Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\mathrm{NaOH}$ is a strong base, thus $$\begin{array}{rl}\left[{\mathrm{OH}}^{-}\right]& =0.0\mathrm{M}=10^{-2}\mathrm{M}\\ \mathrm{pOH}& =-\log \left[{\mathrm{OH}}^{-}\right]=-\log \left(10^{-2}\right)=2\end{array}$$ We know that, $\mathrm{pH}+\mathrm{pOH}=14$ $\therefore \mathrm{pH}=14-2=12$ Thus, option (3) is correct.