b
\(75\,mL\,\frac{M}{5}\,HCl + 25\,ml\,\frac{M}{5}\,NaOH\)

\(25\,mL\,\frac{M}{5}\,NaOH\,\) 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 \(100\,mL\)

\([{H^ + }] = [HCl] = \frac{M}{5} \times \frac{{50}}{{100}} = \frac{M}{{10}}\)

\(pH =  - {\log _{10}}[{H^ + }] =  - {\log _{10}}\frac{M}{{10}} = 1\)