Problem Set 20 – Answers
Part 1: pH and pOH
1)
pH
(a)
7
[H+]
pOH
1 x 10-7 M
7
(b)
9.5
3.4 x 10-10 M
4.5
(d)
14.18
6.7 x 10-15 M
-0.18
(f)
3.8
1.55 x 10-4 M
10.2
(c)
(e)
(g)
4.3
5.0 x 10-5 M
12.7
-1.22
2.0 x 10-13 M
16.6 M
1.3
0.050 M
2) Calculate the pH of the following solutions.
pH = −log[ H + ]
pH = −log[1.00]
pH = 0
[ H + ] = 0.0054 M
b. 0.0054 M HCl pH = −log[0.0054 ]
pH = 2.27
[OH − ] = 0.00071M
pOH = −log[0.00071]
c. 0.00071 M NaOH
pOH = 3.2
pH = 14 − pOH = 10.8 ≈ 11
a. 1.00 M HNO3
d. 0.059 M Ca(OH)2
[OH − ] = (0.059 M) × 2 = 0.12 M
pOH = −log[0.12]
pOH = 0.92
pH = 14 − 0.92 = 13.1
[ H + ] = 2.6 ×10−5 M
e. 2.6 x 10-5M HBr pH = −log[2.6 ×10−5 ]
pH = 4.6
2.9 x 10-5 M
2.0 x 10-10 M
15.22
f.
1 x 10-7 M
9.7
Part 2: pH Calculations with strong acids and bases
[ H + ] = 1.00 M
[OH-]
3.89 x 10-10 M HClO4
1.5 M
6.3 x 10-11 M
6.0 x 10-16 M
[ H + ] = 3.89 ×10−10 M
3.89 ×10−10 M ≤ 10−7
Because the concentration of protons produced
by the perchloric acid is much smaller than the
concentration of protons in solution due to
water dissociation, the [H+] is basically
unaffected by the presence of the acid.
Therefore, pH = 7
Part 3: Simple Weak Acids
3) Calculate the pH of a 0.543 M solution of formic acid.
R
HCOOH (aq) 
HCOO- (aq) +
H+ (aq)
I
0.543 M
0.00 M
0.00 M
C
-x
+x
+x
E
0.543-x ≈ 0.543
+x
+x
Ka = 1.8 ×10−4 =
1.8 ×10−4 =
−
[ H + ][ HCO2 ] ( x)( x)
=
(0.543)
[ HCO2 H ]
[H + ] = x
pH = −log[ x]
pH = −log[0.0098]
pH = 2.0
x2
0.543
x = 0.0098
Note: The approximation here is not perfect. 0.543 – 0.0098 = 0.533. That’s a change of 2%. But I’d still consider that an
acceptable approximation.
4) Calculate the pH of a 0.0664 M solution of aluminum chloride. (Hint: Think about what happens to the aluminum
ions when dissolved in water.)
When AlCl3 dissolves, the ions separate. The Cl- ions are the conjugate base of HCl, a strong acid. Therefore, we can
ignore them. The Al3+ ions get surrounded by water and act as an acid:
Ka=1.4 x 10-5
R
[Al(H2O)6]3+ (aq) 
[Al(H2O)5OH]2+ (aq) +
H+ (aq)
I
0.0664 M
0.00 M
0.00 M
C
-x
+x
+x
E
0.0664-x ≈ 0.0664
+x
+x
Ka = 1.4 ×10−5 =
1.4 ×10−5 =
[ H + ][[ Al( H 2O) 5 OH ]2+ ]
( x)( x)
=
3+
[[ Al( H 2O) 6 ] ]
(0.0664)
x2
0.064
x = 0.000964
[H + ] = x
pH = −log[ x]
pH = −log[0.000964]
pH = 3.0
14.56. (,) Calculate the concentrations of all species present and the pH of a 1.5 M HNO2 solution
(Ka = 4.0 x 10–4).
All quantities in mol/L:
R
HNO2 
H+ +
NO2-
I
1.5
0
0
C
-x
+x
+x
E
≈1.5
x
x
𝐾=
𝑥2
= 4.0 × 10−4
1.5
x = 0.024
[HNO2] = 1.5 M
[H+] = 0.024 M
[NO2-] = 0.024 M
[OH-] = 4.1 x 10-13 M
pH = 1.62
14.68. (,)Calculate the percent dissociation for a 0.22 M solution of chlorous acid
(HClO2, Ka = 1.2 x 10–2).
All quantities in mol/L:
R
HClO2 
H+ +
ClO2-
I
0.22
0
0
C
-x
+x
+x
E
0.22-x
x
x
Note: K is not small enough to use an approximation here.
𝑥2
= 0.012
0.22 − 𝑥
x = 0.0457
𝐾=
𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑜𝑛 =
𝑎𝑚𝑜𝑢𝑛𝑡 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑒𝑑
0.0457 𝑀
=
= 0.21 = 21%
𝑡𝑜𝑡𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡
0.22 𝑀
14.72. (,) Trichloroacetic acid (CCl3CO2H) is a corrosive acid that is used to precipitate proteins. The pH of a
0.050 M solution of trichloroacetic acid is 1.40. Calculate Ka.
[H+] = 10-pH = 10-1.40 = 0.0398 M  This is the concentration of H+ at equilibrium.
All quantities in mol/L:
H+ +
A-
0.050
0
0
C
-x
+x
+x
E
0.050 -xx
x
R
HA
I

Based on the known pH value, x = [H+] = 0.0398
𝑥2
(0.0398)2
=
= 0.155
𝐾𝑎 =
0.050 − 𝑥
0.050 − 0.0398
Based on the Ka you calculated above, is trichloroacetic acid stronger or weaker than acetic acid, CH3CO2H? Can you
explain this difference based on chemical principles?
The chlorine atoms are very electronegative. They essentially pull electron density away from the other atoms in the
molecule. This increases the partial positive charge on the H in the acid, making it more acidic. From the other
perspective, we can say that the partial positive charge induced by the chlorine atoms stabilizes the negative charge on
the conjugate base. A more stable conjugate base makes for a stronger acid. Looking at the Ka value, we can see that
the difference is actually enormous. The Ka for acetic acid is 1.8 x 10-5, almost 10,000 times smaller than the Ka for
trichloroacetic acid.
Part 4: Simple Weak Bases
5) Calculate the pH of a 1.57 M solution of hydrazine.
Kb=3.0 x 10-6
H2NNH3+ (aq) +
OH- (aq)
N/A
0.00 M
0.00 M
-x
N/A
+x
+x
1.57-x ≈ 1.57
N/A
+x
+x
R
H2NNH2 (aq) +
I
1.57 M
C
E
H2O (l) (aq)
+
[OH − ][ H 2 NNH 3 ] ( x)( x)
Kb = 3.0 ×10 =
=
[ H 2 NNH 2 ]
(1.57)
−6
3.0 ×10−6 =
x2

[OH − ] = x
pOH = −log[ x]
pOH = −log[0.00217]
pOH = 2.66
pH = 14 − pOH = 11.34
1.57
x = 0.00217
6) Calculate the pH of a 0.592 M solution of potassium benzoate.
Note that benzoate is the conjugate base of benzoic acid. So potassium benzoate is KC7H5O2. When dissolved in
water, the ions separate. The potassium becomes a spectator, so we ignore it. In order to find Kb for benzoate, we
use the Ka value for benzoic acid.
𝐾𝑏 =
R
C7H5O2- (aq) +
I
10−14
10−14
=
= 1.56 × 10−10
𝐾𝑎
6.4 × 10−5
H2O (l) (aq)
C7H5O2H(aq) +
OH- (aq)
0.592 M
N/A
0.00 M
0.00 M
C
-x
N/A
+x
+x
E
0.592-x ≈ 0.592
N/A
+x
+x
Kb = 1.56 ×10−10 =
−10
1.56 ×10
=
[OH − ][C7 H 5O2 H ] ( x)( x)
=
−
(0.592)
[C7 H 5O2 ]
x2
[OH − ] = 9.61×10−6
pOH = −log[ x]
0.592
−10
1.56 ×10 (0.592) = x 2
pOH = −log[9.61×10−6 ]
pOH = 5.02
pH = 14 − pOH = 8.98 ≈ 9.0
9.25 ×10−11 = x 2
x = 9.61×10−6
14.90a. (,) Calculate [OH–], [H+], and the pH of a 0.200 M solution of the amine aniline.
Part 5: Polyprotic Acids
7) Find the concentration of all species in a 0.0450 M solution of ascorbic acid. (For simplicity’s sake, let’s abbreviate
ascorbic acid as H2Asc). That means you need to find the concentrations of H+, OH-, H2Asc, HAsc-, and Asc2-.
Step 1:
Ka1=7.9 x 10-5
R
H2Asc (aq) 
HAsc (aq) +
H+ (aq)
I
0.0450 M
0.00 M
0.00 M
C
-x
+x
+x
E
0.0450-x
+x
+x
Ka1 = 7.9 ×10−5 =
7.9 ×10−5 =
[ H + ][ HAsc − ]
( x)( x)
=
[ H 2 Asc ]
(0.0450 − x)
x2
0.0450 − x
7.9 ×10 (0.0450 − x) = x 2
−5
3.56 ×10−6 − 7.9 ×10−5 x − x 2 = 0
−b ± b 2 − 4 ac −(−7.9 ×10−5 ) ± (−7.9 ×10−5 ) 2 − 4(−1)(3.56 ×10−6 )
=
2(−1)
2a
x = 0.00185
x=
[ H + ] = [ HAsc − ] = x = 0.00185 M
[ H 2 Asc ] = 0.0450 − x = 0.0432 M
Step 2:
Ka2 = 1.6 x 10-12
R
HAsc- (aq) 
Asc2- (aq) +
H+ (aq)
I
0.00185 M
0.00 M
0.00185 M
C
-x
+x
+x
E
0.00185-x
+x
0.00185 +x
[ H + ][ HAsc − ] (0.00185 + x)( x)
=
[ H 2 Asc ]
(0.00185 − x)
0.00185 + x ≈ 0.00185
0.00185 − x ≈ 0.00185
0.00185 x
=x
1.6 ×10−12 =
0.00185
x = 1.6 ×10−12
Ka 2 = 1.6 ×10−12 =
[ H + ] = 0.00185 + x ≈ 0.00185
[ HAsc − ] = 0.00185 − x ≈ 0.00185 M
[ Asc 2− ] = 1.6 ×10−12 M
[H+]= __0.00185 M_
[OH-]=__5.4 x 10-12 M__
[H2Asc] = __0.0432 M__
[HAsc-]=__0.00185 M___
[Asc2-]=__1.6 x 10-12 M__
14.102. (,) Calculate the pH of a 0.001000 M solution of H2SO4. Note that only the first dissociation is strong.
Ka2 can be found in the book.
14.150. (,) a. The principal equilibrium in a solution of NaHCO3 is
HCO3- + HCO3-  H2CO3 + CO32a. Using the Ka values for carbonic acid, calculate the value of the equilibrium constant for this reaction.