Derive $K_w = K_a \times K_b$ and $pK_w = pK_a + pK_b$ for a weak base $B$ and its conjugate acid $BH^{+}$.

(N/A) For a weak base $B$,the equilibrium in aqueous solution is:
$B_{(aq)} + H_2O_{(l)} \rightleftharpoons BH^{+}_{(aq)} + OH^{-}_{(aq)} \quad \dots (i)$
The base dissociation constant $K_b$ is given by:
$K_b = \frac{[BH^{+}][OH^{-}]}{[B]}$
For the conjugate acid $BH^{+}$,the dissociation equilibrium is:
$BH^{+}_{(aq)} + H_2O_{(l)} \rightleftharpoons B_{(aq)} + H_3O^{+}_{(aq)} \quad \dots (ii)$
The acid dissociation constant $K_a$ is given by:
$K_a = \frac{[B][H_3O^{+}]}{[BH^{+}]}$
Multiplying $K_a$ and $K_b$:
$K_a \times K_b = \left( \frac{[B][H_3O^{+}]}{[BH^{+}]} \right) \times \left( \frac{[BH^{+}][OH^{-}]}{[B]} \right)$
$K_a \times K_b = [H_3O^{+}][OH^{-}] = K_w$
Taking the negative logarithm on both sides:
$-\log(K_a \times K_b) = -\log(K_w)$
$-\log K_a - \log K_b = -\log K_w$
$pK_a + pK_b = pK_w$