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Chapter 15: Applications of Aqueous Equilibria
Problem Solving Center
Self Quiz 2
Self Quiz 2
This activity contains 20 questions.
A pH greater than 7 after the neutralization of equal molar amounts of the given acid and base are produced by
Ca(OH)
2
and HI.
HCl and NaC
2
H
3
O
2
.
HNO
3
and NH
3
.
KOH and HCl.
HNO
2
and LiOH.
The combination that is likely to give a reaction that does
not
proceed as far to completion as the other reactions is
HNO
3
and LiOH.
HCl and NaOH.
HCl and NH
3
.
HBr and Ca(OH)
2
.
HF and CH
3
NH
2
.
The pH of a solution containing 1.0 mole HF and 1.0 mole NaF in 1.0 L of solution will
not
change by
adding NaCl.
adding HCl.
adding NaOH.
adding HF.
adding KF.
The principal proton-transfer reaction in an aqueous solution containing NaF and HF is
HF(
aq
) + Na
+
(
aq
)
NaH
+2
(
aq
) + F
-
(
aq
).
HF(
aq
) + F
-
(
aq
)
F
-
(
aq
) + HF(
aq
).
2H
2
O(
l
)
H
3
O
+
(
aq
) + OH
-
(
aq
).
HF(
aq
) + H
2
O(
l
)
F
-
(
aq
) + H
3
O
+
(
aq
).
Na
+
(
aq
) + H
2
O(
l
)
NaH
2+
(
aq
) + OH
-
(
aq
).
The pH of 1.0 L of a 0.020 M HF-0.030 M NaF buffer solution after adding 0.0020 mole HCl is (K
a
of HF = 3.5 x 10
-4
)
3.56
3.35
3.71
2.70
3.64
The buffer solution that has a buffer capacity greater than that of 100 mL of a 0.020 M HCN - 0.040 M NaCN solution is
100 mL of a 0.040 M HCN - 0.080 M NaCN solution.
50 mL of a 0.010 M HCN - 0.020 M NaCN solution.
200 mL of a 0.010 M HCN - 0.020 M NaCN solution.
50 mL of a 0.020 M HCN - 0.040 M NaCN solution.
100 mL of a 0.010 M HCN - 0.020 M NaCN solution.
The pH of a buffer solution that is 0.30 M H
3
PO
4
and 0.10 M NaH
2
PO
4
is (K
a
of H
3
PO
4
= 7.5 × 10
-3
and K
b
of H
2
PO
4
-
= 1.3 x 10
-12
)
0.70
1.65
0.52
2.60
11.41
The buffer solution that would most likely be used to produce a buffer solution having a pH close to 7 is
0.10 M NaHSO
3
- 0.10 M Na
2
SO
3
, K
a
of HSO
3
-
= 6.3 x 10
-8
.
0.10 M HF - 0.10 M NaF, K
a
of HF = 3.5 x 10
-4
.
0.10 M H
3
BO
3
- 0.10 M NaH
2
BO
3
, K
a
of H
3
BO
3
= 5.8 x 10
-10
.
0.10 M HN
3
- 0.10 M NaN
3
, K
= 1.9 x 10
-5
.
0.10 M HOBr - 0.10 M NaOBr, K
a
of HOBr = 2.0 x 10
-9
.
The solution that is matched
incorrectly
with the given pH is
100 mL of 0.50 M H
2
S-1.0 M NaHS (pK
a
= 7.00) : pH = 7.30.
100 mL of 0.35 M HOCl-1.0 M NaOCl (pK
a
= 7.46) : pH = 7.00.
100 mL of 0.59 M H
3
AsO
3
-1.0 M NaH
2
AsO
3
. (pK
a
= 6.77) : pH = 7.00.
100 mL of 0.38 M H
3
PO
3
-1.0 M NaH
2
PO
3
(pK
a
= 6.58) : pH = 7.00.
100 mL of 1.0 M HNO
2
-0.45 M NaNO
2
(pK
a
= 3.35) : pH = 3.00.
The combination that does
not
represent the equivalence point of a strong acid-strong base titration is
30.0 mL of 0.50 M HClO
4
+ 30.0 mL of 0.40 M LiOH.
10.0 mL of 0.200 M HI + 20.0 mL of 0.100 M RbOH.
30.0 mL of 0.100 M HNO
3
+ 15.0 mL of 0.100 M Sr(OH)
2
.
30.0 mL of 0.20 M HCl + 15.0 mL of 0.20 M Ca (OH)
2
.
30.0 mL of 0.20 M HBr + 60.0 mL of 0.10 M NaOH.
The volume (in mL) of 0.250 M Ca(OH)
2
that is required to reach the point halfway to the equivalence point in the titration of 40.0 mL of 0.200 M HI is
16.0 mL.
10.0 mL.
20.0 mL.
8.00 mL.
32.0 mL.
The pH after the addition of 15.0 mL of 0.0500 M NaOH to 50.0 mL of 0.0400 M ascorbic acid is (K
a
of ascorbic acid = 8.0 x 10
-5
)
3.67
4.40
1.72
3.88
4.32
The weak base that is matched with the correct pH at the point halfway to the equivalence point in the titration of the weak base with a strong acid is
ammonia, K
b
= 1.8 x 10
-5
: pH = 4.74
morphine, K
b
= 1.6 x 10
-6
: pH = 5.80
aniline, K
b
= 4.3 x 10
-10
: pH = 4.63
NaF, K
b
= 2.9 x 10
-11
: pH = 10.54
pyridine, K
b
= 1.9 x 10
-9
: pH = 8.74
Which statement concerning solubility equilibria is true?
The K
sp
value is a constant at all temperatures.
As the amount of the ionic solid increases, the K
sp
value also increases.
The ions from the ionic solid that are in solution will remain in solution.
The rate of dissolution equals the rate of crystallization.
Changing the concentration of one of the ions in the expression will change the value of K
sp
.
The solubility (in grams per liter) of CaF
2
in water at 25 °C is (K
sp
= 1.5 x 10
-10
)
0.020 g.
0.00068 g.
0.041 g.
3.3 x 10
-4
g/L.
0.026 g.
What is [Fe
3+
] if [OH
-
] = 2.3 x 10
-10
? (K
sp
of Fe(OH)
3
= 2.6 x 10
-39
)
3.8 x 10
-30
M
2.1 x 10
-10
M
1.1 x 10
-29
M
7.9 x 10
-12
M
7.7 x 10
-11
M
The compound that shows an increase in solubility as the pH decreases is
Hg
2
Cl
2
.
AgCl.
PbBr
2
.
BaF
2
.
AgI.
The concentration of Cu
2+
in a solution made by adding 0.0500 mole Cu(NO
3
)
2
to 1.00 L of 1.00 M NH
3
is (The reaction Cu
2+
(
aq
) + 4NH
3
(
aq
)
Cu(NH
3
)
4
2+
has a K
f
value of 1.1 x 10
13
.)
0.0 M.
1.1 x 10
-14
M.
5.7 x 10
-15
M.
1.4 x 10
-15
M.
1.3 x 10
12
M.
Which combination of aqueous solutions of Mg(NO
3
)
2
and NaF will
not
produce a precipitate of MgF
2
? (K
sp
of MgF
2
= 7.4 x 10
-11
)
0.0500 L of 4.0 x 10
-4
M Mg(NO
3
)
2
and 0.0500 L of 6.0 x 10
-3
M NaF
0.200 L of 4.0 x 10
-4
M Mg(NO
3
)
2
and 0.300 L of 6.0 x 10
-3
M NaF
0.100 L of 4.0 x 10
-4
M Mg(NO
3
)
2
and 0.200 L of 6.0 x 10
-3
M NaF
0.150 L of 4.0 x 10
-4
M Mg(NO
3
)
2
and 0.0500 L of 6.0 x 10
-3
M NaF
0.200 L of 4.0 x 10
-4
M Mg(NO
3
)
2
and 0.0100 L of 6.0 x 10
-3
M NaF
Which action is
not
characteristic of the selective precipitation method used to separate ions?
Adding water can cause selective precipitation.
Raising the pH can cause selective precipitation.
Adding an anion can cause selective precipitation.
Adding a cation can cause selective precipitation.
Lowering the pH can cause selective precipitation.
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