Answer Key
Chapter 18
Practice Problems
1.
Acid
a.
b.
c.
2. a.
Conjugate
Base
Conjugate
Base
Acid
H2SO3
HSO3
H2O
H3O
H2O
OH
HPO42
H2PO4
HSeO3
SeO32
H2O
H3O
H2CO3(aq)  H2O(l)  H3O(aq)  HCO3(aq)
HCO3(aq)  H2O(l)  H3O(aq)  CO32(aq)
b. H2CrO4(aq)  H2O(l)  H3O(aq)  HCrO4(aq)
HCrO4(aq)  H2O(l)  H3O(aq)  CrO42(aq)
3. a. HF(aq)  H2O(l)  H3O(aq)  F(aq)
[H3O ][F ]
Ka 
[HF]
b. HBrO(aq)  H2O(l)  H3O(aq)  BrO(aq)
Ka 
[H3O ][BrO ]
[HBrO]
4. HSO3(aq)  H2O(l)  H3O(aq)  SO32(aq)
Ka 
[H3O  ][SO32 ]
[HSO3 ]
5. a. C4H9NH2(aq)  H2O(l)  C4H9NH3(aq)  OH(aq)
[C4 H9 NH3 ][OH  ]
[C4 H9 NH 2 ]
3
b. PO4 (aq)  H2O(l)  HPO42(aq)  OH(aq)
Kb 
Kb 
[HPO4 2 ][OH  ]
[PO43 ]
Chemistry: Matter and Change
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Solving Problems: A Chemistry Handbook
Answer Key (continued)
c. HCO3(aq)  H2O(l)  H2CO3(aq)  OH(aq)
[H 2 CO3 ][OH  ]
Kb 
[HCO3 ]
6. a. [H]  1.0108M, basic
b. [OH]  1.0107M, neutral
c. [OH]  1.21012M, acidic
7. a. pH  14.00; pOH  0.00
b. pH  6.75; pOH  7.25
c. pH  2.57; pOH  11.43
d. pH  12.79; pOH  1.21
8. a. [H]  1.6103M; [OH]  6.31012M
b. [H]  6.51014M; [OH]  0.15M
c. [H]  5.8106M; [OH]  1.7109M
9. a. 1.82
c. 3.60
b. 12.13
d. 11.90
2
10. a. 1.6  10
b. 6.4105
11. 0.531M
12. 0.1234M
13. 0.183M
Chapter 18 Review
14. An Arrhenius base is a substance that contains a hydroxide (OH)
group and dissociates in aqueous solution to produce hydroxide
ions. A Brønsted-Lowry base is a hydrogen-ion acceptor that may
or may not contain a hydroxide group.
15. Acid X is a weak acid that ionizes only partially in water. Acid Y
has an even smaller Ka, so it is weaker than X.
16. A decrease in the hydroxide ion concentration is accompanied by an
increase in the hydrogen ion concentration. The solution becomes
more acidic, and the pH decreases.
Chemistry: Matter and Change
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Solving Problems: A Chemistry Handbook
Answer Key (continued)
17. The pH of solution A is 9.0, and the pH of solution B is
14.00  pOH  14.00  3.0  11.0. Thus, both solutions are basic
because their pH values are above 7. Solution A has a lower pH, so
it has a higher concentration of hydrogen ions.
18. Find the hydrogen ion concentration of the HCl solution.
[H]  antilog (pH)  antilog (2.00)  1.0102M  0.010M
HCl is a strong acid, so it ionizes completely and its molarity equals
[H]. Thus, the solution is 0.010M HCl.
19. A large pH change occurs at the equivalence point of an acid-base
titration. This pH change can be detected with a pH meter or an
acid-base indicator.
20. A buffer is a mixture of a weak acid and its conjugate base or a
weak base and its conjugate acid. Sodium formate dissociates in
water to produce sodium ions and formate ions.
NaHCOO(s)  Na(aq)  HCOO(aq)
The HCOO ion is the conjugate base of formic acid, HCOOH, a
weak acid. It is likely that the buffer solution consists of an aqueous
solution of sodium formate and formic acid.
Chemistry: Matter and Change
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Solving Problems: A Chemistry Handbook