Home / AS & A Level Chemistry 23.4 Gibbs free energy change, $\Delta G$: Exam Style Questions Paper 4

AS & A Level Chemistry 23.4 Gibbs free energy change, $\Delta G$: 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 conjugate acid-base pair involves \(HPO_4^{2–}\) (base) and \(H_2PO_4^–\) (acid), as the reaction involves proton transfer between them.

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

Explanation: Buffers maintain pH stability by neutralizing added acids or bases through equilibrium shifts.

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

Explanation: The buffer system works by neutralizing added \(H^+\) (Equation 1) and \(OH^–\) (Equation 2) to maintain pH.

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

Explanation: \(HCO_3^–\) acts as a buffer in blood by equilibrating with \(H^+\) and \(CO_3^{2–}\).

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

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

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

Explanation: Since E is \(X(OH)_2\), it dissociates into 2 \(OH^–\) ions, so \([E] = \frac{[OH^–]}{2} = 0.0615 \, \text{moldm}^{-3}\).

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

Explanation: Moles of E = 0.0615 × 0.25 = 0.0154. Molar mass = \(\frac{2.63}{0.0154} = 171 \, \text{g/mol}\), consistent with \(Ba(OH)_2\).

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

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

(d) \(\Delta H_{sol}\) becomes more exothermic down Group 2.

Explanation: Lattice energy decreases faster than hydration energy, making solubility increase from Mg to Ba.

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