Question

What volume of 0.2 M Acetic acid is to be added to 100ml of 0.4M Sodium acetate so that a Buffer solution of pH equal to 4.94 is obtained? (pK$_a$ of CH$_3$COOH = 4.76)

• A. 132.1 ml
• B. 125.3 ml
• C. 150.2 ml
• D. 110.6 ml

Hint:

For buffer solutions, always remember the Henderson-Hasselbalch equation. The ratio of the concentrations of the acid and its conjugate base controls the pH of the solution.

Answer 1

The Correct Option is A

For a buffer solution, the pH can be calculated using the Henderson-Hasselbalch equation: \[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \] Here, \(\text{pK}_a\) of acetic acid is 4.76, and the pH is given as 4.94. 
We are also given: - Volume of Sodium acetate solution, \( V_{\text{NaOAc}} = 100 \, \text{ml} \) - Molarity of Sodium acetate, \( [\text{NaOAc}] = 0.4 \, \text{M} \) - Molarity of Acetic acid, \( [\text{AcOH}] = 0.2 \, \text{M} \) Let \( V_{\text{AcOH}} \) be the volume of Acetic acid solution to be added. First, apply the Henderson-Hasselbalch equation: \[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{NaOAc}]}{[\text{AcOH}]} \right) \] Substitute the values: \[ 4.94 = 4.76 + \log \left( \frac{0.4 \times 100}{0.2 \times V_{\text{AcOH}}} \right) \] Now simplify the equation: \[ 4.94 - 4.76 = \log \left( \frac{0.4 \times 100}{0.2 \times V_{\text{AcOH}}} \right) \] \[ 0.18 = \log \left( \frac{40}{0.2 \times V_{\text{AcOH}}} \right) \] Now, convert the logarithmic equation into an exponential form: \[ 10^{0.18} = \frac{40}{0.2 \times V_{\text{AcOH}}} \] \[ 1.5 = \frac{40}{0.2 \times V_{\text{AcOH}}} \] Now, solve for \( V_{\text{AcOH}} \): \[ V_{\text{AcOH}} = \frac{40}{1.5 \times 0.2} \] \[ V_{\text{AcOH}} = \frac{40}{0.3} = 132.1 \, \text{ml} \] Thus, the volume of acetic acid required is 132.1 ml.