20 mL of 0.1 M NH4OH is mixed with 40 mL of 0.05 M HCl. The pH of the mixture is nearest to: (Kb(NH4OH) = 1 10-5,\ 2 =…

Step 1 — Moles of each: Moles $\mathrm{NH_4OH}$ = $0.1 \times 0.02 = 2 \times 10^{-3}$ mol Moles HCl = $0.05 \times 0.04 = 2 \times 10^{-3}$ mol

Step 2 — Neutralisation: $\mathrm{NH_4OH + HCl \rightarrow NH_4Cl + H_2O}$ All $\mathrm{NH_4OH}$ is consumed $\rightarrow$ only $\mathrm{NH_4Cl}$ remains (2×10⁻³ mol)

Step 3 — pH of $\mathrm{NH_4Cl}$ solution (salt of weak base + strong acid): Total volume = 60 mL $[\mathrm{NH_4Cl}] = \frac{2 \times 10^{-3}}{0.06} = \frac{1}{30}\ \mathrm{M} \mathrm{pH} = 7 - \frac{1}{2}(\mathrm{pK_b} + \log C) \mathrm{pK_b} = 5, \quad \log C = \log(1/30) = -\log 30 = -(\log 3 + \log 10) = -(0.48 + 1) = -1.48 \mathrm{pH} = 7 - \frac{1}{2}(5 + (-1.48)) = 7 - \frac{3.52}{2} = 7 - 1.76 = 5.24 \approx 5.2$

Answer: Option (3) — 5.2