HSC Exam Practice

Sample response. The ion product Q_sp uses the same mathematical expression as K_sp — the product of ion concentrations raised to stoichiometric powers — but is evaluated using the actual (non-equilibrium) ion concentrations present at any given moment. K_sp is the value of this expression specifically at equilibrium; it is a constant at a given temperature.

Marking notes. 1 mark for correctly defining Q_sp as the product of ion concentrations at any moment (same form as K_sp). 1 mark for distinguishing Q_sp from K_sp: Q_sp uses current/non-equilibrium concentrations; K_sp is the value at equilibrium.

Sample response. (i) Q_sp > K_sp: the solution is supersaturated; ions deposit as solid (precipitate forms) and the equilibrium shifts left until Q_sp decreases to equal K_sp. (ii) Q_sp < K_sp: the solution is unsaturated; no precipitate forms; more solid can dissolve (equilibrium shifts right) until Q_sp increases to equal K_sp. (iii) Q_sp = K_sp: the solution is exactly saturated; dynamic equilibrium exists and there is no net change.

Marking notes. 1 mark per outcome, each requiring both the condition (Q_sp vs K_sp) and the consequence (precipitate / dissolves / no change). Do not award if only the condition is stated without the consequence.

Sample response. When two solutions are mixed, the total volume increases and each ion's concentration decreases proportionally (dilution). If original concentrations are used, Q_sp is overestimated — it may appear that Q_sp > K_sp (precipitate should form) when the actual diluted Q_sp is less than K_sp (no precipitate). In borderline cases this produces the completely wrong conclusion about whether precipitation occurs.

Marking notes. 1 mark for identifying that mixing increases total volume and reduces each ion's concentration (dilution). 1 mark for stating that omitting dilution overestimates Q_sp. 1 mark for describing the consequence — incorrect prediction of precipitation in borderline cases.

Sample response. The dissolution equilibrium is AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). NaCl fully dissociates, releasing Cl⁻ — the common ion (shared with the AgCl equilibrium). This increases [Cl⁻], so Q_sp = [Ag⁺][Cl⁻] rises above K_sp. By Le Chatelier's Principle, the equilibrium shifts left — more AgCl precipitates — decreasing [Ag⁺] until Q_sp returns to K_sp. The molar solubility of AgCl decreases significantly. K_sp is unchanged — only temperature changes K_sp.

Marking notes. 1 mark for identifying Cl⁻ as the common ion from NaCl. 1 mark for stating that [Cl⁻] increases, raising Q_sp above K_sp. 1 mark for Le Chatelier left shift and the consequence (AgCl precipitates / solubility decreases). 1 mark for explicitly stating K_sp is unchanged.

Sample response. Adding a common ion does not change K_sp — it only shifts the equilibrium position (decreases solubility). K_sp is solely a function of temperature. Increasing temperature for a salt with endothermic dissolution shifts the equilibrium right (LCP: equilibrium shifts in the endothermic direction when temperature increases), which increases both K_sp and molar solubility. Common ion addition: solubility decreases, K_sp unchanged. Temperature increase: K_sp increases, solubility increases (for endothermic dissolution).

Marking notes. 1 mark for stating common ion does not change K_sp, only shifts equilibrium position (decreases solubility). 1 mark for correctly stating increasing temperature increases K_sp for endothermic dissolution. 1 mark for the contrast: one changes only position, the other changes the constant itself.

Sample response. Selective precipitation involves adding a reagent ion in a controlled amount so that Q_sp for one salt exceeds its K_sp (that salt precipitates), while Q_sp for the other salt remains below its K_sp (that salt stays dissolved). By filtering off the precipitate, the two ions are separated. The process exploits the large difference in K_sp values between the two salts — the less soluble salt (lower K_sp) precipitates first at a lower reagent concentration. An example is precipitating CuS (K_sp = 6 × 10⁻³⁶) before NiS (K_sp = 1.3 × 10⁻²⁵) by adding low [S²⁻] to a Cu²⁺/Ni²⁺ mixture.

Marking notes. 1 mark for explaining that the reagent is added until Q_sp exceeds K_sp for one salt only. 1 mark for connecting to K_sp difference between the two salts. 1 mark for a correct or illustrative named example (accept any valid pair).

(a) Precipitation from graph (2 marks). Experiments 2 and 3 produce a CaF₂ precipitate because their Q_sp values (1.5 × 10⁻¹⁰ and 8.0 × 10⁻⁹ respectively) are above the K_sp line (3.9 × 10⁻¹¹) — both bars reach above the dashed K_sp line on the graph. Experiments 1 and 4 do not precipitate: Q_sp values (2.0 × 10⁻¹² and 5.0 × 10⁻¹³) are below K_sp, so the solution is unsaturated. [1 mark for correctly identifying experiments 2 and 3 with explanation; 1 mark for correctly identifying 1 and 4 with explanation.]

(b) [F⁻] in Experiment 2 (3 marks). Q_sp(CaF₂) = [Ca²⁺][F⁻]² = 1.5 × 10⁻¹⁰. [1] Rearrange: [F⁻]² = 1.5 × 10⁻¹⁰ / (1.5 × 10⁻⁴) = 1.0 × 10⁻⁶. [1] [F⁻] = √(1.0 × 10⁻⁶) = 1.0 × 10⁻³ mol L⁻¹. [1]

(c) Both precipitate despite different Q_sp (2 marks). Whether a precipitate forms depends only on whether Q_sp exceeds K_sp, not on the absolute value of Q_sp. Both experiments 2 and 3 have Q_sp > K_sp = 3.9 × 10⁻¹¹, so both are supersaturated and both precipitate. [1] The rate of precipitation or the amount of solid formed may differ (experiment 3 is more supersaturated), but the occurrence of precipitation is determined by the Q_sp vs K_sp comparison alone, not by the magnitude of the excess. [1]

(a) Molar solubility in pure water (3 marks). CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq). Let s = molar solubility. [Ca²⁺] = s, [F⁻] = 2s. [1] K_sp = [Ca²⁺][F⁻]² = s(2s)² = 4s³ = 3.9 × 10⁻¹¹. [1] s³ = 9.75 × 10⁻¹². s = (9.75 × 10⁻¹²)^1/3 = 2.14 × 10⁻⁴ mol L⁻¹. [1]

(b) Molar solubility in 0.050 mol L⁻¹ NaF (3 marks). NaF fully dissociates: [F⁻] ≈ 0.050 mol L⁻¹ (dominates). CaF₂ dissolves by s mol L⁻¹, so [Ca²⁺] = s; [F⁻] = 0.050 + 2s ≈ 0.050 (assumption: s << 0.050). [1] K_sp = s(0.050)² = s(2.5 × 10⁻³) = 3.9 × 10⁻¹¹. s = 3.9 × 10⁻¹¹ / 2.5 × 10⁻³ = 1.56 × 10⁻⁸ mol L⁻¹. [1] Verify: 2s = 3.1 × 10⁻⁸ << 0.050 — assumption valid. [1]

(c) Comparison and LCP (1 mark). s in pure water = 2.14 × 10⁻⁴ mol L⁻¹; s in NaF = 1.56 × 10⁻⁸ mol L⁻¹ — approximately 13,700 times less soluble. NaF adds F⁻ (a product of CaF₂ dissolution); by Le Chatelier's Principle this shifts equilibrium left, more CaF₂ precipitates and solubility decreases. K_sp is unchanged — only the equilibrium position shifts. [1 mark for correct comparison with factor; 1 mark for LCP explanation and statement K_sp unchanged — award this mark if included here or distributed across earlier parts.]

(d) Assumption and limitation (1 mark). Assumption: 2s << 0.050 (the contribution of CaF₂ dissolution to [F⁻] is negligible compared to the NaF concentration). Limitation: in real solutions, ion-pairing, activity coefficients (ionic strength effects), or competing equilibria (e.g. CaF⁺ complex formation) would alter the effective ion concentrations, so the calculated s is only an approximation. [1 mark for a valid assumption + a distinct valid limitation.]

Sample response.

The excerpt contains three significant errors. First, it claims that adding any ionic compound raises Q_sp and forces precipitation. This is incorrect. Only adding an ionic compound that releases a common ion — an ion that already appears in the dissolution equilibrium — will affect Q_sp for that dissolution. Adding a salt with no common ion leaves Q_sp unchanged to a first approximation.

Example contrast: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). If NaCl is added to a saturated AgCl solution, it releases Cl⁻ (a common ion), raising [Cl⁻] and thus Q_sp = [Ag⁺][Cl⁻] above K_sp. By Le Chatelier's Principle, the equilibrium shifts left — more AgCl precipitates and [Ag⁺] decreases (solubility decreases). If instead NaNO₃ is added, it releases Na⁺ and NO₃⁻, neither of which appears in the AgCl equilibrium. Q_sp = [Ag⁺][Cl⁻] is not directly increased — no common ion effect, no precipitation.

Second error: the excerpt claims that K_sp increases when more ions are added. This is fundamentally wrong. K_sp is a thermodynamic equilibrium constant that depends only on temperature. Adding ions, regardless of whether they are common or not, does not change K_sp. The common ion effect shifts the equilibrium position (less dissolves, solubility decreases) without altering the equilibrium constant. Confusing the equilibrium position with the equilibrium constant is a critical conceptual error.

Third error: the excerpt implies that “ionic strength raises Q_sp and forces precipitation” as a general rule. In reality, Q_sp = [Ag⁺][Cl⁻] is determined by the actual concentrations of the ions in the equilibrium expression. Adding NaNO₃ may slightly increase the ionic strength, which can affect activity coefficients — in practice this often slightly increases effective solubility (the salting-in effect at low ionic strength), not the reverse. The excerpt's claimed mechanism is incorrect.

Marking criteria.

• 1 mark — Correctly identifies that only a salt with a common ion (not any ionic compound) affects Q_sp for the dissolution equilibrium.
• 1 mark — Correctly contrasts NaCl (common ion Cl⁻, raises Q_sp, causes precipitation) and NaNO₃ (no common ion, Q_sp unchanged, no precipitation) for AgCl.
• 1 mark — Correctly explains the mechanism for NaCl: Cl⁻ raises Q_sp above K_sp → LCP shifts equilibrium left → AgCl precipitates → [Ag⁺] decreases.
• 2 marks — Correctly refutes the claim that K_sp increases when ions are added: K_sp is temperature-dependent only (1 mark); clearly distinguishes between equilibrium position (changes) and equilibrium constant (unchanged) (1 mark).
• 1 mark — Addresses the ionic strength claim: the excerpt's mechanism is wrong — adding a non-common-ion salt does not raise Q_sp via a simple ion-product route; the salting effect operates through activity coefficients, not through Q_sp directly.
• 1 mark — Uses precise lesson terminology throughout (Q_sp, K_sp, common ion, Le Chatelier's Principle, solubility, equilibrium position vs constant) and presents a coherent overall analysis with an explicit judgement about the accuracy of the excerpt.