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, $\left[\mathrm{H}^{+}\right]=10^{-7} \mathrm{M}$ $\begin{aligned} \therefore \text { Total }\left[\mathrm{H}^{+}\right] &=10^{-8}+10^{-7}=10^{-8}(1+10) \\ &=11 \times 10^{-8} \mathrm{M} \\ \therefore \quad \mathrm{pH} &=-\log \left[\mathrm{H}^{+}\right] \\ &=-\log \left(11 \times 10^{-8}\right) \\ &=-\left[\log 11+\log 10^{-8}\right] \\ &=-[1.0414-8]=6.9586 \end{aligned}$