JEE Main · 2019 · Shift-ImediumRDX-023

50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. What is the amount of NaOH in 50…

Redox Reactions · Class 11 · JEE Main Previous Year Question

Question

50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. What is the amount of NaOH in 50 mL of the given sodium hydroxide solution?

Options
  1. a

    2 g

  2. b

    4 g

    ✓
  3. c

    1 g

  4. d

    8 g

Correct Answerb

4 g

Detailed Solution

Step 1 — Reaction of oxalic acid with NaOH:

H2C2O4+2NaOH→Na2C2O4+2H2O\mathrm{H_2C_2O_4 + 2NaOH \rightarrow Na_2C_2O_4 + 2H_2O}

Step 2 — Moles of oxalic acid used:

nH2C2O4=0.5 M×50×10−3 L=0.025 moln_{\mathrm{H_2C_2O_4}} = 0.5\ \mathrm{M} \times 50 \times 10^{-3}\ \mathrm{L} = 0.025\ \mathrm{mol}

Step 3 — Moles of NaOH in 25 mL (stoichiometry 1:2):

nNaOH=2×0.025=0.05 moln_{\mathrm{NaOH}} = 2 \times 0.025 = 0.05\ \mathrm{mol}

Step 4 — Molarity of NaOH solution:

MNaOH=0.05 mol0.025 L=2 MM_{\mathrm{NaOH}} = \frac{0.05\ \mathrm{mol}}{0.025\ \mathrm{L}} = 2\ \mathrm{M}

Step 5 — Amount of NaOH in 50 mL of the same solution:

n=2 M×0.05 L=0.1 moln = 2\ \mathrm{M} \times 0.05\ \mathrm{L} = 0.1\ \mathrm{mol}

m=0.1×40 g mol−1=4 gm = 0.1 \times 40\ \mathrm{g\ mol^{-1}} = \mathbf{4\ \mathrm{g}}

Answer: Option (2) — 4 g

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