JEE Main · 2022 · Shift-ImediumIEQ-041

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 =…

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

20 mL of 0.1 M NH4OH\mathrm{NH_4OH} is mixed with 40 mL of 0.05 M HCl. The pH of the mixture is nearest to:

(Kb(NH4OH)=1×10−5, log⁡2=0.30, log⁡3=0.48, log⁡5=0.69, log⁡7=0.84, log⁡11=1.04)(K_b(\mathrm{NH_4OH}) = 1 \times 10^{-5},\ \log 2 = 0.30,\ \log 3 = 0.48,\ \log 5 = 0.69,\ \log 7 = 0. 84,\ \log 11 = 1. 04)

Options
  1. a

    3.2

  2. b

    4.2

  3. c

    5.2

    ✓
  4. d

    6.2

Correct Answerc

5.2

Detailed Solution

Step 1 — Moles of each: Moles NH4OH\mathrm{NH_4OH} = 0.1×0.02=2×10−30.1 \times 0.02 = 2 \times 10^{-3} mol Moles HCl = 0.05×0.04=2×10−30.05 \times 0.04 = 2 \times 10^{-3} mol

Step 2 — Neutralisation: NH4OH+HCl→NH4Cl+H2O\mathrm{NH_4OH + HCl \rightarrow NH_4Cl + H_2O} All NH4OH\mathrm{NH_4OH} is consumed →\rightarrow only NH4Cl\mathrm{NH_4Cl} remains (2×10⁻³ mol)

Step 3 — pH of NH4Cl\mathrm{NH_4Cl} solution (salt of weak base + strong acid): Total volume = 60 mL [NH4Cl]=2×10−30.06=130 MpH=7−12(pKb+log⁡C)pKb=5,log⁡C=log⁡(1/30)=−log⁡30=−(log⁡3+log⁡10)=−(0.48+1)=−1.48pH=7−12(5+(−1.48))=7−3.522=7−1.76=5.24≈5.2[\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

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