Chemistry • Year 11 • Module 3 • Lesson 3

Precipitation & Solubility Rules

Apply solubility rules to real data and scenarios — predict precipitates, write ionic equations, and interpret quantitative evidence of precipitation reactions.

Apply • Data & Reasoning

Solubility reference (NAGSAG): All nitrates, ammonium, Group 1 soluble. Sulfates mostly soluble (EXCEPT Ba²⁺, Pb²⁺, Ca²⁺). Halides mostly soluble (EXCEPT Ag⁺, Pb²⁺). Carbonates, hydroxides, sulfides mostly insoluble (EXCEPT Group 1 and NH₄⁺). Acetates all soluble.

1. Interpret water-quality monitoring data — Great Artesian Basin bore water

A technician analysed bore water from the Great Artesian Basin. The table below shows dissolved ion concentrations in a sample compared to NSW drinking-water guidelines. Engineers are investigating whether adding sodium carbonate (Na₂CO₃) to the water could selectively remove problematic ions by precipitation. 8 marks

Ion present	Concentration (mg/L)	NSW guideline (mg/L)	Exceeds guideline?
Ca²⁺	320	200	Yes
Ba²⁺	0.8	0.7	Yes
Na⁺	580	500	Yes
SO₄²⁻	140	250	No
Cl⁻	210	250	No

1.1 For each ion that exceeds the guideline, use solubility rules to predict whether adding Na₂CO₃(aq) would precipitate it. Identify the precipitate formula and state symbol if one forms. 3 marks

1.2 Write the net ionic equation for the precipitation of barium ions by carbonate. Include state symbols. 2 marks

1.3 The technician notes that Na⁺ also exceeds the guideline. Explain why adding Na₂CO₃ would not reduce Na⁺ levels and may actually worsen them. 2 marks

1.4 Identify the spectator ions in the Ba²⁺/CO₃²⁻ precipitation reaction when Na₂CO₃(aq) is added to the bore water (assume the main other ions present are Na⁺ and SO₄²⁻). 1 mark

Stuck? Check carbonate solubility for each cation using the NAGSAG table. Na⁺ is Group 1.

2. Interpret a graph — lead removal by precipitation in the Murray–Darling Basin

A water treatment trial added increasing concentrations of Na₂SO₄ to water contaminated with 1.5 mg/L of Pb²⁺ ions. The graph below shows the residual dissolved Pb²⁺ remaining after treatment. 8 marks

Figure 2.1. Residual dissolved Pb²⁺ as a function of Na₂SO₄ concentration added to contaminated water. Hypothetical trial data, after NHMRC (2022) guidelines.

2.1 Describe the trend shown in the graph. Reference at least two data points in your answer. 2 marks

2.2 Estimate the minimum Na₂SO₄ concentration required to reduce Pb²⁺ to below the NSW guideline of 0.01 mg/L. 1 mark

2.3 Use solubility rules to explain why adding Na₂SO₄ causes Pb²⁺ to precipitate. Why is this technique possible — what property of PbSO₄ makes it work? 2 marks

2.4 A student suggests that adding excess Na₂SO₄ (well beyond 2.5 mmol/L) would guarantee zero Pb²⁺ remains. Evaluate this claim using the shape of the graph and your knowledge of Ksp. 3 marks

Stuck? Recall that Ksp for PbSO₄ is very small but non-zero. Check the sulfate solubility exceptions in the NAGSAG table.

3. Predict and justify — which reagent removes which ion?

A Sydney water-treatment plant receives water with excess concentrations of three metal ions: Cu²⁺, Pb²⁺, and Fe³⁺. The engineer has four possible reagents available: NaOH(aq), Na₂CO₃(aq), Na₂SO₄(aq), and NaCl(aq). 6 marks

3.1 For each metal ion below, select the most appropriate reagent and justify your choice using solubility rules. Then write the net ionic equation for the removal reaction (include state symbols). 2 marks each

(a) Cu²⁺ — reagent chosen: __________________

Justification:

Net ionic equation:

(b) Pb²⁺ — reagent chosen: __________________

Justification:

Net ionic equation:

(c) Fe³⁺ — reagent chosen: __________________

Justification:

Net ionic equation:

Stuck? Check what forms an insoluble product with each metal ion. Eliminate any reagent where the precipitate rule does not apply.

4. Cause-and-effect chain — what happens when two ionic solutions are mixed?

Complete the effect boxes using lesson content. Each effect becomes the cause of the next step. 5 marks

Two ionic solutions are mixed in a beaker	→

Four different ion types are now present together	→

One combination of ions forms an insoluble compound	→

A solid lattice forms and sinks	→

Overall outcome (so…): ___________________________________________

Stuck? The four ions present check each possible combination against NAGSAG. Ions that form an insoluble product precipitate; the others remain as spectator ions.

Answers — Do not peek before attempting

Q1 — Great Artesian Basin bore water

1.1 Ca²⁺: CaCO₃(s) precipitates — carbonate rule, Ca²⁺ is not Group 1 [1]. Ba²⁺: BaCO₃(s) precipitates — same rule [1]. Na⁺: Na₂CO₃ is soluble (Group 1 cation on both sides) — no precipitate forms [1].

1.2 Ba²⁺(aq) + CO₃²⁻(aq) → BaCO₃(s) [1 — correct ionic formulas with state symbols]. Award second mark for correct balancing (already balanced 1:1).

1.3 Na₂CO₃ is itself a sodium salt — it introduces additional Na⁺ ions into the water [1]. Na⁺ is Group 1 and forms only soluble salts, so it cannot be removed by precipitation; adding a sodium reagent increases Na⁺ concentration [1].

1.4 Spectator ions: Na⁺(aq) and SO₄²⁻(aq) — they appear on both sides of the full ionic equation unchanged.

Q2 — Graph interpretation (lead removal)

2.1 As Na₂SO₄ concentration increases from 0 to 2.5 mmol/L, residual Pb²⁺ decreases sharply [1]. At 0 mmol/L the starting concentration is 1.5 mg/L; at 1.0 mmol/L it has fallen to approximately 0.05–0.10 mg/L; beyond 1.5 mmol/L the curve plateaus at near 0.02 mg/L [1].

2.2 Approximately 1.5–2.0 mmol/L of Na₂SO₄ is required to reduce Pb²⁺ below 0.01 mg/L (reading from the graph where the curve meets the dashed guideline).

2.3 Pb²⁺ is an exception to the sulfate solubility rule — PbSO₄ is insoluble [1]. When SO₄²⁻ is added, the ion product [Pb²⁺][SO₄²⁻] exceeds Ksp and PbSO₄(s) precipitates, removing Pb²⁺ from solution [1].

2.4 The claim is not fully correct [1]. The graph plateaus near 0.02 mg/L rather than reaching zero — this reflects the very small but non-zero Ksp of PbSO₄ [1]. At equilibrium, a tiny amount of Pb²⁺ always remains in solution; adding excess reagent shifts the equilibrium further toward precipitation (common-ion effect) but cannot reduce [Pb²⁺] to exactly zero [1].

Q3 — Reagent selection for water treatment

(a) Cu²⁺: NaOH(aq). Cu²⁺ is not Group 1, so Cu(OH)₂ is insoluble — pale blue precipitate. Na⁺ and NO₃⁻/SO₄²⁻ are spectator ions. Net ionic: Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s). [2]

(b) Pb²⁺: Na₂CO₃(aq) or Na₂SO₄(aq) both acceptable. With Na₂CO₃: Pb²⁺(aq) + CO₃²⁻(aq) → PbCO₃(s). Pb²⁺ is not Group 1 so PbCO₃ is insoluble. [2]

(c) Fe³⁺: NaOH(aq). Fe³⁺ is not Group 1, so Fe(OH)₃ is insoluble — rust-brown precipitate. Net ionic: Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s). [2]

NaCl is not a valid reagent for any of these ions: Cu²⁺, Pb²⁺, and Fe³⁺ chlorides are all soluble (PbCl₂ is borderline but mainly soluble at low concentrations; not a reliable choice).

Q4 — Cause-and-effect chain

Effect 1: The ions from both solutions are now present together in the same solution (four ion types intermingle freely). Effect 2: Each possible product combination must be checked against the solubility rules to see which are soluble and which are not. Effect 3: A solid precipitate (insoluble ionic lattice) forms as ions that cannot remain dissolved combine. Effect 4: The remaining ions that do not form a precipitate stay dissolved as spectator ions. Overall: A visible solid forms in a previously clear solution — evidence of a chemical reaction; the contaminating ion is removed from solution and can be filtered out.