Derive the equation of the relation between the weak base ionization constant $K_b$ and its conjugate acid ionization constant $K_a$.

(N/A) $NH_3$ (ammonia) is a weak base and its conjugate acid is $NH_4^+$. The following equilibrium is established for $NH_3$:
$(i) NH_{3(aq)} + H_2O_{(l)} \rightleftharpoons NH_{4(aq)}^+ + OH_{(aq)}^-$
$K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}$
$NH_4^+$ is the conjugate acid of $NH_3$. Its equilibrium in aqueous solution is:
$(ii) NH_{4(aq)}^+ + H_2O_{(l)} \rightleftharpoons H_3O_{(aq)}^+ + NH_{3(aq)}$
The ionization constant of the weak acid $NH_4^+$ is $K_a$:
$K_a = \frac{[H_3O^+][NH_3]}{[NH_4^+]}$
Adding reaction $(i)$ and $(ii)$ gives the net reaction:
$(i) + (ii) = 2H_2O_{(l)} \rightleftharpoons H_3O_{(aq)}^+ + OH_{(aq)}^-$
This is the self-ionization of water,where $K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14}$.
Multiplying the equilibrium constants $K_a$ and $K_b$:
$K_a \times K_b = \frac{[H_3O^+][NH_3]}{[NH_4^+]} \times \frac{[NH_4^+][OH^-]}{[NH_3]}$
$K_a \times K_b = [H_3O^+][OH^-] = K_w$
Thus,$K_a \times K_b = K_w$.