Ionization Constants for Weak Acids (ANSWERS)
Answer all of the following questions, showing all work and calculations. Don’t forget to reflect the appropriate
unit and number of significant digits in your final answer.
1. The reaction H2S(g) + H2O(l) → H3O+(aq) + HS-(aq) has a Ka = 1.0 x 10-7.
a. What is [H3O+] in a 0.50 M solution of H2S?
H2 S
H3 O+
0.50 M
0
I
-x
+x
C
0.50 – x
+x
E
Since M = 0.50 = >1000 then 0.50 – x ≈ 0.50
Ka 1.0 x 10-7.
Ka = [H3O+][HS-]
[H2S]
(1.0 x 10-7) = (x)(x)
(0.50)
5.0 x 10-8 = x2
2.2 x 10-4 M = x = [H3O+]
HS0
+x
+x
b. percent dissociation = [H+] / [H2S] x 100 = 2.2 x 10-4 / 0.50 x 100 = 0.044%
2. What is the percent dissociation of a 0.0274 mol/L solution of NH2OH, given that its pH at this
concentration is 9.191? The value of Kb for NH2OH is 8.80 ×10–9.
pOH = 14.000 – pH = 14.000 – 9.191 = 4.809
[OH–] = 10– pOH = 1.55 × 10–5 mol/L
NH2OH(aq) + H2O(l) → NH2OH2+(aq) + OH–(aq)
Percent dissociation =
1.55 105
× 100% = 0.0566 %
0.0274
3. A 0.10 M solution of an unknown monoprotic acid has a pH = 1.96
a. Find Ka for this solution?
[H+] = 10-pH = 10-1.96 =1.10 x 10-2
HX
H+
X0.10 M
0
0
I
-2
-2
- 1.10 x 10
+ 1.10 x 10
+ 1.10 x 10-2
C
0.089
+ 1.10 x 10-2
+ 1.10 x 10-2
E
Ka = [H+][X-]
[HX]
Ka = (1.10 x 10-2)(1.10 x 10-2)
(0.089)
-3
Ka = 1.4 x 10
b. percent dissociation = [H+] / [HX] x 100 = 1.10 x 10-2 / 0.10 x 100 = 11%
2
4. A 0.40 M solution of HClO3 is 0.027% dissociated. Find the Ka for this acid.
[H+] = 0.027 • (0.40) = 1.08 x 10-4 M
100
HClO3(aq) → H+(aq) + ClO3-(aq)
HClO3
H+
ClO30.40 M
0
0
I
-4
-4
- 1.08 x 10
+ 1.08 x 10
+ 1.08 x 10-4
C
0.3999
+ 1.08 x 10-4
+ 1.08 x 10-4
E
Ka = [H+][ClO3-]
[HClO3]
Ka = (1.08 x 10-4)(1.08 x 10-4)
(0.3999)
-8
Ka = 2.9 x 10
5. An unknown base (XOH) has a concentration of 0.18 M and its Kb = 1.6 x 10-6.
a. Write the equilibrium expression to represent this reaction. (1 pt)
Kb = [X+][OH-]
[XOH]
b. Calculate its [OH-] (1 pt)
XOH(aq) → X+(aq) + OH-(aq)
XOH
0.18 M
I
-x
C
0.18 - x
E
X+
0
+x
+x
OH0
+x
+x
Since M = >1000 then 0.18 – x ≈ 0.18
Kb
Kb = [X+][OH-]
[XOH]
(1.6 x 10-6) = (x)(x)
(0.18)
x2 = 2.9 x 10-7
x = [OH-] = 5.4 x 10-4
c. Determine its pOH. (1 pt)
pOH = - log [OH-] = - log [5.4 x 10-4] = 3.27
d. pH = 14 – pOH = 14 – 3.27 = 10.73
e. Determine its percent dissociation. (1 pt)
percent dissociation = [OH-] / [XOH] x 100 = 5.4 x 10-4 / 0.18 x 100 = 0.30%