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Ionic Equilibrium question

2022 · 28 Jun · Shift 2 · Q4
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  5. /2022 · 28 Jun · Shift 2 · Q4

Ionic Equilibrium question

2022 · 28 Jun · Shift 2 · Q4

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio of [CH3CH2COO−][CH3CH2COOH]{{[C{H_3}C{H_2}CO{O^ - }]} \over {[C{H_3}C{H_2}COOH]}}[CH3​CH2​COOH][CH3​CH2​COO−]​ required to make buffer is ‾\underline{\hspace{2cm}}​. Given : Ka(CH3CH2COOH)=1.3×10−5{K_a}(C{H_3}C{H_2}COOH) = 1.3 \times {10^{ - 5}}Ka​(CH3​CH2​COOH)=1.3×10−5
  1. A
    0.03
  2. B
    0.13
  3. C
    0.23
  4. D
    0.33
View written solutionFree

Correct answer: B

  1. Use Henderson–Hasselbalch equation for an acidic buffer:
pH=pKa+log⁡[salt][acid]\text{pH} = \text{p}K_a + \log \frac{[\text{salt}]}{[\text{acid}]}pH=pKa​+log[acid][salt]​

Here,

  • Acid = propanoic acid, CH3CH2COOH\mathrm{CH_3CH_2COOH}CH3​CH2​COOH
  • Salt/base = propanoate ion, CH3CH2COO−\mathrm{CH_3CH_2COO^-}CH3​CH2​COO−

So,

4=pKa+log⁡[CH3CH2COO−][CH3CH2COOH]4 = \text{p}K_a + \log \frac{[\mathrm{CH_3CH_2COO^-}]}{[\mathrm{CH_3CH_2COOH}]}4=pKa​+log[CH3​CH2​COOH][CH3​CH2​COO−]​
  1. Calculate pKa\text{p}K_apKa​

Given:

Ka=1.3×10−5K_a = 1.3 \times 10^{-5}Ka​=1.3×10−5

Therefore,

pKa=−log⁡(1.3×10−5)\text{p}K_a = -\log(1.3 \times 10^{-5})pKa​=−log(1.3×10−5) pKa=5−log⁡1.3\text{p}K_a = 5 - \log 1.3pKa​=5−log1.3

Since log⁡1.3≈0.114\log 1.3 \approx 0.114log1.3≈0.114,

pKa≈5−0.114=4.886\text{p}K_a \approx 5 - 0.114 = 4.886pKa​≈5−0.114=4.886
  1. Substitute into the equation
4=4.886+log⁡[CH3CH2COO−][CH3CH2COOH]4 = 4.886 + \log \frac{[\mathrm{CH_3CH_2COO^-}]}{[\mathrm{CH_3CH_2COOH}]}4=4.886+log[CH3​CH2​COOH][CH3​CH2​COO−]​ log⁡[CH3CH2COO−][CH3CH2COOH]=4−4.886=−0.886\log \frac{[\mathrm{CH_3CH_2COO^-}]}{[\mathrm{CH_3CH_2COOH}]} = 4 - 4.886 = -0.886log[CH3​CH2​COOH][CH3​CH2​COO−]​=4−4.886=−0.886
  1. Find the ratio
[CH3CH2COO−][CH3CH2COOH]=10−0.886\frac{[\mathrm{CH_3CH_2COO^-}]}{[\mathrm{CH_3CH_2COOH}]} = 10^{-0.886}[CH3​CH2​COOH][CH3​CH2​COO−]​=10−0.886 ≈0.13\approx 0.13≈0.13
  1. Match with the options

Thus, the required ratio is:

0.13\boxed{0.13}0.13​

So the correct option is B.

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