1
Ch. 16 Kimball Spring 2019
Discussion Worksheet - Chapter 17
Additional Aspects of Aqueous Equilibria
Dr Cabirac Lecture
1. Determine the pH of
A. 0.20 M NH3 solution
NH3 + H2O ↔ NH4+ + OH- 𝐾𝑏=[𝑁𝐻4+][𝑂𝐻−]
[𝑁𝐻3]
I 0.20M 0 0 𝐾𝑏= 1.8 𝑥 10−5 =(𝑥)(𝑥)
0.20−𝑥 x <<0.20
C - x + x + x
E 0.20 –x x x x = 0.0019 = [OH-] pOH = -log[OH-]
= -log(0.0019) = 2.72
pH = 14 – pOH = 14-2.72 = 11.28
B. a solution that is 0.20 M NH3 and 0.30 M NH4Cl
NH3 + H2O ↔ NH4+ + OH- 𝐾𝑏=[𝑁𝐻4+][𝑂𝐻−]
[𝑁𝐻3]=(𝑥)(𝑥)
0.20−𝑥
I 0.20M 0.30 0 𝐾𝑏= 1.8 𝑥 10−5 =(0.30+𝑥)(𝑥)
0.20−𝑥 x <<0.20
C - x + x + x
E 0.20 –x 0.30+x x x = 1.2x10-5 = [OH-] pOH = -log[1.2x10-5]
= 4.92
pH = 14 – 4.92 = 9.08
or use Henderson-Hasselbalch
𝑝𝐻 = 𝑝𝐾𝑎+ 𝑙𝑜𝑔[𝐵]
[𝐴] = −log(5.6𝑥10−10)+ 𝑙𝑜𝑔(0.20
0.30) = 𝟗.𝟎𝟖 𝐾𝑎=1.0𝑥10−14
1.8𝑥10−5 = 5.6𝑥10−10
2. Which of the following statements has the greatest buffering capacity?
Since all of the following are composed of the same acid/base, the solution with the greatest buffer
capacity will be the one with the greatest concentrations of the acid and the base – this buffer will
be able to react with the most amount of acid or base to resist a change in pH.
A. 0.40 M CH3COONa / 0.20 M CH3COOH
B. 0.40 M CH3COONa / 0.60 M CH3COOH
C. 0.30 M CH3COONa / 0.60 M CH3COOH
3. Which of the following solutions can act as a buffer?
A. KCl/HCl strong acid/salt NO
B. KHSO4 /H2SO4 strong acid/salt NO
C. Na2HPO4 /NaH2PO4weak acid/salt YES
D. KNO2 /HNO3 SA/salt of WA NO
E. H2SO4/Li2SO4SA/salt of WA NO
F. NH3 /NH4NO3 weak base/salt YES
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Ch. 16 Kimball Spring 2019
4. Calculate the pH of the buffer 0.10 M Na2HPO4 / 0.15 M KH2PO4
H2PO4- ↔ H+ + HPO42- 𝐾𝑎=[𝐻+][𝐻𝑃𝑂42−]
[𝐻2𝑃𝑂4−] x <<0.20
I 0.15M 0 0.10M 𝐾𝑎= 6.2 𝑥 10−8 =(𝑥)(−.10+𝑥)
0.15−𝑥 x <<0.10
C - x + x + x
E 0.15 –x x 0.10+x x = 9.3x10-8 = [H+]
pH = -log(9.3x10-8) = 7.03
or use Henderson-Hasselbalch
𝑝𝐻 = 𝑝𝐾𝑎+ 𝑙𝑜𝑔[𝐵]
[𝐴] = −log(6.2𝑥10−8)+ 𝑙𝑜𝑔(0.10
0.15) = 𝟕.𝟎𝟑
5. The diagrams shown below contain one or more of the compounds: H2A, NaHA, Na2A, where H2A is a
weak diprotic acid. (Water molecules and Na+ ions have been omitted for clarity)
A. Which of the solutions can act as a buffer? a, b, c
B. Which of the solutions is the most effective buffer? c greatest amount of weak acid/conj base
6. The pH of blood plasma is 7.40. Assuming the principle buffer system is HCO3- /H2CO3 ,
A. Calculate the ratio [HCO3- ] / [H2CO3 ]
𝑝𝐻 = 𝑝𝐾𝑎+ 𝑙𝑜𝑔[𝐻𝐶𝑂3−]
[𝐻2𝐶𝑂3]
7.40 = −log (4.2𝑥10−7) + 𝑙𝑜𝑔[𝐻𝐶𝑂3−]
[𝐻2𝐶𝑂3]
𝑙𝑜𝑔[𝐻𝐶𝑂3−]
[𝐻2𝐶𝑂3]= 1.02
[𝐻𝐶𝑂3−]
[𝐻2𝐶𝑂3]=10−1.02 = 𝟏.𝟎𝒙𝟏𝟎𝟏
B. Is this buffer more effective against added acid or added base?
[𝑏𝑎𝑠𝑒]
[𝑎𝑐𝑖𝑑] ˃1 so there is more base present than acid, so this buffer is more effective against the
addition of acid
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Ch. 16 Kimball Spring 2019
7. Calculate the pH of 1.0 L of the buffer 1.0 M CH3COONa /1.0 M CH3COOH before and after the addition
of: (assume volume does not change)
Before any additions, pH of buffer:
𝑝𝐻 = 𝑝𝐾𝑎+ 𝑙𝑜𝑔 [𝐶𝐻3𝐶𝑂𝑂−]
[𝐶𝐻3𝐶𝑂𝑂𝐻] = −log(1.8𝑥10−5)+ 𝑙𝑜𝑔1.0
1.0 = 𝟒.𝟕𝟒
A. 0.080 mol NaOH
First, determine moles of HA/A- after A/B rxn HA + OH-→A- + H2O
1.0 mol 0.080mol 1.0 mol
-0.080 -0.080 +0.080
0.92 0 1.08
Now find pH with new concentrations
𝑝𝐻 = 𝑝𝐾𝑎+ 𝑙𝑜𝑔 [𝐶𝐻3𝐶𝑂𝑂−]
[𝐶𝐻3𝐶𝑂𝑂𝐻] = −log(1.8𝑥10−5)+ 𝑙𝑜𝑔1.08
0.92 = 𝟒.𝟖𝟏
B. 0.12 mol HCl
First, determine moles of HA/A- after A/B rxn A-+ H+ →HA
1.0 mol 0.12 mol 1.0 mol
-0.12 -0.12 +0.12
0.88 0 1.12
Now find pH with new concentrations
𝑝𝐻 = 𝑝𝐾𝑎+ 𝑙𝑜𝑔 [𝐶𝐻3𝐶𝑂𝑂−]
[𝐶𝐻3𝐶𝑂𝑂𝐻] = −log(1.8𝑥10−5)+ 𝑙𝑜𝑔0.88
1.12 = 𝟒.𝟔𝟒
8. A 0.2688 g sample of a monoprotic acid neutralizes 16.4 mL of 0.08133 M KOH solution. Calculate the
molar mass of the acid.
HA + OH- → A- + H2O 1:1 stoichiometry for HA and OH-
mol HA = mol OH-
0.0164 𝐿 𝑁𝑎𝑂𝐻(0.08133 𝑚𝑜𝑙
1𝐿 ) = 0.0013 𝑚𝑜𝑙 𝑂𝐻−= 0.0013 𝑚𝑜𝑙 𝐻𝐴
molar mass (HA) = 𝑔 𝐻𝐴
𝑚𝑜𝑙 𝐻𝐴 =0.2688 𝑔
0.0013 𝑚𝑜𝑙 =𝟐𝟎𝟐 𝒈
𝒎𝒐𝒍
9. A student is asked to prepare a buffer solution at pH=8.60, using one of the following
weak acids: HA (Ka = 2.7 x 10-3), HB (Ka = 4.4 x 10-6 ), HC (Ka = 2.6 x 10-9 ). Which
acid would be be the best choice? Why?
HA pKa = -log(2.7 x 10-3) = 2.57
HB pKa = -log(4.4 x 10-6) = 5.36
HC pKa = -log(2.6 x 10-9) = 8.60 for this acid pKa is closest to target pH
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Ch. 16 Kimball Spring 2019
10. The diagrams here represent solutions at various stages in the titration of a weak base B
(such as NH3) with HCl.
art
Identify the solution that corresponds to
A. The initial stage before the addition of HCl c all base particles
B. Halfway to the equivalence point a half B, half BH+
C. The equivalence point d all BH+
D. Beyond the equivalence point b H3O+ present (excess acid)
E. Is the pH>7, pH<7, or pH=7 at the equivalence point?
11. A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH. Calculate
the pH after the following additions of the KOH solution:
A. 0.0 mL Only weak acid present, so determine pH for equilibrium
CH3COOH ↔ H+ + CH3COO - 𝐾𝑎=[𝐻+][𝐶𝐻3𝐶𝑂𝑂−]
[𝐶𝐻3𝐶𝑂𝑂𝐻]
I 0.100M 0 0 1.8 𝑥 10−5 =(𝑥)(𝑥)
0.100−𝑥 x << 0.100
C - x + x +x
E 0.100 –x x x x = 0.0013M = [H+]
pH = -log(0.0013) = 2.87
B. 5.0 mL As a result of A/B rxn, some CB (CH3COO-) has been produced so buffer
𝑚𝑜𝑙 𝐻𝐴 = 0.100𝑚𝑜𝑙
𝐿𝑥0.025𝐿 = 0.0025 𝑚𝑜𝑙
𝑚𝑜𝑙 𝑂𝐻−= 0.200𝑚𝑜𝑙
𝐿𝑥0.005𝐿 = 0.0010 𝑚𝑜𝑙
HA + OH-→A- + H2O𝑝𝐻 = 𝑝𝐾𝑎+ 𝑙𝑜𝑔 [𝐶𝐻3𝐶𝑂𝑂−]
[𝐶𝐻3𝐶𝑂𝑂𝐻]
0.0025 0.0010 0 = −log(1.8𝑥10−5)+𝑙𝑜𝑔0.0010
0.0015 = 𝟒.𝟓𝟕
-0.0010 -0.0010 +0.0010
0.0015 0 0.0010
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Ch. 16 Kimball Spring 2019
C. 10.0 mL As a result of A/B rxn, some CB (CH3COO-) has been produced so buffer
𝑚𝑜𝑙 𝐻𝐴 = 0.100𝑚𝑜𝑙
𝐿𝑥0.025𝐿 = 0.0025 𝑚𝑜𝑙
𝑚𝑜𝑙 𝑂𝐻−= 0.200𝑚𝑜𝑙
𝐿𝑥0.010𝐿 = 0.0020 𝑚𝑜𝑙
HA + OH-→A- + H2O𝑝𝐻 = 𝑝𝐾𝑎+ 𝑙𝑜𝑔 [𝐶𝐻3𝐶𝑂𝑂−]
[𝐶𝐻3𝐶𝑂𝑂𝐻]
0.0025 0.0010 0 = −log(1.8𝑥10−5)+𝑙𝑜𝑔0.0020
0.0005 = 𝟓.𝟑𝟒
-0.0020 -0.0020 +0.0020
0.0005 0 0.0020 0.0010
D. 12.5 mL This is the equivalence pt, so all HA has been neutralized resulting in only A- in
solution 𝑚𝑜𝑙 𝐻𝐴 = 0.100𝑚𝑜𝑙
𝐿𝑥0.025𝐿 = 0.0025 𝑚𝑜𝑙
𝑚𝑜𝑙 𝑂𝐻−= 0.200𝑚𝑜𝑙
𝐿𝑥0.0125 = 0.0025 𝑚𝑜𝑙
HA + OH-→A- + H2O
0.0025 0.0025 0
-0.0025 -0.0025 +0.0025 [𝐶𝐻3𝐶𝑂𝑂−]=0.0025𝑚𝑜𝑙
0.0375𝐿= 0.0667 𝑀
0 0 0.0025
CH3COO - + H2O ↔ CH3COOH + OH- 𝐾𝑏=[𝐶𝐻3𝐶𝑂𝑂𝐻][𝑂𝐻−]
[𝐶𝐻3𝐶𝑂𝑂−]
I 0.0667M 0 0 5.6 𝑥 10−10 =(𝑥)(𝑥)
0.0667−𝑥 x <<0.0667
C - x + x + x
E 0.0667 –x x x x = 6.11x10-6 = [OH-]
pOH = -log[OH-]
= -log(6.11x10-6) = 5.22
pH = 14 – pOH = 14 – 5.22 = 8.78
E. 15.0 mL Beyond eq. point, so excess OH- present
HA + OH-→A- + H2O
0.0025 0.0030 0
-0.0025 -0.0025 +0.0025 [𝑂𝐻−]=0.0005 𝑚𝑜𝑙
0.040 𝐿 = 0.0125𝑀 𝑂𝐻−
0 0.0005 0.0025
strong B weak B pOH = -log(0.0125) = 1.9
pH = 14 – 1.9 = 12.1
Because OH- is strong and CH3COO– weak, the contribution to
pH of CH3COO– is insignificant to that of OH-
Acid
k,
Base
kK
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SMALL
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62x10
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Need
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17
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bases
Pe
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}
Bases
K,>>4
6
Ch. 16 Kimball Spring 2019
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