$?molNaOH=80mgNaOH\times \frac{1gNaOH}{{{10}^{3}}mgNaOH}\times \frac{1molNaOH}{40gNaOH}=0/002molNaOH$  غلظت مولار $\frac{0/002mol}{0/1L}=0/02mol.{{L}^{-1}}$  $\left[ NaOH \right]=\left[ O{{H}^{-}} \right]=0/02mol.{{L}^{-1}}$  $\left[ {{H}_{3}}{{O}^{+}} \right]=\frac{{{K}_{W}}}{\left[ O{{H}^{-}} \right]}=\frac{{{10}^{-14}}}{2\times {{10}^{-2}}}=5\times {{10}^{-13}}mol.{{L}^{-1}}$  $PH=-\log \left[ {{H}_{3}}{{O}^{+}} \right]=-\log 5\times {{10}^{-13}}=-\log \frac{10}{2}\times {{10}^{-13}}=-\log \frac{{{10}^{-12}}}{2}=-(-12\log 10-\log 2)=12\log 10+\log 2=12(1)+0/3=12/3$