$\left[ {{H}_{3}}{{O}^{+}} \right]={{10}^{-PH}}={{10}^{-11}}mol.{{L}^{-1}}$  $\left[ O{{H}^{-}} \right]=\frac{{{K}_{W}}}{\left[ {{H}_{3}}{{O}^{+}} \right]}=\frac{{{10}^{-14}}}{{{10}^{-11}}}={{10}^{-3}}mol.{{L}^{-1}}$  $\left[ O{{H}^{-}} \right]=\left[ KOH \right]={{10}^{-3}}mol.{{L}^{-1}}=0/2LKOH\times \frac{0/001gKOH}{80gKOH}=0/014gKOH$