The pH at the equivalence point of titration of 0.2 $MN{H}_3$ with 0.2M HCl is : $\left(p^{Kb}ofN{H}_3=4.74\right)$

© At the equivalence point, the volume of solution is doubled.

Hence. $\left[N{H}_{4}Cl\right]=\frac{0.2}{2}=0.1M\left[N{H}_4^{+}\right]$

The salt hydrolyses,

$pH=\frac{1}{2}\left(14-4.74-\log 0.1\right)=5.13$