JEE Main · 2022 · Shift-IImediumIEQ-076

A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio…

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio [CH3CH2COO−][CH3CH2COOH]\frac{[\mathrm{CH_3CH_2COO^-}]}{[\mathrm{CH_3CH_2COOH}]} required to make the buffer is:

[Ka(CH3CH2COOH)=1.3×10−5][K_a(\mathrm{CH_3CH_2COOH}) = 1.3 \times 10^{-5}]

Options
  1. a

    0.03

  2. b

    0.13

    ✓
  3. c

    0.23

  4. d

    0.33

Correct Answerb

0.13

Detailed Solution

Step 1 — Henderson-Hasselbalch equation: pH=pKa+log⁡[Salt][Acid]\mathrm{pH} = \mathrm{pK_a} + \log\frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}

Step 2 — Calculate pKa\mathrm{pK_a}: pKa=−log⁡(1.3×10−5)=5−log⁡1.3≈5−0.114=4.886\mathrm{pK_a} = -\log(1.3 \times 10^{-5}) = 5 - \log 1.3 \approx 5 - 0.114 = 4.886

Step 3 — Solve for ratio:

4=4.886+log⁡[Salt][Acid]log⁡[Salt][Acid]=4−4.886=−0.886[Salt][Acid]=10−0.886≈0.134 = 4.886 + \log\frac{[\mathrm{Salt}]}{[\mathrm{Acid}]} \log\frac{[\mathrm{Salt}]}{[\mathrm{Acid}]} = 4 - 4.886 = -0.886 \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]} = 10^{-0.886} \approx 0.13

Answer: Option (2) — 0.13

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