Base Dissociation Constant (Kb) - Chemistry

1. Fundamental Concepts

Definition: The base dissociation constant, $$ K_b $$, is a measure of the strength of a base in solution. It quantifies the extent to which a base dissociates into its constituent ions.
Expression: For a base $$ B $$ that dissociates in water to form $$ BH^+ $$ and $$ OH^- $$, the expression for $$ K_b $$ is given by: $$ K_b = \frac{[BH^+][OH^-]}{[B]} $$
Relationship with $$ K_w $$: The product of the acid dissociation constant ($$ K_a $$) and the base dissociation constant ($$ K_b $$) for a conjugate acid-base pair is equal to the ionization constant of water ($$ K_w $$): $$ K_a \cdot K_b = K_w $$

Strong Bases: $$K_b \gg 1$$

Strong bases completely dissociate in water, resulting in a very large $$ K_b $$.

Weak Bases: $$K_b \ll 1$$

Weak bases only partially dissociate in water, resulting in a small $$ K_b $$.

Application: Used to determine the strength of bases and predict the pH of solutions.

Problem: Calculate the $$ K_b $$ for a 0.1 M solution of ammonia ($$ NH_3 $$) if the concentration of $$ OH^- $$ is $$ 1.8 \times 10^{-5} M $$.

Write the dissociation equation: $$ NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- $$
Set up the $$ K_b $$ expression: $$ K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} $$
Assume the change in concentration of $$ NH_3 $$ is $$ x $$: $$ [NH_4^+] = x = [OH^-] = 1.8 \times 10^{-5} M $$ $$ [NH_3] = 0.1 - x \approx 0.1 M \text{ (since } x \text{ is very small)} $$
Substitute the values into the $$ K_b $$ expression: $$ K_b = \frac{(1.8 \times 10^{-5})(1.8 \times 10^{-5})}{0.1} = \frac{3.24 \times 10^{-10}}{0.1} = 3.24 \times 10^{-9} $$

Validation: The calculated $$ K_b $$ value is consistent with the known $$ K_b $$ for ammonia, which is approximately $$ 1.8 \times 10^{-5} $$.

Problem: Given a 0.2 M solution of methylamine ($$ CH_3NH_2 $$) with a $$ K_b $$ of $$ 4.4 \times 10^{-4} $$, calculate the pH of the solution.

Write the dissociation equation: $$ CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^- $$
Set up the $$ K_b $$ expression: $$ K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]} $$
Let the change in concentration be $$ x $$: $$ [CH_3NH_3^+] = [OH^-] = x $$ $$ [CH_3NH_2] = 0.2 - x \approx 0.2 M \text{ (since } x \text{ is very small)} $$
Substitute the values into the $$ K_b $$ expression: $$ 4.4 \times 10^{-4} = \frac{x^2}{0.2} $$ $$ x^2 = 4.4 \times 10^{-4} \times 0.2 = 8.8 \times 10^{-5} $$ $$ x = \sqrt{8.8 \times 10^{-5}} = 9.38 \times 10^{-3} M $$
Calculate the pOH: $$ pOH = -\log[OH^-] = -\log(9.38 \times 10^{-3}) \approx 2.03 $$
Calculate the pH: $$ pH = 14 - pOH = 14 - 2.03 = 11.97 $$

Validation: The calculated pH is consistent with the expected range for a weak base solution.

ICE Table Method: Use an ICE (Initial, Change, Equilibrium) table to organize the concentrations of reactants and products.
Assumption of Small $$ x $$: For weak bases, assume the change in concentration $$ x $$ is small compared to the initial concentration, simplifying the calculations.
Logarithmic Calculations: Use logarithms to convert between concentrations and pH/pOH values.