JEE Main · 2022 · Shift-IImediumIEQ-030

200 mL of 0.01 M HCl is mixed with 400 mL of 0.01 M H2SO4. The pH of the mixture is

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

200 mL of 0.01 M HCl is mixed with 400 mL of 0.01 M H2SO4\mathrm{H_2SO_4}. The pH of the mixture is

Options
  1. a

    1.14

  2. b

    1.78

    ✓
  3. c

    2.34

  4. d

    3.02

Correct Answerb

1.78

Detailed Solution

Step 1 — Moles of H+\mathrm{H^+}:

From HCl (strong acid, monobasic): nH+=0.01×0.2=2×10−3 moln_{\mathrm{H^+}} = 0.01 \times 0.2 = 2 \times 10^{-3}\ \mathrm{mol}

From H2SO4\mathrm{H_2SO_4} (strong acid, dibasic): nH+=2×0.01×0.4=8×10−3 moln_{\mathrm{H^+}} = 2 \times 0.01 \times 0.4 = 8 \times 10^{-3}\ \mathrm{mol}

Step 2 — Total [H+][\mathrm{H^+}]:

Total volume = 200 + 400 = 600 mL = 0.6 L

[H+]=(2+8)×10−30.6=10×10−30.6=160 M[\mathrm{H^+}] = \frac{(2 + 8) \times 10^{-3}}{0.6} = \frac{10 \times 10^{-3}}{0.6} = \frac{1}{60}\ \mathrm{M}

Step 3 — pH:

pH=−log⁡(160)=log⁡60=log⁡(6×10)=1+log⁡6=1+0.778≈1.78\mathrm{pH} = -\log\left(\frac{1}{60}\right) = \log 60 = \log(6 \times 10) = 1 + \log 6 = 1 + 0.778 \approx 1.78

Answer: Option (2) — pH = 1.78

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