Q14 / 180 · COMEDK 2025 (10 May Shift 1) · Chemistry
What volume of $0.2 \mathrm{M} \mathrm{CH}_3 \mathrm{COOH}$ needs to be added to $100 \mathrm{ml}$ of $0.4 \mathrm{M} \mathrm{CH}_3 \mathrm{COONa}^2$ solution to prepare a buffer of pH equal to 4.91 ? ($\mathrm{pK}_{\mathrm{a}}$ of $\mathrm{CH}_3 \mathrm{COOH}$ is 4.76)
- A213.65 ml
- Correct answerB141.59 ml
- C101.41 ml
- D282.6 ml
Solution
The Henderson-Hasselbalch equation for an acidic buffer is given by:
$\text{pH} = \text{pK}_{a} + \log \left( \dfrac{[\text{Salt}]}{[\text{Acid}]} \right)$
Given $\text{pH} = 4.91$, $\text{pK}_{a} = 4.76$, and the concentration of salt $[\text{CH}_{3}\text{COONa}] = 0.4 \text{ M}$ with volume $V_{s} = 100 \text{ ml}$.
Substituting the values into the equation:
$4.91 = 4.76 + \log \left( \dfrac{n_{\text{salt}}}{n_{\text{acid}}} \right)$
$0.15 = \log \left( \dfrac{0.4 \times 100}{0.2 \times V_{a}} \right)$
Taking the antilog of both sides:
$10^{0.15} = \dfrac{40}{0.2 \times V_{a}}$
$1.4125 = \dfrac{40}{0.2 \times V_{a}}$
$0.2 \times V_{a} = \dfrac{40}{1.4125}$
$0.2 \times V_{a} = 28.3186$
$V_{a} = \dfrac{28.3186}{0.2} = 141.59 \text{ ml}$
Answer: 141.59 ml
$\text{pH} = \text{pK}_{a} + \log \left( \dfrac{[\text{Salt}]}{[\text{Acid}]} \right)$
Given $\text{pH} = 4.91$, $\text{pK}_{a} = 4.76$, and the concentration of salt $[\text{CH}_{3}\text{COONa}] = 0.4 \text{ M}$ with volume $V_{s} = 100 \text{ ml}$.
Substituting the values into the equation:
$4.91 = 4.76 + \log \left( \dfrac{n_{\text{salt}}}{n_{\text{acid}}} \right)$
$0.15 = \log \left( \dfrac{0.4 \times 100}{0.2 \times V_{a}} \right)$
Taking the antilog of both sides:
$10^{0.15} = \dfrac{40}{0.2 \times V_{a}}$
$1.4125 = \dfrac{40}{0.2 \times V_{a}}$
$0.2 \times V_{a} = \dfrac{40}{1.4125}$
$0.2 \times V_{a} = 28.3186$
$V_{a} = \dfrac{28.3186}{0.2} = 141.59 \text{ ml}$
Answer: 141.59 ml
Hence, the correct answer is option: 'B'.