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

First, determine the moles of ethanoic acid and ethanoate before the addition of NaOH:

• Moles of ethanoic acid = $0.260 \text{ mol dm}^{-3} \times \frac{500 \text{ cm}^3}{1000} = 0.130 \text{ mol}$
• Moles of ethanoate = $0.121 \text{ mol dm}^{-3} \times \frac{500 \text{ cm}^3}{1000} = 0.0605 \text{ mol}$

After adding NaOH, which reacts with ethanoic acid:

• Moles of NaOH added = $7.00 \times 10^{-3} \text{ mol}$

Reaction:

• Remaining moles of ethanoic acid = $0.130 - 0.007 = 0.123 \text{ mol}$
• Moles of ethanoate = $0.0605 + 0.007 = 0.0675 \text{ mol}$

Now, we can calculate the new concentrations:

• Concentration of ethanoic acid: $[CH_3COOH] = \frac{0.123}{0.500} = 0.246 \text{ mol dm}^{-3}$
• Concentration of ethanoate: $[CH_3COO^-] = \frac{0.0675}{0.500} = 0.135 \text{ mol dm}^{-3}$

Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[CH_3COO^-]}{[CH_3COOH]}$

Where:

• $pK_a = -\log(1.74 \times 10^{-5}) \approx 4.76$

Substituting values: $pH = 4.76 + \log \frac{0.135}{0.246} \approx 4.76 - 0.2081 \approx 4.55$

Thus, the pH of the buffer solution after the sodium hydroxide was added is approximately 4.55.