Determine the degree of ionization and $pH$ of a $0.05 \, M$ ammonia solution. The ionization constant of ammonia $(K_{b})$ is $1.77 \times 10^{-5}$. Also,calculate the ionization constant of the conjugate acid of ammonia.

(N/A) $NH_{3} + H_{2}O \rightleftharpoons NH_{4}^{+} + OH^{-}$
We use the expression for the ionization constant of a weak base:
$[OH^{-}] = c \alpha = 0.05 \alpha$
$K_{b} = \frac{c \alpha^{2}}{1 - \alpha}$
Since $\alpha$ is small,we assume $1 - \alpha \approx 1$,so $K_{b} \approx c \alpha^{2}$.
$\alpha = \sqrt{\frac{K_{b}}{c}} = \sqrt{\frac{1.77 \times 10^{-5}}{0.05}} = 0.0188$
$[OH^{-}] = c \alpha = 0.05 \times 0.0188 = 9.4 \times 10^{-4} \, M$
$[H^{+}] = \frac{K_{w}}{[OH^{-}]} = \frac{10^{-14}}{9.4 \times 10^{-4}} = 1.064 \times 10^{-11} \, M$
$pH = -\log[H^{+}] = -\log(1.064 \times 10^{-11}) = 10.97$
For the conjugate acid-base pair,$K_{a} \times K_{b} = K_{w}$.
$K_{a} = \frac{K_{w}}{K_{b}} = \frac{10^{-14}}{1.77 \times 10^{-5}} = 5.65 \times 10^{-10}$