Home / AS & A Level Chemistry 24.2 Standard electrode potentials $E^{\ominus}$: Exam Style Questions Paper 4

AS & A Level Chemistry 24.2 Standard electrode potentials $E^{\ominus}$: Exam Style Questions Paper 4

A Level Chemistry Paper 4 - All Chapters
Question

(a) Disodium phosphate, \((Na^+)_2(HPO_4^{2–})\), reacts with an acid to form monosodium phosphate, \(Na^+(H_2PO_4^–)\).
(i) Identify the ions that are a conjugate acid–base pair in this reaction, using the formulae of the species involved.
(ii) Define buffer solution.
(iii) Write two equations to show how a mixture of \((Na^+)_2(HPO_4^{2–})\) and \(Na^+(H_2PO_4^–)\) can act as a buffer solution.
(iv) Identify one inorganic ion that acts as a buffer in blood.

(b) Compound E is the hydroxide of a Group 2 element. Compound E is a strong alkali. 2.63g of E is dissolved in water to make 250 cm³ of solution F. Solution F has a pH of 13.09 at 298K.
(i) Show that the concentration of hydroxide ions in solution F is 0.123 moldm⁻³.
(ii) Explain why the concentration of compound E in solution F is 0.0615moldm⁻³.
(iii) Use the concentration given in (ii) to identify compound E.

(c) Compound E is much more soluble than magnesium hydroxide. A saturated solution of magnesium hydroxide in water has a concentration of 1.40 × 10⁻⁴ moldm⁻³ at 298K. Calculate the solubility product, \(K_{sp}\), of magnesium hydroxide. Include units.

(d) Explain why compound E is much more soluble than magnesium hydroxide.

▶️ Answer/Explanation
Solution

(a)(i) Conjugate acid: \(H_2PO_4^–\), Conjugate base: \(HPO_4^{2–}\).

Explanation: The pair \(HPO_4^{2–}/H_2PO_4^–\) are conjugate base and acid respectively, as they differ by one proton.

(a)(ii) A buffer solution resists changes in pH when small amounts of acid or base are added.

Explanation: It maintains pH stability through equilibrium shifts between its weak acid and conjugate base components.

(a)(iii)
Equation 1: \(HPO_4^{2–} + H^+ \to H_2PO_4^–\) (neutralizes added acid)
Equation 2: \(H_2PO_4^– + OH^– \to HPO_4^{2–} + H_2O\) (neutralizes added base)

Explanation: The buffer system uses both reactions to maintain pH by consuming added \(H^+\) or \(OH^–\).

(a)(iv) \(HCO_3^–\) (hydrogen carbonate).

Explanation: In blood, \(HCO_3^–/H_2CO_3\) acts as the primary buffer system to regulate pH.

(b)(i) \([OH^–] = 0.123 \, \text{moldm}^{-3}\).

Explanation: Given pH = 13.09, pOH = 14 – 13.09 = 0.91. Thus, \([OH^–] = 10^{-0.91} = 0.123 \, \text{moldm}^{-3}\).

(b)(ii) Concentration of E is 0.0615 moldm⁻³.

Explanation: Since E is \(X(OH)_2\), it dissociates completely to give 2 \(OH^–\) ions per formula unit. Thus, \([E] = \frac{[OH^–]}{2} = \frac{0.123}{2} = 0.0615 \, \text{moldm}^{-3}\).

(b)(iii) Barium hydroxide, \(Ba(OH)_2\).

Explanation:
– Moles of E in 250 cm³ = 0.0615 × 0.25 = 0.0154 mol
– Molar mass = \(\frac{2.63}{0.0154} ≈ 171 \, \text{g/mol}\)
– This matches \(Ba(OH)_2\) (Ba = 137, O = 16, H = 1 → 137 + 2×(16+1) = 171).

(c) \(K_{sp} = 1.10 \times 10^{-11} \, \text{mol}^3\text{dm}^{-9}\).

Explanation:
For \(Mg(OH)_2\): \(K_{sp} = [Mg^{2+}][OH^–]^2\)
Given \([Mg^{2+}] = 1.4 \times 10^{-4} \, \text{moldm}^{-3}\), \([OH^–] = 2.8 \times 10^{-4} \, \text{moldm}^{-3}\)
Thus, \(K_{sp} = (1.4 \times 10^{-4}) \times (2.8 \times 10^{-4})^2 = 1.10 \times 10^{-11}\).

(d) Three key reasons:

Explanation:
1. Lattice energy (\(\Delta H_{latt}\)) decreases more than hydration energy (\(\Delta H_{hyd}\)) down Group 2.
2. The change in \(\Delta H_{latt}\) dominates, making \(\Delta H_{sol}\) more exothermic for larger ions like Ba²⁺.
3. \(Ba(OH)_2\) has weaker ionic bonds (due to larger ion size) compared to \(Mg(OH)_2\), enhancing solubility.

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