fied. The prediction based upon the empirical definition of a buffer was accurate. It appears that we can continue to
have confidence in this concept.
CHAPTER 8 SUMMARY
MAKE A SUMMARY
(Page 630)
(Answers may vary, but should include a page for each of the following six sections: The Nature of Acid–Base Equilibria;
Weak Acids and Bases; Acid–Base Properties of Salt Solutions; Acid–Base Titration; Buffers; and The Science of Acid
Deposition.)
CHAPTER 8 SELF-QUIZ
(Page 631)
1. False. The stronger a Brønsted–Lowry acid is, the weaker its conjugate base.
2. False. Group I metal ions produce neutral solutions.
3. True
4. True
5. False. A solution of the bicarbonate ion is basic.
6. False. The pH of water would be less than 7.
7. True
8. False. Most dyes that act as acid–base indicators are weak acids.
9. True
10. (b)
11. (b)
12. (e)
13. (a)
14. (b)
15. (c)
16. (e)
17. (a)
18. (b)
19. (a)
CHAPTER 8 REVIEW
(Page 632)
Understanding Concepts
8.50 g
1. nNaOH 40.00 g/mol
nNaOH 0.2125 mol
(extra digits carried)
0.2125 mol
[OH–] 0.500 L
[OH–] 0.425 mol/L
pOH –log 0.425
pOH –0.372
The pOH of sodium hydroxide is –0.372.
342
Chapter 8
Copyright © 2003 Nelson
3.05 104 g
2. (a) nHCl 36.46 g/mol
nHCl 836.5 mol
(extra digits carried)
836.5 mol
[H] 806 L
[H] 1.038 mol/L
pH –log 1.038
pH 0.016
pOH 14 – 0.016
pOH 13.984
The pH and pOH of the hydrochloric acid are 0.016 and 13.984, respectively.
(b) Assumptions are that the temperature remains constant, and that water does not contribute a significant amount
–
of H
(aq) or OH(aq).
H
–log[H]
pH
3. H2O e
OH–
–log[OH–]
pOH
Kw
–log Kw
14
–
4. HF(aq) e H
(aq) F(aq)
–
[H
(aq)][F(aq)]
Ka [HF(aq)]
ICE Table for the Dissociation of Hydrofluoric Acid
HF(aq) e
H+(aq)
–
+ F(aq)
Initial concentration (mol/L)
2.0
0.0
0.0
Change in concentration (mol/L)
–x
+x
+x
Equilibrium concentration (mol/L)
2.0 – x
x
x
–
[H
(aq)][F(aq)]
Ka [HF(aq)]
Ka 6.6 10–4
Predict whether a simplifying assumption is justified:
[HA]initial
2.0
Ka
6.6 10–4
[HA]initial
3000
Ka
Since 3000 > 100, we may assume that 2.0 – x 2.0.
The equilibrium expression becomes
x2
6.6 10–4
2.00
x 0.036
The 5% rule justifies the assumption that 2.0 – x 2.0.
[H
(aq)]
[F–(aq)] 0.036 mol/L
The concentrations of hydrogen and fluoride in a 2.0 mol/L hydrofluoric acid solution are 0.036 mol/L.
Copyright © 2003 Nelson
Acid–Base Equilibrium
343
5.
Kw
[H] [OH–]
1.0 10–14
2.5 10–7
[H] 4.0 10–8 mol/L
pH –log 4.0 10–8
pH 7.40
The hydrogen ion concentration and pH of blood are 4.0 10–8 mol/L and 7.40, respectively.
60 000 g
6. nHC H O 2 3 2(aq)
60.06 g/mol
nHC H O
999 mol
CHC H O
999 mol
1250 L
CHC H O
0.799 mol/L
2 3 2(aq)
2 3 2(aq)
2 3 2(aq)
ICE Table for the Dissociation of Acetic Acid
HC2H3O2(aq) e
+
H(aq)
+
C2H3O2–(aq)
0.799
0.0
0.0
Change in concentration (mol/L)
–x
+x
+x
Equilibrium concentration (mol/L)
0.799 – x
x
x
Initial concentration (mol/L)
–
[H
(aq)][C2H3O2 (aq)]
1.8 10–5
[HC2H3O2(aq)]
x2
1.8 10–5
0.799 – x
Predicting whether 0.799 – x 0.799 – x …
[HA]initial
0.799 mol/L
Ka
1.8 10–5
[HA]initial
44400
Ka
Since 44400 > 100, we assume that 0.799 – x 0.799
x2
1.8 10–5
0.799
x 1.8 10–5
(extra digits carried)
The 5% rule justifies the assumption.
[H
(aq)]
3.795 10–3 mol/L
pH – log [H
(aq)]
– log [3.795 10–3]
pH 2.421
pOH 14 – 2.421
pOH 11.579
The pH and pOH of the acetic acid solution are 2.421 and 11.579.
344
Chapter 8
Copyright © 2003 Nelson
7. (a)
ICE Table for the Ionization of ASA
HC10H7CO4(aq) e
+
H(aq)
+
–
C10H7CO4(aq)
0.018
0.0
0.0
Change in concentration (mol/L)
–x
+x
+x
Equilibrium concentration (mol/L)
0.018 – x
x
x
Initial concentration (mol/L)
–
[H
(aq)][C10H7CO4 (aq)]
3.27 10–4
[HC10H7CO4(aq)]
x2
3.27 10–4
0.018 – x
Predicting whether 0.018 – x 0.018 …
[HA]initial
0.018 mol/L
3.27 10–4
Ka
[HA]initial
55
Ka
Since 55 < 100, we cannot assume that 0.018 – x 0.018.
x2
3.27 10–4
0.018
x2
3.27 10–4
0.018 – x
(0.018 – x)(3.27 10–4) x2
5.89 10–6 – 3.27 10–4x x2
x2 3.27 10–4x – 5.89 10–6 0
10–4)2(–5.89
– 4(1)
10–6)
–3.27 10–4 (3.27
x 2(1)
–3.27 10–4 4.86 10–3
2
x 2.27 10–3 or –8.13 10–4
Since negative concentrations are meaningless,
x 2.27 10–3
–3
[H
(aq)] 2.27 10 mol/L
pH – log[H
(aq)]
pH – log[2.27 10–3]
pH 2.644
The pH of the ASA solution is 2.644.
(b) An increase in temperature shifts the autoionization equilibrium of water to the right, producing more H
(aq). As
a result, the pH of the solution would decrease.
8. (a)
7.9 10–3%
–
HCN(aq) e H
(aq) CN(aq)
–
[H
(aq)][CN(aq)]
6.2 10–10
[HCN(aq)]
Copyright © 2003 Nelson
Acid–Base Equilibrium
345
–
(b) HCN(aq) e H
(aq) CN(aq)
[H(aq)][CN–(aq)]
6.2 10–10
[HCN(aq)]
x2
6.2 10–10
0.10 – x
Predicting whether 0.10 – x 0.10 …
[HA]initial
0.10 mol/L
Ka
6.2 10–10
[HA]initial
109
Ka
Since 109 > 100, we assume that 0.10 – x 0.10.
x2
6.2 10–10
(0.10)
x 7.9 10–6
The 5% rule justifies the assumption.
[H
(aq)]
7.9 10–6 mol/L
pH – log[H
(aq)]
– log[7.9 10–6]
pH 5.10
The pH of the hydrocyanic acid is 5.10.
Calculating percent ionization …
p
[H
[H(aq)]
(aq)] 100
[H
(aq)]
p 100%
[H(aq)]
7.9 10–6 mol/L
100%
0.100 mol/L
7.9 10–3 %
The percent ionization of hydrocyanic acid is 7.9 10–3 %.
– H O
–
9. HC6H6O2(aq)
2 (l) e H2C6H6O2(aq) OH(aq)
pH 8.65
pOH 5.35
–5.35
[OH
(aq)] 10
4.47 10–6 mol/L (extra digits carried)
[OH
(aq)] [H2C6H6O2(aq)]
ICE Table for the Ionization of Ascorbate
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
346
Chapter 8
HC6H6O2–(aq) +
H2O(l) e
H2C6H6O2(aq) +
–
OH(aq)
0.15
–
0.00
0.00
–4.47 10–6
0.15 – (4.47 10–6)
–
4.47 10–6
4.47 10–6
–
4.47 10–6
4.47 10–6
Copyright © 2003 Nelson
– ]
[H2C6H6O2(aq)][OH(aq)
Kb
– ]
[HC6H6O2 (aq)
(4.47 10–6)2
Kb
0.15 – 4.47 10–6
Assume that 0.15 – 4.47 10–6 0.15.
(4.47 10–6)2
1.3 10–10
0.15
Kb for the ascorbate ion is 1.3 10–10.
10. (a)
acid
conjugate base
acid
conjugate base
(C6H5)3CH
(C6H5)C–
C4H4NH
C4H4N–
HC2H3O2
C2H3O2–
H2S
HS–
(C6H5)3CH
(C6H5)C–
OH–
O2–
C4H4NH
C4H4N–
H2S
HS–
(b) (strongest acid) HC2H3O2; H2S; C4H4NH; (C6H5)3CH (weakest acid)
11. The final solution is basic because of the hydrolysis of the methanoate ion:
– H O e HCH O
–
CH3O(aq)
2 (l)
3 (aq) OH(aq)
12.
Ionic
Molecular
neutral
acidic
basic
neutral
acidic
basic
CaCl2
NH4Cl
NaOH
C6H12O6
HCl
NH3
NaNO3
Al(NO3)3
Na2CO3
O2
H2S
N2H4
13. (a) yellow
(b) red
(c) yellow-green
(d) colourless
(e) yellow
14. H3PO4(aq) 2 NaOH(aq) → Na2HPO4(aq) 2 H2O(l)
25.0 mL
C
17.9 mL
1.5 mol/L
nNaOH
1.50 mol
17.9 mL 1L
nNaOH
26.9 mmol
(aq)
(aq)
1
26.9 mmol 13.4 mmol
2
13.4 mmol
CH PO 3
4(aq)
25.0 mL
nH PO
3
4(aq)
CH PO
3
4(aq)
0.537 mol/L
The concentration of the phosphoric acid solution is 0.537 mol/L.
15. (a) HCl(aq) NH3(aq) → H2O(l) NH4Cl(aq)
nNH
3(aq)
CNH
3(aq)
VNH
3(aq)
0.10 mol/L 10.0 mL
nNH
3(aq)
1.0 mmol
Copyright © 2003 Nelson
Acid–Base Equilibrium
347
Final solution volume 20.0 mL
– , NH , H O
Entities remaining in solution at the equivalence point: Cl(aq)
4 (aq)
2 (l)
– does not hydrolyze, the pH of the solution is determined by NH .
Since Cl(aq)
4(aq)
e NH
NH4(aq)
3(aq) H (aq)
is present at the equivalence point.
Since 1.0 mmol of NH3(aq) was present initially, 1.0 mmol of NH4(aq)
1.0 mmol
]
[NH4(aq)
20.0 mL
] 0.050 mol/L
[NH4(aq)
e NH
NH4(aq)
3(aq) H (aq)
[H
(aq)][NH3(aq)]
5.8 10–10
]
[NH4(a
q)
ICE Table for the Hydrolysis of NH4+(aq)
NH4+(aq) e
+
H(aq)
+
NH3(aq)
0.050
0.000
0.000
Change in concentration (mol/L)
–x
+x
+x
Equilibrium concentration (mol/L)
0.050 – x
x
x
Initial concentration (mol/L)
e NH
NH4(aq)
3(aq) H (aq)
[H
(aq)][NH3(aq)]
5.8 10–10
]
[NH4(a
q)
x2
5.8 10–10
0.050 – x
Applying the hundred rule …
[NH4(aq)]initial
0.050
> 100
Ka
5.8 10–10
The equilibrium simplifies to
x2
5.8 10–10
0.050
x 5.385 10–6
(extra digits carried)
The 5% rule verifies the assumption.
[H
(aq)]
5.385 10–6 mol/L
pH – log [5.385 10–6]
pH 5.27
The pH of the solution at the equivalence point is 5.27.
–
[NH4
(aq)][OH(aq)]
(b)
Kb [NH3(aq)]
Kb 1.8 10–5
x2
1.8 10–5
0.10 – x
348
Chapter 8
Copyright © 2003 Nelson
Predicting the validity of the assumption …
0.10
100
1.8 10–5
we many assume that 0.10 – x 0.10
The equilibrium expression becomes
x2
1.8 10–5
0.10
x 1.342 10–3
(extra digits carried)
The 5% rule verifies the simplification assumption.
– ] 1.342 10–3 mol/L
[OH(aq)
pOH – log[1.342 10–3]
pOH 2.872
pH 14 – pOH
14.0 – 2.872
pH 11.13
The initial pH of the ammonia solution is 11.13.
(c)
VHCl
5.00 mL
CHCl
0.100 mol/L
nHCl
VHCl
(aq)
(aq)
(aq)
(aq)
CHCl
(aq)
5.00 mL 0.100 mol/L
nHCl
0.50 mmol
VNH
10.00 mL
CNH
0.100 mol/L
nNH
VNH
nNH
10.00 mL 0.100 mol/L
nNH
1.00 mmol
nNH
remaining …
(aq)
3(aq)
3(aq)
3(aq)
3(aq)
3(aq)
3(aq)
3(aq)
CNH
3(aq)
nNH
1.00 mmol – 0.50 mmol
nNH
0.50 mmol
3(aq)
3(aq)
Total volume 5.00 mL 10.00 mL 15.00 mL
CNH
0.50 mmol
15.00 mL
CNH
0.0333 mol/L
3(aq)
3(aq)
(extra digits carried)
0.50 mmol
CNH 4 (aq)
15.00 mL
CNH 0.0333 mol/L
4 (aq)
(extra digits carried)
Let x represent the changes in concentration that occur as the system reestablishes equilibrium.
Copyright © 2003 Nelson
Acid–Base Equilibrium
349
ICE Table for the Ionization of Ammonia
NH3(aq)
+ H2O(l) e
–
OH(aq)
+
NH4+(aq)
0.0333
–
0.00
0.0333
Change in concentration (mol/L)
–x
–
+x
+x
Equilibrium concentration (mol/L)
0.0333 – x
–
x
0.0333 + x
Initial concentration (mol/L)
][OH– ]
[NH4(a
q)
(aq)
Kb
[NH3(aq)]
x(0.0333 x)
1.8 10–5
0.0333 – x
Applying the hundred rule …
[NH3(aq)]initial
0.0333
Ka
1.8 10–5
[NH3(aq)]initial
1850
Ka
Since 1850 > 100, we can assume that 0.0333 x 0.0333 and that 0.0333 – x 0.0333
The equilibrium simplifies to
x(0.0333)
1.8 10–5
0.0333
x 1.8 10–5
The 5% rule validates the assumption.
– ] 1.8 10–5 mol/L
[OH(aq)
pOH – log [1.8 10–5]
4.7445
(extra digits carried)
pH 9.26
The pH after adding 5.0 mL of HCl is 9.26.
–
(d) Entities at the equivalence point are: H2O(l), NH4
(aq), Cl(aq)
(e) The equivalence point is reached after 10.0 mL of HCl(aq) is added. The pH of this solution is 5.27. (See calculation in (a).)
(f)
nHCl
(aq)
VHCl
(aq)
CHCl
(aq)
15.0 mL 0.10 mol/L
nHCl
1.5 mmol
nNH
VNH
nNH
10.0 mL 0.10 mol/L
(aq)
3(aq)
3(aq)
3(aq)
(extra digits carried)
CNH
3(aq)
1.0 mmol
nHCl
(aq)
remaining 1.5 mmol 1.0 mmol
0.5 mmol
nHCl
(aq)
CHCl (aq)
VHCl
(aq)
0.5 mmol
25 mL
350
Chapter 8
Copyright © 2003 Nelson
CHCl
(aq)
0.020 mol/L
[H
(aq)] 0.020 mol/L
pH –log 0.020
pH 1.70
(g)
16. (a)
(b)
(c)
(d)
After 15.0 mL of acid is added, the pH of the solution is 1.70.
A suitable indicator for this titration would be methyl red (pH range 4.4–6.2).
NH4
NH2–
2 NH3 e NH4 NH2–
pNH 4
Titration Curve for Titrating NH2– with NH4+
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
5
10
15
20
35
25
30
Volume of added (mL)
40
45
50
17. The pH would be between 5.0 and 5.4. An estimate of the hydrogen ion concentration is 1 10–5 mol/L.
>50%
– HCO –
2–
18. OH(aq)
3(aq) e H2O(l) CO3(aq)
<50%
–
H3O
(aq) HCO3(aq) e H2O(l) H2CO3(aq)
Applying Inquiry Skills
19. (i) Test the solutions with a pH meter. The strongest bases have the largest pH values.
(ii) Test the solutions with universal indicator. Compare the indicator colour with a colour chart to determine the
approximate pH.
20. (a) (i)
neutral
(ii) basic
(iii) basic
(iv) <7
(v)
<7
(vi) basic
(vii) <7
(b) The predictions could be tested with a pH meter. (The meter should be carefully calibrated, and rinsed between
solutions.)
21. (Sample answer—there are many possible correct solutions to this problem.) If the solutions are tested with a pH
meter, and the pH values are ordered from smallest to largest, then the solutions are sulfuric acid, hydrochloric acid,
acetic acid, ethanediol, ammonia, sodium hydroxide, and barium hydroxide, respectively.
Copyright © 2003 Nelson
Acid–Base Equilibrium
351
Diagnostic Tests on the Unlabelled Solutions
Litmus
Conductivity
Acid/Base titration
Analysis
red
low
one volume
CH3COOH(aq)
blue
very high
two volumes
Ba(OH)2(aq)
blue
low
one volume
NH3(aq)
no change
none
not applicable
C2H4(OH)2(aq)
red
higher
two volumes
H2SO4(aq)
red
high
one volume
HCl(aq)
blue
high
one volume
NaOH(aq)
22. (a) There is no third pH endpoint and therefore, no third equivalence point in this titration.
(b) Removing the hydroxide ions by precipitation will cause the equilibrium to shift to the right, producing more
hydroxide ions.
(c) Hydrochloric acid is not a primary standard.
(d) Both reactants and products form basic solutions and litmus cannot be used to distinguish among basic solutions.
(e) Cobalt chloride paper is used in a diagnostic test for the production of water in a strong acid–strong base reaction.
(f) The strength of an acid is determined by titration.
Making Connections
23. (a) Mg(OH)2(s) 2 HCl e 2 H2O(l) MgCl2(aq)
(b) Alkalosis is based on the carbonic acid/bicarbonate equilibrium:
–
H2CO3(aq) e H
(aq) HCO3(aq)
During alkalosis, an increase in blood levels of the bicarbonate ion removes hydrogen ions, thereby increasing
blood pH. Symptoms of alkalosis include vomiting, headache, and nausea.
(c) Hydroxide-based antacids are very common.
Extension
24. H2SO4(aq) Ba(OH)2(aq) e 2 H2O(l) BaSO4(s)
Place an accurately measured volume of Ba(OH)2(aq) into a beaker and immerse the leads from a conductivity meter
in the solution. Measure and record the initial conductivity reading. Titrate with standardized H2SO4(aq) and record
the conductivity readings. A sudden drop in the conductivity serves as the endpoint of the titration. This drop in
conductivity corresponds to the removal of most of the ions through the production of water and insoluble barium
sulfate. When the H2SO4(aq) is added in excess, the conductivity increases due to the ions present in the acid.
–
25. H2SO4(aq) e H
(aq) HSO4(aq)
From the first ionization:
–2
Since sulfuric acid is a strong acid [H
(aq)] 1.0 10 mol/L
From the second ionization:
– e H SO 2–
HSO4(aq)
(aq)
4(aq)
–
[H
(aq)][HSO4(aq)]
Ka [HSO4(–aq)]
Ka 1.0 10–2
ICE Table for the Ionization of the Hydrogen Sulfate Ion
HSO4–(aq) e
+
H(aq)
+
2–
SO4(aq)
0.010
0.00
0.00
Change in concentration (mol/L)
–x
+x
+x
Equilibrium concentration (mol/L)
0.010 – x
x
x
Initial concentration (mol/L)
352
Chapter 8
Copyright © 2003 Nelson
– e H SO 2–
HSO4(aq)
4(aq)
(aq)
–
[H
(aq)][HSO4(aq)]
Ka 1.0 10–2
[HSO4(–aq)]
x2
1.0 10–2
0.010 – x
Predict whether a simplifying assumption is justified …
[HA]initial
0.010
Ka
1.0 10–2
1
Therefore, we may not assume that 0.010 – x 0.010.
x2
1.0 10–2
0.010 – x
x2 0.01x – 0.0001 0
–0.01 (0.01)2
–4(–0
.0001)
x 2
The only positive root is x 6.18 10–3
Total
[H
(aq)]
6.18 10–3
(extra digits carried)
mol/L 1.0 10–2 mol/L
1.618 10–2 mol/L
pH 1.79
The pH of the sulfuric acid solution is 1.79.
Copyright © 2003 Nelson
Acid–Base Equilibrium
353