$NH _{3}+ H _{2} O \rightleftharpoons NH _{4}^{+}+ OH ^{-}$

The ionization constant of $NH _{3}$

$K_{ b }=$ antilog $\left(- pK _{ b }\right)$ i.e.

$K_{b}=10^{-4.75}=1.77 \times 10^{-5} \,M$

$NH _{3}+ H _{2} O \rightleftharpoons NH _{4}^{+}+ OH ^{-}$

Initial concentration $(M)$

$0.10$                              $0.20$         $0$

Change to reach equilibrium $(M)$

$-x$                                  $+x$            $+x$

At equilibrium $(M)$

$0.10-x$                       $0.20+x$           $x$

$K_{ b }=\left[ NH _{4}^{+}\right]\left[ OH ^{-}\right] /\left[ NH _{3}\right]$

$=(0.20+x)(x) /(0.1-x)=1.77 \times 10^{-5}$

As $K_{ b }$ is small, we can neglect $x$ in comparison to $0.1 \,M$ and $0.2\, M$. Thus,

$\left[ OH ^{-}\right]= x =0.88 \times 10^{-5}$

Therefore, $\left[ H ^{+}\right]=1.12 \times 10^{-9}$

$pH =-\log \left[ H ^{+}\right]=8.95$