A $50mL$ of $0.02MN{H}_{4}OH$ solution is titrated with $0.02MHCl$. Find the $pH$ of the solution when $50.1mL$ of $HCl$ is added.
Given: $K_b\left(N{H}_{4}OH\right)=1.8\times {10}^{-5}$

The correct option is B 4.65
mmol of $N{H}_{4}OH=50\times 0.02=1$
mmol of $HCl=50.1\times 0.02=1.002$

$N{H}_{4}OH\left(aq\right)+HCl\left(aq\right)\to N{H}_{4}Cl\left(aq\right)+H_{2}O\left(l\right)Initially:11.00200Final:00.00211$
$\text{Final Concentration}=\frac{\text{Moles}}{\text{Total Volume}}$
Taking Total volume $\approx 100mL$ for calculation purposes.
$\left[HCl\right]=\frac{0.002}{100}=2\times {10}^{-5}M\left[N{H}_{4}Cl\right]=\frac{1}{100}=0.01M$
We will take $H^{+}$ from both $HCl$ and hydrolysis of $N{H}_{4}Cl$
$N{H}_{4}Cl$ will further hydrolyse as

$N{H}_4^{+}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons N{H}_{4}OH\left(aq\right)+H^{+}\left(aq\right)Equilibrium:C(1-h)ChCh$
where $h$ is the degree of hydrolysis:
$h=\sqrt{\frac{K_w}{K_b\times C}}h=\sqrt{\frac{{10}^{-14}}{1.8\times {10}^{-5}\times 0.01}}h=\sqrt{5.55\times {10}^{-8}}h=2.35\times {10}^{-4}$
$\left[H^{+}\right]=Ch=0.01\times 2.35\times {10}^{-4}\left[H^{+}\right]=2.35\times {10}^{-6}$
$[H^{+}{]}_{total}=[H^{+}{]}_{HCl}+\left[H^{+}{]}_{N{H}_{4}Cl}\right[H^{+}{]}_{total}=(2\times {10}^{-5})+(2.35\times {10}^{-6})=2.235\times {10}^{-5}pH=-log\left[H^{+}\right]pH=-log(2.235\times {10}^{-5})=(5-0.35)pH=4.65$