Qsp, Precipitation & Common Ion Effect

Q1 — Sample Band 6 response (8 marks), annotated

Calculations (4 values):

• Patient A, before: Q_sp = (2.5 × 10⁻³)(1.8 × 10⁻³) = 4.5 × 10⁻⁶ >> K_sp = 2.3 × 10⁻⁹ — highly supersaturated; stones form.
• Patient A, after: Q_sp = (2.8 × 10⁻³)(5.0 × 10⁻⁴) = 1.4 × 10⁻⁶ > K_sp — still supersaturated, but Q_sp has fallen to ~31% of baseline. Stone risk reduced but not eliminated.
• Patient B, before: same as A before: Q_sp = 4.5 × 10⁻⁶.
• Patient B, after: Q_sp = (8.0 × 10⁻³)(1.5 × 10⁻³) = 1.2 × 10⁻⁵ >> K_sp — Q_sp has increased nearly threefold compared to baseline. Stone risk worsens.

Marking criteria.

• 2 marks — All four Q_sp values correct (1 mark for Patient A before and after; 1 mark for Patient B before and after), each compared to K_sp with a statement of whether the solution is supersaturated.
• 1 mark — Patient A: the common ion is Ca²⁺ (added modestly via food); the main effect is that dietary Ca²⁺ binds gut oxalate, lowering urinary [C₂O₄²⁻] substantially — this is the dominant factor reducing Q_sp. The slight rise in urinary [Ca²⁺] is outweighed by the large fall in [C₂O₄²⁻].
• 1 mark — Patient B: the common ion is Ca²⁺; supplemental calcium increases urinary [Ca²⁺] dramatically (from 2.5 × 10⁻³ to 8.0 × 10⁻³ mol L⁻¹). Although [C₂O₄²⁻] decreases modestly, the much larger increase in [Ca²⁺] raises Q_sp overall — the product of two concentrations worsens when one term triples and the other falls only slightly.
• 1 mark — Accounts for why Patient B's intervention worsens risk: Q_sp is a product of both ion concentrations; tripling [Ca²⁺] while [C₂O₄²⁻] falls only 17% raises Q_sp from 4.5 × 10⁻⁶ to 1.2 × 10⁻⁵. Ksp is unchanged — only the equilibrium position is relevant.
• 1 mark — Explicit justified conclusion: Patient A's dietary calcium intervention is more chemically sound because Q_sp decreases (from 4.5 × 10⁻⁶ to 1.4 × 10⁻⁶), reducing supersaturation; Patient B's supplement strategy increases Q_sp, worsening stone risk despite lowering [C₂O₄²⁻].
• 1 mark — Uses precise lesson terminology throughout: Q_sp, K_sp, supersaturated, common ion effect, Le Chatelier's Principle (optional but valued), solubility.

Q2 — Sample Band 6 response (7 marks), annotated

Overall judgement: The claim contains one correct idea embedded in three significant errors.

Correct element: Adding a soluble salt with a common ion does decrease the solubility of a sparingly soluble salt. The reasoning for Sydney Water's use of Na₂CO₃ is directionally correct — CO₃²⁻ is a common ion for the CaCO₃ dissolution equilibrium (CaCO₃(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq)). [1 mark]

Error 1 — “any soluble salt”: Only salts that release a common ion (an ion already present in the dissolution equilibrium) decrease solubility. Adding NaCl to a CaCO₃ solution introduces Na⁺ and Cl⁻, neither of which appears in the CaCO₃ equilibrium. Q_sp = [Ca²⁺][CO₃²⁻] is unaffected; solubility of CaCO₃ is unchanged (to a first approximation). [1 mark]

Error 2 — Ksp decreases: Ksp is a thermodynamic equilibrium constant that depends only on temperature. Adding a common ion shifts the equilibrium position (left shift — less dissolves) but does not change the energy difference between dissolved and solid states. K_sp(CaCO₃) is the same whether Na₂CO₃ is present or absent, at constant temperature. Only changing temperature changes K_sp. [1 mark]

Correct mechanism for Sydney Water: Na₂CO₃ fully dissociates, releasing CO₃²⁻ (a common ion). This raises Q_sp = [Ca²⁺][CO₃²⁻] above K_sp. By Le Chatelier's Principle, the equilibrium shifts left — CaCO₃ precipitates, reducing dissolved [Ca²⁺] (hardness). K_sp is unchanged; only the equilibrium position shifts. [1 mark]

Calculation / example: Suppose [Ca²⁺] = 2.0 × 10⁻³ mol L⁻¹ and K_sp(CaCO₃) = 3.4 × 10⁻⁹. Molar solubility in pure water s = √(3.4 × 10⁻⁹) = 5.8 × 10⁻⁵ mol L⁻¹. If Na₂CO₃ is added to give [CO₃²⁻] = 0.010 mol L⁻¹, new s(Ca²⁺) = K_sp/[CO₃²⁻] = 3.4 × 10⁻⁹/0.010 = 3.4 × 10⁻⁷ mol L⁻¹ — ~170 times less soluble. K_sp = 3.4 × 10⁻⁹ throughout. [1 mark]

Defensible reformulation: “The common ion effect occurs only when the added salt releases an ion that also appears in the dissolution equilibrium of the sparingly soluble salt. Adding such a common ion raises Q_sp above K_sp, causing the equilibrium to shift left and reducing solubility. Ksp is unchanged — only the equilibrium position changes. Sydney Water's addition of Na₂CO₃ exploits this: CO₃²⁻ is the common ion, precipitating CaCO₃ and reducing water hardness. A salt with no common ion (e.g. NaCl added to a CaCO₃ system) has no meaningful common ion effect on CaCO₃ solubility.” [1 mark for defensible reformulation addressing all three errors]

Marking criteria.

• 1 mark — Identifies the one correct element (common ion does decrease solubility of a sparingly soluble salt that shares that ion).
• 1 mark — Correctly refutes “any soluble salt” — must be a salt with a common ion; adds NaCl or other non-common-ion example.
• 1 mark — Correctly refutes Ksp changing — K_sp is temperature-dependent only; common ion changes equilibrium position, not the constant.
• 1 mark — Accurately explains the correct mechanism for Sydney Water's hardness treatment using CO₃²⁻ as the common ion, LCP, and Q_sp > K_sp.
• 1 mark — Provides a supporting calculation or quantitative example showing solubility decrease while K_sp is constant.
• 2 marks — Reformulates the claim into a scientifically accurate statement that addresses all three errors (common ion specificity, K_sp constancy, Sydney Water mechanism) and uses precise lesson terminology. Award 1 mark if only two of three elements are addressed.