Chemistry - Ionic Equilibrium Question with Solution | TestHub

$75\mathrm{mL}\frac{\mathrm{M}}{5}\mathrm{HCl}+25\mathrm{mL}\frac{\mathrm{M}}{5}\mathrm{NaOH}$ $25\mathrm{mL}\frac{\mathrm{M}}{5}\mathrm{NaOH}$ will neutralise $25\mathrm{mL}\frac{\mathrm{M}}{5}\mathrm{HCl}$ $75-25=50\mathrm{mL}\frac{\mathrm{M}}{5}\mathrm{HCl}$ will remain. Total volume will be $75+25=100\mathrm{mL}$ $50\mathrm{mL}\frac{\mathrm{M}}{5}\mathrm{HCl}$ is diluted to $100\mathrm{mL}$ $\left[{\mathrm{H}}^{+}\right]=[\mathrm{HCl}]=\frac{\mathrm{M}}{5}\times \frac{50}{100}=\frac{\mathrm{M}}{10}$ $$\mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right]=-{\log }_{10}\frac{\mathrm{M}}{10}=1$$