The \(\mathrm{pH}\) of \(10^{-8} \mathrm{M} \mathrm{HCl}\) solution is

As \(\mathrm{HCl}\) solution is acidic in nature and the \(\mathrm{pH}\) of an acidic solution cannot exceed more than 7 . Thus, the answer must be \(6.9586\).
We can also find this answer through calculations. From acid, \(\left[\mathrm{H}^{+}\right]=10^{-8} \mathrm{M}\)
But the \(\left[\mathrm{H}^{+}\right]\)ions from water, i.e., \(\left[\mathrm{H}^{+}\right]=10^{-7} \mathrm{M}\) cannot be neglected in comparison to \(10^{-8} \mathrm{M}\). The \(\mathrm{pH}\) can be calculated as follows
From acid, \(\left[\mathrm{H}^{+}\right]=10^{-8} \mathrm{M}\)
From water, \(\left[\mathrm{H}^{+}\right]=10^{-7} \mathrm{M}\)
\(\begin{aligned}
\therefore \text { Total }\left[\mathrm{H}^{+}\right] &=10^{-8}+10^{-7}=10^{-8}(1+10) \\
&=11 \times 10^{-8} \mathrm{M} \\
\therefore \quad \mathrm{pH} &=-\log \left[\mathrm{H}^{+}\right] \\
&=-\log \left(11 \times 10^{-8}\right) \\
&=-\left[\log 11+\log 10^{-8}\right] \\
&=-[1.0414-8]=6.9586
\end{aligned}\)