CHM 1046. Chapter 15 Homework Solutions.
Problems: 4, 6, 8, 24, 34, 36, 38, 39, 42, 52, 56a, 62, 68, 70, 74, 80a, 89, 93, 100, 103a,
116, 117, 124
4)
An Arrhenius acid is a substance that produces H+ ions when added to water. An
Arrhenius base is a substance that produces OH- ions when added to water.
6)
A Bronstead acid is a proton donor, and forms a conjugate base after donating a
proton. A Bronsted base is a proton acceptor, and forms a conjugate acid after accepting
a proton.
8)
A substance is amphoteric if it can behave as either a Bronstead acid or a
Bronsted base, depending on the particular reaction taking place. A good example (and
the most important amphoteric substance) is water.
HF(aq) + H2O()  H3O+(aq) + F-(aq)
water acts as a Bronsted base
NH3(aq) + H2O()  OH-(aq) + NH4+(aq)
water acts as a Bronsted acid
24)
Cations will act as weak acids if they are small and have multiple positive
charges. Examples are Al3+, Be2+, and Cu2+ ion. See Appendix II-C, page A-12, for
other examples. (Note - the problem meant to ask about cations as weak acids, not as
weak bases.)
34)
a) NaOH is an Arrhenius base
NaOH(aq)  Na+(aq) + OH-(aq)
b) H2SO4 is an Arrhenius acid
H2SO4(aq)  H+(aq) + HSO4-(aq)
c) HBr is an Arrhenius acid
HBr(aq)  H+(aq) + Br-(aq)
d) Sr(OH)2 is an Arrhenius base
Sr(OH)2(aq)  Sr2+(aq) + 2 OH-(aq)
36)
a)
HI is a Bronsted acid, I- is the conjugate base
H2O is a Bronsted base, H3O+ is the conjugate acid
b)
H2O is a Bronsted acid, OH- is the conjugate base
CH3NH2 is a Bronsted base, CH3NH3+ is the conjugate acid
1
38)
c)
H2O is a Bronsted acid, OH- is the conjugate base
CO32- is a Bronsted base, HCO3- is the conjugate acid
d)
HBr is a Bronsted acid, Br- is the conjugate base
H2O is a Bronsted base, H3O+ is the conjugate acid
The conjugate acid forms when a proton is added to the substance.
a) NH4+
39)
b) HClO4
c) H2SO4
d) HCO3-
As an acid
H2PO4-(aq) + NH3(aq)  HPO42-(aq) + NH4+(aq)
As a base
H2PO4-(aq) + CH3COOH(aq)  H3PO4(aq) + CH3COO-(aq)
42)
HF(aq) + H2O()  H3O+(aq) + F-(aq)
a) HF is a weak acid
Ka = [H3O+] [F-]
[HF]
b) HCHO2 is a weak acid
HCHO2(aq) + H2O()  H3O+(aq) + HCO2-(aq)
Ka = [H3O+] [HCO2-]
[HCHO2]
c) H2SO4 is a strong acid
H2SO4(aq) + H2O()  H3O+(aq) + HSO4-(aq)
d) H2CO3 is a weak acid
H2CO3(aq) + H2O()  H3O+(aq) + HCO3-(aq)
Ka = [H3O+] [HCO3-]
[H2CO3]
52)
To do this problem we use relationships
[H3O+] [OH-] = 1.0 x 10-14
pH = - log10[H3O+]
[H3O+] = 10-pH
Also, if pH < 7.00 we have an acid solution, while if pH > 7.00 we have a basic
solution.
2
The values given in the problem are shown in parentheses.
56)
[H3O+]
[OH-]
pH
Acidic or Basic
(3.5 x 10-3)
2.9 x 10-12
2.46
Acidic
2.6 x 10-8
(3.8 x 10-7)
7.58
Basic
(1.8 x 10-9)
5.6 x 10-6
8.74
Basic
7.1 x 10-8
1.4 x 10-7
(7.15)
Basic
a) HI is a strong acid
HI(aq) + H2O()  H3O+(aq) + I-(aq)
[H3O+] = 0.048 mol HI 1 mol H3O+ = 0.048 M
1L
1 mol HI
pH = - log10(0.048) = 1.32
62)
Formic acid is HCOOH, Ka = 1.8 x 10-4 (from Table 15.5, page 672)
HCOOH(aq) + H2O()  H3O+(aq) + HCOO-(aq)
Ka = [H3O+] [HCOO-] = 1.8 x 10-4
[HCOOH]
H3O+
HCOOHCOOH
Initial
Change
Equilibrium
0
0
0.200
x
x
-x
x
x
0.200 - x
(x) (x) = 1.8 x 10-4
(0.200 - x)
Assume x << 0.200. Then
x2 = 1.8 x 10-4 ; x2 = (0.200) (1.8 x 10-4) = 3.6 x 10-5
0.200
x = (3.6 x 10-5)1/2 = 6.0 x 10-3
Is 6.0 x 10-3 << 0.200 ? YES. (We say small if at least 10 times smaller).
Then [H3O+] = x = 6.0 x 10-3 M ; pH = - log10(6.0 x 10-3) = 2.22
3
68)
HA is a weak acid, and so
HA(aq) + H2O()  H3O+(aq) + A-(aq)
Ka = [H3O+] [A-]
[HA]
[H3O+] = 10-pH = 10-3.29 = 5.1 x 10-4 M
H3O+
AHA
Initial
Change
Equilibrium
0
0
0.115
x
x
-x
x = 5.13 x 10-4
x = 5.13 x 10-4
0.115 - x = 0.1145
Ka = (5.13 x 10-4) (5.13 x 10-4) = 2.3 x 10-6
0.1145
70)
Benzoic acid is C6H5COOH, Ka = 6.5 x 10-5 (from Table 15.5, page 672)
C6H5COOH(aq) + H2O()  H3O+(aq) + C6H5COO-(aq)
Ka = [H3O+] [C6H5COO-] = 6.5 x 10-5
[C6H5COOH]
H3O+
C6H5COOC6H5COOH
Initial
Change
Equilibrium
0
0
0.225
x
x
-x
x
x
0.225 - x
(x) (x) = 1.8 x 10-4
(0.225 - x)
Assume x << 0.225. Then
x2 = 6.5 x 10-5 ; x2 = (0.225) (6.5 x 10-5) = 1.46 x 10-5
0.225
x = (1.46 x 10-5)1/2 = 3.82 x 10-3
Is 3.82 x 10-3 << 0.225 ? YES. (We say small if at least 10 times smaller).
Now, percent ionization = concentration of ionized acid x 100%
initial concentration of acid
So % ionization = 3.82 x 10-3 . 100 % = 1.7 %
0.225
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74)
HA(aq) + H2O()  H3O+(aq) + A-(aq)
Ka = [H3O+] [A-]
[HA]
Percent ionization = concentration of ionized acid x 100%
initial concentration of acid
Here the % ionization is 0.59 %.
S0 [A-] = % ionization [HA]initial = 0.59 % (0.085) = 5.0 x 10-4 M
100 %
100 %
Based on the above reaction we can say [H3O+] = [A-] = 5.0 x 10-4 M
[HA] = 0.085 - 5.0 x 10-4 = 0.085 M
Ka = (5.0 x 10-4) (5.0 x 10-4) = 2.9 x 10-6
0.085
80)
a)
LiOH(aq)  Li+(aq) + OH-(aq)
[OH-] = 8.77 x 10-3 mol LiOH
1L
1 mol OH- = 8.77 x 10-3 M OH1 mol LiOH
pOH = - log10(8.77 x 10-3) = 2.057
pH = 14.00 - pOH = 14.00 - 2.057 = 11.94
[H3O+] = 10-pH = 10-11.94 = 1.1 x 10-12 M
89)
M(C8H10N4O2) = 194.2 g/mol
So the concentration of the caffeine solution is
455 x 10-3 g 1 mol = 2.34 x 10-3 M
1L
194.2 g
pKb = 10.4 , so Kb = 10-pKb = 10-10.4 = 4.0 x 10-11
Let C8H10N4O2 = B = weak base. We may then write the reaction as
5
B(aq) + H2O()  HB+(aq) + OH-(q)
Kb = [HB+] [OH-] = 4.0 x 10-11
[B]
OHHB+
B
Initial
Change
Equilibrium
0
0
2.34 x 10-3
x
x
-x
x
x
2.4 x 10-3 - x
(x) (x)
= 4.0 x 10-11
(2.34 x 10-3 - x)
Assume x << 2.34 x 10-3. Then
x2
= 4.0 x 10-11 ; x2 = (2.34 x 10-3) (4.0 x 10-11) = 9.36 x 10-14
-3
2.34 x 10
x = (9.36 x 10-14)1/2 = 3.06 x 10-7
Is 3.06 x 10-7 << 2.34 x 10-3 ? YES. (We say small if at least 10 times smaller).
So pOH = - log10(3.06 x 10-7) = 6.51
pH = 14.00 - pOH = 14.00 - 6.51 = 7.49
93)
a) Br- is the conjugate base of HBr, a strong acid, and so Br- does not act as a
weak base in solution.
b) ClO- is the conjugate base of HClO, a weak acid, and so ClO- is a weak base.
ClO-(aq) + H2O()  HClO(aq) + OH-(aq)
c) CN- is the conjugate base of HCN, a weak acid, and so CN- is a weak base.
CN-(aq) + H2O()  HCN(aq) + OH-(aq)
d) Cl- is the conjugate base of HCl, a strong acid, and so Cl- does not act as a
weak base in solution.
100)
a) Al(NO3)3(aq)  Al3+(aq) = 3 NO3-(aq)
NO3- is the conjugate base of HNO3, a strong acid, and so NO3- does not act as a
weak base in solution. Al3+ is a small cation with a multiple positive charge, and so acts
like a weak acid in solution. Therefore the salt solution will be acidic.
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b) C2H5NH3NO3(aq)  C2H5NH3+(aq) + NO3-(aq)
NO3- is the conjugate base of HNO3, a strong acid, and so NO3- does not act as a
weak base in solution. C2H5NH3+ is the conjugate acid of C2H5NH2, a weak base, and so
C2H5NH3+ is a weak acid. Therefore the salt solution will be acidic.
c) K2CO3(aq)  2 K+(aq) + CO32-(aq)
K+ has a small positive charge and so is not a weak acid. CO32- is the conjugate
base of HCO3-, a weak acid, and so CO32- is a weak base. Therefore the salt solution is
basic.
d) RbI(aq)  Rb+(aq) + I-(aq)
Rb+ has a small positive charge and so is not a weak acid. I- is the conjugate base
of HI, a strong acid, and so I- does not act as a weak base in solution. Therefore the salt
solution is neutral.
e) NH4ClO(aq)  NH4+(aq) + ClO-(aq)
NH4+ is the conjugate acid of NH3, a weak base, and so NH4+ is a weak acid.
ClO is the conjugate base of HClO, a weak acid, and so ClO- is a weak base. So this acts
like a solution with equal amounts of a weak acid and a weak base. Therefore, the
solution is approximately neutral. (In fact, since Kb for NH3 is larger than Ka for HClO,
the solution will be slightly basic.)
-
103)
a)
NH4Cl(aq)  NH4+(aq) + Cl-(aq)
NH4+(aq) + H2O()  NH3(aq) + H3O+(aq)
Ka = [NH3] [H3O+]
[NH4+]
Now, Kb(NH3) = 1.76 x 10-5 (from table 15.8, page 688)
Since Ka Kb = 1.0 x 10-14 for a base/conjugate acid pair,
Ka = 1.0 x 10-14 = 1.0 x 10-14 = 5.7 x 10-10
Kb
1.76 x 10-5
NH3
H3O+
NH+
Initial
Change
Equilibrium
0
0
0.10
x
x
-x
x
x
0.10 - x
7
(x) (x) = 5.7 x 10-10
(0.10 - x)
Assume x << 0.10. Then
x2 = 5.7 x 10-10 ; x2 = (0.10) (5.7 x 10-10) = 15.7 x 10-11
0.10
x = (5.7 x 10-11)1/2 = 7.5 x 10-6
Is 7.5 x 10-6 << 0.10 ? YES. (We say small if at least 10 times smaller).
So pH = - log10(7.5 x 10-6) = 5.12
116)
Based on the periodic trends...
H2Te is stronger than H2S (acid strength increases from top to bottom within a
group).
HI is stronger than H2Te (acid strength increases from left to right within a row).
H2S is stronger than NaH (acid strength increases from left to right within a row).
So NaH < H2S < H2Te < HI
117) a) H2SO4 > H2SO3. For a ternary acid with the same third element, the more
oxygen atoms the stronger the acid.
b) HClO2 > HClO. For a ternary acid with the same third element, the more
oxygen atoms the stronger the acid.
c) HClO > HBrO. For a ternary acid with the same number of oxygen atoms, acid
strength increases from bottom to top within a group.
d) CCl3COOH > CH3COOH. We did not talk about this sort of prediction. I
would argue that the CCl3COO- ion is more stable than the CH3COO- ion, since it has
several electronegative Cl atoms, and so will be more easy to form.
124) A Lewis acid is an electron pair acceptor, and a Lewis base is an electron pair
donor.
a) Ag+ is a Lewis acid, NH3 is a Lewis base (Ag+ is accepting electron pairs from
the N atom in NH3).
b) AlBr3 is a Lewis acid, and NH3 is a Lewis base (Al is accepting an electron pair
from the N atom in NH3).
c) BF3 is a Lewis acid, and F- is a Lewis base (B is accepting an electron pair
from F )
Note that if a substance is a Bronsted base it is usually going to also be a Lewis
base.
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