Chemistry • Year 12 • Module 5 • Lesson 18

Qsp, Precipitation & Common Ion Effect

Apply Qsp vs Ksp reasoning and the common ion effect to data tables, graphs, a real-world scenario, and a predict-and-justify task.

Apply · Data & Reasoning

1. Interpret precipitation data — Sydney Water hardness control

Sydney Water adds sodium carbonate (Na₂CO₃) to hard water supplies to precipitate calcium carbonate (CaCO₃) and reduce water hardness. A technician mixes samples of hard water (containing Ca²⁺) with Na₂CO₃ solution at different concentrations and records which samples produce a precipitate. K_sp(CaCO₃) = 3.4 × 10⁻⁹ at 25 °C. 7 marks

Trial	[Ca²⁺] after mixing (mol L⁻¹)	[CO₃²⁻] after mixing (mol L⁻¹)	Precipitate observed?
1	2.0 × 10⁻⁴	1.0 × 10⁻⁵	No
2	2.0 × 10⁻⁴	2.0 × 10⁻⁵	No
3	2.0 × 10⁻⁴	2.0 × 10⁻⁴	Yes
4	4.0 × 10⁻³	2.0 × 10⁻³	Yes
5	1.0 × 10⁻⁵	1.0 × 10⁻⁵	No

1.1 Calculate Qsp for Trial 1 and Trial 3. Show full working. 2 marks

1.2 Explain why no precipitate forms in Trial 2 but a precipitate forms in Trial 3, despite the same [Ca²⁺] in both. 3 marks

1.3 A student claims Trial 5 shows no precipitate because the concentrations are “too small to matter.” Use a Qsp calculation to evaluate this claim precisely. 2 marks

Stuck? For each trial, calculate Qsp = [Ca²⁺][CO₃²⁻] and compare to K_sp = 3.4 × 10⁻⁹.

2. Interpret graph — common ion effect on BaSO₄ solubility

The graph below shows the molar solubility of barium sulfate (BaSO₄, K_sp = 1.1 × 10⁻¹⁰) as a function of the concentration of Na₂SO₄ added to the solution. K_sp(BaSO₄) is constant throughout (temperature unchanged). 7 marks

Figure 2. Molar solubility of BaSO₄ vs [Na₂SO₄] added at 25 °C. Logarithmic scale on y-axis. Adapted from illustrative equilibrium modelling data.

2.1 Describe the trend shown in the graph. What happens to the molar solubility of BaSO₄ as [Na₂SO₄] increases? 2 marks

2.2 Using the graph, estimate by what factor the solubility of BaSO₄ decreases when [Na₂SO₄] increases from 0 to 0.10 mol L⁻¹. Show how you estimated this from the graph. 2 marks

2.3 Explain the shape of the curve using Le Chatelier's Principle and the concept of Qsp. Why does the solubility fall so steeply at first and then level off at higher [Na₂SO₄]? 3 marks

Stuck? Read the log scale carefully: each major division is one power of ten. LCP: adding SO₄²⁻ (a product) shifts dissolution left.

3. Case study — forensic soil analysis for lead contamination

Forensic scientists investigating an industrial site test soil leachate for lead contamination. They add sulfate ions (SO₄²⁻) to the leachate and measure whether PbSO₄ precipitates. K_sp(PbSO₄) = 1.6 × 10⁻⁸. A soil sample gives a leachate in which [Pb²⁺] = 3.2 × 10⁻³ mol L⁻¹ after dilution. 5 marks

Context. Soil from former lead-battery recycling sites in Western Sydney has been found to contain elevated Pb²⁺ leaching into groundwater. Comparing Qsp to K_sp allows scientists to determine whether precipitation would naturally reduce dissolved Pb²⁺ at a given sulfate concentration.

3.1 Calculate the minimum [SO₄²⁻] required to initiate precipitation of PbSO₄ from this leachate. Show full working. 3 marks

3.2 If natural sulfate levels in the soil water are only 1.0 × 10⁻⁶ mol L⁻¹, will PbSO₄ precipitate spontaneously? Justify your answer by calculating Qsp. 2 marks

Stuck? For minimum [SO₄²⁻], set Qsp = K_sp and solve for [SO₄²⁻].

4. Predict and justify — copper/nickel selective precipitation in mining

BHP's Olympic Dam hydrometallurgical process separates Cu²⁺ and Ni²⁺ ions from a mixed solution by adding sulfide ions (S²⁻) in a controlled way. Relevant K_sp values: K_sp(CuS) = 6.0 × 10⁻³⁶; K_sp(NiS) = 1.3 × 10⁻²⁵. Both [Cu²⁺] and [Ni²⁺] are initially 0.10 mol L⁻¹. 5 marks

Predict which ion precipitates first as [S²⁻] is gradually increased from zero, and calculate the [S²⁻] at which each sulfide begins to precipitate. Use your answers to explain how selective precipitation of Cu²⁺ from Ni²⁺ is achievable. Justify your reasoning using Qsp and K_sp.

Hint: Precipitation begins when Qsp = K_sp. Calculate [S²⁻] required to initiate CuS precipitation and NiS precipitation separately, then compare.

Answers — Do not peek before attempting

Q1.1 — Qsp for Trial 1 and Trial 3 (2 marks)

Trial 1: Q_sp = [Ca²⁺][CO₃²⁻] = (2.0 × 10⁻⁴)(1.0 × 10⁻⁵) = 2.0 × 10⁻⁹. Since Q_sp = 2.0 × 10⁻⁹ < K_sp = 3.4 × 10⁻⁹, no precipitate forms. [1 mark]

Trial 3: Q_sp = (2.0 × 10⁻⁴)(2.0 × 10⁻⁴) = 4.0 × 10⁻⁸. Since Q_sp = 4.0 × 10⁻⁸ > K_sp = 3.4 × 10⁻⁹, a precipitate forms. [1 mark]

Q1.2 — Why no precipitate in Trial 2 but precipitate in Trial 3 (3 marks)

Trial 2: Q_sp = (2.0 × 10⁻⁴)(2.0 × 10⁻⁵) = 4.0 × 10⁻⁹ < K_sp = 3.4 × 10⁻⁹ — solution is unsaturated, so no precipitate forms. [1]

Trial 3: Q_sp = 4.0 × 10⁻⁸ >> K_sp, so the solution is supersaturated and CaCO₃ precipitates until Qsp returns to K_sp. [1]

The key difference is [CO₃²⁻]: increasing it 10-fold (from 2.0 × 10⁻⁵ to 2.0 × 10⁻⁴) increases Qsp 10-fold, crossing the K_sp threshold. [1]

Q1.3 — Trial 5 Qsp calculation (2 marks)

Q_sp = (1.0 × 10⁻⁵)(1.0 × 10⁻⁵) = 1.0 × 10⁻¹⁰. Since Q_sp = 1.0 × 10⁻¹⁰ < K_sp = 3.4 × 10⁻⁹, the solution is unsaturated — no precipitate. [1]

The student is wrong to say it is just “too small to matter” — the reason is precise: Qsp is less than K_sp by a factor of ~34. A Qsp comparison, not vague size judgment, determines whether precipitation occurs. [1]

Q2.1 — Trend description (2 marks)

As [Na₂SO₄] increases from 0 to 0.10 mol L⁻¹, the molar solubility of BaSO₄ decreases dramatically — from approximately 1.05 × 10⁻⁵ mol L⁻¹ in pure water to approximately 1.1 × 10⁻⁹ mol L⁻¹ at [Na₂SO₄] = 0.10 mol L⁻¹. [1 mark for direction of change; 1 mark for quoting approximate values or scale.]

Q2.2 — Factor of decrease (2 marks)

From the graph: s in pure water ≈ 1.05 × 10⁻⁵ mol L⁻¹; s at 0.10 mol L⁻¹ Na₂SO₄ ≈ 1.1 × 10⁻⁹ mol L⁻¹. Ratio = (1.05 × 10⁻⁵) / (1.1 × 10⁻⁹) ≈ 9,500-fold decrease. [1 mark for reading both values off the graph; 1 mark for computing ratio.]

Q2.3 — Shape of curve using LCP and Qsp (3 marks)

The dissolution equilibrium is BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq). Adding Na₂SO₄ increases [SO₄²⁻], making Q_sp > K_sp. By Le Chatelier's Principle, the equilibrium shifts left — more BaSO₄ precipitates — reducing [Ba²⁺] (i.e. solubility). [1]

The curve falls steeply at first because even small additions of SO₄²⁻ to an already-dilute system dramatically change the ion product. [1]

At high [Na₂SO₄], the curve flattens because [SO₄²⁻] ≈ [Na₂SO₄] >> [Ba²⁺], so further additions of SO₄²⁻ change the ratio [Ba²⁺][SO₄²⁻] proportionally less. K_sp is constant throughout — only solubility changes. [1]

Q3.1 — Minimum [SO₄²⁻] to initiate PbSO₄ precipitation (3 marks)

Precipitation begins when Q_sp = K_sp. Write K_sp(PbSO₄) = [Pb²⁺][SO₄²⁻] = 1.6 × 10⁻⁸. [1]

Solve for [SO₄²⁻]: [SO₄²⁻] = K_sp / [Pb²⁺] = (1.6 × 10⁻⁸) / (3.2 × 10⁻³). [1]

[SO₄²⁻] = 5.0 × 10⁻⁶ mol L⁻¹ — any concentration above this will cause PbSO₄ to precipitate. [1]

Q3.2 — Will PbSO₄ precipitate at [SO₄²⁻] = 1.0 × 10⁻⁶ mol L⁻¹? (2 marks)

Q_sp = [Pb²⁺][SO₄²⁻] = (3.2 × 10⁻³)(1.0 × 10⁻⁶) = 3.2 × 10⁻⁹. [1]

Since Q_sp = 3.2 × 10⁻⁹ < K_sp = 1.6 × 10⁻⁸, the solution is unsaturated — PbSO₄ does not precipitate spontaneously at this natural sulfate level. [1]

Q4 — Selective precipitation of Cu²⁺ from Ni²⁺ (5 marks)

Initiation of CuS precipitation: Q_sp(CuS) = [Cu²⁺][S²⁻] = K_sp(CuS). [S²⁻] = K_sp/[Cu²⁺] = 6.0 × 10⁻³⁶ / 0.10 = 6.0 × 10⁻³⁵ mol L⁻¹. [1]

Initiation of NiS precipitation: [S²⁻] = K_sp(NiS)/[Ni²⁺] = 1.3 × 10⁻²⁵ / 0.10 = 1.3 × 10⁻²⁴ mol L⁻¹. [1]

CuS begins to precipitate at [S²⁻] = 6.0 × 10⁻³⁵ mol L⁻¹, which is approximately 2 × 10¹⁰ times lower than the [S²⁻] needed to precipitate NiS. [1]

This enormous difference means that by controlling [S²⁻] to a value between 6 × 10⁻³⁵ and 1.3 × 10⁻²⁴ mol L⁻¹, Cu²⁺ is quantitatively precipitated as CuS while Ni²⁺ remains in solution. [1]

This is selective precipitation: Qsp for CuS exceeds K_sp(CuS) while Qsp for NiS remains below K_sp(NiS). Filtering the precipitate separates the two metal ions — exploited industrially at Olympic Dam to refine mixed ore leachates. [1]