One ${\mathrm{dm}}^3$ solution of ${10}^{-4}\mathrm{M} \mathrm{HCl}$ solution is diluted 1000 times. Its pH will be

Since HCl solution has a pH of 5,

$\left[{\mathrm{H}}^{+}\right]={10}^{-5}$

It is twice diluted $\Rightarrow {\mathrm{V}}_2=1000{\mathrm{V}}_1$

In terms of millimetres, ${\mathrm{M}}_1{\mathrm{V}}_1={\mathrm{M}}_2{\mathrm{V}}_2$

${\mathrm{M}}_2={10}^{-8}$ but since $\left[{\mathrm{H}}^{+}\right]={10}^{-8}\text{. }$

pH=8 (cannot be 8 for an acid)

Therefore, we must take into account the concentration of ${10}^{-6}\mathrm{H}$ when the concentration of HCl is smaller than  ${\mathrm{H}}^{+}={10}^{-7}$ via water

So, total $\left[{\mathrm{H}}^{+}\right]={10}^{-7}+{10}^{-8}$

$\begin{array}{l}\left[{\mathrm{H}}^{+}\right]=\left(1.1\right)\times {10}^{-7}\\ \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]\\ =7-\log \left[\mathrm{H}\right]\\ =7.004\\ =6.96\cong 7\end{array}$

HCl would have a pH of 7.

Hence the option B is correct.