**Answer: 8.78**

*$pK_a(NH^+_4) = 9.26$*

*$\therefore pK_b (NH_3) = pK_w - pK_a(NH^+_4)$ = 14 - 9.26 = 4.74*

*On adding 15 m mol (50 mL of 0.3 M) HCl to 20 m mol (50 mL of 0.4 M) $NH_3$ the resulting solution contains 15 m mol (in 100 mL) of $NH_4Cl$ and 5 m mol (in 100 mL) of $NH_3$.*

*$\therefore [Salt] = 15 \times 10^{-2} M , [Base] = 5 \times 10^{-2} M$*

*$pOH = pK_b + log \frac{[Salt]}{[Base]} = 4.74 + log \frac{15 \times 10^{-2}}{5 \times 10^{-2}} = 4.74 + log 3 = 4,74 + 0.48 = 5.22$*

*pH = 14 - pOH = 14 - 5.22 = 8.78*