Following four solutions are prepared by mixing different volumes of NaOH and HCl of different concentrations, pH of which one of them will be equal to 1?

$$\begin{gathered} 75\mathrm{mL}\frac{\mathrm{M}}{5}\mathrm{HCl}+25\mathrm{mL}\frac{\mathrm{M}}{5}\mathrm{NaOH} \\ 25\mathrm{mL}\frac{\mathrm{M}}{5}\mathrm{NaOH} \mathrm{will} \mathrm{neutralise} 25\mathrm{mL}\frac{\mathrm{M}}{5}\mathrm{HCl} \end{gathered}$$

$$\begin{gathered} 75-25=50\mathrm{mL}\frac{\mathrm{M}}{5}\mathrm{HCl} \mathrm{will} \mathrm{remain}. \\ \mathrm{Total} \mathrm{volume} \mathrm{will} \mathrm{be} 75+25=100 \mathrm{mL} \\ \mathrm{When} 50\mathrm{mL}\frac{\mathrm{M}}{5}\mathrm{HCl} \mathrm{is} \mathrm{diluted} \mathrm{to} 100 \mathrm{mL}, \mathrm{molarity} \mathrm{will} \\ \mathrm{reduce} \mathrm{to} \mathrm{half}, \mathrm{i}.\mathrm{e}., \frac{\mathrm{M}}{10} \\ \left[{\mathrm{H}}^{+}\right]=\left[\mathrm{HCl}\right]=\frac{1}{10} \\ \mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right]=-{\log }_{10}\frac{1}{10}= \end{gathered}$$