To 1L solution containing 0.1 mol each of $N{H}_3$  and $N{H}_{4}Cl,$ 0.05 mol of NaOH is added .The change in pH will be $\left(p{K}_{b}forN{H}_3=4.74\right)$  

$N{H}_{4}Cl+NaOH\stackrel{}{\to }NaCl+N{H}_3+H_{2}O$

Moles of $N{H}_4^{+}$ left =0.1- 0.05 = 0.05

Total moles of $N{H}_3$ = 0.1+0.05 = 0.15

${\left(pOH\right)}_1=p{K}_b+\log \frac{\left[salt\right]}{\left[base\right]}=p{K}_b+\log \frac{0.1}{0.1}=p{K}_b$

${\left(pOH\right)}_2=p{K}_b+\log \frac{0.05}{0.15}=p{K}_b-\log 3$

Change in $pOH={\left(pOH\right)}_2-{\left(pOH\right)}_1=-\log 3$

Change in pH = log 3 = 0.48