Every time a water treatment plant removes lead or mercury from drinking water, it's using a precipitation reaction — chemistry that turns dissolved poisons into harmless solids that can be filtered out. Clear solution in, clean water out.
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A water treatment plant receives water contaminated with dissolved lead ions from old pipes. The lead is invisible — the water looks perfectly clear. Engineers add a chemical to the water and within seconds a white solid forms and sinks to the bottom, taking the lead with it.
Here's the puzzle: how can adding one clear solution to another clear solution suddenly produce a solid? And how would you know in advance which chemical to add? Write down your prediction before reading on.
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
📚 Core Content
Rather than memorising every compound individually, use the NAGSAG rules to predict solubility. Apply them in order of priority:
| Ion | Solubility | Insoluble Exceptions |
|---|---|---|
| NO₃⁻ (nitrate) | All soluble | None |
| NH₄⁺ (ammonium) | All soluble | None |
| Group 1 metal ions | All soluble | None |
| SO₄²⁻ (sulfate) | Mostly soluble | Ba²⁺, Pb²⁺, Ca²⁺ |
| Cl⁻, Br⁻, I⁻ (halides) | Mostly soluble | Ag⁺, Pb²⁺ |
| CO₃²⁻ (carbonate) | Mostly insoluble | Group 1, NH₄⁺ |
| OH⁻ (hydroxide) | Mostly insoluble | Group 1, Ba²⁺ |
There are three levels of equation for precipitation reactions. Using the reaction of Pb(NO₃)₂(aq) with KI(aq) as the example:
The precipitation reactions you've just learned are the same reactions municipal water engineers use to remove dissolved toxic metals from contaminated water supplies.
Heavy metal ions such as Pb²⁺, Hg²⁺, and Cd²⁺ dissolve from corroded pipes and industrial runoff. They are colourless and tasteless — contaminated water looks identical to clean water. Treatment adds a reagent that triggers precipitation:
🧮 Worked Examples
🧪 Activities
1 Aqueous silver nitrate AgNO₃(aq) is mixed with aqueous sodium chloride NaCl(aq). Does a precipitate form?
2 Aqueous potassium carbonate K₂CO₃(aq) is mixed with aqueous calcium chloride CaCl₂(aq). Does a precipitate form?
3 Aqueous sodium sulfate Na₂SO₄(aq) is mixed with aqueous potassium nitrate KNO₃(aq). Does a precipitate form?
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| Target ion to remove | Other ions present | Suggested reagent | Precipitate formed | Net ionic equation |
|---|---|---|---|---|
| Pb²⁺ | Na⁺, Cl⁻ | Your answer | Your answer | Your answer |
| Cu²⁺ | Na⁺, NO₃⁻ | Your answer | Your answer | Your answer |
| Fe³⁺ | K⁺, SO₄²⁻ | Your answer | Your answer | Your answer |
Complete the table and answer the question below:
Complete the table and answer the question in your workbook.
Earlier you were asked: How can two clear solutions suddenly produce a solid? And how do you know in advance which chemical to add?
The key insight: when two ionic solutions mix, you effectively have four ions in the same beaker. If any combination of those ions produces a compound that is insoluble according to the solubility rules, it will immediately precipitate as a solid. You predict this in advance using the solubility rules — the solid forms because the new ionic combination cannot stay dissolved. This is exactly how water treatment engineers remove lead and mercury: they choose a reagent whose ion pairs with the target metal to form an insoluble product.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Wrong: All ionic compounds dissolve in water because water is polar.
Right: Water dissolves many ionic compounds through ion-dipole interactions, but not all. Solubility depends on the balance between lattice energy and hydration energy. Compounds with very high lattice energy (e.g., AgCl, BaSO₄) are insoluble despite water's polarity.
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
8. Explain what a spectator ion is and why it does not appear in the net ionic equation. Use the reaction between potassium iodide and lead(II) nitrate as your example. Write the full ionic equation and identify the spectator ions. 4 MARKS
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Answer in your workbook.
9. Predict and describe what happens when aqueous iron(III) chloride (FeCl₃) is mixed with aqueous sodium hydroxide (NaOH). (a) Use solubility rules to predict whether a precipitate forms, and identify it with its colour. (1 mark) (b) Write the balanced molecular equation with state symbols. (1 mark) (c) Write the net ionic equation. (1 mark) (d) Identify the spectator ions. (1 mark) 4 MARKS
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10. A water treatment facility receives water contaminated with dissolved mercury(II) ions (Hg²⁺). An engineer proposes adding sodium sulfide (Na₂S) to the water to remove the mercury. (a) Write the net ionic equation for the removal of Hg²⁺ using Na₂S. (1 mark) (b) Use solubility rules to justify why mercury(II) sulfide (HgS) precipitates. (1 mark) (c) Evaluate one advantage and one limitation of using precipitation to remove heavy metals from water supplies. (2 marks) (d) The engineer adds excess Na₂S to ensure complete mercury removal. Assess the risk this creates. (1 mark) 5 MARKS
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Answer in your workbook.
1. AgNO₃ + NaCl: Ions: Ag⁺, NO₃⁻, Na⁺, Cl⁻. Products: AgCl (Ag⁺ + Cl⁻ — insoluble, white precipitate), NaNO₃ (soluble). Net ionic: Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
2. K₂CO₃ + CaCl₂: Ions: K⁺, CO₃²⁻, Ca²⁺, Cl⁻. Products: CaCO₃ (insoluble, white precipitate), KCl (Group 1 → soluble). Net ionic: Ca²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s)
3. Na₂SO₄ + KNO₃: Ions: Na⁺, SO₄²⁻, K⁺, NO₃⁻. All possible products (NaNO₃, K₂SO₄) involve Group 1 → all soluble. No precipitate forms.
Row 1 (Pb²⁺, with Na⁺ Cl⁻): Add Na₂CO₃ or Na₂SO₄. With Na₂CO₃: PbCO₃(s) white precipitate forms. Net ionic: Pb²⁺(aq) + CO₃²⁻(aq) → PbCO₃(s)
Row 2 (Cu²⁺, with Na⁺ NO₃⁻): Add NaOH. Cu(OH)₂(s) pale blue precipitate forms (Cu²⁺ + OH⁻; copper is not Group 1 so Cu(OH)₂ insoluble). Net ionic: Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s)
Row 3 (Fe³⁺, with K⁺ SO₄²⁻): Add KOH or NaOH. Fe(OH)₃(s) rust brown precipitate (Fe³⁺ + OH⁻; iron not Group 1 so Fe(OH)₃ insoluble). Net ionic: Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s)
Why not NaCl for Pb²⁺? The Cl⁻ ion is already present in solution. Adding NaCl increases the Cl⁻ concentration but PbCl₂ is only slightly soluble (not completely insoluble like PbSO₄ or PbCO₃) — significant Pb²⁺ would remain in solution. Additionally, the Cl⁻ ion already present means no driving force for additional precipitation from NaCl.
1. C — PbI₂: Pb²⁺ is an exception for iodides → insoluble. The others: Na₂CO₃ (Group 1 → soluble), (NH₄)₂SO₄ (ammonium → soluble), KOH (Group 1 → soluble).
2. B — The precipitate is AgCl(s). Na⁺ and NO₃⁻ remain dissolved unchanged — spectator ions.
3. B — CaCO₃ is an insoluble ionic solid (not free ions). It must be written as CaCO₃(s) in any ionic equation.
4. D — Cu²⁺ + 2OH⁻ → Cu(OH)₂(s). Spectator ions Na⁺ and SO₄²⁻ are removed. Options A and B are molecular/full ionic equations, not net ionic. Option C is incorrect (CuSO₄ is soluble).
5. A — A white solid forming and persisting (cannot be redissolved) confirms new substance PbSO₄(s) produced — strongest evidence of chemical change. Temperature change alone is insufficient; cloudiness could be physical.
6. C (Band 5) — PbCl₂ is only slightly soluble (it sits at the borderline of the solubility exceptions), meaning Pb²⁺ concentration would not drop to safe levels. Also, Cl⁻ ions added to drinking water have taste/health implications.
7. B (Band 6) — This is a nuanced question. Both BaSO₄ and CaSO₄ are insoluble sulfates, so Na₂SO₄ would precipitate both. Na₂CO₃ would also precipitate both (BaCO₃ and CaCO₃ are both insoluble). The only way to selectively precipitate Ba²⁺ is to use very carefully controlled concentrations of Na₂SO₄, since BaSO₄ (Ksp ≈ 1.1×10⁻¹⁰) is about 10,000 times less soluble than CaSO₄ (Ksp ≈ 4.9×10⁻⁵). This is a Band 6 question requiring understanding beyond simple solubility rules. Accept option B as the best available answer with this reasoning.
Q8 (4 marks): A spectator ion is an ion that appears unchanged (same formula, same state) on both sides of the full ionic equation — it does not participate in the reaction [1]. It is excluded from the net ionic equation because it plays no role in forming the product [1]. Full ionic equation for Pb(NO₃)₂ + 2KI: Pb²⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) + 2K⁺(aq) + 2NO₃⁻(aq) [1]. Spectator ions: K⁺(aq) and NO₃⁻(aq) — they appear unchanged on both sides [1].
Q9 (4 marks): (a) Fe(OH)₃ — rust brown precipitate. Fe³⁺ with OH⁻: iron is not Group 1 → hydroxide insoluble; Na⁺ with Cl⁻ → NaCl soluble [1]. (b) FeCl₃(aq) + 3NaOH(aq) → Fe(OH)₃(s) + 3NaCl(aq) [1 — balanced with state symbols]. (c) Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s) [1]. (d) Spectator ions: Na⁺(aq) and Cl⁻(aq) [1].
Q10 (5 marks): (a) Hg²⁺(aq) + S²⁻(aq) → HgS(s) [1]. (b) HgS is a sulfide with a non-Group 1, non-ammonium cation → falls under "generally insoluble" rule for sulfides; very low solubility confirms precipitation [1]. (c) Advantage: converts dissolved invisible ions into filterable solids, removing them effectively and at relatively low cost [1]. Limitation: not all heavy metal compounds are sufficiently insoluble to reduce contamination below safe limits, or may require precise stoichiometry [1]. (d) Excess Na₂S introduces S²⁻ ions into the treated water, which is itself toxic/harmful and constitutes secondary contamination — negating the purpose of treatment [1].
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