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This question is about sulfuric acid and its salts - AQA - A-Level Chemistry - Question 2 - 2019 - Paper 3
Question 2
This question is about sulfuric acid and its salts. **2.1** Draw the displayed formula of a molecule of H₂SO₄. **2.2** In aqueous solution, sulfuric acid acts as a... show full transcript
Worked Solution & Example Answer:This question is about sulfuric acid and its salts - AQA - A-Level Chemistry - Question 2 - 2019 - Paper 3
Step 1
Step 2
Step 3
Describe the method that the student should use to make this solution.
Answer
To prepare the solution, follow these steps:
- Weighing the Solid: Accurately weigh the amount of sodium hydrogensulfate (NaHSO₄) required to achieve the desired concentration in 250 cm³.
- Dissolving the Solid: Transfer the solid to a beaker and add a small amount of distilled water to dissolve it, stirring until fully dissolved.
- Transferring to a Volumetric Flask: Pour the dissolved solution into a 250 cm³ volumetric flask, rinsing the beaker with water to ensure all solute is transferred.
- Making Up to Volume: Fill the flask with distilled water up to the 250 cm³ mark and mix thoroughly by inverting the flask several times.
Step 4
Calculate the value of Kₐ for the hydrogen sulfate ion (HSO₄⁻) that is behaving as a weak acid.
Answer
To calculate Kₐ for HSO₄⁻, we start by determining the concentration of hydrogen ions [H⁺] from pH:
ightarrow [H^+] = 10^{-1.72} = 0.0183 ext{ mol/dm}^3$$ Next, calculate the concentration of HSO₄⁻ in 605 mg of NaHSO₄: $$605 ext{ mg} = 0.605 ext{ g}$$ $$\text{Molar mass of NaHSO₄} = 120.06 ext{ g/mol}$$ $$ ext{Moles of NaHSO₄} = \frac{0.605}{120.06} = 0.00504 ext{ mol}$$ In 100 cm³ (or 0.1 dm³) of solution, the concentration is: $$\text{Concentration} = \frac{0.00504}{0.1} = 0.0504 ext{ mol/dm}^3$$ Using the equation for weak acid dissociation: $$K_a = \frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$$ Assuming [H⁺] = [SO₄²⁻], we have: $$K_a = \frac{(0.0183)(0.0183)}{0.0504 - 0.0183} \\ = \frac{(0.00033489)}{0.0321} = 0.01043 ext{ (to three significant figures: } 0.0104)$$Step 6
Explain why this increases the pH of the solution.
Answer
The addition of sodium sulfate (Na₂SO₄) introduces more sulfate ions (SO₄²⁻) into the solution. This increase in sulfate ions will shift the equilibrium:
According to Le Chatelier's Principle, adding SO₄²⁻ reduces the concentration of HSO₄⁻, leading to the consumption of H⁺ ions and thus reducing their concentration in solution. As a result, the pH of the solution increases.