The \(\mathrm{pH}\) of a \(10^{-8}\) molar solution of \(\mathrm{HCl}\) in water is

\(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log 10^{-8}=8\)
It is not possible for acid, so it is \(\left[\mathrm{H}^{+}\right]\), the \(\left[\mathrm{H}^{+}\right]\) of water is also added. Total \(\left[\mathrm{H}^{+}\right]\) in solution
\(=\left[\mathrm{H}^{+}\right] \text {of } \mathrm{HCl}+\left[\mathrm{H}^{+}\right] \text {of water } \)
\( =\left(1 \times 10^{-8}+1 \times 10^{-7}\right) \mathrm{M} \)
\( =(1+10) \times 10^{-8}=11 \times 10^{-8} \mathrm{M} \)
\( \therefore \quad \mathrm{pH} =-\log \left[\mathrm{H}^{+}\right] \)
\( =-\log 11 \times 10^{-8} \)
\( =-\log 11+8 \log 10 \)
\( =-1.0414+8=6.9586\)