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

Using the formula for the acid dissociation constant:

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

We know:

• $K_a = 1.74 \times 10^{-5}$
• $[CH_3COO^-] = 0.136 \text{ mol dm}^{-3}$
• We can find $[H^+]$ using the pH:

$pH = -\log[H^+] \Rightarrow [H^+] = 10^{-pH} = 10^{-3.87} \approx 1.35 \times 10^{-4} \text{ mol dm}^{-3}$

Now, substituting in the K_a equation:

$1.74 \times 10^{-5} = \frac{(0.136)(1.35 \times 10^{-4})}{[CH_3COOH]}$

Rearranging gives:

$[CH_3COOH] = \frac{(0.136)(1.35 \times 10^{-4})}{1.74 \times 10^{-5}} \approx 1.06 \text{ mol dm}^{-3}$

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