3.2 - The Base Ionization Constant, Kb
Module 8
Lesson 3.2 The Base Ionization Constant, Kb
Key Concepts
You learned in Lesson 1.2 that bases vary in strength (similar to acids). The Base Ionization Constant, Kb, describes the extent to which weak bases produce hydroxide ions in water.
Consider the following reactions of the cyanide ion with water:
\( \mathrm { CN^-(aq) + H_2O(l) \leftrightarrows HCN(aq) + OH^-(aq) } \)
The equilibrium law expression for this reaction is written as follows:
\( \mathrm { K_b = \dfrac{[HCN(aq)] [OH^-(aq)] }{ [CN^-(aq)] } } \)
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Note that the concentration of liquid water is omitted from the Kb expression. Assume that the value remains virtually constant. |
The Table of Relative Strengths of Aqueous Acids and Bases does not provide Kb values. Fortunately, we can derive the Kb value for a weak acid easily by using the following formula:
\( \mathrm { K_w = K_a \times K_b } \)
Therefore, \( \mathrm { K_b = \dfrac{K_w}{K_a} } \)
To find the Kb for any weak base, identify its conjugate acid on the table of Relative Strengths of Aqueous Acids and Bases. Then divide Kw (1.0 x 10-14) by the Ka value for that particular conjugate acid.
To calculate empirical Kb values and amount concentrations for weak bases, use ICE tables. Remember the "Rule of Thumb" that states that the initial concentration of the base will remain unchanged at equilibrium if the initial concentration of the base is at least 1000 times greater than its Kb value.
Watch
Read pages 744 to 749 of the textbook and work through the Sample problems and Communication examples.
Check Your Understanding
Complete Practice Questions 10 to 13 on page 746 of the textbook. Also, complete Section 16.3 Questions 1, and 2 on page 750. Check your answers by clicking the link below.
Page 746 Practice Question 10
| Empirical Property | Expected Result with a Strong Base | Expected Result with a Weak Base |
| Conductivity | strong electrolyte | weak electrolyte |
| pH | high | lower, but above 7 |
| Rate of reaction | higher | lower |
Page 746 Practice Question 11
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\( \mathrm { CN^-(aq) + H_2O(l) \leftrightarrows HCN(aq) + OH^-(aq) } \)
\( \mathrm { K_b = \dfrac{[HCN(aq)] [OH^-(aq)] }{ [CN^-(aq)] } } \)
-
\( \mathrm { SO_4^{2-}(aq) + H_2O(l) \leftrightarrows HSO_4^- (aq) + OH^-(aq) } \)
\( \mathrm { K_b = \dfrac{[HSO_4^- (aq)] [OH^-(aq)] }{ [SO_4^{2-}(aq)] } } \)
Page 746 Practice Question 12
| Amount Concentration |
[C2H5COO-(aq)]
(mol/L) |
[C2H5COOH(aq)]
(mol/L) |
[OH-(aq)]
(mol/L) |
| Initial | 0.157 | 0 | 0 |
| Change | - 1.1 x 10-5 | + 1.1 x 10-5 | + 1.1 x 10-5 |
| Equilibrium |
0.157
(= 0.157 x 10-5 = 0.1576989, an insignificant change) |
1.1 x 10-5 | 1.1 x 10-5 |
\( \mathrm { K_b = \dfrac{[C_2H_5COOH(aq)] [OH^-(aq)] }{ [C_2H_5COO^-(aq)] } } \)
\( \mathrm { K_b = \dfrac{[1.1 \times 10^{-5} \frac{mol}{L}] [1.1 \times 10^{-5} \frac{mol}{L}] }{ [0.157 \frac{mol}{L}] } = 7.7 \times 10^{-10 } } \)
Page 746 Practice Question 13
\( \mathrm { pOH = 14.00 - 8.81 = 5.19 } \)
\( \mathrm { [OH^-(aq)] = 10^{-5.19} = 6.5 \times10^{-6} \frac{mol}{L} } \)
| Amount Concentration |
[C6H5NH2(aq)]
(mol/L) |
[C6H5NH3(aq)]
(mol/L) |
[OH-(aq)]
(mol/L) |
| Initial | 0.10 | 0 | 0 |
| Change | - 6.5 x 10-6 | + 6.5 x 10-6 | + 6.5 x 10-6 |
| Equilibrium |
0.10
(an insignificant change) |
6.5 x 10-6 | 6.5 x 10-6 |
\( \mathrm { K_b = \dfrac{[6.5 \times 10^{-6} \frac{mol}{L}] [6.5 \times 10^{-6} \frac{mol}{L}] }{ [0.10 \frac{mol}{L}] } = 4.2 \times 10^{-10 } } \)
Page 750 Section 16.3 Question 1
\( \mathrm { Cod(aq) + H_2O(l) \leftrightarrows Cod^+(aq) + OH^-(aq) } \)
Approximation:
\( \mathrm { \dfrac{[Cod(aq)] }{ K_b } = \dfrac{0.020 \frac{mol}{L} }{ 1.73 \times 10^{-6} }= 11560 } \), significantly higher than 1000; therefore, the assumption holds, and the change in concentration in codeine is insignificant.
\( \mathrm { K_b = \dfrac{ [Cod^+(aq)] [OH^-(aq)] }{ [Cod(aq)] } } \)
Let \( \mathrm { x = [OH^-(aq)] = [Cod^+(aq)] } \)
\( \mathrm { 1.73 \times 10^{-6} = \dfrac{[x] [x] }{ [0.020 \frac{mol}{L}] } } \)
\( \mathrm { x = \sqrt{1.73 \times 10^{-6} ~[0.020 \frac{mol}{L}]} } \)
\( \mathrm { x = 1.9 \times 10^{-4} \frac{mol}{L} } \)
\( \mathrm { [OH^-(aq)] = 1.9 \times 10^{-4} \frac{mol}{L} } \)
\( \mathrm { pOH = -log(1.9 \times 10^{-4} \frac{mol}{L}) = 3.72 } \)
Page 750 Section 16.3 Question 2
\( \mathrm { CN^-(aq) + H_2O(l) \leftrightarrows HCN(aq) + OH^-(aq) } \)
\( \mathrm { K_{b_{(cyanide~ion)}} = \left(\dfrac{K_w}{K_a}\right)_{HCN} = \dfrac{1.00 \times 10^{-14}}{6.2 \times 10 ^{-5}} } \)
Approximation: \( \mathrm { \dfrac{0.18 \frac{mol}{L} }{ 1.61 \times 10^{-5} }= 11160 } \), which is greater than 1000; therefore, the change in concentration of CN- is not significant.
\( \mathrm { K_b = \dfrac{ [HCN(aq)] [OH^-(aq)] }{ [CN^-(aq)] } } \)
Let \( \mathrm { x = [OH^-(aq)] = [HCN(aq)] } \)
\( \mathrm { 1.61 \times 10^{-5} = \dfrac{[x] [x] }{ 0.18 \frac{mol}{L} } } \)
\( \mathrm { x = \sqrt{ 1.61 \times 10^{-5} ~ \left(0.18 \frac{mol}{L}\right)} } \)
\( \mathrm { x = 0.0017\frac{mol}{L} } \)
\( \mathrm { [OH^-(aq)] = 0.0017\frac{mol}{L} } \)
\( \mathrm { pOH = -log(0.0017\frac{mol}{L}) = 2.77 } \)
\( \mathrm { pH = 14.00 -2.77 = 11.23 } \)