⇒Ionic Equilibrium

Percent dissociation $=1.42\mathrm{\%}$

$\therefore \alpha =0.0142$

For a weak monoacidic base,

$\begin{array}{rl}{\mathrm{K}}_{\mathrm{b}}& ={\alpha }^2\mathrm{C}\\ & =(0.0142{)}^2\times (0.05)\\ & =1.0082\times {10}^{-5}\end{array}$

$\underset{0.02M}{NaOH}\stackrel{}{⟶}N{a}^{+}+\underset{0.02M}{O{H}^{-}}$

$\begin{array}{rl}{P}^{\mathrm{OH}}& =-{\log }_{10}\left[{\mathrm{OH}}^{-}\right]=-{\log }_{10}[0.02]\\ & =-{\log }_{10}(2\times {10}^{-2})=2-0.3010\\ & {\mathrm{P}}^{\mathrm{OH}}=1.6990\\ \therefore & {\mathrm{P}}^{\mathrm{H}}=14-{\mathrm{P}}^{\mathrm{H}}=14-1.6990=12.301\end{array}$