Derive the equation $K_a \times K_b = K_w$.

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