On mixing final volume $=255 \mathrm{ml}$ $$ \begin{aligned} & {\left[\mathrm{NH}_4 \mathrm{OH}\right]=\frac{5 \times 0.1}{255},\left[\mathrm{NH}_4 \mathrm{Cl}\right]=\frac{250 \times 0.1}{255}} \\ & \mathrm{p}^{\mathrm{OH}}=\mathrm{pk}_{\mathrm{b}}+\log \frac{\left[\mathrm{NH}_4 \mathrm{Cl}\right]}{\mathrm{NH}_4 \mathrm{OH}} \\ & \mathrm{p}^{\mathrm{OH}}=4.74+\log \frac{250 \times 0.1}{5 \times 0.1} \\ & \mathrm{P}^{\mathrm{H}}=14-\mathrm{P}^{\mathrm{OH}}=14-6.44 \end{aligned} $$