JEE Main · 2019 · Shift-IhardIEQ-040

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\…

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

20 mL of 0.1 M H2SO4\mathrm{H_2SO_4} solution is added to 30 mL of 0.2 M NH4OH\mathrm{NH_4OH} solution. The pH of the resultant mixture is:

[pKb of NH4OH=4.7][\mathrm{pK_b\ of\ NH_4OH} = 4.7]

Options
  1. a

    5.0

  2. b

    5.2

  3. c

    9.4

  4. d

    9.0

    ✓
Correct Answerd

9.0

Detailed Solution

Step 1 — Moles of each: Moles H2SO4\mathrm{H_2SO_4} = 0.1×0.02=2×10−30.1 \times 0.02 = 2 \times 10^{-3} mol →H+\rightarrow \mathrm{H^+} = 4×10−34 \times 10^{-3} mol Moles NH4OH\mathrm{NH_4OH} = 0.2×0.03=6×10−30.2 \times 0.03 = 6 \times 10^{-3} mol

Step 2 — Neutralisation: NH4OH+H+→NH4++H2O\mathrm{NH_4OH + H^+ \rightarrow NH_4^+ + H_2O} H⁺ consumed = 4×10−34 \times 10^{-3} mol NH4OH\mathrm{NH_4OH} remaining = 6×10−3−4×10−3=2×10−36 \times 10^{-3} - 4 \times 10^{-3} = 2 \times 10^{-3} mol NH4+\mathrm{NH_4^+} formed = 4×10−34 \times 10^{-3} mol

Step 3 — Buffer solution (weak base + its salt): Total volume = 50 mL [NH4OH]=2×10−30.05=0.04 M[NH4+]=4×10−30.05=0.08 M[\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: pOH=pKb+log⁡[Salt][Base]=4.7+log⁡0.080.04=4.7+log⁡2=4.7+0.3=5.0pH=14−5.0=9.0\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

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