This question is about sulfuric acid and its salts - AQA - A-Level Chemistry - Question 2 - 2019 - Paper 3

Equation 1: $\text{H}_2\text{SO}_4 \rightarrow \text{HSO}_4^- + \text{H}^+$

Equation 2: $\text{HSO}_4^- \rightarrow \text{SO}_4^{2-} + \text{H}^+$

To prepare a 250 cm³ solution of sodium hydrogen sulfate (NaHSO₄):

1. Weigh the Solid: Accurately weigh out the appropriate mass of NaHSO₄ using a balance.
2. Dissolve the Solid: Transfer the NaHSO₄ into a volumetric flask or a beaker containing distilled water, and stir until completely dissolved.
3. Transfer to Volumetric Flask: Pour the solution into a 250 cm³ volumetric flask, ensuring no solid remains in the beaker.
4. Make Up to Mark: Add distilled water to the flask until the bottom of the meniscus is level with the 250 cm³ mark. Cap the flask and invert to mix thoroughly.

To find the dissociation constant Kₐ for the hydrogen sulfate ion (HSO4−):

1. Start with Dissociation Equation: $\text{HSO}_4^- \rightleftharpoons \text{H}^+ + \text{SO}_4^{2-}$
2. Calculate Concentrations: Given the initial moles of NaHSO₄ is calculated from 605 mg: $\mathrm{Molar\ mass\ of\ NaHSO}_4 \approx 120\text{g/mol}$ $moles = \frac{605\text{mg}}{1000}\div 120\approx 0.00504\text{mol}$ Initial concentration in 100 cm³: $C_{HSO_4^-} = 0.00504\text{mol}/0.1\text{L} = 0.0504\text{mol/L}$ After dissociation, since it is a weak acid, we can assume all HSO4− partially dissociates into H+ and SO4^2−.
3. Calculate pH: Given pH = 1.72, $[H^+] = 10^{-1.72} ext{ M} \approx 0.01884 ext{ M}$
4. Use Kₐ Expression: $K_a = \frac{[H^+][SO_4^{2-}]}{[HSO_4^-]} \approx \frac{(0.01884)(0.01884)}{(0.0504 - 0.01884)} \\ \approx \frac{0.000355}{0.03156} \approx 0.0113$
5. Final Answer: Thus, to three significant figures, Kₐ ≈ 0.0113. Units: The units of Kₐ are mol/dm³.

Adding sodium sulfate (Na₂SO₄) does not provide additional H+ ions, as it is a neutral salt. When dissolved, it dissociates into Na+ and SO₄²−. The increased concentration of sulfate ions can shift the equilibrium of the dissociation of HSO4− back to the left:
$\text{HSO}_4^- \rightleftharpoons \text{H}^+ + \text{SO}_4^{2-}$ This reduces the concentration of H+ ions in the solution, thereby increasing the pH of the solution.