At certain temperature, ${\mathrm{p}}^{\mathrm{H}}$ of 0.01N $\mathrm{Ba}{\left(\mathrm{OH}\right)}_2$ is 11.699. Ionic product of water a that at temperature is

$\begin{array}{l}\left[{\mathrm{OH}}^{-}\right]={10}^{-2}\mathrm{M};\\ {\mathrm{p}}^{\mathrm{OH}}=2;\\ {\mathrm{p}}^{\mathrm{H}}=11.699;\\ {\mathrm{p}}^{\mathrm{H}}+{\mathrm{p}}^{\mathrm{OH}}=13.699;\\ {\mathrm{p}}^{{\mathrm{k}}_{\mathrm{W}}}=13.699;\\ {\mathrm{p}}^{{\mathrm{k}}_{\mathrm{W}}}=-{\mathrm{logk}}_{\mathrm{W}}\\ \log {\mathrm{k}}_{\mathrm{w}}=-13-0.699\\ \log {\mathrm{k}}_{\mathrm{w}}=-13-0.699+1-1\\ \log {\mathrm{k}}_{\mathrm{w}}=-14+0.301\\ {\mathrm{k}}_{\mathrm{w}}=\mathrm{Antilog}\left(\overline{14}.3010\right)\\ {\mathrm{k}}_{\mathrm{w}}=2\times {10}^{-14}\end{array}$