## General Chemistry 10th Edition

We know that $H_3O^+=\frac{K_w}{OH^-}$ $H_3O^+=\frac{1.0\times 10^{-14}}{0.050}=2\times 10^{-3}M$ $PH=-log(H_3O^+)$ $PH=-log (2\times 10^{-13})=12.698$