The acid dissociation constant, K_a, for ethanoic acid is given by the expression $$K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$$ The value of K_a for ethanoic acid is 1.74 × 10^-5 mol dm^-3 at 25 °C A buffer solution with a pH of 3.87 was prepared using ethanoic acid and sodium ethanoate - AQA - A-Level Chemistry - Question 2 - 2017 - Paper 1

Start with the expression for the acid dissociation constant:

$K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$

Rearranging gives:

$[CH_3COOH] = \frac{[CH_3COO^-][H^+]}{K_a}$

Use the provided values:

• $K_a = 1.74 \times 10^{-5} \ \text{mol dm}^{-3}$
• $[CH_3COO^-] = 0.136 \ \text{mol dm}^{-3}$
• Use the formula for $[H^+]$ derived from pH:

$[H^+] = 10^{-pH} = 10^{-3.87} = 1.3498 \times 10^{-4} \ \text{mol dm}^{-3}$

Substitute into the rearranged equation:

$[CH_3COOH] = \frac{(0.136)(1.3498 \times 10^{-4})}{1.74 \times 10^{-5}}$

Calculate the concentration:

$[CH_3COOH] = \frac{1.834 \times 10^{-8}}{1.74 \times 10^{-5}} \approx 1.06 \ \text{mol dm}^{-3}$

Thus, the concentration of the ethanoic acid in the buffer solution is approximately 1.06 mol dm^-3.