Chapter 15
Applications of
Aqueous Equilibria
QUESTION
Suppose the weak acid HNO2 (Ka = 4.0 10–4) is added to a
solution of NaNO2. If the concentration of acid is 0.10 M and
the salt concentration is 0.060 M, what is the [H+]?
1.
2.
3.
4.
2.4  10–4 M
6.7  10–4 M
2.0  10–4 M
4.0  10–5 M
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CRS Question, 15–2
ANSWER
Choice 2 represents the approximate [H+]. The Ka expression
may be used with the assumption that the dissociation of the
acid is very slight. The presence of NO2– (from the salt) does
reduce the amount of dissociation of the acid. The assumption
that any NO2– from the dissociation of the acid can be ignored
is used in the calculation so that [H+] = Ka (0.10)/(0.060).
Section 15.1: Solutions of Acids or Bases Containing a
Common Ion
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CRS Question, 15–3
QUESTION
The dissociation constant of propanoic acid (HC3H5O2) is 1.3 
10–5. What is the pH of a buffer made with 0.50 M propanoic
acid and 0.25 M NaC3H5O2 (the sodium salt of propanoic acid)?
1.
2.
3.
4.
4.59
5.19
7.00
I’m not getting any of these answers.
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CRS Question, 15–4
ANSWER
Choice 1 is the correct pH of the buffer. There is an
assumption being made here about the concentration of the
salt to be only from the salt added. This ignores the small
amount of propionate [C3H5O2–] that may be produced from
the acid dissociation.
Section 15.2: Buffered Solutions
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CRS Question, 15–5
QUESTION
A certain chemical reaction runs very slow at high pH values
and very fast at lower pH values. To study the reaction, a
chemist needs to buffer the solution at a basic pH. Determine
the pH of a buffer solution that had the following
concentrations: 0.42 M NH4Cl and 0.75 M NH3; Kb of NH3 =
1.8  10–5 at 25°C.
1.
2.
3.
4.
9.00
4.49
9.51
11.57
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CRS Question, 15–6
ANSWER
Choice 3 provides the correct basic pH of the solution. Using
the Kb value of ammonia allows the calculation of [OH–] with
the given NH3 and NH4Cl concentrations. Assuming that no
significant NH4+ comes from the NH3 + H2O reaction, the
[OH–] could be approximated via [OH–] = Kb (NH3)/(NH4+);
the pH can then be determined from the pOH.
Section 15.2: Buffered Solutions
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CRS Question, 15–7
QUESTION
A 0.50 L buffer solution containing 0.42 M NH4Cl and 0.75 M
NH3 (Kb of NH3 = 1.8  10–5 ) has a pH of 9.51 at 25°C. The
solution receives 0.010 moles of HCl from an outside source.
Assuming no significant change in volume of the solution,
what is the pH of the solution after the addition of the HCl at
25°C?
1.
2.
3.
4.
9.48
9.51
9.34
8.76
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CRS Question, 15–8
ANSWER
Choice 1 correctly incorporates the new concentrations of base
and salt after the addition of H+. The concentrations of both base
and its salt can be converted to amounts by using V  M = moles.
Then the moles of NH3 are reduced by 0.010 moles of H+ from
the HCl. This means that the NH4+ amount is increased by the
same 0.010 moles of H+. The new amounts are then divided by
the solution volume (0.50 L) to determine the new
concentrations. These are then used to solve for the [OH–],
which is converted to pOH and finally to pH.
Section 15.2: Buffered Solutions
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CRS Question, 15–9
QUESTION
Although buffers resist change in pH they are still affected by the
addition of a new base or acid. A 0.50 L buffer solution
containing 0.42 M NH4Cl and 0.75 M NH3;(Kb of NH3 = 1.8 
10–5 ) has a pH of 9.51 at 25°C. What is the new pH of the
buffer after it is adjusted by the addition of 0.010 moles of
NaOH. Assume no change in the solution volume.
1.
2.
3.
4.
8.97
9.51
12.00
9.54
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CRS Question, 15–10
ANSWER
Choice 4 provides the newly adjusted pH. The addition of
0.010 moles of new OH– causes the NH4+ concentration to
decrease and the NH3 concentration to increase. The molarity
of both is first changed to moles (V  M = moles;) then the
0.010 moles of OH– is added and subtracted appropriately.
The new number of moles of each are divided by the solution
volume to obtain molarity. The system can now be treated
with the Kb expression using the new molarities of the NH4+
and NH3 to solve for [OH–], which can then be used to obtain
pOH and pH.
Section 15.2: Buffered Solutions
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CRS Question, 15–11
QUESTION
Two separate buffer solutions are prepared using propanoic
acid and calcium propanoate (also used in some food
preservative applications). If buffer “A” was 0.10 M in both
acid and salt while buffer “B” was 0.20 M in both acid and
salt, which of the following would be true?
1. Both solutions would have the same pH and would have
the same buffer capacity.
2. Solution “B” would would have a lower pH and both
would have the same buffer capacity.
3. Solution “B” would have a lower pH and a larger buffer
capacity.
4. Both solutions would have the same pH; “B” would have
a larger buffer capacity.
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CRS Question, 15–12
ANSWER
Choice 4 offers the accurate comparison of the pH and buffer
capacity for the two solutions. As the salt and acid have the
same concentration, the pH = pKa relationship is the same.
However, because solution “B” has a higher molarity for
both acid and salt any additional acid or base additions will
have to be much larger before the ratio changes significantly.
Therefore, solution “B” has a higher buffering capacity.
Section 15.3: Buffering Capacity
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CRS Question, 15–13
QUESTION
To culture a certain bacteria a microbiologist would like to
buffer the media at a pH of 3.75. To maximize the efficiency
of the system a 1:1 ratio of acid to salt will be used. Which of
the following acids would make the best choice for the buffer?
1.
2.
3.
4.
Acetic acid; Ka = 1.8  10–5
Propanoic acid; Ka = 1.3  10–5
Formic acid; Ka = 1.8  10–4
Nitrous acid; Ka = 4.0  10–4
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CRS Question, 15–14
ANSWER
Choice 3 represents the best selection for this buffer. If the
acid:salt ratio is to be 1:1, the pH will equal pKa. At a pH of
3.75 the closest pKa would be from formic acid: 1.3  10–4.
Section 15.3: Buffering Capacity
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CRS Question, 15–15
QUESTION
If 25.0 mL of 0.250 M HCl were titrated with 0.500 M NaOH,
what are the initial pH, pH at the equivalence point and
volume of the total solution at the equivalence point?
1.
2.
3.
4.
1.000; 7.00; 25.0 mL
0.602; 7.00; 50.0 mL
0.602; 7.00; 37.5 mL
1.000; 7.00; 12.5 mL
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CRS Question, 15–16
ANSWER
Choice 3 provides the correct response for all three questions.
The pH of a strong acid is taken directly from the molarity.
(–log 0.250). The pH of the equivalence point for a strong
acid and strong base titration is 7.00. The moles of HCl
initially is 25.0 mL  0.250 M, or 6.25 millimoles. Therefore,
the base must provide 6.25 millimoles for neutralization. The
volume of 0.500 M NaOH that would provide this could be
calculated from V  0.500 M = 6.25; V = 6.25/0.500 = 12.5
mL.
Section 15.4: Titrations and pH Curves
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CRS Question, 15–17
QUESTION
What is the pH of the equivalence point of the titration of
30.0 mL of 0.100 M benzoic acid, C6H5CO2H
(Ka = 6.4  10–5), with 0.100 M NaOH at 25°C?
1.
2.
3.
4.
8.60
5.55
8.45
I must be missing a step, I don’t get any of these.
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CRS Question, 15–18
ANSWER
Choice 3 correctly accounts for the formation of the moles of
the salt of benzoic acid during the titration (volume acid 
molarity of the acid), molarity of the salt (moles / total
volume), determination of the Kb of the conjugate salt (Kw/Ka),
and finally the pH (14 – pOH).
Section 15.4: Titrations and pH Curves
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CRS Question, 15–19
QUESTION
50.0 mL of an acid solution (HA) contains 0.0200 mole of the
acid. The solution is titrated with 0.200 M NaOH. Half way
to the equivalence point the pH is 5.25. What is the total
volume of the solution at that point and what is the Ka of the
acid?
1.
2.
3.
4.
0.100 L; Ka = 5.6  10–6
0.150 L; Ka = 5.6  10–6
0.100 L; Ka = 1.8  10–9
0.150 L; Ka = 1.8  10–9
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CRS Question, 15–20
ANSWER
Choice 1 provides both the correct volume of the solution and
correct Ka value for the acid. 50.0 mL of acid containing
0.0200 mole would require 0.100 L of 0.200 M NaOH to
neutralize, so half way to that value would be 50.0 mL. This
is then added to the original 50.0 mL of acid to obtain the
solution volume of 0.100 L half way to the equivalence point.
At this point the concentration of HA is approximately that of
A–, so Ka = [H+] or pKa = pH.
Section 15.4: Titrations and pH Curves
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CRS Question, 15–21
QUESTION
Suppose you have 50.0 mL each of 0.10 M solutions of
propanoic acid (Ka = 1.3  10–5); acetic acid
(Ka = 1.8  10–5); HCl (all monoprotic acids). Which would
require the LEAST volume of 0.10 M NaOH to neutralize?
1.
2.
3.
4.
Propanoic acid
Acetic acid
HCl
I think they all require the same.
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CRS Question, 15–22
ANSWER
Choice 4 is correct. As all the acids have the same volume
and molarity, they have the same number of moles. NaOH is
a strong base and will react with all the moles of each acid.
Therefore, each would require the same moles of NaOH for
neutralization.
Section 15.4: Titrations and pH Curves
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CRS Question, 15–23
QUESTION
Most acid-base indicators are weak acids. In a titration of
0.50 M acetic acid (at 25°C, Ka = 1.8  10–5) with KOH,
which indicator would best indicate the pH at the
equivalence point? The approximate Ka for each choice is
provided.
1.
2.
3.
4.
Bromophenol blue; Ka ~ 1  10–4
Methyl red; Ka ~ 1  10–5
Bromothymol blue; Ka ~ 1  10–7
Thymolphthalein; Ka ~ 1  10–10
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CRS Question, 15–24
ANSWER
Choice 4 provides the best choice although there may also be
better choices available than these four. The equivalence
point pH should be as close as possible to the pKa of the
indicator. As acetic acid is a fairly weak acid and NaOH is a
strong base, the pH at the equivalence point will be above 7.
The only choice above 7 in the list was thymolphthalein.
Without a more detailed calculation, this would be the best
choice.
Section 15.5: Acid–Base Indicators
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CRS Question, 15–25
QUESTION
The acid-base indicator bromcresol purple has an interesting
yellow-to-purple color change. If the approximate Ka of this
indicator is 1.0  10–6, what would be the ratio of purple [A–]
to yellow [HA] at a pH of 4.0?
1.
2.
3.
4.
100:1
1:100
1:1
This choice indicates that I don’t know.
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CRS Question, 15–26
ANSWER
Choice 2 shows the [A–]/[HA] ratio at pH 4.0 for bromcresol
purple. The pH can be converted to [H+] and divided into the
Ka value to reveal the [A–]/[HA] ratio at pH 4.0.
Ka/[H+] = [A–]/[HA].
Section 15.5: Acid–Base Indicators
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CRS Question, 15–27
QUESTION
Lead (II) iodide is used in some camera batteries. PbI2 has a
Ksp of 1.4  10–8. What is the molar solubility of this
compound?
1.
2.
3.
4.
1.9  10–3 M
2.4  10–3 M
1.5  10–3 M
8.4  10–5 M
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CRS Question, 15–28
ANSWER
Choice 3 correctly uses the stoichiometry and equilibrium
expression to determine the molar solubility of PbI2. Since
the dissociation of PbI2 shows Pb2+(aq) + 2I–(aq), the Ksp
expression would appear as: Ksp = [X][2X]2. Solving for X
shows the relationship for PbI2 dissociating in a 1:1 ratio
between PbI2 and Pb2+.
Section 15.6: Solubility Equilibria and the Solubility Product
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CRS Question, 15–29
QUESTION
Calcium phosphate (Ca3(PO4)2) is only slightly soluble. In
fact, it is a common ingredient in phosphate rock and is a
major source of phosphate fertilizer. If the molar solubility is
2  10–7, what is the value of the Ksp?
1.
2.
3.
4.
3  10–34
2  10–10
3  10–32
I don’t get any of those values.
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CRS Question, 15–30
ANSWER
Choice 3 correctly incorporates the stoichiometry and
equilibrium expression for calcium phosphate. Ksp =
[Ca2+]3[PO43–]2. Since the molar solubility represents the
dissociation of Ca3(PO4)2, the ratio of that to the ions would
be 1:3 and 1:2 respectively. So, the expression becomes:
Ksp = [3  (2  10–7)]3[2  (2  10–7)]2.
Section 15.6: Solubility Equilibria and the Solubility Product
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CRS Question, 15–31
QUESTION
The Ksp of Ag2CrO4 is 9.0  10–12. What would be the
solubility of this compound in a solution that was already
0.10 M in potassium chromate?
1.
2.
3.
4.
9.0  10–13 M
9.5  10–6 M
3.0  10–6 M
4.7  10–6 M
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CRS Question, 15–32
ANSWER
Choice 4 provides the correct solubility of the compound in a
solution with a common ion. Ksp = [Ag+]2[CrO42–]; The
[Ag+] = 2  the molar solubility. Proper substitution of
known values produces:
9.0  10–12 = [Ag+]2[~ .10]
Section 15.6: Solubility Equilibria and the Solubility Product
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CRS Question, 15–33
QUESTION
Will a precipitate of BaSO4 form when 10.0 mL of 0.0010 M
barium nitrate are mixed with 20.0 mL of 0.000020 M of
sodium sulfate? The Ksp of barium sulfate is 1.5  10–9.
Prove your answer by reporting the calculated value of Q.
1.
2.
3.
4.
Yes; Q = 2.0  10–8
Yes; Q = 4.4  10–9
No; Q = 7.0  10–10
No; Q = 3.0  10–17
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CRS Question, 15–34
ANSWER
Choice 2 provides the correct prediction. Once the molarity
of each ion (Ba2+ and SO42–) is determined they are
multiplied to obtain Q. Since Q is larger than Ksp a
precipitate will form.
Section 15.7: Precipitation and Qualitative Analysis
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CRS Question, 15–35