A 0.100 mol dm⁻³ solution of sodium hydroxide was gradually added to 25.0 cm³ of a solution of a weak acid, HX, in the presence of a suitable indicator - AQA - A-Level Chemistry - Question 4 - 2017 - Paper 3

The expression for the acid dissociation constant, Ka, is given by the formula:

$$K_a = \frac{[H^+][X^-]}{[HX]}$$

To find the concentration of HX, we first note that 20.0 cm³ of 0.100 mol dm⁻³ NaOH was needed to reach the first pH change.

The amount of NaOH used can be calculated as:

$$\text{Amount NaOH} = \text{Concentration} \times \text{Volume} = (0.100 \, \text{mol dm}^{-3}) \times (20.0 \, \text{cm}^3 \times \frac{1 \, \text{dm}^3}{1000 \, \text{cm}^3}) = 0.00200 \, \text{mol}$$

Since NaOH completely reacts with HX in a 1:1 ratio, the same amount of HX was present:

Thus, the initial concentration of HX is given by:

$$\text{Concentration of HX} = \frac{0.00200 \, \text{mol}}{0.025 \, \text{dm}^3} = 0.0800 \, \text{mol dm}^{-3}$$

Before any sodium hydroxide is added, we can calculate the pH of the weak acid solution.

Given the concentration of HX is 0.0800 mol dm⁻³.

Using the formula:

K_a = 2.62 imes 10^{-5} = \frac{[H^+]^2}{[HX]}\n$$ Letting [H⁺] = x, we have:

2.62 imes 10^{-5} = \frac{x^2}{0.0800 - x} \approx \frac{x^2}{0.0800}

$$Solving for x gives:$$

x = \sqrt{(2.62 imes 10^{-5}) \times 0.0800} \approx 0.0014 , \text{mol dm}^{-3}

$$Finally, the pH is:$$

pH = -\log(0.0014) \approx 2.85

$$$$

When half of the acid has reacted, the concentrations of HX and its conjugate base (X⁻) are equal, making it a buffer solution.

Using the Henderson-Hasselbalch equation:

$$pH = pK_a + \log \frac{[A^-]}{[HA]}$$

Where:

• $pK_a = -\log(2.62 imes 10^{-5}) \approx 4.58$
• $[A^-] = [HX]$ at half equivalence = 0.0400 mol dm⁻³ (since half has reacted into 0.0400)
• $[HA] = [HX] = 0.0400 mol dm⁻³$

Thus,

$$pH = 4.58 + \log(1) = 4.58$$