Chemistry • Year 12 • Module 5 • Lesson 17

Solubility Product Ksp

Synthesise Ksp theory with quantitative data and real-world contexts to produce extended, evidence-based responses at Band 5–6 level.

Master • Band 5–6

Question 1 — Sydney Water hardness removal: evaluate the strategy

Scenario

Sydney Water treats bore water from the Hawkesbury–Nepean catchment to reduce ‘hardness’ — excess dissolved Ca2+ and Mg2+ ions that cause scale build-up in pipes and reduce detergent effectiveness. One approved method is lime softening: adding Ca(OH)2 (hydrated lime) to raise pH, which precipitates Mg2+ as Mg(OH)2 and reduces Ca2+ as CaCO3.

A junior engineer proposes an alternative: adding Na2SO4 to precipitate Ca2+ as CaSO4 instead of CaCO3, arguing it is cheaper and equally effective.

Table 1. Ksp data for relevant salts at 25°C
Salt Dissolution equation Formula type Ksp
Mg(OH)2 Mg(OH)2 ⇌ Mg2+ + 2OH− 1:2 5.6 × 10−12
CaCO3 CaCO3 ⇌ Ca2+ + CO32− 1:1 3.3 × 10−9
CaSO4 CaSO4 ⇌ Ca2+ + SO42− 1:1 4.9 × 10−5
BaSO4 BaSO4 ⇌ Ba2+ + SO42− 1:1 1.1 × 10−10

Extended response prompt 8 marks

Using Table 1 and your knowledge of Ksp, evaluate the junior engineer’s proposal that Na2SO4 addition would be equally effective at removing Ca2+ from Sydney Water.

In your response you must:

  1. Calculate the molar solubility (s) of CaSO4 and CaCO3 from the Ksp data, showing all steps.
  2. Compare the effectiveness of CaSO4 precipitation versus CaCO3 precipitation at removing Ca2+ from water.
  3. Identify and explain one additional criterion (other than Ksp) that would affect the engineer’s decision.
  4. Reach a clear, evidence-based judgement on whether the proposal is equally effective.
Worked answer in the key below. Do not peek until you have drafted a full response.

Question 2 — Multi-step Ksp investigation: lead contamination in old NSW housing

Scenario

Lead-based paints used in NSW homes before 1970 contain compounds including PbCO3 (white lead) and PbSO4. NSW Health reports that even small dissolved Pb2+ concentrations pose neurological risk to children. A community group is assessing which paint compound poses the greater long-term dissolution hazard in garden soil.

Relevant Ksp data (25°C): PbCO3 = 7.4 × 10−14 (1:1 type); PbSO4 = 2.5 × 10−8 (1:1 type).

7 marks

(a) Write the Ksp expression for each compound and calculate the molar solubility of PbCO3 and PbSO4 in pure water. Show all working. 3 marks

(b) Using your calculated molar solubilities, convert each to g/L (Mr: PbCO3 = 267.2 g/mol; PbSO4 = 303.3 g/mol). Identify which compound releases more dissolved Pb2+ per litre of rainwater seeping through garden soil and calculate the ratio of the two Pb2+ concentrations. 2 marks

(c) The NSW Health action level for dissolved lead in drinking water is 10 µg/L (= 4.83 × 10−8 mol/L). Based on your calculations, state whether either compound would exceed this action level when dissolved in pure water, and identify one assumption this calculation makes that may not hold in real soil conditions. 2 marks

Stuck? Revisit lesson Cards 2 and 4. Convert mol/L to g/L by multiplying by Mr.
Answers — Do not peek before attempting

Q1 — Sydney Water hardness removal

Step 1 — Molar solubility calculations:
CaSO4 (1:1): Ksp = s2 = 4.9 × 10−5. s = √(4.9 × 10−5) = 7.0 × 10−3 mol/L.
CaCO3 (1:1): Ksp = s2 = 3.3 × 10−9. s = √(3.3 × 10−9) = 5.7 × 10−5 mol/L.

Step 2 — Comparison:
CaSO4 has s = 7.0 × 10−3 mol/L — roughly 120 times more soluble than CaCO3 (s = 5.7 × 10−5 mol/L). This means adding Na2SO4 would form CaSO4, but a far greater [Ca2+] would remain dissolved in solution at equilibrium — the treatment would remove far less Ca2+ than the lime-softening method (CaCO3 precipitation).

Step 3 — Additional criterion:
Any one of: cost of Na2SO4 vs lime; introduction of SO42− ions to drinking water (health limit 500 mg/L); pH adjustment requirements; filter efficiency for different precipitate particle sizes; sludge disposal.

Step 4 — Judgement:
The proposal is NOT equally effective. CaSO4 is approximately 120× more soluble than CaCO3, so precipitation of Ca2+ as CaSO4 would leave far more dissolved Ca2+ in the treated water, failing to adequately reduce hardness.

Marking notes: 1 mark each calculation (2 total) • 2 marks comparison (quantitative with ratio) • 2 marks additional criterion (identify + explain) • 2 marks judgement (evidence-based, clear direction).

Q2 — Lead contamination

(a)
PbCO3: Ksp = [Pb2+][CO32−] = s2 = 7.4 × 10−14. s = √(7.4 × 10−14) = 2.72 × 10−7 mol/L.
PbSO4: Ksp = [Pb2+][SO42−] = s2 = 2.5 × 10−8. s = √(2.5 × 10−8) = 1.58 × 10−4 mol/L.
(1 mark Ksp expressions correctly written for both; 1 mark each molar solubility.)

(b)
PbCO3 in g/L: 2.72 × 10−7 mol/L × 267.2 g/mol = 7.27 × 10−5 g/L = 0.0727 mg/L.
PbSO4 in g/L: 1.58 × 10−4 mol/L × 303.3 g/mol = 4.79 × 10−2 g/L = 47.9 mg/L.
PbSO4 releases far more dissolved Pb2+. Ratio: 1.58 × 10−4 / 2.72 × 10−7 ≈ 581 — PbSO4 releases ~580× more lead into solution. (1 mark correct g/L conversions; 1 mark correct ratio and identification.)

(c)
PbCO3: s = 2.72 × 10−7 mol/L. NSW Health limit = 4.83 × 10−8 mol/L. PbCO3 exceeds the limit by ~5.6×. PbSO4 exceeds the limit by ~3300×. Both compounds exceed the action level in pure water.
Assumption: the calculation assumes equilibrium in pure water at 25°C. Real soil conditions include: other dissolved ions (common ion effect), pH variation (acidic soil increases dissolution), organic ligands complexing Pb2+, and temperature variation — all of which can increase effective Pb2+ concentrations. (1 mark both exceed stated clearly; 1 mark valid assumption identified and explained.)

HSCScience • Chemistry • Year 12 • Module 5 • Lesson 17 • Worksheet 3 (Master)