$\left[ {{H}_{3}}{{O}^{+}} \right]\left[ O{{H}^{-}} \right]={{K}_{W}}\Rightarrow {{10}^{-14}}=\left( 4\times {{10}^{8}}\times \left[ O{{H}^{-}} \right] \right)\left[ O{{H}^{-}} \right]\Rightarrow {{\left[ O{{H}^{-}} \right]}^{2}}=\frac{{{10}^{-14}}}{4\times {{10}^{8}}}=0/25\times {{10}^{-22}}=2/5\times {{10}^{-23}}$  $_{PH=-\log \left[ {{H}_{3}}{{O}^{+}} \right]=-\log 2\times {{10}^{-3}}=-\log 2+3\log 10=-0/3+3=2/7}^{\left[ {{H}_{3}}{{O}^{+}} \right]=\frac{{{K}_{W}}}{\left[ O{{H}^{-}} \right]}=\frac{{{10}^{-14}}}{0/5\times {{10}^{-12}}}=2\times {{10}^{-3}}mol.{{L}^{-1}}}$