HSC Exam Practice

Sample response. The solubility product K_sp is the equilibrium constant expression for the dissolution of a sparingly soluble ionic compound in water, equal to the product of the ion concentrations each raised to the power of their stoichiometric coefficients. The solid is excluded because a pure solid has constant activity (= 1 by convention in thermodynamics), so including it would only multiply the expression by a constant without changing its value — it is standard practice to absorb the activity of the solid into the value of K_sp.

Marking notes. 1 mark — K_sp correctly identified as an equilibrium constant for dissolution; 1 mark — expression involves ion concentrations raised to stoichiometric powers; 1 mark — solid excluded due to constant activity (= 1).

(a) Ag₂CrO₄: Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq); K_sp = [Ag⁺]²[CrO₄²⁻]. (1 mark balanced equation with states; 1 mark correct K_sp with powers.)

(b) Mg(OH)₂: Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq); K_sp = [Mg²⁺][OH⁻]². (1 mark balanced equation with states; 1 mark correct K_sp with powers.)

Sample response. K_sp is the equilibrium constant for dissolution — a fixed value at a given temperature that describes how far the dissolution equilibrium proceeds. Molar solubility (s) is the amount (in mol/L) that actually dissolves per litre of solution to reach saturation. K_sp is used to compare sparingly soluble salts of the same formula type and to predict whether a precipitate will form, while molar solubility expresses the dissolved quantity in physically measurable concentration units and is required to compare salts of different formula types.

Marking notes. 1 mark — K_sp correctly defined as an equilibrium constant (not solubility); 1 mark — molar solubility defined as mol/L dissolved at saturation; 1 mark — appropriate distinction of how each is used (formula-type context / comparison / prediction).

Sample response. CaF₂ is a 1:2 type salt (dissolves to give Ca²⁺ and 2F⁻), while AgCl is a 1:1 type. Their K_sp-to-solubility relationships differ: for AgCl, K_sp = s², giving s = √K_sp = 1.34 × 10⁻⁵ mol/L; for CaF₂, K_sp = 4s³, giving s = (K_sp/4)^1/3 = 2.14 × 10⁻⁴ mol/L. Despite a smaller K_sp, CaF₂ is actually more soluble than AgCl. Direct K_sp comparison is only valid for salts with the same formula type — i.e. the same stoichiometric ion ratio.

Marking notes. 1 mark — identifies different formula types (1:2 vs 1:1) as the reason the comparison is invalid; 1 mark — shows or states the correct molar solubility relationship for each formula type; 1 mark — demonstrates or states that CaF₂ is MORE soluble despite the smaller K_sp (counter-intuitive result).

Sample response. Tooth enamel is mainly hydroxyapatite, Ca₁₀(PO₄)₆(OH)₂, with K_sp ≈ 2.3 × 10⁻⁵⁹. When acid (H⁺ from bacteria) is present, OH⁻ and PO₄³⁻ ions are consumed, removing products of dissolution and shifting the equilibrium right (Le Chatelier) — the enamel dissolves. Fluoride ions replace OH⁻ in the enamel structure to form fluorapatite, Ca₁₀(PO₄)₆F₂, which has K_sp ≈ 1 × 10⁻¹²¹ — many orders of magnitude smaller. A far lower K_sp means a much lower molar solubility, so fluorapatite is far more resistant to dissolution even when H⁺ is present.

Marking notes. 1 mark — K_sp(fluorapatite) << K_sp(hydroxyapatite) stated or implied; 1 mark — Le Chatelier explanation: acid removes a product of dissolution, shifting equilibrium right; 1 mark — links smaller K_sp to lower solubility / greater resistance to dissolution.

Sample response. Dissolution equation: Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq). Ion concentrations in terms of s: [Ag⁺] = 2s = 2(6.5 × 10⁻⁵) = 1.30 × 10⁻⁴ mol/L; [CrO₄²⁻] = s = 6.5 × 10⁻⁵ mol/L. K_sp = [Ag⁺]²[CrO₄²⁻] = (1.30 × 10⁻⁴)²(6.5 × 10⁻⁵) = (1.69 × 10⁻⁸)(6.5 × 10⁻⁵) = 1.1 × 10⁻¹².

Marking notes. 1 mark — correct dissolution equation with [Ag⁺] = 2s and [CrO₄²⁻] = s identified; 1 mark — correct K_sp value (accept 1.0–1.2 × 10⁻¹² for rounding).

(a) Molar solubility calculations:
CaCO₃ (1:1): K_sp = [Ca²⁺][CO₃²⁻] = s² = 3.3 × 10⁻⁹. s = √(3.3 × 10⁻⁹) = 5.74 × 10⁻⁵ mol/L.
CaSO₄ (1:1): K_sp = [Ca²⁺][SO₄²⁻] = s² = 4.9 × 10⁻⁵. s = √(4.9 × 10⁻⁵) = 7.00 × 10⁻³ mol/L.
(1 mark each K_sp expression, 1 mark each correctly calculated s — total 4 marks.)

(b) Which dissolves more readily: CaSO₄ (gypsum) is ~122× more soluble than calcite (CaCO₃). Exposed to rainwater, CaSO₄ dissolves far more readily. For precipitation to form a cave mineral, the ion product must exceed K_sp: because CaSO₄ is more soluble (larger K_sp), it is harder for evaporation or CO₂ loss to raise concentrations above its K_sp threshold — therefore calcite (CaCO₃) precipitates more readily to form stalactites/stalagmites. (2 marks: 1 for quantitative comparison, 1 for correct conclusion about calcite precipitating more readily.)

(c) Ba²⁺ conversion of CaSO₄ to BaSO₄: CaSO₄ (K_sp = 4.9 × 10⁻⁵) has a far larger K_sp than BaSO₄ (K_sp = 1.1 × 10⁻¹⁰). The net reaction CaSO₄(s) + Ba²⁺(aq) → BaSO₄(s) + Ca²⁺(aq) is favoured because BaSO₄ is far less soluble — the system converts the more soluble sulphate to the less soluble sulphate, releasing Ca²⁺ into solution. K_net = K_sp(CaSO₄)/K_sp(BaSO₄) = 4.9 × 10⁻⁵/1.1 × 10⁻¹⁰ ≈ 4.5 × 10⁵ ≫ 1 — thermodynamically highly favourable. (3 marks: 1 for K_sp comparison, 1 for direction/net reaction, 1 for K_net calculation or equivalent argument.)

Sample response. K_sp alone is sufficient to rank solubility only when the compounds being compared have the same formula type (the same ion ratio). For compounds with different formula types, K_sp alone is not sufficient and can be actively misleading. Consider AgCl (1:1 type, K_sp = 1.8 × 10⁻¹⁰) and CaF₂ (1:2 type, K_sp = 3.9 × 10⁻¹¹). A naive comparison suggests CaF₂ is less soluble because its K_sp is smaller. Calculating molar solubility reveals the opposite: for AgCl (1:1), K_sp = s², so s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol/L. For CaF₂ (1:2), [Ca²⁺] = s and [F⁻] = 2s, so K_sp = s(2s)² = 4s³ = 3.9 × 10⁻¹¹; s = (9.75 × 10⁻¹²)^1/3 = 2.14 × 10⁻⁴ mol/L. CaF₂ is approximately 16 times more soluble than AgCl despite its smaller K_sp, because producing two F⁻ ions per formula unit amplifies the contribution of each mole dissolved to the ion product. The claim is therefore partially correct: K_sp alone is sufficient for comparing salts of the same formula type but is insufficient and unreliable for different formula types, where molar solubility must be calculated for each salt and compared directly.

Marking notes. 1 mark — identifies that K_sp comparison is valid only for same formula type. 1 mark — names two appropriate compounds with different formula types (e.g. AgCl and CaF₂). 1 mark — correct K_sp expression for the 1:2 salt (K_sp = 4s³). 1 mark — correct molar solubility calculation for at least one compound, showing all working. 1 mark — correct molar solubility for the second compound. 1 mark — explicitly states which compound is more soluble and reconciles this with the K_sp values (counter-intuitive result). 1 mark — clear evaluative judgement: K_sp sufficient for same formula type, insufficient across different formula types.