Chemistry 122
Mines, Spring 2014
Answer Key, Problem Set 7 (With explanations)
1. 16.6 (Write out net ionic equations that represent what happens in each case. Represent the weak acid as HA and its conjugate base as
A-); 2. 16.50(ab) Assume 25°C! ; 3. 16.54; 4. 16.150 & 16.59(ac); 5. 16.61 & 16.11; 6. 16.62(abc); 7. 16.65; 8. 16.67**; 9.
16.71; 10. 16.75 Assume equal initial volumes! ; 11. 16.78; 12. 16.81**; 13. 16.100; 14. NT1; 15. 15.111 (“all species” H2SO3,
HSO3-, H3O+, OH-, and SO32-. Assume T = 25°C. Also calculate the pH.); 16. 15.27 & 15.116 (except omit H2S from the problem);
17. 15.117
-----------------------------------1. 16.6 Suppose that a buffer contains equal amounts of a weak acid and its conjugate base. (a) What happens to the relative
amounts of the weak acid and conjugate base when a small amount of strong acid is added to the buffer? (b) What
happens when a small amount of strong base is added? (Write out net ionic equations in each case.)
Answer: (a) The amount of conjugate base goes down a bit and the amount of weak acid goes up.
(b) The amount of conjugate acid goes down and the amount of conjugate base goes up.
Explanation:
(a) When a bit of strong acid (e.g., HNO3) is added to a buffer (weak acid  HA; weak base  A ), it reacts
with the “best base around”, which is A . The reaction that occurs can be represented by:
-
-
HNO3 + A
(OR
H3O+ + A-
100%


 HA + NO3
-
100%

 HA + H2O [If you allow the HNO3 to react with water first to make H3O+])
This reaction goes “until somebody runs out”.
I.e., it goes “to completion”, which is to say, it has a K value that is extremely large. This can be seen in two ways: 1) The
first reaction (above) must be more product favored than the reaction of HNO3 with H2O since A- is a better base than H2O,
and 2) the second reaction (above) is the exact reverse of the acid ionization reaction equation, and thus has
K
1
1 . Since HA is,by definition, weak (or else you would not have a buffer!),
K a  1 
 1
Ka
Ka
-
As such, the net result is that A is converted into HA. If the moles of strong acid added are less than
the moles of A present, the strong acid is the limiting reactant, and only a fraction of the original A will
be converted into HA. Thus a buffer will remain after the reaction occurs (but the pH will be a bit
lower). The reaction can be visualized as follows:
+
HNO3
H3O+ + A-  HA + H2O
HA
-
A
HA
-
A
(b) When a bit of strong base (e.g., NaOH) is added to a buffer, it reacts with the “best acid around”, which
is HA. The reaction that occurs can be represented by:
100%
OH + HA 

 A + H2O [If you allow the NaOH to dissociate first to make OH-]
-
-
This reaction goes “until somebody runs out”.
I.e., it goes “to completion”, which is to say, it has a K value that is extremely large. This can be seen by recognizing that
the reaction (above) is the exact reverse of the base ionization reaction equation, and thus has K 
definition, weak (or else you would not have a buffer!), K b  1 
1 . Since A- is, by
Kb
1
 1
Kb
-
As such, the net result is that HA is converted into A . If the moles of strong base added are less than
the moles of HA present, the strong base is the limiting reactant, and only a fraction of the original HA
will be converted into A . Thus a buffer will remain after the reaction occurs (but the pH will be a bit
higher). The reaction can be visualized as follows:
+
-
OH
OH- + HA  A- + H2O
HA
-
A
HA
PS7-1
-
A
Answer Key, Problem Set 7
2. 16.50(ab) Assume 25°C! ; For each solution, calculate the initial and final pH after adding 0.010 mol of NaOH.
(a) 250.0 mL of pure water
(b) 250.0 mL of a buffer solution that is 0.195 M in HCHO2 and 0.275 M in KCHO2
Answers: (a) 7.00 before, 12.60 after; (b) 3.89 before, 4.05 after
Strategy:
pH’s before addition of NaOH:
(a) Pure water at 25°C has [H3O ]  1.0 x 10 M and pH  7.00 (see PS5 and PS6)
+
-7
(b) As discussed in PS6, the pH of a buffer with known “initial” (prepared) concentrations can be
determined “straight away” (because the small x approximation applies to both HA and A-) using:
[HA]  moles HA 
[A  ]  moles A  
+
 or

[H3O ]  Ka x
OR pH  pKa + log

 
 
[HA]  moles HA 
[A ]  moles A 
(or you could do a longer ICE approach with small x approximations)
NOTE: Find Ka(HCHO2) in Table 15.5.
pH’s after addition of NaOH:
Always consider dissociation of soluble ionic compounds first. Thus, the first thing to occur (in either
+
case) is that NaOH will dissociate into Na (a negligible acid) and OH .
For (a). The only other process that occurs is a shifting in the water autoionization equilibrium. Thus,
to calculate pH, find the [OH ] after the dissociation, and assume that is the equilibrium
+
value. Then use the water autoionization equilibrium condition (Kw  [H3O ][OH ]) to find
+
[H3O ] and then pH.
-
For (b) (a buffer). The primary process that occurs after the dissociation is that OH will react with
HA (see Q1 above) until somebody runs out. So first calculate the moles of HCHO 2 in the
solution and determine whether it is more or less than the 0.010 mol of NaOH added.
1) If it is greater, then the resulting solution will be a buffer since there will still be some acid
and base component remaining (use an “ICF” table to find final moles of HA and A-**). You
moles HA
+
can determine the pH of the buffer using [H3O ] = Ka x
(or the HH equation)
moles A 
**NOTE: You must “let” KCHO2 dissociate first into K+ (ignore) and CHO2- (“A-“). Then find moles CHO2- and
use in the “ICF” calculation.
-
2) If it is smaller, then find the concentration of excess OH and proceed as in (a).
Execution of Strategy:
(a) pH before addition of NaOH: 7.00 (neutral at 25°C)
pH after:
0.010 mol NaOH in 250.0 mL  0.010 mol Na and 0.010 mol OH in 250.0 mL ( 0.250 L)
+
-
-
 [OH ] 
0.010 mol OH1.0 x 10-14
 2.5 x 10-13 M
 0.040M  [H3 O  ] 
0.040
0.250L
 pH = -log(2.5 x 10 ) = 12.602 = 12.60
-13
(b) pH before NaOH:
[H3O ]  Ka x
+
[HA]
0.195M
 1.8 x 10-4 x
 1.276.x 10-4  pH  -log(1.276 x 10-4)  3.894  3.89

0.275M
[A ]
Answer is reasonable. pKa here is –log(1.8 x 10-4)
= 3.74. This buffer has more A- than HA  pH
should be a bit higher than 3.74 and it is.
PS7-2
Answer Key, Problem Set 7
pH after NaOH added:
0.195 mol HCHO 2
 0.04875 mol
L
(initial) moles HCHO2 = 0.2500 L x

-
(initial) moles CHO2 = 0.2500 L x
0.275 mol CHO 2
 0.06875 mol
L
-
-
Reaction that occurs: OH + HCHO2 → CHO2 + H2O (OH- will react with the strongest acid around.)
-
-
in moles
OH (moles)
HCHO2 (mol)
CHO2 (mol)
I
0.010 (LR)
0.04875
0.06875
C
-0.010
-0.010
+ 0.010
F
0
0.03875
0.07875
Decent amounts of HCHO2 and CHO2 are present after this reaction  a buffer!
-
[H3O ]  Ka x
+
moles HCHO 2
moles CHO 2

 1.8 x 10-4 x
0.03875
 8.857..x 10-5 M
0.07875
 pH  -log(8.857..x 10 )  4.052..  4.05
-5
Answer is reasonable. There is more CHO2- and
less HCHO2 here than in the original buffer. Thus,
the pH should be a bit higher than 3.89 and it is.
NOTE: The increase in pH in pure water was 12.60 – 7.00  5.60 units.
The increase in pH in the buffer was only 4.05 - 3.89  0.16 units. The buffer “worked”!
3. 16.54. Determine whether or not the mixing of the two solutions will result in a buffer.
(a) 75.0 mL of 0.10 M HF; 55.0 mL of 0.15 NaF
(b) 150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl
(c) 165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH
(d) 125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl
(e) 105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl
Answers: (a), (c), (d), and (e) will result in buffers. (b) will not
Strategy:
NOTE: A buffer must contain some of a weak acid and some of its conjugate base. So if those two
components are mixed together directly, clearly a buffer will result. However, one must realize that
if a strong acid or strong base is one of the substances being mixed a reaction may occur to
generate one of the components of the buffer that is not “there” initially. So you must be on the
lookout for this kind of mixture. As will be described below, in such an instance, a calculation needs
to be done to see whether or not a buffer will result.
1) Look to see what two kinds of substances are being mixed. If the following substances are mixed,
you can make the conclusion right away without doing a calculation:
a) If two acids are mixed, it won’t be a buffer.
(You would neither have or generate the weak
base component.)
b) If two bases are mixed, it won’t be a buffer.
(You would neither have or generate the weak
base component.)
c) If a weak acid and a Gp I salt of its conjugate base are mixed (e.g., HA and NaA or KA), it will be
a buffer.
d) If a weak molecular base and a salt of its conjugate acid with a negligible base anion (e.g., RNH2
and RNH3Cl, where “R” represents C (with other atoms bound to it) or H), it will be a buffer.
2) In the following two cases, a calculation (an “ICF” type) will be needed:
PS7-3
Answer Key, Problem Set 7
d) If a weak acid and strong base are mixed, it may be a buffer because they will react with one
another until one of them runs out, producing some of the conjugate base of the weak
acid. The question becomes, “What will be left over?” So do an ICF type calculation (with
moles) and assess the result:
i) If there is excess weak acid (i.e., the SB is limiting), the resulting solution is a buffer.
ii) If there is excess strong base (i.e., the WA is limiting), the resulting solution will not be a buffer
(in effect, a buffer was initially created but then so much SB was around that the buffer capacity was exceeded).
iii) If there is no excess of either, the resulting solution will not be a buffer. The solution contains
only the conjugate base of the weak acid. (Solution of a weak base.)
e) If a weak base salt (i.e., Gp I salt of a base anion) and a strong acid are mixed, it may be a
buffer because they will react with one another until one of them runs out, producing
some of the conjugate acid of the weak base. The question becomes, “What will be left
over?” So do an ICF type calculation (with moles) and assess the result:
i) If there is excess weak base (i.e., the SA is limiting), the resulting solution is a buffer.
ii) If there is excess strong acid (i.e., the WB is limiting), the resulting solution will not be a buffer
(in effect, a buffer was initially created but then so much SA was around that the buffer capacity was exceeded).
iii) If there is no excess of either, the resulting solution will not be a buffer. The solution contains
only the conjugate acid of the weak base. (Solution of a weak acid.)
Execution of Strategy:
(a) 75.0 mL of 0.10 M HF; 55.0 mL of 0.15 NaF
-
HF and NaF are mixed. HF and F will both be present. This will make a buffer.
The amounts are basically irrelevant since neither one is ridiculously small.
(b) 150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl
-
HF and HCl are both acids. No F will be present initially nor generated by any reaction.
This will NOT make a buffer.
(c) 165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH
~100%
HF is mixed with KOH. They will react to form F : HF + OH 
 F + H2O
-
-
-
 Do an ICF type calculation to see who’s limiting.
moles HF: 0.1650 L x 0.10 mol/L  0.01650 mol HF
moles OH : 0.1350 L x 0.050 mol/L  0.00675 mol OH
-
-
L.R.  This will make a buffer.
NOTE: You need not actually complete a full ICF table to make this conclusion, but just to “prove” it
to you (and for completeness), I will show the results of that here:
in moles
OH-
F-
HF
I
0.00675 (LR)
0.01650
0
C
- 0.00675
- 0.00675
+ 0.00675
F
0
0.00975
0.00675
A buffer!
(d) 125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl
+
CH3NH2 (a weak base) and CH3NH3 (its conjugate) will be present. This will make a buffer.
(e) 105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl
PS7-4
Answer Key, Problem Set 7
+
CH3NH2 is mixed with HCl. They will react to form CH3NH3 :
~100%
CH3NH2 + HCl 
 CH3NH3 + Cl
+
-
 Do an ICF type calculation to see who’s limiting.
moles CH3NH2: 0.1050 L x 0.15 mol/L  0.01575 mol CH3NH2
moles HCl: 0.0950 L x 0.10 mol/L  0.00950 mol HCl
L.R.  This will make a buffer.
NOTE: You need not actually complete a full ICF table to make this conclusion, but just to “prove” it
to you (and for completeness), I will show the results of that here:
in moles
CH3NH2
HCl
CH3NH3+
I
0.01575
0.00950 (LR)
0
C
- 0.00950
- 0.00950
+ 0.00950
F
0.00625
0
0.00950
A buffer!
4. 16.150 & 16.59(ac). Determine whether or not each addition exceeds the capacity of the buffer.
16.150: Buffer contains 0.10 mol of a weak acid and 0.10 mol of its conjugate base in 1.0 L of solution.
(a) 0.020 mol of NaOH
(b) 0.020 mol of HCl
(c) 0.10 mol of NaOH
(d) 0.010 mol of HCl
16.59: A 500.0 mL buffer is 0.100 M in HNO2 and 0.150 M in KNO2.
(a) 250 mg NaOH
(b) 350 mg KOH
(c) 1.25 g HBr
(d) 1.35 g HI
Answers: 16.150: Only (c) exceeds the capacity;
16.59: None exceed the capacity!
Strategy:
(Perhaps the authors meant the initial volume of the buffer to
be 50.0 mL? This is quite silly to me. Sorry I didn’t catch it
earlier)
“Exceeding the buffer capacity” means “reacting away all of a buffer component”. Thus, compare:
moles of strong acid added to the moles of weak base component initially present
OR
moles of strong base added to the moles of weak acid component initially present
IF:
moles SA  moles WB (initial)  all the WB will run out (L.R.) and buffer capacity will be exceeded.
OR
moles SB  moles WA (initial)  all the WA will run out (L.R.) and buffer capacity will be exceeded.
Otherwise, the buffer capacity will not be exceeded.
Note the similarity of this problem to the prior one. The key idea is that whether a SA (or SB) is added to a buffer
or a solution of a WB (or WA), an “reaction to completion” will occur, and you may end up with a buffer, or you may
end up with excess SA (or SB) [obviously not a buffer].
Execution of Strategy:
16.150: Buffer contains 0.10 mol of a weak acid and 0.10 mol of its conjugate base in 1.0 L of solution.
(a) 0.020 mol of NaOH
-
0.020 mol OH < 0.10 mol HA
 OH runs out; some HA left over
-
 capacity not exceeded
(b) 0.020 mol of HCl
-
0.020 mol HCl < 0.10 mol A
 HCl runs out; some A left over
-
 capacity not exceeded
(c) 0.10 mol of NaOH
-
0.10 mol OH = 0.10 mol HA
PS7-5
 NaOH and HA will run out
Answer Key, Problem Set 7
 capacity is exceeded
(d) 0.010 mol of HCl
 HCl runs out; some A left over
-
-
0.010 mol HCl < 0.10 mol A
 capacity not exceeded
16.59: A 500.0 mL buffer is 0.100 M in HNO2 and 0.150 M in KNO2.
moles HNO2: 0.5000 L x 0.100 mol/L = 0.0500 mol HNO2
-
-
moles NO2 : 0.5000 L x 0.150 mol/L = 0.0750 mol NO2
(Let KNO2 dissociate into K+ + NO2-)
(a) 250 mg NaOH
moles NaOH: 250 mg NaOH x
1g
1mol
x
 0.00625mol NaOH
1000 mg (22.99  16.00  1.01) g
 0.00625 mol OH (after dissociation) < 0.0500 mol HNO2
-
 OH runs out; some HNO2 left over
-
 capacity not exceeded (not even close!)
(b) 350 mg KOH [Removed formally from set on 3/14/12]
moles KOH: 350 mg NaOH x
1g
1mol
x
 0.00623..mol KOH
1000 mg (39.10  16.00  1.01) g
 0.00623.. mol OH- (after dissociation) < 0.0500 mol HNO2
 OH- runs out; some HNO2 left over
 capacity not exceeded (sorry for the repetition; this is nearly identical to (a)!)
(c) 1.25 g HBr
moles HBr: 1.25 g HBr x
1mol
 0.0154..mol HBr
(1.01 79.90)g
 0.0154.. mol HBr < 0.0750 mol NO2
-
 HBr runs out; some NO2 left over
-
 capacity not exceeded (still not very close!)
(d) 1.35 g HI [Removed formally from set on 3/14/12]
moles HBr: 1.35 g HI x
1mol
 0.0105..mol HI
(1.01 126.90)g
 0.0155.. mol HI < 0.0750 mol NO2 HI runs out; some NO2- left over
 capacity not exceeded (sorry for the repetition)
PS7-6
Answer Key, Problem Set 7
5. 16.61 & 16.11.
16.61. The graphs labeled (a) and (b) show the titration curves for two equal-volume samples of monoprotic acids, one weak
and one strong. Both titrations were carried out with the same concentration of strong base.
14
14
(a)
12
10
pH
(b)
12
10
x
8
pH
8
6
6
4
4
2
2
0
x
0
0
20
40
60
80
100
Volume of base added (mL)
0
20
40
60
80
100
Volume of base added (mL)
(i) What is the approximate pH at the equivalence point of each curve?
Answers: (a) ~8.5;
(b) ~7.0 (we cannot say “exactly” 7.00 because the T was not specified as 25°C)
(ii) Which curve corresponds to the titration of the strong acid and which one to the titration of the weak acid?
Answer: The curve in (b) is the strong acid; in (a) is the weak acid
Reasons:
(i) The equivalence point of a titration of a monoprotic acid is the point at which moles of added
base = initial moles of acid. This point will also be the point on the curve that is the steepest
point, which is also called an “inflection point” (it is roughly in the middle of the steep segment
that appears nearly linear). These points are shown by an “x” on each plot.
(ii) At the equivalence point, by definition, all the HA has been reacted away and turned into A
100%
(because the titration reaction is: OH + HA 

 A + H2O
-
-
-
-
)
+
Thus, the solution at the equivalence point is a solution of A (with Na around as well):
-
If HA is a weak acid, then A is a weak base (PS5 ideas), and the solution will be basic.
-
If HA is a strong acid, A will be a negligible base (PS5), and the solution will be neutral.
Since the equivalence point pH is well above 7.0 (basic) in (a) it must be the one with the weak
acid. The one in (b), with a pH of ~7.0 (~neutral) must contain the strong acid (assume T ~25°C).
16.11 The pH at the equivalence point of the titration of a strong acid with a strong base is 7.0 (AT 25°C!!). However, the pH
at the equivalence point of the titration of a weak acid with a strong base is above 7.0. Why?
-
-
Answer: At the e.p., the solution is the same as a solution of A . Thus if HA is weak, a solution of A ,
a weak base, will be basic. But if HA is strong, a solution of A , a negligible base, will be
neutral. See above reason for (ii) for more details.
6. 16.62(abc) Two 25.0-mL samples, one 0.100 M HCl and the other 0.100 M HF, were titirated with 0.200 M KOH.
(a) What is the volume of added base at the equivalence point for each titration?
Answer: 12.5 mL (for both)
Work / Reasoning:
1) At the equivalence point (ep), moles of base added = moles of initial acid (monoprotic case)
 moles NaOH added  moles HCl  25.0 mL x
1L
0.100 mol HCl
x
0.00250 mol
1000 mL
L
2) Knowing moles of NaOH added, along with the molarity of the NaOH solution, allows calculation
of the volume of the NaOH solution:
PS7-7
Answer Key, Problem Set 7
0.00250mol NaOH x
1L
1000 mL
x
 12.5 mL
0.200 mol NaOH
L
3) No need to recalculate for HF. Answer will be the same because the molarity and volume of the
HF solution is identical to that of the HCl.
(b) Predict whether the pH at the equivalence point for each titration will be acidic, basic, or neutral.
Answers: neutral for the HCl titration; basic for the HF titration
Reasoning: At ep, HCl titration is a solution of Cl (negligible base)  neutral.
-
At ep, HF titration is a solution of F (weak base)  basic. See #5 (prior problem) for details
-
(c) Predict which titration curve will have the lower initial pH.
Answer: HCl
Reasoning: Initial pH refers to the solutions before any NaOH is added (i.e., the acid solutions). If
+
initial concentrations are the same, the stronger acid will have a greater [H3O ]eq and thus a lower pH.
7. 16.65. Consider the curve for the titration of a weak monoprotic acid with a strong
14
base and answer each question.
(a)
12
What is the pH and what is the volume of added base at the equivalence
point?
10
(b) At what volume of added base is the pH calculated by working an
equilibrium problem based on the initial concentration and Ka of the weak
acid?
(c)
6
x
4
At what volume of added base does pH = pKa?
2
(d) At what volume of added base is the pH calculated by working an
equilibrium problem based on the concentration and Kb of the conjugate
base?
(e)
x
8
pH
0
0
20
40
60
80
Volume of base added (mL)
Beyond what volume of added base is the pH calculated by focusing on the
amount of excess strong base added?
Answers:
(a) pH ~ 8.7 (between 8.5 and 9.0); Vep ~ 30 mL
(b) 0 mL of added base (i.e., the initial point)
(c) At ~15 mL (the half equivalence point [i.e., half of the Vep])
Because here moles HA  moles
A-  ½ initial moles of HA
-
(d) At ~30 mL (the equivalence point, because here the solution is a solution of A )
(e) Again, 30 mL. That is, after the equivalence point volume (because then there is excess base, and
+
that dominates the [OH ]eq and thus [H3O ]eq.
8. 16.67. Consider the titration of a 35.0 mL sample of 0.175 M HBr with 0.200 M KOH. Determine each quantity:
(a) the initial pH
(b) the volume of added base required to reach the equivalence point
(c) the pH at 10.0 mL of added base
(d) the pH at the equivalence point
(e) the pH after adding 5.0 mL of base beyond the equivalence point
Answers: (a) 0.76; (b) 30.6 mL; (c) 1.04; (d) 7.00 (assuming T = 25°C); (e) 12.15
Strategy:
As noted on the PS7 sheet: “In such a titration, one simply does an ICF type calculation, and then
determines whether there is excess HCl, NaOH, or neither (e.p.). There are no buffer calculations or
weak acid or weak base calculations because there is never a buffer nor weak acid nor weak base!”
More specifically, the titration reaction is represented by (this problem involves HBr (not HCl) and KOH (not NaOH):
~100%
 KBr + H2O
HBr + KOH 
-
(OR after “dissociation”: HBr + OH
1) Calculate initial moles from the volume and molarity.
PS7-8
~100%

 H2O + Br -)
Answer Key, Problem Set 7
2) Calculate the moles of KOH added, when appropriate. Use stoichiometry (ICF, if you like) to
determine who is limiting and who is left over (and how much, in moles).
3) Taking into account the total volume (i.e., initial volume + volume added), calculate the
+
concentration of (left over) H3O or OH .
-
4) If OH is left over, assume T = 25°C (because they didn’t specify, and if you don’t assume this, you can’t find the
+
-14
pH), and determine [H3O ] from [OH ] as “usual” (using Kw = 1.0 x 10
at 25°C).
5) For the part where you are asked to calculate Vep, do as was done in problem #6 on this set
(basically a “stoichiometry with molarity and volume” problem).
Execution of Strategy:
Initial moles of HBr: 0.0350 L x 0.175 mol/L = 0.006125 mol HBr
(a) Initial [HBr]  0.175 M  [H3O ]  0.175 M  pH  -log(0.175)  0.7569… 0.757 (3 SF)
[0.76 is OK]
(b) At ep, moles KOH added  moles HBr initial  0.006125 mol (see above)
+
0.006125 mol KOH x
1L
1000 mL
x
 30.625 mL  30.6 mL (of 0.200 M KOH)
0.200mol KOH
L
(c) After 10.0 mL ( 0.0100 L) of KOH added:
mol KOH added  0.0100 L x 0.200 mol/L  0.00200 mol
Set up an ICF table (not necessary, but helpful to do / show stoichiometry):
-
in moles
HBr
OH
I
0.006125
0.00200 (LR)
C
-0.00200
-0.00200
F
0.004125
0
Excess HBr  [HBr]excess 
Br
–
(not relevant)
0.004125 mol
 0.09166 M HBr
 1L

(35.0 mL  10.0 mL)

 1000 mL 
VHBr, initial
VKOH
added
 pH  -log(0.09166)  1.0377..  1.038 (3 SF) [1.04 is OK]
+
-
(d) At the equivalence point, the solution must be neutral (solution of K and Br ). Thus, assuming T =
25°C, pH = 7.00
(e) After 5.00 mL of KOH beyond the equivalence point (i.e., at 35.6 mL of KOH added), the amount of
excess KOH can be calculated directly (because they told you it was 5.00 mL (= 0.00500 L) past the
equivalence point!):
-
moles excess OH = 0.00500 L x 0.200 mol/L = 0.00100 mol OH
[OH ]excess 
0.00100mol
-
 1L

(35.0 mL  30.6 mL  5.00 mL)

 1000 mL 
VHBr, initial
[H3O ] 
+
-
 0.01416 M OH-
VKOH added to reach ep VKOH added after ep
1.0 x 10-14
-13
 7.06..x 10-13 M  pH  -log(7.06..x 10 )  12.151  12.15
0.01416
PS7-9
Answer Key, Problem Set 7
9. 16.71 Consider the titration of a 20.0-mL sample of 0.105 M HC2H3O2 with 0.125 M NaOH. Determine each
quantity:
(a) the initial pH
(b) the volume of added base required to reach the equivalence point
(c) the pH at 5.0 mL of added base
(d) the pH at one-half the equivalence point
(e) the pH at the equivalence point
(f) the pH after adding 5.0 mL of base beyond the equivalence point
Answers: (a) 2.86; (b) 16.8 mL; (c) 4.37; (d) 4.74; (e) 8.75; (f) 12.17
Strategy:
a) The initial pH is determined by doing a standard “weak acid” problem. Find Ka from Table 15.5 and
use the given initial concentration to set up an ICE table and “problem”. Solve for x, which will be
+
equal to the [H3O ] and then calculate pH.
b) Same as in prior problem. Finding the equivalence point volume (of added base) is a stoichiometry
problem with molarity and volumes.
c) For points between the initial and the equivalence point, you have a buffer. Use an ICF table to find
moles of HA remaining and moles of A generated. Then find the pH as discussed earlier:
+
[H3O ]  Ka x [HA]  

moles HA 

[A ]  moles A 



pH  pKa + log [A ]  or moles A 


OR
[HA] 
moles HA 
(or you could do a longer ICE approach with small x approximations)
d) At half the equivalence point, pH  pKa ([H3O ]  Ka)
+
-
-
e) At the equivalence point, you must calculate the concentration of A by dividing the moles of A (=
initial moles of HA) by the total volume (Vinitial acid solution + Vbase added), in L. Then calculate Kb from Ka
and do an ICE type of problem. Remember, however, that “x” here is going to be equal to [OH ], not
+
+
[H3O ], so you’ll need to calculate [H3O ] before finding pH.
-
f) After the equivalence point, the only thing that matters is how much excess OH is left in the
solution. The calculation is analogous to part (e) of the prior problem.
Execution of Strategy:
(a) Ka(HC2H3O2)  1.8 x 10 (at 25°C)
-5

C2H3O2 + H3O ; K a 
-
HC2H3O2 + H2O
+
-
[C2H3O2 ]eq [H3O  ]eq
[HC2H3O2 ]eq
+
[HC2H3O2] (M)
[C2H3O2 ] (M)
[H3O ] (M)
I
0.105
0
~0
C
-x
+x
+x
E
0.105- x
x
x

Ka 
[C2H3O2 ]eq [H3O  ]eq
[HC2H3O2 ]eq
 1.8 x 10-5 
 1.8 x 10-5
x x   1.8 x 10-5
0.105  x 
Try the “small x” approximation (borderline, but let’s hope!):
x x   1.8 x 10-5 
x x   1.8 x 10-5
Assume 0.105 - x  0.105

0.105  x 
0.105
 x 2  0.105(1.8x 10-5 )  1.89 x 10-6
 x  1.89 x 10-6   1.374.. x 10-3
A negative x here does not make sense so use the (+) one.
PS7-10
Answer Key, Problem Set 7

 Check assumption :



1.374.. x 10-3
x 100  1.3% (  5% so OK! ) 
0.105

[H3O ]  x  1.374 x 10
+
-3
 pH  -log(1.374 x 10 )  2.861..  2.86
-3
(b) At ep, moles NaOH added  moles HC2H3O2 initial  0.0200L x
0.00210mol NaOH x
0.105mol HC 2H3O2
 0.00210mol
1L
1L
1000 mL
x
 16.8 mL
0.125mol KOH
L
(c) pH after 5.0 mL of 0.125 M KOH added:
(initial) moles HC2H3O2  0.00210 (from (b))
(initial) moles OH  0.00500 L x 0.125 mol/L  0.000625 mol
-
-
-
Reaction that occurs: OH + HC2H3O2 → C2H3O2 + H2O (OH- will react with the strongest acid around.)
in moles
OH
I
-
-
HC2H3O2
C2H3O2
0.000625 (LR)
0.00210
0
C
- 0.000625
- 0.000625
+ 0.000625
F
0
0.001475
0.000625
Decent amounts of HC2H3O2 and C2H3O2 are present after this reaction  a buffer!
-
[H3O ]  Ka x
+
moles HC 2H3 O2
moles C2H3O2

 1.8 x 10-5 x
0.001475
 4.248 x 10-5 M
0.000625
Answer is reasonable. pKa here is –log(1.8 x 10-5)
 4.74. This buffer has more HA than A-  pH
should be a bit lower than 4.74 and it is.
 pH  -log(4.248 x 10 )  4.371..  4.37
-5
(d) At the half equivalence point, pH  pKa  -log(1.8 x 10 )  4.74
-5
(e) At the e.p., moles NaOH added  moles HC2H3O2  0.00210 (part (b))
(You should not need an ICF table here because you “know” you’re at the equivalence point, but I will include one so that
you can see how this would work out if you had not calculated the e.p. first.)
-
-
OH + HC2H3O2 → C2H3O2 + H2O
in moles
OH
I
-
-
HC2H3O2
C2H3O2
0.00210
0.00210
~0
C
- 0.00210
- 0.00210
+ 0.00210
F
0
0
0.00210
No HC2H3O2 left  not a buffer! Just a solution of
a weak base.
----------------“pH of a weak base” problem:
Kw
1.0 x 10-14

 5.56 x 10-10
Need Kb and [C2H3O2 ]. Kb 
K a (HC 2H3O2 ) 1.8 x 10-5
C2H3O2 ,
And since the total volume is now 20.0 mL (original acid) + 16.8 mL (added KOH) = 36.8 mL = 0.0368 L,
-
[C2H3O2 ] =
0.00210mol
 0.057065..M
0.0368L
PS7-11
Answer Key, Problem Set 7
HC2H3O2 + OH ; K b 
-
-
C2H3O2 + H2O
-
[HC2H3O2 ]eq [OH- ]eq
-
[C2H3O2 ]eq
-
[C2H3O2 ] (M)
[HC2H3O2] (M)
[OH ] (M)
I
0.057065..
0
~0
C
-x
+x
+x
E
0.057065..- x
x
x
K b  5.56 x 10-10 
 5.56 x 10-10
( x )( x )
x2

 x  (0.057065..) x (5.56 x 10-10 )  5.630.. x 10-6
(0.057065..  x ) 0.057065..
-6
(Assumption is clearly valid 5 x 10 <<
0.057)
(Assume that 0.05706 – x 
0.05706..)
[OH ]  x  5.630.. x 10 M; [H3O ] 
-
-6
+
1.0 x 10-14
-9
 1.77.. x 10-9 M  pH  -log(1.77.. x 10 )  8.75
-6
5.630.. x 10
(f) After 5.00 mL of NaOH beyond the equivalence point, the amount of excess NaOH can be calculated
directly (because they told you it was 5.00 mL ( 0.00500 L) past the equivalence point!):
-
moles excess OH = 0.00500 L x 0.125 mol/L = 0.000625 mol OH
[OH ]excess 
0.000625mol
-
 1L

(20.0 mL  16.8 mL  5.00 mL)

1000
mL


VHBr, initial
[H3O ] 
+
-
 0.01495 M OH-
VKOH added to reach ep VKOH added after ep
1.0 x 10-14
-13
 6.688..x10-13 M  pH  -log(6.688..x 10 )  12.174  12.17
0.01495
10. 16.75 Consider the titration curves for two weak acids, both titrated with 0.100 M NaOH. Assume equal initial volumes!
14
14
(a)
12
10
pH
(b)
12
10
8
pH
x
8
6
6
4
4
2
2
0
x
0
0
20
40
60
80
100
Volume of base added (mL)
(i) Which acid solution is more concentrated?
(ii) Which acid has the larger Ka?
0
20
40
60
80
100
Volume of base added (mL)
Answers:
(a)
(b)
Reasoning:
(i) Assuming equal initial volumes, the one with greater concentration of HA has the greater number of
moles of HA. Since solution (a) takes more moles of NaOH (more mL of solution--~38 mL vs ~30
mL) to reach the equivalence point, it must contain more initial moles of HA. Don’t be fooled by the
initial pH! That value depends on both Ka and [HA]. Although the initial pH in (a) is higher (meaning
PS7-12
Answer Key, Problem Set 7
+
the initial [H3O ] is lower) that does not mean the initial [HA] is lower—it it not! The reason the
+
[H3O ] is lower in (a) is because the acid is considerably weaker (see part (ii)).
(ii) To determine the Ka of each acid, look at the half equivalence points (“x’s” on the plots), since the
pH there equals pKa and Ka  10-pKa . The higher the pKa, the smaller the Ka, so clearly (a) (with pKa
 7.5) has the acid with lower Ka and (b) (with pKa  5.5) has the acid with the greater Ka. Acid (a) is
weaker (which is why the pH is higher initially despite the
14
higher initial [HA] (see part (i)).
12
11. 16.78 A 0.446-g sample of an unknown monoprotic acid is titrated
with 0.105 M KOH. The resulting titration curve is shown
here. Determine the molar mass and pKa of the acid.
10
pH
8
6
Answers: MM = 120 g/mol; pKa  4.3  0.2 (at least)
4
Reasoning:
2
x
To determine molar mass, you need two things: grams of a
0
0
20
40
60
80
100
sample, and moles of that sample. The grams of this sample is
Volume of base added (mL)
given. To find moles of HA here, find the moles of KOH needed
to react with all of the HA (i.e., the moles of KOH needed to
reach the equivalence point). This can be done using the equivalence point volume and the molarity of
the KOH solution:
Vep (from curve) is about 36 mL (see vertical line on plot)  0.036 L
0.036 L x 0.105 mol/L  0.00378 moles KOH  moles HA (initial) because it’s a 1:1 stoichiometry.
Molar Mass 
0.446 g HA
 117.9..  120 g/mol
0.00378 mol HA
The reading of the plot (Vep) limits precision here
to about 2 SF in my estimation. Also, my plot (in
the key) is not exactly the same as the one in the
book (it is not a copy), so there will be some
variability in answers to greater than 2 SF, to be
sure.
The pKa is the pH at the half equivalence point (“x” on plot;
~18 mL). I estimate a value of about 4.3 from my plot,
although it looks a bit lower than that (maybe 4.1 or 4.2-ish?) on the plot in the text itself. Clearly there is
uncertainly to at least 0.1 and probably a bit more.
12. 16.81
Methyl red has a pKa of 5.0 and is red in its acid form and yellow in its basic form. (a) If several
drops of this indicator are placed in a 25.0-mL sample of 0.100 M HCl, what color with the solution
appear? (b) If 0.100 M NaOH is slowly added (i.e., added in small increments) to the HCl sample, in
what pH range will the indicator change color?
Answers:
(a) red; (b) from about 4.0 – 6.0
Reasoning:
-
Let’s call the acid form of methyl red HIn and the base form In . The pH of a 0.100 M HCl solution is
1.0 (–log(0.100) = 1.00). This is 4 pH units below methyl red’s pKa. Thus, the ratio of HIn / In will be
10000 : 1 (see mathematical “proof” of this at the end of this solution*), meaning nearly all of the indicator units are
in the HIn (acid) form, which is red (see problem info). Thus the solution will appear red.
-
Now as the pH increases, the ratio of HIn / In will get smaller and smaller, meaning that there will be
fewer of the In species in the HIn (red) form and more in the In (yellow) form. When it is one unit below
pKa (here, that would be 4), the ratio will be 10 : 1, so 10 out of every 11 “In” species will be in the HIn
form. That’s over 90%, so the color will still appear mostly red. But nearly 10% of the In species will
now be yellow and the color will thus start to change. By the time the pH reaches pKa (here, 5), there
will be equal amounts of the red form and the yellow form, so the color will be orange. And by the time
the pH reaches one unit above pKa (here, 6), the ratio of HIn / In will be 1 : 10, meaning over 90% of
the In species will be in the yellow form, and the solution will appear primarily yellow. In general, then,
for any indicator, the pH range in which its color changes is its own “buffer range”: i.e., pKa  1. In this
problem, that is 5.0  1.0 or 4.0 – 6.0.
PS7-13
Answer Key, Problem Set 7
* Proof of this: From a calculational point of view, think of a solution of HIn and In- as a buffer (even though it won’t “act” like one
in this solution because there is such a small amount in the system relative to the amount of HCl and/or NaOH (later)). If pKa  5.0, then Ka
 10-5. At pH 1.0, [H3O+]  10-1 M and we have:
[H3O ]eq  K a x
[HIn]
[In- ]

[HIn] [H3O ]eq 101

 5  104  10000
Ka
[In ]
10

Obviously, you could see this with the HH equation as well (Tro does this), but as you already know, I prefer the “modified
Ka expression” approach to buffers (it’s easier to rederive if you forget it [just one step from the Ka expression equation).
13. 16.100. Determine whether or not each compound will be more soluble in acidic solution than in pure water. Explain.
(a) Hg2Br2
(b) Mg(OH)2
(c) CaCO3
(d) AgI
Answers: (b) and (c) will be more soluble in acidic solution; (a) and (d) will not be.
Reasoning:
The issue here is whether or not the anion of the ionic compound will react with added acid (i.e., if it is
a non-negligible base). If it does, then adding acid will have the effect of decreasing the concentration
of the base anion, which will shift the solubility equilibrium to the right (i.e., toward dissolution of the
solid). If, however, the anion is one of the negligible bases (conjugate to a strong acid), then there will
be no such effect, and the solubility will not increase when acid is added.
-
-
In this case, (a) and (d) have Br and I , both negligible bases, as the anion. Thus solubility will be
2unaffected. However, (b) has OH and (c) has CO3 , both non-negligible bases. Thus these
compounds’ solubility will increase in acidic solution.
14. NT1. You have two salts, AgX and AgY, with very similar Ksp values. You know that the Ka value for HX is much greater
than the Ka value for HY. Which salt is more soluble at low pH? Explain.
Answer: AgY is more soluble in acidic solution.
Reasoning:
As noted in the prior problem, solubility will increase as pH decreases (i.e, as acid is added or solution
is acidified) if the anion of the salt is a non-negligible base. This is because the added acid will react
with the anion (A ) to form its conjugate acid HA thereby lowering the concentration of A (a product)
and “coaxing” more forward reaction (dissolution) to occur. This can be thought of in Le Châtelier’s
terms as a “stress” in which a product species’ concentration is decreased—forward reaction (in this
case, dissolution) will occur to help “alleviate the stress”. The stronger of a base the anion is, the
+
greater will be the extent to which it reacts with a given concentration of H3O , and so the greater will
be the increase in solubility.
-
In this case, since Ka for HX is “much greater”, Kb for X must be much smaller (the stronger the acid,
the weaker its conjugate base; or KaKb = Kw), and so AgX should be less soluble at low pH than AgY (which
will be affected more by a lowering in pH).
One could look at this mathematically as follows:
Kb 
[HA]eq [OH- ]eq
[A - ]eq

[HA]eq
[A - ]eq

[H3O ]eq K b
Kb
Kb


[OH- ]eq Kw /[H3O ]eq
Kw
+
This equation shows that for a given [H3O ] (and thus a given pH), the greater the Kb, the greater is the
[HA]eq
ratio
, which means a greater fraction of A ‘s are protonated and thus “taken out of solution”.
[A - ] eq
PS7-14
Answer Key, Problem Set 7
15. 15.111. Calculate the concentration of all species in a 0.500 M solution of H2SO3. (By “all species”, let’s assume they
mean H2SO3, HSO3-, H3O+, OH-, and SO32-). Assume T = 25°C. Also calculate the pH.
Answers: [H2SO3]  0.418 M; [HSO3 ]  [H3O ]  8.2 x 10 M; pH = 1.10;
-
[OH ]  1.2 x 10
-
+
-2
M; [SO3 ]  6.4 x 10 M
-13
2-
-8
Strategy:
1) Recognize that H2SO3 is a diprotic acid, and thus there will be two acid ionization equilibria involved.
-2
-8
Look up Ka1 and Ka2 for H2SO3 in Table 15.10 (1.6 x 10 and 6.4 x 10 respectively).
2) Although not necessary at this moment, it is helpful to write out the two acid ionization equations to
help you see what is “going on”:
-
H2SO3(aq) + H2O(l)
+
-2
HSO3 (aq) + H3O (aq)
-
2-
HSO3 (aq) + H2O(l)
Ka1 = 1.6 x 10
+
-8
SO3 (aq) + H3O (aq)
Ka2 = 6.4 x 10
nd
3) Now recognize that since the 2 ionization is much less product favored than the first, we can
+
consider the first equilibrium (alone) to determine [H2SO3]eq, [HSO3 ]eq, [H3O ]eq (and thus pH and
[OH ]). That is, the only purpose for doing any calculation involving the second ionization will be to
2determine [SO3 ]eq!
4) Do an weak acid ionization problem (ICE table setup) considering the first ionization only. This will
+
result in determining [H2SO3]eq, [HSO3 ]eq, and [H3O ]eq.
5) Calculate pH and [OH ]eq from [H3O ]eq using [H3O ][OH ]  Kw ( 1.0 x 10
-
+
+
[HSO3 ]eq
-
-14
at 25°C)
+
6) Use the equilibrium values for
and [H3O ]eq already determined as the initial
concentrations in the second ionization / ICE type problem. Apply the “small x” approximation and
2solve for x, which here will equal [SO3 ].
Execution of Strategy:
[HSO3 ]eq [H3 O  ]eq
-
-
H2SO3(aq) + H2O(l)
HSO3 (aq)
-
[H2SO3 ]eq
[HSO3 ] (M)
[H3O ] (M)
I
0.500
0
~0
C
-x
+x
+x
E
0.500 – x
x
x
[HSO3 ]eq [H3O  ]eq
 1.6 x 10-2 
[H2SO3 ]eq
 1.6 x 10-2
+
[H2SO3] (M)
-
K a1 
K a1 
+
+ H3O (aq)
( x )( x )
 x 2  1.6 x 10-2 0.500  x 
(0.500  x )
 x2  8.00 x 10-3  1.6 x 10-2 x  x2  1.6 x 10-2 x  8.00 x 10-3  0
 x
 1.6 x 10-2 
1.6 x 10 
-2 2

 4(1)  8.00 x 10-3
2(1)
   1.6 x 10
-2
 0.1795..
2
 8.179..x 10-2 ()root
(Negative root is not meaningful for this physical system.)


(8.179..x 10-2 )(8.179..x 10-2 )
 Check (+) root :
 1.599.. x 10-2 (matchesK a ) 
-2

(0.500

8.
1
79
..
x
10
)


Thus:
[H2SO3]eq  0.500 – x  0.500 – 0.08179..  0.4182  0.418 M
[HSO3 ]eq  [H3O ]eq  x  8.1799.. x 10 M
-
+
-2
PS7-15
 pH  -log(8.1799.. x 10 )  1.087  1.09
-2
Answer Key, Problem Set 7
[OH ]eq  1.0 x 10 /0.0818  1.22.. x 10
-
-14
-13
 1.2 x 10
-13
M
[SO3 ]eq [H3O  ]eq
2-
HSO3 (aq)
2SO3 (aq)
+ H2O(l)
+ H3O (aq)
-
-
[HSO3 ]eq
[HSO3 ](M)
[SO3 ] (M)
2-
[H3O ] (M)
I
0.0818
0
0.0818
C
-x
+x
+x
E
0.0818 – x
x
0.0818 + x
[SO3 ]eq [H3O  ]eq
2-
K a2 
K a2 
+
-
[HSO3 ]eq
 6.4 x 10-8 
 [SO3 ]  x  6.4 x 10 M
2-
-8
 6.4 x 10-8
+
( x )(0.0818  x ) ( x )(0.0818)

 x (Assuming x << 0.0818)
(0.0818  x )
(0.0818)
(Clearly the assumption was good: 6 x 10-8 <<
0.08)
16. 15.27 & 15.116 (except omit H2S from the problem)
15.27. For a binary acid, H—Y, what factors affect the relative ease with which the acid ionizes?
Answer:
The polarity and bond strength of the H-Y bond both play a role.
All other things being equal:
The greater the polarity of the bond, with the H being more positive, the stronger the acid will be.
The greater the bond strength of the H-Y bond, the weaker the acid will be.
15.116. Based on their molecular structure, arrange the binary compounds in order of increasing acid strength. Give your
reasoning.
H2Te, HI, NaH
Answer: NaH < H2Te < HI
Reasoning:
These are binary acids. As such, the bond polarity and bond strength play a role (see above). The
electronegativities of Na, Te, and I are in the order Na < Te < I. You can rationalize this as follows (i.e., you
need not have the actual values of electronegativity available). Na is a metal on the far left of the periodic table. Its
electronegativity will be smaller than any nonmetal. Te and I are nonmetals on the far right, but I is to the right of Te. Since EN
tends to increase as you move right on the periodic table, I’s EN should be greater than Te’s.
The greater the EN, the more an atom will pull bonding electrons to itself. Thus, the H-I bond should
be most polar (with H being partially positive), making HI the best acid. The H-Na bond should be the
most polar (with H being partially negative), making NaI the worst acid. and H2Te is in the middle.
NOTE: This argument ignores bond strength.
17. 15.117. Based on their molecular structure, pick the stronger acid from each pair of oxyacids. Give reasoning.
Answers
(a) H2SO4 or H2SO3
H2SO4
(b) HClO2 or HClO
HClO2
(c) HClO or HBrO
HClO
Reasoning
In both of these cases, the chosen acid has one more O bonded to the “X”
atom (S or Cl, here). That pulls more electron density from “X”, which, in turn
will pull more electron density from the O bonded to an acidic H, making the
H-O bond more polar, the H more positive, and thus the acid stronger. See
picture below.
The number of O’s is the same, but in one case “X” is Cl and in the other, it is
Br. Since Cl has a greater EN, it will pull more electron density from O
bonded to the H, making the H-O bond more polar and the acid stronger. See
picture below.
PS7-16
Answer Key, Problem Set 7
(d) CCl3COOH or
CH3COOH
CCl3COOH
Cl has a greater EN than H. So the three Cl’s will pull electron density from
the C (bound to the C bound to the O) to a greater extent than the 3 H’s on
the other structure. As noted above, such an “inductive effect” will make the
H-O bond more poloar, the H more positive, and the acid stronger. See
picture below (although imagine the picture “flipped” (since the H is on the
right in these formulas rather than on the left).
H—O—X—O
“Acidic hydrogen” (i.e., a
H that will be donated as
+
H when the acid acts
like a BL acid)
PS7-17
Anything that causes electrons
to move in the direction of the
arrows will make the H more
partially positive and thus easier
+
to be given away as “H ” (i.e.,
makes the acid stronger).