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