Chapter 11 - Solutions
CH3COO–(aq) + H3O+ (aq)
1. a. CH3COOH(aq) + H2O (l)
G = RT ln Ka
Ka =
 - G 


e  RT 
G = 369.31 kJ + (237.13 kJ)  (389.9 kJ) – (237.13 kJ)
= +20.59 kJ
3
-1


- 20.59 x 10 J·mol




-1 -1
8.3145 J·mol ·K )(298 K) 

Ka = e
Ka = 2.46  104
[CH3COO ][H  ]
[CH3COOH]
[CH3COO ]
pH = pKa + log
[CH3COOH]
b. Ka 
4.8 = log (2.46  104) + log
[CH3COO ]
[CH3COOH]
[CH 3COO ]
= 1.19
[CH 3COOH]
[CH3COO–] = (1.19)(1 M) = 1.19 M
n (CH3COO–) = (2.5 L)(1.19 M) = 2.98 moles
m (CH3COONa) = 2.98 moles 
82.04 g
= 244 g
1 mol
3. a. n (OH)= (0.052 L)(0.350 mol·L-1) = 0.0182 mol
n (OH) = n HA (equivalence point)
M = molar mass
4.25 g
M (HA) =
= 234 g·mol-1
0.0182 mol
[A  ] 

b. pH = pKa + log 
 in “buffer region”
[HA]




V (OH-) = 0.026 L, 50% of the HA has been converted to A–, so [A–] = [HA]
pKa = pH = 3.82
Ka = 10pKa = 103.82 = 1.51  104
5. Ksp (AgI) at 25C = 8  1017.
Determine H of
AgI(s)
Ag+(aq) + I (aq)
H = 105.58 kJ + (55.19 kJ) – (61.84 kJ) = +112.23 kJ
 K  ΔH
ln  1  
R
 K2 
 1

T
 2

1 
T1 
 8  10 17  112000  1
1 






K 2  8.314  358 298 

ln 
 8  10 17 
  7.58

K
2



17
8  10
ln 
K2
= e7.58 = 5.1  104
K2 = 1.6  1013 = Ksp at 85C
Ksp = [Ag+][I] = (s)2 = 1.6  1013
s = 4  107 M
[base]
[acid]

because HCO3 is acting as the base and CO2 is the acid, this equation can be
manipulated into the following form:
7. a. pH = pKa + log
 [CO ] 
2 
 [HCO  ] 
3 

pH = pK - log 





6
 (5.5 x 10
M) 



[CO 2 ]

0 = -log 

 (5.5 x 10  6 M) 




[CO 2 ]

10 0 = 1 = 

 (5.5 x 10  6 M) 


6
[CO 2 ] = 5.5 10 M
n (CO 2 in 1.0 L) = 5.5 10 - 6 mol
6.1 = 6.1 - log 
[CO 2 ]
b. Assume that all of the added H+ reacts completely with HCO3 to form CO2. That
is, every mole of H+ depletes the amount of HCO3 by one mole and increases the
amount of CO2 by one mole.
n (HCO3 originally present) = n (CO2 originally present)
n (HCO3 originally present) = (5.5x106 M)(1.0 L) = = 5.5  106 mol
n (HCO3 after addition of acid) = 5.5  106 mol  6.5  107 mol
n (HCO3 after addition of acid) = 4.9  106 mol HCO3
n (CO2 after addition of acid) = 5.5  106 mol + 6.5  107 mol
n (CO2 after addition of acid) = 6.2  106 mol CO2
Because V = 1.0 L,
[CO2] = 6.2  106 M
[HCO3] = 4.9  106 M
pH = pK - log
pH = 6.1 - log
[CO 2 ]
[HCO 3  ]
(6.2 10 - 6 M)
(4.9 10 - 6 M)
= 6.0
pH = 6.1  6.0 = 0.1
9. a. n (H3PO3) = n (H2PO3) = n (NaOH) (between the equivalence points)
n (NaOH) = 0.0375 L  0.10 M = 0.00375 mol OH
n (H3PO3)= 0.00375 mol
0.00375 mol
c (H3PO3) =
= 0.15 M
0.025 L
H2PO3 (aq) + OH (aq) Ka2 for H2PO3 = 2.6  107
b. HPO32 (aq) + H2O (l)
Kw
1 x 10 -14
=
= 3.8  108
7
K a2
2.6 x 10
0.00375 mol
[H2PO32] =
= 0.0375 M
0.025 L  0.0375 L  0.0375 L)
Kb2 =
Kb2 =
[OH  ][H 2 PO 3 ]
[HPO 32  ]

(x)(x)
= 3.8  108
0.0375 M
x = 3.8  105 M = H2PO3
[HPO32] = 0.0375 – x = 0.0375 – 3.8  105 = 0.0375 M