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Chemistry, Student Solutions Manual
Chapter 14
Chapter 14 Aqueous Acid-Base Equilibria
Solutions to Problems in Chapter 14
14.1 HBr is a strong acid that will transfer its hydrogen atom to a water molecule,
generating a hydronium cation and a bromide anion:
14.3 HClO4 is a strong acid that dissolves in water to generate H3O+ cations and ClO4–
anions. Since there are no bases present for the hydronium ion to react with, the only
equilibrium occurring is the proton transfer reaction between water molecules:
H 2O  H 2O
OH   H3O 
K w  1.00  1014
To determine the concentrations of hydroxide and hydronium ions, set up a
concentration table, write the equilibrium expression, and solve for the final
concentrations.
The initial concentration of hydronium ions is the same as the concentration of
perchloric acid, [H3O+] = 1.25 × 10–3 M. Here is the concentration table:
Reaction:
H2O +
H2O
H3O+
Initial concentration (M)
Solvent
Solvent
1.25 × 10–3
Change in concentration
(M)
Solvent
Solvent
Final concentration (M)
Solvent
Solvent
K w  [H 3O + ][OH  ]
+
OH–
0
+x
+x
1.25 × 10–3 + x
x
1.00  1014  [0.00125  x][x]
Assume that x << 0.00125 M:
1.00  1014  [0.00125][x]
x  8.00  1012 M  [OH  ]
The assumption that x << 0.00125 is valid.
[H 3O  ]  (1.25  103 M)  (8.00  10 12 M) = 1.25  10 3 M
14.5 We are asked to determine the final concentrations of all the ions in a final solution.
Begin by analyzing the chemistry. HCl is a strong acid that dissolves in water to
generate H3O+ cations and Cl– anions. Any water solution always has OH– and H3O+
ions with the equilibrium:
H 2O  H 2O
OH   H3O 
K w  1.00  1014
Therefore, the ions present in this solution are H3O+, OH–, and Cl–.
The process involves dilution, so the first step is to determine the concentration of
HCl in the flask after the dilution:
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Chapter 14
MiVi = MfVf
Mf 
M iVi (12.1 M)(1.00 mL)

 0.121 M HCl in final solution.
Vf
100. mL
Since Cl– is a spectator ion, its concentration is the same as that of HCl: [Cl–] = 0.121
M. The rest of the ion concentrations are determined by the equilibrium. Set up a
concentration table, write the equilibrium expression, and solve for the final
concentrations.
H3O+ +
OH–
Reaction:
H2O +
Initial concentration (M)
Solvent
Solvent
0.121
0
Change in concentration (M)
Solvent
Solvent
+x
+x
Final concentration (M)
Solvent
Solvent
0.121 + x
x
K w  [H 3O + ][OH  ]
H2O
1.00  1014  x(0.121  x)
1.00  1014  x(0.121)
Assume that x << 0.0121:
x  [OH  ]  8.26  1014 M
The assumption is valid.
[H 3O  ]  (0.121 M)  (8.26  10 14 M) = 0.121 M
Here are the final concentrations:
[Cl ]  [H3O+ ]  0.121 M
[OH  ]  8.26  1014 M
14.7 We are asked to determine the concentrations of hydroxide and hydronium ions in the
solution. Begin by analyzing the chemistry. HCl is a strong acid that dissolves in
water to generate H3O+ cations and Cl– anions. Any water solution always has OH–
and H3O+ ions with the equilibrium:
H 2O  H 2O
OH   H3O 
K w  1.00  1014
The first step is to determine the initial number of moles of HCl gas and convert that
to concentration of HCl dissolved in the solution:
MHCl = 1.008 g/mol + 35.453 g/mol = 36.461 g/mol
 1 mol 
0.488 g 
  0.01338 mol HCl dissolved
 36.461 g 
0.01338 mol
[HCl] 
 0.0412 M
0.325 L
To determine the concentrations of the ions, construct a concentration table, write the
equilibrium expression, and solve for the final concentrations. The initial
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Chapter 14
concentration of hydronium ions will be the same as the concentration of HCl,
[H3O+] = 0.0412 M.
H2O
H3O+ +
OH–
Solvent
Solvent
0.0412
0
Change in concentration (M) Solvent
Solvent
+x
+x
Solvent
Solvent
0.0412 + x
x
Reaction:
H2O +
Initial concentration (M)
Final concentration (M)
K w  [H 3O + ][OH  ]
1.00  1014  (0.0412  x)x
Assume that x << 0.0412:
1.00  1014  (0.0412)x
x  2.43  1013 M  [OH  ]
The assumption is valid.
[H 3O  ]  (0.0412 M)  (2.43  1013 M)  0.0412 M
14.9 Conversion from hydronium ion molarity to pH is accomplished by taking the
logarithm to base ten and changing the sign, pH = – (log[H3O+]):
(a) – 0.60; (b) 5.426; (c) 2.32; and (d) 3.593
14.11 Because pH + pOH = 14.00, pH = 14.00 – pOH. Convert from hydroxide ion
molarity to pOH by taking the logarithm to base ten and changing the sign, pOH = –
(log[OH–]). Thus, pH = 14 + log[OH–]:
(a) 14.60; (b) 8.574; (c) 11.68; and (d) 10.407
14.13 Take 10–pH to convert pH into hydronium ion concentration:
(a) 0.22 M; (b) 1.4 × 10–8 M; (c) 2.1 × 10–4 M; and (d) 4.7 × 10–15 M
14.15 Use pH + pOH = 14.00 to convert pH to pOH. Then take 10–pOH to convert pOH into
hydroxide ion concentration:
(a) pOH = 13.34, [OH–] = 4.6 × 10–14 M; (b) pOH = 6.15, [OH–] = 7.1 × 10–7 M; (c)
pOH = 10.32, [OH–] = 4.8 × 10–11 M; and (d) pOH = –0.33, [OH–] = 2.1 M
14.17 To calculate the pH of a solution, it is necessary to determine either the hydronium
ion concentration or the hydroxide ion concentration.
Determine the nature and initial concentration of each species, construct a
concentration table, write the equilibrium expression, and solve for the
concentrations:
(a) Weak base, carry out an equilibrium calculation to determine [OH–]:
Reaction:
H2O +
Initial concentration (M)
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C5H5N
1.5
C5H5NH+ +
OH–
0
0
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Chapter 14
Change in concentration (M)
Final concentration (M)
K b  1.7  109 
[C5 H 5 NH  ]eq
[C5 H 5 N]eq

–x
+x
+x
1.5 – x
x
x
x2
; assume that x   
1.5  x
2
x
1.5
2
x  2.55  109 , so x  5.0 105 ; the assumption is valid.
   
[OH  ] = 5.0  105 M
pOH =  log(5.0  10 5 ) = 4.30 pH = 14.00  4.30 = 9.70
(b) Weak base, carry out an equilibrium calculation to determine [OH-]:
Reaction:
H2O +
NH2OH
NH3OH+ +
OH–
Initial concentration (M)
1.5
0
0
Change in concentration (M)
–x
+x
+x
1.5 – x
x
x
Final concentration (M)
K b  8.7  109 
[NH3OH  ]eq [OH  ]eq
[NH 2 OH]eq

x2
; assume that x << 1.5:
1.5  x
x2
1.5
2
x  1.3  108 , so x  1.1  104 ; the assumption is valid.
8.7  109 
[OH  ]  1.1  104 M
pOH  log (1.1  10 4 )  3.96
pH = 14.00  3.96  10.04
(c) Weak acid, carry out an equilibrium calculation to determine [H3O+]:
H2O+
Reaction:
HCH2H
HCO2– +
H3O+
Initial concentration (M)
1.5
0
0
Change in concentration (M)
–x
+x
+x
1.5 – x
x
x
Final concentration (M)
K a  1.8  104 
[HCH 2  ]eq [H3O  ]eq
[HCO 2 H]eq

x2
; assume that x << 1.5:
1.5  x
x2
1.5
2
x  2.7  104 , so x  1.6  102 ; the assumption is valid.
1.8  104 
[H 3O  ]  1.6  102 M
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pH  log (1.6  10 2 )  1.80
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Chapter 14
14.19 HONH2 is a weak base that will accept a proton from water to form a hydroxide ion:
HCO2H is a weak acid that will donate a proton to water to form a hydronium ion:
14.21 Follow standard procedures for dealing with equilibrium problems:
(a) HN3 is a weak acid. Major species: HN3 and H2O; minor species: N3–, H3O+, and
OH–.
(b) Construct a concentration table, write the equilibrium expression, and solve for
the concentrations:
Reaction:
H2O +
N3–
HN3
H3O+
+
Initial concentration (M)
1.50
0
0
Change in concentration (M)
–x
+x
+x
1.50 – x
x
X
Final concentration (M)
K a  2.5  10 5 
[N 3 ]eq [H 3O  ]eq
[HN 3 ]eq

x2
; assume that x << 1.50:
1.50  x
x2
1.50
2
x  3.75  10 5 , so x  6.1  10 3 ; the assumption is valid.
2.5  10 5 
[H 3O  ]  [N 3 ]  6.1  10 3 M, and [HN 3 ]  (1.50 M)  (6.1  10 3 M)  1.50 M
[OH  ] =
1.0  1014
 1.6  10 12 M
6.1  10 3
(c) pH = –log (6.1 × 10–3) = 2.21
(d) The dominant equilibrium is proton transfer from HN3 to water:
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Chapter 14
14.23 Determine the percent ionization using Equation 14-3 in your textbook:
 H 3O + 
eq
%HA ionized  100%
HA
 initial
The equilibrium concentration of hydronium ions was calculated in Problem
14.21, and the initial concentration of the weak acid is given in the problem:
 6.1  103M 
%HA ionized  100% 
  0.41%
 1.50 M 
14.25 Follow standard procedures for dealing with equilibrium problems:
(a) N(CH3)3 is a weak base. Major species: N(CH3)3, H2O; minor species:
HN(CH3)3+,OH–, and H3O+.
(b) Construct a concentration table, write the equilibrium expression, and solve for
the concentrations:
Reaction:
H2O +
HN(CH3)3+
0.350
0
0
–x
+x
+x
0.350 – x
x
x
Initial concentration (M)
Change in concentration (M)
Final concentration (M)
K b  6.5  105 
[HN(CH 3 )3 ]eq [OH  ]eq
[N(CH 3 )3 ]eq
OH–
+
N(CH3)3

x2
; assume that x << 0.350:
0.350  x
x2
0.350
2
x  2.28  105 , so x  4.8  10 3 ; the assumption is valid.
6.5  105 
[OH  ]  [HN(CH 3 )3 ]  4.8  10 3 M, and [N(CH 3 )3 ]  (0.350 M)  (4.8  10 3 M)  0.345 M
[H 3O  ] =
1.0  1014
 2.1  10 12 M
4.8  103
(c) pH = –log(2.1 × 10–12) = 11.68
(d) The dominant equilibrium is proton transfer from water to trimethylamine:
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Chapter 14
14.27 Determine the percent ionization using Equation 14-3 in your textbook:
 H 3O + 
eq
%HA ionized  100%
 HA initial
Follow standard procedures for dealing with equilibrium problems:
Acetic acid is a weak acid. Major species: CH3CO2H and H2O.
Construct a concentration table, write the equilibrium expression, and solve for the
concentration of hydronium ions:
Reaction:
H2O +
CH3CO2H
CH3CO2– +
H3O+
Initial concentration (M)
0.75
0
0
Change in concentration (M)
–x
+x
+x
0.75 – x
x
x
Final concentration (M)
Find Ka in Appendix E of your textbook:
5
K a  1.8  10 
[CH3CO 2 ]eq [H 3O + ]eq
[CH 3CO 2 H]eq

x2
; assume that x << 0.75:
0.75  x
x2
1.8  10 
0.75
2
x  1.35  105 , so x  3.7  103 ; the assumption is valid.
5
[H3O  ]  3.7  103 M
Substitute this result into Equation 14-3:
 3.7 103 M 
%HA ionized  100% 
  0.49%
 0.75 M 
14.29 Examine the chemical formula to determine the nature of a compound: (a) weak
base; (b) weak acid; (c) weak acid; and (d) strong base.
14.31 Identify each substance using the standard colour code. The first is C2H5CO2H,
propanoic acid, a carboxylic acid. The second is H3PO4, phosphoric acid. The third is
CH3NH2, methyl amine, a base. Draw the conjugate base of an acid by removing an
acidic H+, leaving a negative charge. Draw the conjugate acid of a base by adding H+
to an atom that has a lone pair:
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14.33 The conjugate base will have one less H and one lower charge; a conjugate acid will
have one more H and one higher charge:
C5 H5 N conjugate acid is C5H5 NH+
HONH2 conjugate acid is HONH3+
HCO2 H conjugate base is HCO2 
14.35 Conjugate pairs are connected. Any aqueous solution has OH– and H3O+ ions with
the equilibrium:
(a) NH3 is a weak base
(b) HCNO is a weak acid
(c) HClO is a weak acid
(d) Ba(OH)2 is a strong base
14.37 Follow standard procedures for dealing with equilibrium problems:
(a) The compound is a salt, so major species are Na+, SO32–, and H2O.
(b) The species with acid–base properties are SO32– (a weak base) and H2O, so the
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Chapter 14
dominant equilibrium is H 2O (l ) + SO32 (aq)
HSO3 (aq) + OH  (aq).
(c) Construct a concentration table, write the equilibrium expression, and carry out an
equilibrium calculation to determine [OH–]:
Reaction:
HSO3– +
SO32–
H2O +
OH–
Initial concentration (M)
0.45
0
0
Change in concentration (M)
–x
+x
+x
0.45 – x
x
x
Final concentration (M)
The equilibrium reaction is a weak base proton transfer. HSO3– is the species
resulting from the gain of a proton by SO32–, so use Ka2 to evaluate Kb:
K a2  6.3  108
K b  1.6  107 
Kb 
[HSO3 ]eq [OH  ]eq
[SO32 ]eq
K W 1.0  1014

 1.6  10 7
8
K a 2 6.3  10
=
x2
; assume x <<0.45:
0.45  x
x2
0.45
2
x  7.2  108 , so x  2.7  10 4 ; the assumption is valid.
1.6  107 
[OH  ]  2.7  104 M
pOH  log(2.7  104 )  3.57
pH  14.00  3.57  10.43
14.39 Follow standard procedures for dealing with equilibrium problems:
(a) The compound is a salt, so the major species are NH4+, NO3–, and H2O.
(b) The species with acid–base properties are NH4+ (a weak acid) and H2O, so the
dominant equilibrium is H2O (l) + NH4+ (aq)
NH3 (aq) + H3O+ (aq).
(c) Carry out an equilibrium calculation to determine [H3O+]:
Reaction:
H2O +
Initial concentration (M)
Change in concentration (M)
Final concentration (M)
NH4+
NH3
0.0100
–x
0.0100 – x
+
H3O+
0
0
+x
+x
x
x
The equilibrium reaction is a weak acid proton transfer. NH3 is the species resulting
from the loss of a proton from NH4+ so use Kb to determine Ka:
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K b  1.5  105
K a  5.6  1010 
Ka 
Chapter 14
K W 1.0  1014

 5.6  1010
K b 1.8  105
[NH 4  ]eq [H3O  ]eq
[NH3 ]eq
=
x2
; assume that x <<0.0100:
0.0100  x
5.6 x 1010 
x 2  5.6  1012 , so x  2.37  106 ; the assumption is valid.
[H3O  ]  2.37  106 M
pH  log(2.37  106 )  5.63
14.41 To determine the acid–base properties of a salt solution, examine the species for
their acidic or basic character:
(a) Species are H2O (acid and base), Na+ (neither), and HS– (conjugate base of H2S);
the solution is basic with pH determined by
H 2 O (l )  HS (aq)
OH  (aq)  H 2S (aq)
(b) Species are H2O (acid and base), Na+ (neither), and OI– (conjugate base of HOI);
the solution is basic with pH determined by
H 2O (l )  OI  (aq)
OH  (aq)  HOI (aq)
(c) Species are H2O (acid and base), Li+ (neither), and ClO4– (conjugate base of a
strong acid, thus neither); the solution is neutral, with pH determined by the water
equilibrium:
H 2 O (l )  H 2 O (l )
OH  (aq)  H 3O  (aq)
(d) Species are H2O (acid and base), HC5H5N+ (conjugate acid of pyridine, a weak
base), and Cl– (conjugate base of a strong acid, thus neither); the solution is acidic
with pH determined by
H 2O (l )  HC5H 5 N  (aq)
H 3O  (aq)  C5H 5 N (aq)
14.43 The pH of an aqueous solution of a salt is determined by the acid–base
characteristics of the cation and anion. Because Na+ has no acid–base tendencies, the
anions in these compounds determine the pH of their solutions. Solution pH increases
with the strength of the basic anion, which in turn is inversely proportional to the
strength of the parent weak acid. Here is the order for these compounds:
NaI (neutral) < NaF (Ka = 6.3 × 10–4) < NaC6H5CO2 (Ka = 6.3 × 10–5)
< Na3PO4 (Ka3 = 4.8 × 10–13) < NaOH (strong base)
14.45 (a) H2SO4 is stronger because anions are poorer proton donors than neutral species.
(b) HClO is stronger because Cl is a more electronegative atom than I. A higher
electronegativity means that Cl attracts more of the electron density around it than I,
weakening the H–X bond and making it easier to break (hence a better proton donor).
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0.0100
Chemistry, Student Solutions Manual
Chapter 14
(c) HClO2 is stronger. O atoms are highly electronegative and attract electron density
around them. Having two O atoms, HClO2 will have less electron density in the H–O
bond than HClO, and thus making the bond easier to break.
14.47 Use arrows of different sizes to show differences in electron density shifts.
14.49 The phenolate anion can be drawn with several resonance structures, placing the
negative charge at different locations around the benzene ring instead of on the
oxygen atom:
Acid strength increases as the stability of the conjugate base increases. The Lewis
structures show that the phenolate anion can distribute its negative charge around the
benzene ring, increasing its stability compared with that of the localized O– that
results from removal of a proton from ethanol. That is why phenol is a weak acid,
whereas alcohols such as ethanol are not acidic.
14.51 This problem asks for the concentration of all ionic species present in the solution.
Begin by analyzing the chemistry. NaC2H3O2 is a salt that dissolves in solution to
form Na+ and C2H3O2–. The acetate anion is a weak base:
H 2O  C2 H 3O 2
C2 H 3O 2 H  OH 
K b  5.6  10 10
Every aqueous solution has the water equilibrium:
H 2O  H 2O
H3O   OH 
K w  1.0  1014
Thus, the ionic species present in the solution are Na+, C2H3O2–, H3O+, and OH–. Na+
is a spectator ion and will have the same concentration as the initial salt:
[Na+] = 0.250 M
For the remaining ions, set up concentration tables, write the equilibrium
expressions, and solve for the ionic concentrations.
Reaction:
H2O +
Initial concentration (M)
Change in concentration (M)
© John Wiley and Sons Canada, Ltd.
C2H3O2H +
OH–
0.250
0
0
–x
+x
+x
C2H3O2–
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Chapter 14
0.250 – x
Final concentration (M)
K b  6.5  1010 
x
x
x2
; assume that x << 0.250:
0.250  x
x2
5.6  10 
0.250
2
10
x  1.4  10 , so x  1.2  105 ; the assumption is valid.
10
[C2 H 3O 2 H]  [OH  ]  1.2  105 M
[C2 H 3O 2  ]  (0.250 M)  (1.2  105 M)  0.250 M
Use a second concentration table to determine the concentrations of hydronium ions:
H2O
H3O+ +
OH–
Initial concentration (M)
Solvent
0
1.2 × 10–5
Change in concentration (M)
Solvent
+x
+x
Final concentration (M)
Solvent
x
1.2 × 10–5 + x
Reaction: H2O
+
K w  1.00  1014  (x)(1.2  105  x); assume that x << 1.2  105 :
1.00  1014  (x)(1.2  105 )
x  8.3  1010 M  [H 3O  ]; the assumption is valid.
[OH  ]  1.2  105 M
The ionic concentrations are
[Na  ]  [C2 H 3O 2  ]  0.250 M
[OH  ]  1.2  105 M
[H3O+ ]  8.3  1010 M
14.53 This problem asks for the concentration of all ionic species present in the solution.
Begin by analyzing the chemistry. H2CO3 is a diprotic acid with equilibria:
H 2O  H 2CO3
HCO3  H 3O 
K a1  4.5  107
H 2O  HCO3
CO32  H 3O 
K a2  4.7  1011
Every aqueous solution has the water equilibrium:
H 2O  H 2O
H3O   OH 
K w  1.00  1014
Thus, the ionic species present in the solution are HCO3–, CO32–, H3O+, and OH–.
Set up concentration tables, write equilibrium expressions, and solve for the ionic
concentrations.
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Chemistry, Student Solutions Manual
Reaction: H2O +
Chapter 14
HCO3– +
H2CO3
1.55 × 10–2
0
0
–x
+x
+x
1.55 × 10–2 – x
x
x
Initial concentration (M)
Change in concentration (M)
Final concentration (M)
H3O+
x2
K a1  4.5  10 
; assume that x << 1.55  10 2 :
2
1.55  10  x
x2
4.5  107 
1.55  102
x 2  6.98  109 , so x  8.4  10 5 ; the assumption is valid.
7
[H 3O  ]  [HCO3 ]  8.4  10 5 M
[H 2CO 3 ]  1.55  102 M
Set up a concentration table for the second equilibrium:
Reaction: H2O +
HCO3–
H3O+
8.4 × 10–5
0
8.4
–x
+x
+x
8.4 × 10–5 – x
x
8.4 + x
Initial concentration (M)
Change in concentration (M)
Final concentration (M)
CO32– +
x(8.4  105  x)
; assume that x << 8.4  10 5:
5
8.4  10  x
M  [CO32 ]
K a2  4.7  1011 
x  4.7  1011
Use Kw to determine the concentration of hydroxide ions:
K w  1.00  1014  (8.4  105 )[OH  ]
[OH  ]  1.2  1010 M
Ionic concentrations:
[H3O ]  [HCO3 ]  8.4  105 M
[CO32 ]  4.7  1011 M
[OH ]  1.2  1010 M
14.55 In order to determine concentrations of all ions, we need to consider more than one
equilibrium. This is done in stages, starting with the dominant equilibrium. The
problem asks us for the concentrations of the ions in aqueous sodium carbonate, in
which the major species are Na+, CO32–, and H2O. The sodium ion is a spectator ion.
The carbonate anion undergoes proton transfer with equilibrium constant Kb2:
© John Wiley and Sons Canada, Ltd.
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Chemistry, Student Solutions Manual
H2O (l )  CO3
2
(aq)
Chapter 14


OH (aq)  HCO3 (aq)
Kb2 =
[OH ]eq [HCO3 ]eq
[CO32  ]eq
Table 14-5 provides the value of the acid equilibrium constant: K a 2  4.7  1011
K b2 =
K w 1.0  1014

 2.1  104
11
K a2 4.7  10
This is much larger than Kw, so this is the dominant equilibrium, which we use to
begin our calculations.
Because carbonic acid is diprotic, a second proton transfer equilibrium has an effect
on the ion concentrations, and the water equilibrium also plays a secondary role. Now
we are ready to organize the data and the unknowns and do the calculations. The
spectator ion is the easiest to deal with: [Na+] = 2 [CO32–] = 2(0.055 M) = 0.11 M.
There are multiple equilibria affecting ion concentrations, so we must work with
more than one concentration table, starting with the dominant equilibrium. Set up a
concentration table to determine concentrations of the ions generated by this
reaction:
Reaction: H2O
+
OH– +
HCO3−
0.055
0
0
–x
+x
+x
0.055 − x
x
x
CO32–
Initial concentration (M)
Change in concentration (M)
Final concentration (M)
Substitute the equilibrium concentrations into the equilibrium constant expression
and solve for x. We cannot make an approximation, so use the quadratic formula:
x2
0.055  x
2
4
x = (2.1  10 )(0.055  x) = (1.16  10 5   (2.1  104 x)
K b2 = 2.1  104 =
0 = x 2 + (2.1  10 4 x)  (1.16  10 5 )
b ± b 2  4ac (2.1  104 ) ± (2.1  10 4 ) 2  4(  1.16  10 5 )
x=
=
= 3.2  103
2a
2
[OH−] = [HCO3−] = 3.2 × 10−3 M, and [CO32−] = 0.055 − 0.0032 = 0.052 M
Take into account the proton transfer equilibrium involving HCO3– with a second
concentration table, using concentrations calculated for the first equilibrium:
Reaction: H2O +
© John Wiley and Sons Canada, Ltd.
HCO3–
OH– +
H2CO3
263
Chemistry, Student Solutions Manual
Initial concentration (M)
Change in concentration (M)
Final concentration (M)
Chapter 14
3.2 × 10–3
3.2 × 10–3
0
–x
+x
+x
3.2 × 10–3 – x
3.2 × 10–3 + x
x
Substitute equilibrium concentrations into the equilibrium constant expression and
solve for x, making the approximation that x << 3.2 × 10–3.
[OH  ]eq [H 2CO3 ]eq
K w 1.0  1014
8
K b1 =
=
= 2.2  10 
K a1 4.5  107
[HCO3 ]
2.2  108 =
(x)(3.2  103  x) (x)(3.2  103 )

=x
3.2  103  x
3.2  103
This value is too small to cause a measurable change in the concentrations already
calculated, but it does tell us the concentration of carbonic acid in the solution:
[H2CO3] = 2.2 × 10–8 M
One more ion remains, H3O+, generated from the water equilibrium. Apply the water
equilibrium expression directly:
[H3O+ ] =
Kw
(1.0  1014 )
=
= 3.1  1012 M

3
[OH ] (3.2  10 )
14.57 (a) An acid–base equilibrium reaction involves proton transfer, in this case from
boric acid to water:
(b) To calculate the pH of a solution, follow the standard procedure for equilibrium
calculations:
Reaction: H2O +
H3BO3
Final concentration (M)
© John Wiley and Sons Canada, Ltd.
H3O+
3.2 × 10–3
0
–x
+x
+x
0.050 − x
x
x
Initial concentration (M)
Change in concentration (M)
H2BO3– +
264
Chemistry, Student Solutions Manual
K a = 5.4  10
10
=
Chapter 14
[H 2 BO3 ]eq [H 3O + ]eq
[H 3BO3 ]eq
=
x2
; assume that x << 0.050:
0.050  x
x2
5.4  10 =
0.050
2
11
x = 2.7  10 , so x = 5.2  10 6 ; the assumption is valid.
10
[H 3O + ] = 5.2  106 M
pH =  log(5.2  106 ) = 5.28
14.59 There is much interesting chemical information provided in the statement of this
problem, but the calculation is a straightforward equilibrium determination for a
solution of a weak base. Follow the standard procedures to determine the pH. Begin
by constructing a concentration table, write the equilibrium expression, and solve for
the concentration of hydroxide ions.
Reaction: H2O
+
LSDH+ +
LSD
OH–
Initial concentration (M)
0.55
0
0
Change in concentration (M)
–x
+x
+x
0.55 − x
x
x
Final concentration (M)
K b = 7.6  10
7
=
[LSDH + ]eq [OH  ]eq
[LSD]eq
=
x2
; assume that x << 0.55:
0.55  x
x2
7.6  10 =
0.55
2
7
x = 4.2  10 , so x = 6.5  10 4 ; the assumption is valid.
7
[OH  ] = 6.5  104 M
pOH =  log(6.5  104 ) = 3.19
pH = 14.00  3.19 = 10.81
14.61 Follow the standard procedure for dealing with equilibrium calculations: the major
species are Na+, A–, and H2O, and the acid–base equilibrium is
A  (aq) + H 2O (l )
HA (aq) + OH  (aq)
Set up a concentration table and use it to determine Kb:
pOH = 14.00  pH = 3.00
Reaction: H2O
+
Initial concentration (M)
Change in concentration (M)
© John Wiley and Sons Canada, Ltd.
[OH  ] = 103.00 = 1.00  103 M
A–
HA +
OH–
0.0100
0
0
– 0.00100
+ 0.00100
+ 0.00100
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Chemistry, Student Solutions Manual
Chapter 14
Final concentration (M)
Kb =
Ka =
[HA]eq [OH  ]eq
[A  ]eq
0.0090
=
0.00100
0.00100
(1.00  103 ) 2
= 1.1  10 4
0.0090
K w 1.00  1014
=
= 9.1  10 11
K b 1.1  104
14.63 Determine the percent ionization using Equation 14-3 in your textbook:
%HA ionized = 100%
[H 3O + ]eq
[HA]initial
The equilibrium concentration of hydronium ions was calculated in Problem 14.57:
[H3O+] = 5.2 × 10–6 M, and the initial concentration of the weak acid is given in the
problem, 0.050 M:
 5.2  106 M 
%HA ionized = 100% 
 = 0.010%
 0.050 M 
14.65 There are two amine groups, one at either end of the molecule, each of which can
accept a proton from a water molecule. Convert the line structure to a Lewis structure
using the standard procedures, and then show the transfer of one proton to each N
atom:
14.67 Follow the standard procedure for solving equilibrium problems:
1. Major species: H2O, Na+, and F–
2. Dominant acid–base equilibrium: H 2O + F
3. From Table 14-2, Ka = 6.3  104 ; K b =
© John Wiley and Sons Canada, Ltd.
HF + OH 
K w 1.0  1014
=
= 1.6  1011
4
Ka 6.3  10
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Chemistry, Student Solutions Manual
Chapter 14
4. Set up a concentration table:
Reaction: H2O +
F–
Initial concentration (M)
0.250
Change in concentration (M)
–x
Final concentration (M)
0.250 – x
Kb = 1.6  10
11
=
[HF]eq [OH ]eq
[F ]eq
OH–
HF +
0
0
+x
+x
x
x
x2
=
; assume that x << 0.250:
0.250  x
x2
0.250
2
x = 4.0  1012 , so x = 2.0  10 6 ; the assumption is valid.
1.6  1011 =
[OH ] = 2.0  106 M
pOH =  log(2.0  106 ) = 5.70
pH = 14.00  5.70 = 8.30
14.69 To determine concentrations of species in a solution, follow the standard procedure:
1. This is a strong acid. Major species are H2O, H3O+, and HSO4–.
2. The dominant acid–base equilibrium is H 2O + HSO4 
SO4 2 + H3O + .
3. From Table 14-5, Ka2 = 1.0 × 10–2.
4. In this solution, there are hydronium ions from the strong acid present initially:
HSO4–
Reaction: H2O +
2.00
Initial concentration (M)
–x
Change in concentration (M)
2.00 – x
Final concentration (M)
K a2 = 1.0  10
1.0  102 =
2
=
[SO 4 2 ]eq [H 3O + ]eq

[HSO 4 ]eq
=
SO42– +
H3O+
0
2.00
+x
+x
x
2.00 + x
x(2.00 + x)
; assume that x << 2.00:
2.00  x
(2.00)(x)
, so x = 1.0  102 ; the assumption is valid.
(2.00)
[H 3O + ] = 2.00 + 0.010 = 2.01 M
[SO 4 2 ] = 1.0  102 M
[HSO 4  ] = 2.00  x = 1.99 M
14.71 (a) H2SO4 is a strong acid, so the major species in solution are H2O, HSO4–, and
H3O+. The hydrogen sulphate ion is a weak acid, so the equilibrium reaction that
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Chapter 14
determines the pH is H 2O + HSO 4 
SO 4 2 + H3O + .
(b) Na2SO4 is a salt, so the major species in solution are H2O, SO42–, and Na+. The
sulphate ion is the conjugate base of a weak acid, so the equilibrium reaction that
HSO 4  + OH  .
determines pH is SO 4 2 + H 2O
(c) NaHSO4 is a salt, so the major species in solution are H2O, HSO4–, and Na+. The
hydrogen sulphate ion is a weak acid, so the equilibrium reaction that determines the
SO 4 2 + H 3O + .
pH is H 2 O + HSO 4 
(d) NH4Cl is a salt, so the major species in solution are H2O, Cl–, and NH4+. The
ammonium ion is the conjugate acid of a weak base, so the equilibrium reaction that
NH 3 + H3O + .
determines pH is H 2O + NH 4 +
14.73 Tabulated equilibrium constants for acid–base reactions always refer to reactions in
which H2O is one of the reactants. The reaction in this problem is the reverse of a
base reaction:
HPO4 2 (aq) + OH  (aq)
PO 43 (aq) + H 2O (l )
Table 14-5 lists Ka values for phosphoric acid:
HPO4 2 (aq) + H 2 O (l )
PO43 (aq) + H3O+ (aq)
K a3 = 4.8  1013
Ka and Kb for a conjugate acid–base pair are related through Ka Kb = Kw:
1.0  10 14
Kb =
= 2.1  10 2
13
4.8  10
1
K eq =
= 48
Kb
14.75 Molecular pictures must show the correct relative numbers of the various species in
the solution. From the starting condition (six molecules of oxalic acid), make
appropriate changes and then draw new pictures:
(a) Hydroxide ions react with oxalic acid to form water and hydrogen oxalate ions:
H 2C2O4 + OH 
H 2O + HC2O 4  . The picture shows two molecules of oxalic
acid and four each of water and hydrogen oxalate:
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Chemistry, Student Solutions Manual
Chapter 14
(b) When all oxalic acid has reacted, hydroxide ions react with hydrogen oxalate ions
H 2 O + C2O 4 2 . The picture
to form water and oxalate ions: HC2 O 4  + OH 
shows four hydrogen oxalate ions, eight water molecules, and two oxalate ions:
(c) NH3, a weak base, accepts a proton from oxalic acid, a weak acid:
H 2 C2O 4 + NH 3
NH 4 + + HC2O 4 
The picture shows two oxalic acid molecules and four each of ammonium and
hydrogen oxalate ions:
14.77 The chemical reaction that occurs is
P4O10 + 6 H2O → 4 H3PO4
(a) The major species present are H2O and H3PO4.
(b) To determine the ranking of the minor species, consider what reactions generate
them. H3PO4 undergoes proton transfer with water to form H2PO4– and H3O+ in equal
concentrations, but H2PO4– undergoes further proton transfer with water to form
HPO42– and H3O+. HPO42–, in turn, generates a tiny amount of PO43–, and the water
equilibrium generates a tiny amount of OH–. The minor species are (in order of
highest concentration to lowest concentration): H3O+, H2PO4–, HPO42–, OH– , and
PO43–.
(c) The dominant equilibrium that determines the pH is
H3PO 4 + H 2 O
H 2 PO 4  + H3O +
Set up a concentration table, solve for hydronium ion concentration, and then
calculate the pH. Determine the initial concentration using standard
stoichiometric procedures:
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Chemistry, Student Solutions Manual
Chapter 14
 1 mol   4 mol H 3PO 4 
nH3PO4 = 3.5 g P4O10 
 = 0.0493 mol

 283.88 g   1 mol P4O10 
0.0493 mol
[H3PO 4 ] =
= 0.033 M
1.50 L
Reaction: H2O +
H3PO4
Initial concentration (M)
H2PO4− +
H3O+
0
0
0.033
Change in concentration (M)
–x
0.033 − x
Final concentration (M)
+x
+x
x
x
Now substitute into the equilibrium constant expression and solve for x:
K a1 = 0.0069 =
[H 2 PO 4  ]eq [H3O + ]eq
[H3 PO 4 ]eq
x = [H3O + ] = 0.012 M
x2

; solve by the quadratic equation.
0.033  x
pH =  log(0.012) = 1.92
14.79 Proton transfer occurs from the carboxylic acid O–H (shown screened below) and
the amino nitrogen atom:
14.81 Net ionic equations show only the reacting species. Remember that strong acids
generate H3O+ in solution and react to completion with weak bases, and strong bases
generate OH– in solution and react to completion with weak acids:
(a) Strong base reacting with weak acid: OH− + C6H5CO2H
(b) Strong acid reacting with weak base: H3O+ + (CH3)3N
H2O + C6H5CO2
H2O + (CH3)3NH+
(c) Weak base reacting with weak acid: SO42− + CH3CO2H
HSO4– + CH3CO2−
HSO4–, pKa = 1.99; CH3CO2H, pKa = 4.75; HSO4– is stronger, so this reaction
proceeds to a small extent.
(d) Strong base reacting with weak acid: OH− + NH4+
(e) Weak base reacting with weak acid: HPO42− + NH3
H2O +NH3
PO43− + NH4+
HPO42–, pKa = 12.32; NH4+, pKa = 9.25; NH4+ is stronger, so this reaction proceeds to
a small extent.
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Chemistry, Student Solutions Manual
Chapter 14
14.83 Follow the standard procedure for dealing with equilibrium calculations:
(a) The major species are Na+, HCO3–, and H2O.
There are two equilibria involving major species:
HCO3  aq  + H 2O  l 
CO32  aq  + H3O+  aq  pK eq  pK a 2  10.33
HCO3  aq  + H 2O  l 
H 2CO3  aq  + OH   aq 
pK b1  pK w  pKa1  14.00  6.35  7.65
The equilibrium with the larger Keq (smaller pKeq) dominates, making this solution
basic.
(b) Set up a concentration table, solve for hydroxide ion concentration, and then
calculate the pH. Determine the initial concentration using standard stoichiometric
procedures:
n 0.0228 mol
 HCO3  = =
= 0.152 M
V
0.150 L
Reaction: H2O +
HCO3
H2CO3 +
OH–
Initial concentration (M)
0.152
0
0
Change in concentration (M)
–x
+x
+x
Final concentration (M)
0.152 – x
x
x
Now substitute into the equilibrium constant expression and solve for x:
Keq = 10–7.65 = 2.2 × 10–8
 H 2CO3 eq OH   eq
x2
8
2.2  10 =
=
; assume that x << 0.152:
0.152  x
 HCO3 
eq
x 2 = 3.3  109 , from which x = 5.8  105 ; the assumption is valid.
[OH–] = 5.8 × 10–5 M
pOH  log  5.8  105  = 4.24
pH = 14.00  4.24 = 9.76
14.85 (a) HBr is a strong acid, so when this gas bubbles through water, it generates
hydronium ions. The major species are H2O, H3O+, Br–, Ca2+, and OH–, and the
reaction that goes to completion is H3O+ + OH–
2O.
(b) The major species are H2O, Na+, HSO4–, and OH–, and the reaction that goes to
2–.
completion is HSO4– + OH–
2O + SO4
+ –
2+
(c) The major species are H2O, NH4 , I , Pb , and NO3–, and the reaction that goes to
completion is formation of PbI2 precipitate: Pb2+ + 2 I–
2 (s).
14.87 (a) H 2SO 2 (l )  H 2SO 2 (l )
(b) Lewis structure:
© John Wiley and Sons Canada, Ltd.
H 3SO 4  (solvated) + HSO 4  (solvated)
271
Chemistry, Student Solutions Manual
Chapter 14
Either doubly-bonded oxygen atom can accept a proton.
(c) H 2SO2 (l )  HClO 4 (l )
© John Wiley and Sons Canada, Ltd.
H 3SO 2  (solvated)  ClO 4  (solvated)
272