Only litre buffer solution was prepared by adding 0.10 mol each of $$NH_3$$ and $$NH_4Cl$$ in deionised water. The change in pH on addition of 0.05 mol of HCl to the above solution is _____ $$\times 10^{-2}$$.(Nearest integer)
(Given : $$pK_b$$ of $$NH_3$$ = 4.745 and $$\log_{10}3 = 0.477$$)

For a basic buffer the Henderson-Hasselbalch form is
$$\mathrm{pOH}=pK_b+\log\left(\frac{[\text{salt}]}{[\text{base}]}\right)$$

Case 1: Before adding HCl
Number of moles of $$NH_3 = 0.10$$ mol, number of moles of $$NH_4Cl = 0.10$$ mol in $$1\,$$L, so
$$[\text{base}]=0.10\ \text{M},\;[\text{salt}]=0.10\ \text{M}$$
$$\frac{[\text{salt}]}{[\text{base}]}=1 \;\Longrightarrow\; \log 1 = 0$$
$$\therefore \mathrm{pOH}_1 = pK_b + 0 = 4.745$$
$$\mathrm{pH}_1 = 14 - \mathrm{pOH}_1 = 14 - 4.745 = 9.255$$

Case 2: After adding $$0.05$$ mol HCl
The strong acid reacts completely: $$NH_3 + H^+ \rightarrow NH_4^+$$
Moles after reaction:
$$NH_3: 0.10 - 0.05 = 0.05\ \text{mol}$$
$$NH_4^+: 0.10 + 0.05 = 0.15\ \text{mol}$$
With total volume still $$1\,$$L,
$$[\text{base}]=0.05\ \text{M},\;[\text{salt}]=0.15\ \text{M}$$
$$\frac{[\text{salt}]}{[\text{base}]} =\frac{0.15}{0.05}=3,\;\; \log 3 = 0.477$$
$$\mathrm{pOH}_2 = pK_b + 0.477 = 4.745 + 0.477 = 5.222$$
$$\mathrm{pH}_2 = 14 - 5.222 = 8.778$$

Change in pH
$$\Delta\mathrm{pH} = \mathrm{pH}_1 - \mathrm{pH}_2 = 9.255 - 8.778 = 0.477 \approx 0.48$$

Expressed as $$\Delta\mathrm{pH}=48 \times 10^{-2}$$.