Explanation

(a)
Given, $K_{b} = 1.8 \times 10^{- 5},\ \left\lbrack NH_{3} \right\rbrack = 0.1\ M$
$NH_{3} + H_{2}O \rightleftharpoons NH_{4}^{+} + OH^{-}$
Let concentration of $\left\lbrack NH_{4}^{+} \right\rbrack = \left\lbrack OH^{-} \right\rbrack = x$
$K_{b} = \frac{\left\lbrack NH_{4}^{+} \right\rbrack\left\lbrack OH^{-} \right\rbrack}{\left\lbrack NH_{3} \right\rbrack}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{x \times x}{0.1}$
or $1.8 \times 10^{- 5} = \frac{x^{2}}{0.1}$
or $x^{2} = 1.8 \times 10^{- 6}$
$\therefore$ $x = 1.35 \times 10^{- 3}$
$\therefore$ $\left\lbrack OH^{-} \right\rbrack = 1.35 \times 10^{- 3}$
$pOH = - \log{\lbrack OH^{-}\rbrack}$
$= - \log\left( 1.35 \times 10^{- 3} \right)$
$= 2.87$
$pH = 14 - pOH = 14 - 2.87 = 11.13$