50$\mathrm{mL}$ of $0.2\mathrm{M}$ ammonia solution is treated with $25 \mathrm{mL}$ of $0.2\mathrm{MHCl}$. If ${\mathrm{pK}}_{\mathrm{b}}$ of ammonia solution is $4.75.\mathrm{pH}$ of the mixture will be

 Number of millimoles of  ${\mathrm{NH}}_3=50 \mathrm{mL}\times 0.2\mathrm{M}=10\mathrm{mmol}$

Number of millimoles of $\mathrm{HCl}=25 \mathrm{mL}\times 0.2\mathrm{M}=5\mathrm{mmol}$

$5{\mathrm{mmolNH}}_3$ will react with $5\mathrm{mmolHCl}$ to form $5{\mathrm{mmolNH}}_4\mathrm{Cl}$

$10-5=5{\mathrm{mmolNH}}_3$ will remain.

$\mathrm{pOH}$=${\mathrm{pK}}_{\mathrm{b}}-\log \frac{\left[{\mathrm{NH}}_4\mathrm{Cl}\right]}{\left[{\mathrm{NH}}_3\right]}$

$\mathrm{pOH}=4.75-\log \frac{5/\mathrm{V}}{5/\mathrm{V}}$

$\mathrm{pOH}=4.75$

$\mathrm{pH}=14-\mathrm{pOH}$=$14-4.75=9.25$

Therefore, the correct option is (C).