Explanation

(b)
Given, aqueous solution of $NH_{3}\ (weak\ base)C = 0.01$ M
$\alpha = 5\% = \frac{5}{100}$
$\left\lbrack OH^{-} \right\rbrack = C\alpha = 0.01 \times \frac{5}{100} = 5 \times 10^{- 4}$
$pOH = - \log{\lbrack OH^{-}\rbrack}$
$= \log\left( 5 \times 10^{- 4} \right)$
$= 4\log{10 - \log 5}$
$= 4 - 0.6989$
$= 3.3010$
$\therefore$ $pH = 14 - pOH = 14 - 3.3010 = 10.6990$