A litre of buffer solution contains 0.1 mole of each of NH$$_3$$ and NH$$_4$$Cl. On the addition of 0.02 mole of HCl by dissolving gaseous HCl, the pH of the solution is found to be _____ $$\times 10^{-3}$$ (Nearest integer)
Given: pK$$_b$$(NH$$_3$$) = 4.745
log 2 = 0.301
log 3 = 0.477
T = 298 K

Find the pH of a buffer solution containing 0.1 mole each of $$NH_3$$ and $$NH_4Cl$$ in 1 L after adding 0.02 mole of HCl.

We first determine the moles after HCl addition. HCl reacts with $$NH_3$$: $$NH_3 + HCl \rightarrow NH_4Cl$$. Moles of $$NH_3$$ remaining = $$0.1 - 0.02 = 0.08$$ mol and moles of $$NH_4^+$$ formed = $$0.1 + 0.02 = 0.12$$ mol.

Next, we apply the Henderson-Hasselbalch equation for a basic buffer: $$pOH = pK_b + \log\frac{[NH_4^+]}{[NH_3]}$$, which yields $$pOH = 4.745 + \log\frac{0.12}{0.08}$$.

Now the logarithm is computed as $$\log\frac{0.12}{0.08} = \log\frac{3}{2} = \log 3 - \log 2 = 0.477 - 0.301 = 0.176$$.

Therefore, $$pOH = 4.745 + 0.176 = 4.921$$ and $$pH = 14 - 4.921 = 9.079$$. Expressing this in the required form gives $$pH = 9.079 = 9079 \times 10^{-3}$$, so the answer is 9079.