Explanation

(b)
$CH_{3}NH_{2}$ (base) on reaction with $HCl$ (acid) to give a salt of weak base and strong acid as $CH_{3}NH_{3}^{+}Cl^{-}$
$CH_{3}NH_{2} + HCl \rightarrow \ CH_{3}NH_{3}^{+}Cl^{-}$
At=0 0.1 mol 0.8 mol 0.08 mol
After reaction (0.1-0.08)
=0.02 mol
So, it acts as basic buffer solution due to presence of base and salt in solution of 1 L.
$pOH = - \log_{10}K_{b} + \log_{10}\frac{\lbrack salt\rbrack}{\lbrack base\rbrack}$
$pOH = - \log_{10}K_{b} + \log_{10}\frac{\lbrack CH_{3}NH_{3}^{+}Cl^{-}\rbrack}{\lbrack CH_{3}NH_{2}\rbrack}$
$= - \log_{10}{5 \times 10^{- 4}} + \log_{10}\frac{\lbrack 0.08\rbrack}{\lbrack 0.02\rbrack}$
$= - \log_{10}{5 +}4\log_{10}10 + \log_{10}\frac{\lbrack 0.08\rbrack}{\lbrack 0.02\rbrack}$
$= - 0.699 + 4 + 0.602 = 3.903$
$pH = 14 - pOH$
$= 14 - 3.903$
$= 10.097 = - \log_{10}{\lbrack H^{+}\rbrack}$
$\therefore$ $\left\lbrack H^{+} \right\rbrack = 8.0 \times 10^{- 11}$