Derive the expression for the ionization constant $(K_b)$ of a weak base.

Consider a weak base $MOH$ that ionizes in water as follows:
$MOH_{(aq)} \rightleftharpoons M^+_{(aq)} + OH^-_{(aq)}$
Let $C$ be the initial concentration of the base and $\alpha$ be the degree of ionization.
At equilibrium:
$[MOH] = C(1 - \alpha)$
$[M^+] = C\alpha$
$[OH^-] = C\alpha$
The ionization constant $K_b$ is given by the law of mass action:
$K_b = \frac{[M^+][OH^-]}{[MOH]}$
Substituting the equilibrium concentrations:
$K_b = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)}$
$K_b = \frac{C^2\alpha^2}{C(1 - \alpha)}$
$K_b = \frac{C\alpha^2}{1 - \alpha}$
For a very weak base,$\alpha \ll 1$,so $(1 - \alpha) \approx 1$. Thus,the expression simplifies to:
$K_b \approx C\alpha^2$
$\alpha = \sqrt{\frac{K_b}{C}}$
$[OH^-] = C\alpha = \sqrt{K_b \cdot C}$