20 mL of 0.1 M H2SO4 solution is added to 30 mL of 0.2 M NH4OH solution. The pH of the resultant mixture is: [pKb\ of\…

Step 1 — Moles of each: Moles $\mathrm{H_2SO_4}$ = $0.1 \times 0.02 = 2 \times 10^{-3}$ mol $\rightarrow \mathrm{H^+}$ = $4 \times 10^{-3}$ mol Moles $\mathrm{NH_4OH}$ = $0.2 \times 0.03 = 6 \times 10^{-3}$ mol

Step 2 — Neutralisation: $\mathrm{NH_4OH + H^+ \rightarrow NH_4^+ + H_2O}$ H⁺ consumed = $4 \times 10^{-3}$ mol $\mathrm{NH_4OH}$ remaining = $6 \times 10^{-3} - 4 \times 10^{-3} = 2 \times 10^{-3}$ mol $\mathrm{NH_4^+}$ formed = $4 \times 10^{-3}$ mol

Step 3 — Buffer solution (weak base + its salt): Total volume = 50 mL $[\mathrm{NH_4OH}] = \frac{2 \times 10^{-3}}{0.05} = 0.04\ \mathrm{M} [\mathrm{NH_4^+}] = \frac{4 \times 10^{-3}}{0.05} = 0.08\ \mathrm{M}$

Step 4 — pOH using Henderson-Hasselbalch for base: $\mathrm{pOH} = \mathrm{pK_b} + \log\frac{[\mathrm{Salt}]}{[\mathrm{Base}]} = 4.7 + \log\frac{0.08}{0.04} = 4.7 + \log 2 = 4.7 + 0.3 = 5.0 \mathrm{pH} = 14 - 5.0 = 9.0$

Answer: Option (4) — 9.0