JEE Main · 2023 · Shift-IIeasyRDX-046

The volume of 0.02 M aqueous HBr required to neutralize 10.0 mL of 0.01 M aqueous Ba(OH)2 is (Assume complete…

Redox Reactions · Class 11 · JEE Main Previous Year Question

Question

The volume of 0.02 M aqueous HBr required to neutralize 10.0 mL of 0.01 M aqueous Ba(OH)2\mathrm{Ba(OH)_2} is (Assume complete neutralization)

Options
  1. a

    2.5 mL

  2. b

    5.0 mL

  3. c

    10.0 mL

    ✓
  4. d

    7.5 mL

Correct Answerc

10.0 mL

Detailed Solution

Step 1 — Reaction: Ba(OH)2+2HBr→BaBr2+2H2O\mathrm{Ba(OH)_2 + 2HBr \rightarrow BaBr_2 + 2H_2O}

Step 2 — Moles of Ba(OH)2\mathrm{Ba(OH)_2}: nBa(OH)2=0.01 M×10.0×10−3 L=1.0×10−4 moln_{\mathrm{Ba(OH)_2}} = 0.01\ \mathrm{M} \times 10.0 \times 10^{-3}\ \mathrm{L} = 1.0 \times 10^{-4}\ \mathrm{mol}

Step 3 — Moles of HBr required (stoichiometry 1:2): nHBr=2×1.0×10−4=2.0×10−4 moln_{\mathrm{HBr}} = 2 \times 1.0 \times 10^{-4} = 2.0 \times 10^{-4}\ \mathrm{mol}

Step 4 — Volume of HBr: V=nM=2.0×10−40.02=0.01 L=10.0 mLV = \frac{n}{M} = \frac{2.0 \times 10^{-4}}{0.02} = 0.01\ \mathrm{L} = 10.0\ \mathrm{mL}

Answer: Option (3) — 10.0 mL

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