Chemistry 122 [Tyvoll]
Spring 2008
Chapter 16 Homework Solutions for Moore
Problems # – 1, 7, 12, 20, 22, 32, 36, 42, 49, 55, 59, 63, 65, 67, 72, 75, 89, 99, 112
1. A Brønsted-Lowry acid is a substance capable of donating H+ ion. A Brønsted-Lowry base is a
substance capable of accepting H+ ion.
7. (a) HNO3(aq) + H2O(l) ' H3O+(aq) + NO3–(aq)
acid
base
conj. acid conj. base
(b) NH4+(aq) + CN–(aq) ' NH3(aq) + HCN(aq)
acid
base
conj. acid
conj. base
12. An H+(aq) ion is transferred from the acid to the water molecule to make H3O+(aq)
and the conjugate base of the acid.
(a) HCO3 –(aq) + H2O(l) ' H3O+(aq) + CO32–(aq)
(b) HCl (aq) + H2O (l) ' H3O+(aq) + Cl–(aq)
(c) CH3COOH(aq) + H2O(l) ' H3O+(aq) + CH3COO–(aq)
(d) HCN (aq) + H2O (l) ' H3O+(aq) + CN–(aq)
20. Two relationships between acids can be used to help us decide: If the central atoms
are the same, then the acid with more O atoms is stronger. If the numbers of O atoms are the
same, then the acid with the more electronegative central atom is stronger. (Electronegativities
are found in Figure 8.6.)
(a) H2SO4 is stronger than H2SO3 because it has more O atoms, and H2SO3 is stronger than
H2CO3 because it has a more electronegative central atom (ENS > ENC), so H2SO4 is
stronger than H2CO3.
(b) HNO3 is stronger than HNO2 because it has more O atoms.
(c) HClO4 is stronger than H2SO4 because it has a more electronegative atom (ENCl > ENS).
(d) HClO3 is stronger than H3PO4 because it has a more electronegative atom (ENCl > ENP).
(e) H2SO4 is stronger than H2SO3 because it has more O atoms.
22. (a) HI is a Brønsted-Lowry acid. Its conjugate base is I–, iodide ion.
(b) NO3– is a Brønsted-Lowry base. Its conjugate acid is HNO3, nitric acid.
(c) CO32– is a Brønsted-Lowry base. Its conjugate acid is HCO3–, hydrogen carbonate ion.
(d) H2CO3 is a Brønsted-Lowry acid. Its conjugate base is HCO3–, hydrogen carbonate ion.
(e) HSO4– as a Brønsted-Lowry base, has a conjugate acid of H2SO4, sulfuric acid. HSO4– as a
Brønsted-Lowry acid, has a conjugate base of SO42–, sulfate ion.
(f) SO32– is a Brønsted-Lowry base. Its conjugate is HSO3–, hydrogen sulfite ion.
32. When pH = 3.30, [H3O+] = 10–3.30 = 5.0 × 10–4 M. The solution is acidic.
36. HNO3 is a strong acid, and completely ionizes to form NO3– and H3O+: so [H3O+] = 0.0013 M,
pH = – log[H3O+] = 2.89
pOH = 14.00 – pH = 14.00 – 2.89 = 11.11
42. (a) pH = 1.00 [H3O+] = 10–pH = 1.0 × 10–1 M
acidic solution (pH < 10–7)
pOH = 14 – pH = 13
[OH–] = 10–pOH = 1.0 × 10–13 M
(b) pH = 10.5 [H3O+] = 10–pH = 3 × 10–11 M
basic solution (pH > 10–7)
–
–pOH
pOH = 14 – pH = 3.5 [OH ] = 10
= 3 × 10–4 M
(c) [H3O+] = 1.8 × 10–4 M pH = –log[H3O+] = 3.74
acidic solution (pH < 10–7)
pOH = 14 – pH = 10.26 [OH–] = 10–pOH = 5.6 × 10–11 M
(d) [OH–] =2.3 × 10–5 M pOH = –log[OH–] = 4.64
pH = 14 – pOH = 9.36 [H3O+] = 10–pH = 4.3 × 10–10 M basic solution (pH > 10–7)
49. (a) F– (aq) + H2O (l) ' HF (aq) + OH– (aq)
Kb = [HF][OH−]/[F−]
(b) NH3 (aq) + H2O (l) ' NH4+ (aq) + OH– (aq)
Kb =[NH4+][OH−]/[NH3]
(c) H2CO3 (aq) + H2O (l) ' HCO3– (aq) + H3O+ (aq)
Ka = [HCO3−][H3O+]/[H2CO3]
(d) H3PO4 (aq) + H2O (l) ' H2PO4– (aq) + H3O+ (aq)
Ka = [H2PO4–][H3O+]/[H3PO4]
(e) CH3COO– (aq) + H2O (l) ' CH3COOH (aq) + OH– (aq)
Kb = [CH3COOH][OH–]/[CH3COO–]
55. HCNO is cyanic acid. The equation for the equilibrium and the equilibrium expression are:
HCNO (aq) + H2O (l) ' H3O+ (aq) + CNO– (aq)
Ka = [H3O+][CNO–]/[HCNO] = ?
At equilibrium, pH = 2.67, so [H3O+] = 10–pH = 10–2.67 = 2.1 × 10–3 M. Since the H3O+ ions are
produced from the decomposition of HCNO, an equal quantity of CNO– is formed, and the
concentrations change in the following way, where x = [H3O+] = [CNO1-] = 2.1 × 10–3 M:
Setting up the “ICE” table:
I
C
E
[H3O+] (aq)]
0
+x
x
[HCNO (aq)]
0.015
-x
0.015 - x
[CNO1- ] (aq)]
0
+x
x
0.015 - x
x
x
HCNO(aq) + H2O ' H3O+(aq) + CNO1-(aq)
Ka = [H3O+][CNO–]/[HCNO] = (2.1 × 10–3)(2.1 × 10–3)/(0.015−2.1 × 10–3)
Ka = 3.6 x 10−4
59. The equation for the equilibrium and the equilibrium expression are:
C6H5COOH (aq) + H2O (l) ' H3O+ (aq) + C6H5COO– (aq)
Ka = [H3O+][C6H5COO–]/[C6H5COOH] = 6.3 x 10−5
Setting up the “ICE” table:
I
C
E
[C6H5COOH (aq)]
0.050
−x
0.050 − x
[H3O+] (aq)]
0
+x
x
[C6H5COO– ] (aq)]
0
+x
x
0.050 − x
x
x
C6H5COOH (aq) + H2O ' H3O+ (aq) + C6H5COO– (aq)
Ka = (x)(x) / (0.050 − x ) = 6.3 x 10−5
Ka ≅ (x)(x) / (0.050) ≅ 6.3 x 10−5
(assume x << 0.050, i.e. “neglect” x)
x ≅ (6.3 x 10−5)(0.050)1/2 = 1.8 x 10−3 M
Check assumption: x/C = 1.8 x 10−3/0.050 = 0.036 > 0.01 so technically have to use quadratic.
However, text accepts this as “OK”, so pH = −log (1.8 x 10−3) = 2.74
And % ionization = x/C x 100 = 1.8 x 10−3/0.050 = 3.6 % ionization
63. The equation for the equilibrium and the equilibrium expression are:
CH3NH2 (aq) + H2O (l) ' CH3NH3+ (aq) + OH– (aq)
Kb = [CH3NH3+][OH–]/[CH3NH2] = 5.0 x 10–4
I
C
E
[OH–] (aq)]
0
+x
x
[CH3NH2 (aq)]
0.024
−x
0.024 − x
[CH3NH2 ] (aq)]
0
+x
x
Kb = (x)(x)/(0.024 − x) = 5.0 x 10–4
x2 = (5.0 x 10–4)(0.024 – x)
x2 + 4.2 x 10–4x – 1.01 × 10–5 = 0
(requires quadratic since x not << 0.024)
x = 3.0 x 10–3 M = [OH–]
pOH = – log[OH–] = – log(3.0 x 10–3) = 2.72
pH = 14.00 – pOH = 14.00 – 2.72 = 11.28
65. (a) Pyridine reacts with water, undergoing hydrolysis to form a basic solution.
The equation for the equilibrium is: C5H5N(aq) + H2O(l) ' C5H5NH+(aq) + OH–(aq)
(b) Kb = KW/Ka = 1.0 x 10–14/6.5 x 10–6 = 1.5 x 10–9 = [C5H5NH+][OH–]/[C5H5N]
[OH– ] = 10–pOH = 10–(14.00– pH) = 10 (pH– 14.00) = 10 (8.5– 14.00) = 10–5.5 = 3 x 10–6 M
I
C
E
[C5H5N] (aq)]
0.2
−x
0.2 − x
[C5H5NH+] (aq)]
0
+x
x
Kb = (3 x 10–6)(x)/(0.2 − x) = 1.5 x 10–9
x = 0.0001
[C5H5N] = 0.2 – x = 0.2 – 0.0001 = 0.2 M
[OH– ] (aq)]
?
+x
3 x 10–6
67. Find the initial concentration of the HC9H7O4:
C(HC9H7O4) = (2 tablets/200 mL)(0.325 g HC9H7O4/tablet)
x (1 mol HC9H7O4/180.16 g HC9H7O4)(1000 mL/L) = 0.0180 M
HC9H7O4 (aq) + H2O (l) ' H3O+ (aq) + C9H7O4– (aq)
Kb = [H3O+][C9H7O4–]/[HC9H7O4] = 3.27 x 10–4
I
C
E
[HC9H7O4] (aq)]
0.0180
−x
0.0180 − x
[H3O+] (aq)]
0
+x
x
[C9H7O4–] (aq)]
0
+x
x
Kb = (x)(x)/(0.0180 − x) = 3.27 x 10–4
x2 = (3.27 x 10–4)(0.0180 – x)
x2 + 3.27 x 10–4x – 5.89 x 10–6 = 0 (x not << 0.0180 so must solve using quadratic)\
x = 2.27 x 10–3 M = [H3O+]
pH = – log[H3O+] = – log(2.27 x 10–3) = 2.644
72. Compare the reactant acid to the product acid and identify which is stronger and which is
weaker. Do the same with the bases. Equilibrium favors the weaker species in the reaction.
(a) CN– (aq) + HSO4 – (aq) ' HCN (aq) + SO42– (aq) The reaction is product-favored.
(b) H2S (aq) + H2O (l) ' H3O+ (aq) + HS– (aq) The reaction is reactant-favored.
(c) H– (aq) + H2O (l) ' OH– (aq) + H2 (g) (aq) The reaction is product-favored.
75. (a) NaHSO4 (s) ' Na+ (aq) + HSO4 – (aq) The Na+ ions do not affect the pH of a water solution.
HSO4– is a weak acid and a weak base, so we must compare the size of Ka and Kb:
HSO4– (aq) + H2O (l) ' H3O+ (aq) + SO42– (aq)
Ka = 1.2 x 10–2
HSO4– (aq) + H2O (l) ' H2SO4 (aq) + OH– (aq)
Kb = very small
∴HSO4– is a stronger acid than it is a base, so we predict pH less than 7.
(b) NH4Br (s) ' NH4+ (aq) + Br– (aq).
Br– ions do not affect the pH of a water solution. NH4+ is a weak acid, so we predict pH less
than 7.
75. (c) KClO4 (s) K+ (aq) + ClO4– (aq). (100%)
Neither of these ions affects the pH of a water solution, because KOH is a strong base and
HClO4 is a strong acid, so we predict pH equal to 7.
89. The Lewis model focuses on what electron pairs are doing. The substance capable
of donating the electron pair to form a new bond is called a Lewis base. The substance capable
of accepting an electron pair is a Lewis acid.
(a) NH3 has a lone pair of electrons that can form a new bond, so NH3 can be a Lewis base. Its
H atoms could interact with lone pairs on other strong Lewis bases, so NH3 can also be a
Lewis acid. (see p. 386 for structure of NH3)
(b) BeCl2 has a central Be atom with only four electrons, so it can interact with lone pairs on
Lewis bases, so BeCl2 can be a Lewis acid.
(c) BCl3 has a central B atom with only six electrons, so it can interact with lone pairs on
Lewis bases, so BCl3 can be a Lewis acid. (e. g. see p. 385 for structure of BF3)
99. (a) The strongest acid of the three has the largest Ka value, and that is HF. The weakest acid of
the three has the smallest Ka value, and that is HS–.
(b) The strongest acid has the weakest conjugate base. F– is the weakest of the three bases, so
HF is the acid with the weakest conjugate base.
(c) The weakest acid has the strongest conjugate base. S2– is the strongest of the three bases, so
HS– is the acid with the strongest conjugate base.
112. Conjugates in an acid-base pair must differ by only one H+ ion. The acids and bases
were identified correctly, but the conjugates were not. HCO3− is the conjugate of H2CO3 and
HSO4− is the conjugate of SO42–.