurations result for all of the hexaaqua complexes. The
electron configurations for the metal ions in the high-spin hexaaqua complexes and their
associated crystal field stabilization energies (CFSE) are provided in the table below
Metal Ion Configuration t2g eg number of unpaired e CFSE(o)
Ca2+
3d 0
0 0
0
0
2+
1
Sc
3d
1 0
1
2/ 5
Ti2+
3d 2
2 0
2
4/ 5
V2+
3d 3
3 0
3
6/ 5
Cr2+
3d 4
3 1
4
3/ 5
2+
5
Mn
3d
3 2
5
0
Fe2+
3d 6
4 2
4
2/ 5
Co2+
3d 7
5 2
3
4/ 5
Ni2+
3d 8
6 2
2
6/ 5
2+
9
Cu
3d
6 3
1
3/ 5
2+
10
Zn
3d
6 4
0
0
2+
2+
Thus, the crystal field stabilization energy is zero for Ca . Mn and Zn2+.
(c) The lines drawn between those ions that have a CFSE = 0 show the trend for the enthalpy of
hydration after the contribution from the crystal field stabilization energy has been subtracted
from the experimental values. The Ca to Mn and Mn to Zn lines are quite similar. Both line
have slopes that are negative and are of comparable magnitude. This trend shows that as one
proceeds from left to right across the periodic table, the energy of hydration for dications
becomes increasingly more negative. The hexaaqua complexes become progressively more
stable because the Zeff experienced by the bonding electrons in the valence shell of the metal
ion steadily increases as we move further and further to the right. Put another way, the Zeff
climbs steadily as we move from left to right and this leads to the positive charge density on
the metal becoming larger and larger, which results in the water ligands steadily being pulled
closer and closer to the nucleus. Of course, the closer the approach of the water ligands to the
metal, the greater is the energy released upon successful coordination of the ligand.
1191
Chapter 24: Complex Ions and Coordination Compounds
(d) Those ions that exhibit crystal field stabilization energies greater than zero have heats of
hydration that are more negative (i.e. more energy released) than the hypothetical heat of
hydration for the ion with CFSE subtracted out. The heat of hydration without CFSE for
a given ion falls on the line drawn between the two flanking ions with CFSE = 0 at a
position directly above the point for the experimental hydration energy. The energy
difference between the observed heat of hydration for the ion and the heat of hydration
without CFSE is, of course, approximately equal to the CFSE for the ion.
(e) As was mentioned in the answer to part (c), the straight line drawn between manganese
and zinc (both ions with CFSE = 0) on the previous plot, describes the enthalpy trend
after the ligand field stabilization energy has been subtracted from the experimental
values for the hydration enthalpy. Thus, o for Fe2+ in [Fe(H2O)6]2+ is approximately
equal to 5/2 of the energy difference (in kJ/mol) between the observed hydration energy
for Fe2+(g) and the point for Fe2+ on the line connecting Mn2+ and Zn2+, which is the
expected enthalpy of hydration after the CFSE has been subtracted out. Remember that
the crystal field stabilization energy for Fe2+ that is obtained from the graph is not o, but
rather 2/5o, since the CFSE for a 3d6 ion in an octahedral field is just 2/5 of o.
Consequently, to obtain o , we must multiply the enthalpy difference by 5/2. According
to the graph, the high-spin CFSE for Fe2+ is 2843 kJ/mol (2782 kJ/mol) or 61
kJ/mol. Consequently, o = 5/2(61 kJ/mol) = 153 kJ/mol, or 1.5 102 kJ/mol is the
energy difference between the eg and t2g orbital sets.
(f) The color of an octahedral complex is the result of the promotion of an electron on the
metal from a t2g orbital eg orbital. The energy difference between the eg and t2g orbital
sets is o. As the metal-ligand bonding becomes stronger, the separation between the t2g
and eg orbitals becomes larger. If the eg set is not full, then the metal complex will exhibit
an absorption band corresponding to a t2g eg transition. Thus, the [Fe(H2O)6]2+complex
ion should absorb electromagnetic radiation that has Ephoton = o. Since o 150 kJ/mol
(calculated in part (e) of this question) for a mole of [Fe(H2O)6]2+(aq),
1mol [Fe(H 2 O)6 ]2+
1.5 102 kJ
1000 J
2+
23
2+
1mol[Fe(H 2 O)6 ]
6.022 10 [Fe(H 2 O)6 ]
1kJ
19
= 2.5 10 J per ion
E
2.5 1019 J
= =
= 3.8 1014 s1
h 6.626 1034 J s
Ephoton =
=
c 2.998 108 m s -1
=
= 7.8 107 m (780 nm)
h
3.8 1014 s -1
So, the [Fe(H2O)6]2+ ion will absorb radiation with a wavelength of 780 nm, which is
red light in the visible part of the electromagnetic spectrum.
1192
Chapter 24: Complex Ions and Coordination Compounds
SELF-ASSESSMENT EXERCISES
67.
(E) (a) Coordination number is the number of ligands coordinated to a transition metal
complex.
(b) 0 is the crystal field splitting parameter which depends on the coordination geometry
of a transition metal complex.
(c) Ammine complex is the complex that contains NH3 ligands.
(d) An enantiomer is one of two stereoisomers that are mirror images of each other that
are "non-superposable" (not identical).
68.
(E) (a) A spectrochemical series is a list of ligands ordered on ligand strength and a list
of metal ions based on oxidation number, group and identity.
(b) Crystal field theory (CFT) is a model that describes the electronic structure of
transition metal compounds, all of which can be considered coordination complexes.
(c) Optical isomers are two compounds which contain the same number and kinds of
atoms, and bonds (i.e., the connectivity between atoms is the same), and different spatial
arrangements of the atoms, but which have non-superimposable mirror images.
(d) Structural isomerism in accordance with IUPAC, is a form of isomerism in which
molecules with the same molecular formula have atoms bonded together in different
orders.
69.
(E) (a) Coordination number is the number of ligands coordinated to a transition metal
complex. Oxidation number of a central atom in a coordination compound is the charge
that it would have if all the ligands were removed along with the electron pairs that were
shared with the central atom.
(b) A monodentate ligand has one point at which it can attach to the central atom.
Polydentate ligand has many points at which it can coordinate to a transition metal.
(c) Cis isomer has identical groups on the same side. In trans isomer, on the other hand,
identical groups are on the opposite side.
(d) Dextrorotation and levorotation refer, respectively, to the properties of rotating plane
polarized light clockwise (for dextrorotation) or counterclockwise (for levorotation). A
compound with dextrorotation is called dextrorotary, while a compound with levorotation
is called levorotary.
(e) When metal is coordinated to ligands to form a complex, its "d" orbital splits into high
and low energy groups of suborbitals. Depending on the nature of the ligands, the energy
difference separating these groups can be large or small. In the first case, electrons of the
d orbital tend to pair in the low energy suborbitals, a configuration known as "low spin".
If the energy difference is low, electrons tend to distribute unpaired, giving rise to a "high
spin" configuration. High spin is associated with paramagnetism (the property of being
attracted to magnetic fields), while low spin is associated to diamagnetism (inert or
repelled by magnets).
70.
(E) (d)
71.
(E) (e)
1193
Chapter 24: Complex Ions and Coordination Compounds
72.
(E) (b)
73.
(E) (a)
74.
(E) (d)
75.
(E) (c)
76.
(E) (b)
77.
(M) (a) pentaamminebromocobalt(III)sulfate, no isomerism.
(b) hexaamminechromium(III)hexacyanocobaltate(III), no isomerism.
(c) sodiumhexanitrito-N-cobaltate(III), no isomerism
(d) tris(ethylenediamine)cobalt(III)chloride, two optical isomers.
78.
(M) (a) [Ag(CN)2]-; (b) [Pt(NO2)(NH3)3]+; (c) [CoCl(en)2(H2O)]2+; (d) K4[Cr(CN)6]
79.
(M)
-
Cl
Cl
Pt
H2N
(a)
Cl
Cl
H3N
Fe
Cl
O
Cl
NH3
Cr
NH2
H3N
H2N
Cl
80.
O
Fe
Cl
Cl
O
Cl
H2
N
2-
+
Cl
NH3
OH
O
(b)
(c)
(d)
(M) (a) one structure only (all positions are equivalent for the H2O ligand; NH3 ligands
attach at the remaining five sites).
+3
NH3
H2O
NH3
Co
H3N
NH3
NH3
(b) two structures, cis and trans, based on the placement of H2O.
+3
+3
NH3
H2O
NH3
NH3
H2O
Co
H3N
NH3
Co
OH2
H2O
NH3
NH3
NH3
trans
cis
1194
Chapter 24: Complex Ions and Coordination Compounds
(c) two structures, fac and mer.
+3
+3
NH3
H3N
NH3
NH3
H2O
Co
H2O
OH2
Co
OH2
H2O
OH2
NH3
NH3
fac
mer
(d) two structures, cis and trans, based on the placement of NH3.
+3
+3
NH3
H2O
OH2
OH2
H3N
Co
H2O
OH2
Co
OH2
H 3N
NH3
OH2
OH2
trans
cis
81.
(M) (a) coordination isomerism, based on the interchange of ligands between the
complex cation and complex anion.
(b) linkage isomerism, based on the mode of attachment of the SCN- ligand (either –
SCN- or –NCS-).
(c) no isomerism
(d) geometric isomerism, based on whether Cl- ligands are cis or trans.
(e) geometric isomerism, based on whether the NH3 or OH- ligands are fac or mer.
82.
(M) (a) geometric isomerism (cis and trans) and optical isomerism in the cis isomer.
(b) geometric isomerism (cis and trans), optical isomerism in the cis isomer, and linkage
isomerism in the thiocyanate ligand.
(c) no isomers.
(d) no isomers for this square-planar complex.
(e) two geometric isomers, one with the tridentate ligand occupying meridional positions
and the other with the tridentate ligand occupying facial positions.
83.
(M) Because ethylenediamine (en) is a stronger field ligand than H2O, more energy must
be absorbed by [Co(en)3]3+ than by [Co(H2O)6]3+ to simulate an electronic transition.
This means that [Co(en)3]3+(aq) absorbs shorter wavelength light and transmits longer
wavelength light than does [Co(H2O)6]3+(aq). Thus [Co(en)3]3+(aq) is yellow and
[Co(H2O)6]3+(aq) is blue.
1195
Source Exif Data:
File Type : PDF File Type Extension : pdf MIME Type : application/pdf PDF Version : 1.6 Linearized : Yes XMP Toolkit : Adobe XMP Core 4.0-c321 44.398116, Tue Aug 04 2009 14:24:39 Modify Date : 2010:11:22 17:21:27-08:00 Create Date : 2010:11:22 17:21:27-08:00 Metadata Date : 2010:11:22 17:21:27-08:00 Creator Tool : PScript5.dll Version 5.2.2 Format : application/pdf Title : Microsoft Word - 14_Petrucci10e_CSM.doc Creator : UCampL2 Document ID : uuid:e33eb2d6-57ac-47c1-b4eb-4cd3bc5faebb Instance ID : uuid:2b6f55fd-8147-47b8-9b6a-6107753c230a Producer : Acrobat Distiller 9.0.0 (Windows) Page Count : 486 Author : UCampL2EXIF Metadata provided by EXIF.tools