Question

Calculate the pH of ${10}^{-8}$ molar HCl solution.

Answer 1

Water has ${10}^{-7}$ M $H^{+}$ and ${10}^{-7}$ M $O{H}^{-}$.

When ${10}^{-8}$ M $HCl$ is added the concentration of proton in water is not neglected.

So, Total $H^{+}$ = ${10}^{-7}+{10}^{-8}$

pH=$-\log \left(H^{+}\right)=-\log \left({10}^{-8}\left(11\right)\right)=8-\log 11=6.598$