Following four solutions are prepared by mixing different volumes of $\mathrm{NaoH}$and $\mathrm{HCl}$ of different concentrations, $\mathrm{pH}$ of which one of them will be equal to 1 ?

 $\mathrm{pH}$ of $75 \mathrm{mL}\frac{\mathrm{M}}{5}\mathrm{HCl}+25 \mathrm{mL}\frac{\mathrm{M}}{5}\mathrm{NaOH}$ will be equal to 1 .

Total volume $=75 \mathrm{mL}+25 \mathrm{mL}=100 \mathrm{mL}$

Number of $\mathrm{mmol}$ of $\mathrm{HCl}=75 \mathrm{mL}\times \frac{\mathrm{M}}{5}=15\mathrm{mmol}$

Number of $\mathrm{mmol}$ of $\mathrm{NaOH}=25 \mathrm{mL}\times \frac{\mathrm{M}}{5}=5\mathrm{mmol}$

$5\mathrm{mmol}$ of \$\mathrm{NaOH}$ will neutralise $5\mathrm{mmol}$ of $\mathrm{HCl}$.

$15-5=10\mathrm{mmol}$ of $\mathrm{HCl}$ will remain.

$\left[{\mathrm{H}}^{+}\right]=\left[\mathrm{HCl}\right]=\frac{10\mathrm{mmol}}{100 \mathrm{mL}}=0.1\mathrm{M}$

$\mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right]$

$\mathrm{pH}=-{\log }_{10}0.1\mathrm{M}=1$

Therefore, the correct option is (B).