JEE Main · 2020 · Shift-IeasyIEQ-013

An acidic buffer is obtained on mixing:

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

An acidic buffer is obtained on mixing:

Options
  1. a

    100 mL of 0.1 M CH3COOH\mathrm{CH_3COOH} and 100 mL of 0.1 M NaOH

  2. b

    100 mL of 0.1 M HCl and 200 mL of 0.1 M NaCl

  3. c

    100 mL of 0.1 M CH3COOH\mathrm{CH_3COOH} and 200 mL of 0.1 M NaOH

  4. d

    100 mL of 0.1 M HCl and 200 mL of 0.1 M CH3COONa\mathrm{CH_3COONa}

    ✓
Correct Answerd

100 mL of 0.1 M HCl and 200 mL of 0.1 M CH3COONa\mathrm{CH_3COONa}

Detailed Solution

Acidic buffer = weak acid + its conjugate base (salt), with pH < 7.

Step 1 — Analyse each option:

(1) 100 mL 0.1M CH3COOH\mathrm{CH_3COOH} + 100 mL 0.1M NaOH: Moles: acid = 0.01, base = 0.01 →\rightarrow complete neutralisation →\rightarrow only CH3COONa\mathrm{CH_3COONa} (basic salt solution, not a buffer) ✗

(2) 100 mL 0.1M HCl + 200 mL 0.1M NaCl: HCl is strong acid, NaCl is neutral salt →\rightarrow not a buffer ✗

(3) 100 mL 0.1M CH3COOH\mathrm{CH_3COOH} + 200 mL 0.1M NaOH: Moles: acid = 0.01, base = 0.02 →\rightarrow NaOH in excess →\rightarrow basic solution ✗

(4) 100 mL 0.1M HCl + 200 mL 0.1M CH3COONa\mathrm{CH_3COONa}: Moles: HCl = 0.01, CH3COONa\mathrm{CH_3COONa} = 0.02 HCl reacts with CH3COONa\mathrm{CH_3COONa}: HCl+CH3COONa→CH3COOH+NaCl\mathrm{HCl + CH_3COONa \rightarrow CH_3COOH + NaCl} After reaction: CH3COOH\mathrm{CH_3COOH} = 0.01 mol, CH3COONa\mathrm{CH_3COONa} = 0.01 mol →\rightarrow acidic buffer ✓

Answer: Option (4)

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