Calculate the pH of a 0.150 mol dm⁻³ solution of ethanoic acid at 25 °C - AQA - A-Level Chemistry - Question 4 - 2022 - Paper 1

Using the expression for the dissociation of ethanoic acid:

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

Assuming that the dissociation is small, we can approximate:

$[CH_3COOH] \approx 0.150$

Let ([H^+] = x), then the expression becomes:

$K_a = \frac{x^2}{0.150 - x} \approx \frac{x^2}{0.150}$

Therefore, we get:

$x^2 = 1.74 \times 10^{-5} \times 0.150$

Calculating gives:

$x^2 \approx 2.61 \times 10^{-6}$

Taking the square root, we find:

$[H^+] \approx 1.61 \times 10^{-3}$.