${\mathrm{p}}^{\mathrm{H}}$ of 0.005M Barium hydroxide solution at 298 K will be

$$\begin{gathered} \left[{\mathrm{OH}}^{-}\right]=\mathrm{Normality}\mathrm{of}\mathrm{Ba}{\left(\mathrm{OH}\right)}_2\text{\hspace{0.17em};} \\ \left[{\mathrm{OH}}^{-}\right]=2\times 5\times {10}^{-3}; \\ \left[{\mathrm{OH}}^{-}\right]={10}^{-2}; \\ {\mathrm{p}}^{\mathrm{OH}}=2; \\ {\mathrm{p}}^{\mathrm{H}}=12 \end{gathered}$$