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

Equation 1:
$\mathrm{H_2SO_4 \rightarrow HSO_4^- + H^+}$
Equation 2:
$\mathrm{HSO_4^- \rightleftharpoons SO_4^{2-} + H^+}$

1. Weigh out the exact mass of sodium hydrogensulfate (NaHSO₄) required for 250 cm³ of solution.
2. Transfer the weighed sodium hydrogensulfate into a clean, dry beaker.
3. Add a small volume of distilled water to the beaker to dissolve the solid completely. Stir the solution until all the solid has dissolved.
4. Once dissolved, pour the solution into a 250 cm³ volumetric flask.
5. Rinse the beaker with distilled water and add the rinse water to the volumetric flask.
6. Fill the volumetric flask with distilled water until the bottom of the meniscus is at the 250 cm³ mark.
7. Stopper the flask and shake gently to ensure thorough mixing.

To find Kₐ for HSO₄⁻, we can use the formula:

1. Calculate the initial concentration of the solution using the mass of NaHSO₄:

   Initial mass of NaHSO₄ = 605 mg
   Molar mass of NaHSO₄ = 120.06 g/mol
   Amount in moles = [ \frac{605 mg}{1000} \times \frac{1 g}{1000 mg} \times \frac{1 mol}{120.06 g} \approx 0.00504 mol ]

   Therefore, concentration in 100 cm³ = [ \frac{0.00504 mol}{0.1 L} = 0.0504 mol/L ]

2. Using the pH given:
   pH = 1.72

   Hence, [ [H^+] = 10^{-1.72} = 0.0183 mol/L ]

3. Next, calculate [HSO₄⁻]:

   Initial concentration of HSO₄⁻ is approximately equal to that of NaHSO₄, which is 0.0504 mol/L, and at equilibrium:

   [ [HSO_4^-] \approx 0.0504 - x ] where ( x = [H^+] = 0.0183 )

4. So, [ [HSO_4^-] \approx 0.0504 - 0.0183 = 0.0321 mol/L ]

5. Now apply the expression for Kₐ:

   [ K_a = \frac{[H^+][SO_4^{2-}]}{[HSO_4^-]} \approx \frac{(0.0183)(0.0183)}{0.0321} ]

6. Calculation yields:
   [ K_a \approx 0.01027 \Rightarrow 1.03 \times 10^{-2} ]

7. Therefore, the value of Kₐ is:
   [ K_a = 1.03 \times 10^{-2} ]
   and the units are mol/L.

The addition of sodium sulfate (Na₂SO₄) introduces more sulfate ions (SO₄²⁻) into the solution. This results in the common ion effect, which shifts the equilibrium of the dissociation of HSO₄⁻:

$\mathrm{HSO_4^- \rightleftharpoons H^+ + SO_4^{2-}}$

The presence of additional SO₄²⁻ ions decreases the concentration of H⁺ ions, which raises the pH of the solution. Additionally, as more sulfate ions are introduced, they compete with HSO₄⁻ for H⁺ ions, further reducing the acidity of the solution.