When 0.1 mole solid NaOH is added to 1 L of 0.1 M ${\mathrm{NH}}_3$ (aq ), then which statement is wrong? 

$\left(K_b=2\times {10}^{-5},\log 2=0.3\right)$

$\begin{array}{l}{\mathrm{K}}_{\mathrm{b}}=2\times {10}^{-5}\left({\mathrm{NH}}_3\right)\\ {\mathrm{NH}}_3\to {\mathrm{NH}}_4+{\mathrm{OH}}^{-}\\ {\mathrm{Ca}}^2=\mathrm{Ka}\\ \alpha =\sqrt{\frac{\mathrm{Ka}}{\mathrm{c}}}=\sqrt{\frac{2\times {10}^{-5}}{0.1}}=\sqrt{2}\times {10}^{-3}\\ \left[{\mathrm{OH}}^{-}\right]=0.1\times \sqrt{2}\times {10}^{-3}=\sqrt{2}\times {10}^{-4}\\ {\mathrm{NaOH}}_{\mathrm{a}}\to \mathrm{Na}+{\mathrm{OH}}^{-}\\ \left[{\mathrm{OH}}^{-}\right]=0.1\mathrm{M}\\ {\mathrm{P}}^{\mathrm{H}}=14-\log 0.1\\ =10\end{array}$

$\begin{array}{r}{\mathrm{P}}^{\mathrm{H}}\end{array}$due to ${\mathrm{NH}}_3=14-\log \sqrt{2}\times {10}^{-4}$

$\begin{array}{l}=14\left(\frac{1}{2}{\log }^2+\log {10}^{-4}\right)\\ =14-(-4+0.15)=10.15\end{array}$

(A) degree of dissociation approaches to zero

$\alpha \to 0$

(B) Change in ${\mathrm{P}}^{\mathrm{H}}$be $13-10.15=2.85$

$\left(\mathrm{C}\right)\left({\mathrm{Na}}^{+}\right)=0.1\mathrm{M}{\mathrm{NH}}_3\simeq 0.1\mathrm{M}$

$\left[{\mathrm{OH}}^{-}\right]=0.1+\sqrt{2}\times {10}^{-4}\simeq 0.1$.