Chapter 16: 3, 5, 7, 15, 17, 21, 23, 31, 35, 41, 45, 55, 73, 87, 93, 105
3.
A Bronsted acid is a proton donor and a Bronsted base is a proton acceptor
H3O+ an acid
NH2− a base
HBr an acid
H2O an acid and a base
NH3 a base
NO3− a base
HCN an acid
a.
CH3COO−
base
b.
HCO3−
base
+
HCO3−
acid
c.
H2PO4−
acid
+
NH3
base
d.
HClO
acid
+
CH3NH2
base
e.
CO32−
base
+
H2O
base
f.
CH3COO−
base
7.
a.
d.
g.
j.
m.
CH2ClCOO−
HPO42−
SO42−
NH3
ClO−
15.
a.
[H3O+] = 10−2.42 = 3.8 x 10−3 M
b.
[H3O+] = 10−11.21 = 6.2 x 10−12 M
c.
[H3O+] = 10−6.96 = 1.1 x 10−7 M
d.
[H3O+] = 10−15.00 = 1.0 x 10−15 M
a.
pH = −log[H3O+] = −log(0.0010) = 3.00
b.
Kw
1.0 x 1014

 1.3 x 1014 M
[H 3O ] 

[OH ]
0.76
5.
17.
b.
e.
h.
OH− a base
NH4+ an acid
CO32− a base
a.
d.
g.
j.
+
+
HCN
acid
c.
f.
i.
CH3COOH
conj. acid
H2CO3
conj. acid
H2O
acid
+
CH3COOH
conj. acid
b.
e.
h.
k.

pH = −log(1.3 x 10−14) = 13.89
NH4+
conj. acid
+
CH3NH2+
conj. acid
HCO3−
conj. acid
CO32−
conj. base
+
HPO42−
conj. base
IO4−
PO43−
HCOO−
HS−
CN−
conj. base
+
ClO−
conj. base
+
OH−
conj. base
+
OH−
conj. base
c.
f.
i.
l.
H2PO42−
HSO4−
SO32−
S2−
c.
Ba2+
Ba(OH)2
+
2OH−
1.0 x 1014
 1.8 x 1011 M ; pH = −log(1.8 x 10−11) = 10.74
[H 3O ] 
4
2(2.8 x 10 )

d.
21.
pH = −log(5.2 x 10−4) = 3.28
pH = 14.00 – pOH = 14.00 – 9.40 = 4.60
1 mol HCl
36.46 g HCl
0.662 L
18.4 g HCl 
23.
[HCl]  [H 3O  ] 
 0.762 M
pH = −log(0.762) = 0.118
HNO3 is a strong acid
H2SO4 is a strong acid (1st ionization only)
H2CO3 is a weak acid
HCl is a strong acid
HNO2 is a weak acid
b.
d.
f.
h.
HNO3 is a weak acid
HSO4− is a weak acid
HCO3− is a weak acid
HCN is a weak acid
31.
a.
c.
e.
g.
i.
35.
The direction should favor formation of F(aq) and H2O(ℓ). Hydroxide ion is a stronger
base than fluoride ion, and hydrofluoric acid is a stronger acid than water.
41.
initial
change
equilibrium
K a  4.9 x 1010 
HCN(aq)
HCN
0.15
−x
0.15 − x
+
H3O+(aq)
H3O+
~0.0
+x
x
H2O(ℓ)
H2O
-
[H 3O  ][CN  ]
(x)(x)

[HCN]
0.15  x

+ CN−(aq)
CN−
0.0
+x
x
x2
0.15  x
If you assume that x << 0.15, then 0.15 – x ≈ 0.15
x2
x2

 4.9 x 1010 ; x2 = (0.15)(4.9 x 10–10) = 7.35 x 10–11;
0.15  x
0.15
x = 8.6 x 10–6
[H3O+] = [CN−] = 8.6 x 10–6 M and [HCN] = 0.15 M
Kw
1.0 x 1014

[OH ] 

 1.2 x 109 M

6
[H 3O ]
8.6 x 10
45.
initial
change
equilibrium
K a  7.1 x 104 
HF(aq) +
HF
0.060
−x
0.060 − x
H3O+(aq)
H3O+
~0.0
+x
x
H2O(ℓ)
H2O
-
[H 3O  ][F ]
(x)(x)

[HF]
0.060  x

+ F−(aq)
F−
0.0
+x
x
x2
0.060  x
If you assume that x << 0.060, then 0.060 – x ≈ 0.060
x2
x2

 7.1 x 104 ; x2 = 4.3 x 10–5; x = 6.6 x 10–3 (assumpt not
0.060  x
0.060
good!)
If you do not make the assumption, then you need to solve the quadratic equation
x2 + Kax – CHAKa = x2 + 7.1 x 10–4x – 4.3 x 10–5 = 0
x = 6.2 x 10–3 or –6.9 x 10–3 (2nd root is physically impossible); [H3O+] = 6.2 x 10–3 M
pH = −log(6.2 x 10−3) = 2.21
55.
a.
initial
change
equilibrium
K b  1.8 x 105 
NH3(aq)
NH3
0.10
−x
0.10 − x
+
NH4+(aq)
H2O(ℓ)
H2O
-
[NH 4  ][OH  ]
(x)(x)

[NH 3 ]
0.10  x
NH4+
0.0
+x
x

+ OH−(aq)
OH−
~0.0
+x
x
x2
0.10  x
If you assume that x << 0.10, then 0.10 – x ≈ 0.10
x2
x2

 1.8 x 105 ; x2 = 1.8 x 10–6; x = 1.3 x 10–3 = [OH−]
0.10  x
0.10
pOH = −log(1.3 x 10−3) = 2.89;
pH = 14.00 – 2.89 = 11.11
b.
initial
change
equilibrium
K b  1.7 x 109 
C5H5N +(aq) + OH−(aq)
C5H5N +
OH−
0.0
~0.0
+x
+x
x
x
C5H5N(aq) + H2O(ℓ)
C5H5N
H2O
0.050
−x
0.050 − x
[C5 H 5 N  ][OH  ]
(x)(x)

[C5 H 5 N]
0.050  x

x2
0.050  x
If you assume that x << 0.050, then 0.050 – x ≈ 0.050
x2
x2

 1.7 x 109 ; x2 = 8.5 x 10–11; x = 9.2 x 10–6 = [OH−]
0.050  x
0.050
pOH = −log(9.2 x 10−6) = 5.04;
73.
CH3COONa(aq)
pH = 14.00 – 5.04 = 8.96
CH3COO−(aq)
+
Na+(aq)
Na+ is a neutral cation and CH3COO− is a weak base anion
CH3COO−(aq)
CH3COO−
initial
0.36
change
−x
equilibrium 0.36 − x
+
H2O(ℓ)
H2O
-
CH3COOH(aq) + OH−(aq)
CH3COOH
OH−
0.0
~0.0
+x
+x
x
x
Ka x Kb = Kw for acid/conjugate base pair; Ka for CH3COOH is 1.8 x 10–5
Kb 
[CH 3COOH][OH  ]
1.0 x 1014
(x)(x)
10
5.6
x
10



5

1.8 x 10
[CH 3COO ]
0.36  x

x2
0.36  x
If you assume that x << 0.36, then 0.36 – x ≈ 0.36
x2
x2

 5.6 x 1010 ; x2 = 2.0 x 10–10; x = 1.4 x 10–5 = [OH−]
0.36  x
0.36
pOH = −log(1.4 x 10–5) = 4.85;
87.
pH = 14.00 – 4.85 = 9.15
a.
NH3/NH4+ are a base/conjugate acid pair and NH3/NH2− are an acid/conjugate
base pair
b.
H+ (H3O+) corresponds to and OH− corresponds to NH2−; the condition for a
neutral solution is [NH4+] = [NH2−]
93.
CH3COO−(aq)
CH3COONH4(aq)
+
NH4+(aq)
NH4+ is a weak acid cation and CH3COO− is a weak base anion
Ka x Kb = Kw for acid/conjugate base pair
Ka for CH3COOH is 1.8 x 10–5 ; K b (CH 3COO  ) 
1.0 x 1014
 5.6 x 1010
1.8 x 105
1.0 x 1014
Kb for NH3 is 1.8 x 10 ; K a (NH 4 ) 
 5.6 x 1010
5
1.8 x 10
–5

Since Kb (CH3COO−) = Ka (NH4+) the solution is neutral and pH = 7.00
105.
a.
NH2−
base
N3−
base
b.
+
+
H2O
acid
3H2O
acid
NH3
conj. acid
NH3
conj. acid
+
+
N3− is the stronger base since it produces 3 OH− ions.
OH−
conj. base
3OH−
conj. base