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

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

Before adding 0.1 M NaOH ${\mathrm{p}}^{\mathrm{H}}$  of weak base 0.1M ${\mathrm{NH}}_3$ is

${\mathrm{p}}^{\mathrm{H}}=\frac{1}{2}\left[{\mathrm{p}}^{{\mathrm{k}}_{\mathrm{b}}}-{\log }_{10}\mathrm{C}\right]=\frac{1}{2}\left[4.7+1\right]=2.85$

When 0.1 M NaOH is added due to common ion effect dissociation of ${\mathrm{NH}}_3$ decreases.

$$\begin{gathered} {\mathrm{NH}}_3+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{NH}}_4^{+}+{\mathrm{OH}}^{-} \\ {\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\mathrm{NH}}_4^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{NH}}_3\right]}=\frac{0.1\left[{\mathrm{OH}}^{-}\right]}{0.1+\mathrm{x}}=2\times {10}^{-5} \\ \mathrm{x}=\left[{\mathrm{OH}}^{-}\right] \mathrm{combining} \mathrm{with} \left[{\mathrm{NH}}_4^{+}\right] \mathrm{to} \mathrm{form} \mathrm{ammonia} \mathrm{solution} \mathrm{in} \mathrm{backward} \mathrm{direction} \\ \left[{\mathrm{OH}}^{-}\right]=2\times {10}^{-5} \left[\mathrm{since} 0.1+\mathrm{x}\cong 0.1\right] \\ {\mathrm{p}}^{\mathrm{OH}}=4.7 \\ \mathrm{Change} \mathrm{in} {\mathrm{p}}^{\mathrm{OH}}=4.7-2.85=1.85 \\ \mathrm{For} \mathrm{a} \mathrm{given} \mathrm{basic} \mathrm{solution} \mathrm{change} \mathrm{in} {\mathrm{p}}^{\mathrm{OH}} \mathrm{is} \mathrm{also} \mathrm{equal} \mathrm{to} \mathrm{change} \mathrm{in} {\mathrm{p}}^{\mathrm{H}} \end{gathered}$$