d
\(N{H_3} + HCl \to N{H_4}Cl\)

moles of \(HCl = 0.2\,M \times 25 \times {10^{ - 3}}\,L = 0.005\)

moles \(HCl\) (total consumed)

moles of \(N{H_3} = 0.2\,M \times 50 \times {10^{ - 3}}\,L = 0.01\)

moles \(HCl\)

excess \(N{H_3} = 0.01 - 0.005 = 0.005\) moles

\(1\) mole ammonia \(= 1 \) mole \(N{H_4}Cl\)

\(0.005\,N{H_3} = 0.005\,N{H_4}Cl\)

Total volume 

\( = {V_{HCl}} + {V_{N{H_3}}} = 25 + 50\, = 75\,ml\)

\([N{H_3}] = [N{H_4}Cl] = \frac{{0.005\,mole}}{{75 \times {{10}^{ - 3}}\,L}} = 0.066\,M\)

\(pOH = p{K_b} + \log \frac{{[N{H_4}Cl]}}{{[N{H_3}]}}\)

\(pOH = 4.75 + \log \frac{{[0.066]}}{{[0.066]}}\)

\(pOH = 4.75\)

\(pH = 14 - pOH \Rightarrow pH = 9.25\)