Year 11 Chemistry Module 3 ⏱ ~35 min Lesson 3 of 12

Precipitation & Solubility Rules

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.

🌊
Printable worksheet

Download this lesson's worksheet

Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.

Think First

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.

Write your initial thinking in your book
Saved
📐

Key Tools — This Lesson

$\text{X}^{n+}_{(aq)} + \text{Y}^{m-}_{(aq)} \rightarrow \text{XY}_{(s)}\downarrow$  (net ionic)
Solubility rules (NAGSAG): used to predict whether an ionic compound will dissolve (soluble) or precipitate (insoluble) Net ionic equation: full ionic equation minus spectator ions — ions that appear identically on both sides Precipitation indicator: cloudy or coloured solid ($\downarrow$) appearing in a previously clear solution
📖 Know

Key Facts

  • The NAGSAG solubility rules and their exceptions
  • Common precipitates and their colours
  • What spectator ions are
💡 Understand

Concepts

  • Why mixing two clear solutions can produce a solid
  • The difference between molecular, full ionic, and net ionic equations
  • How precipitation is used in water treatment
✅ Can Do

Skills

  • Apply solubility rules to predict if a precipitate forms
  • Write molecular, full ionic, and net ionic equations
  • Identify spectator ions in a reaction
Key Terms — scan these before reading
Solubility rules (NAGSAG)The maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature.
Net ionic equationExplain what a spectator ion is and why it does not appear in the net ionic equation.
Precipitation indicatorA weak acid whose acid and conjugate base forms have different colours, used to signal the endpoint of a titration.
Synthesis reactionA reaction where two or more reactants combine to form a single product.
Decomposition reactionA reaction where a single compound breaks down into simpler substances.
Precipitation reactionA reaction in which an insoluble solid forms when two solutions are mixed.
📋

Solubility Rules — The NAGSAG Framework

Rather than memorising every compound individually, use the NAGSAG rules to predict solubility. Apply them in order of priority:

NAGSAG — SOLUBILITY RULE MEMORY AID N Nitrates ALL soluble No exceptions A Ammonium ALL soluble No exceptions G Group 1 ALL soluble Li⁺ Na⁺ K⁺ ... S Sulfates MOSTLY soluble EXCEPT Ba²⁺ Pb²⁺ Ca²⁺ A Acetates ALL soluble No exceptions G Generally insoluble CO₃²⁻ PO₄³⁻ OH⁻ S²⁻ UNLESS with Group 1 or ammonium
Ion Solubility Insoluble Exceptions
NO₃⁻ (nitrate)All solubleNone
NH₄⁺ (ammonium)All solubleNone
Group 1 metal ionsAll solubleNone
SO₄²⁻ (sulfate)Mostly solubleBa²⁺, Pb²⁺, Ca²⁺
Cl⁻, Br⁻, I⁻ (halides)Mostly solubleAg⁺, Pb²⁺
CO₃²⁻ (carbonate)Mostly insolubleGroup 1, NH₄⁺
OH⁻ (hydroxide)Mostly insolubleGroup 1, Ba²⁺
Apply rules to both possible products: When two solutions are mixed, check both possible product combinations separately — one may be soluble, the other insoluble.
Calcium sulfate exception: CaSO₄ is insoluble — students frequently forget this because calcium compounds are often soluble. Memorise it: Ca²⁺ is an exception for sulfates.
Medical insight — BaSO₄: Barium sulfate is so insoluble it is used as a "barium meal" in medical X-rays of the digestive system. Patients swallow it without absorbing any into the bloodstream — its insolubility is what makes it safe as a contrast agent.
02

Writing Full Ionic and Net Ionic Equations

There are three levels of equation for precipitation reactions. Using the reaction of Pb(NO₃)₂(aq) with KI(aq) as the example:

Level 1 — Molecular equation (complete formulas):
Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)
Level 2 — Full ionic equation (split all soluble ionic compounds into ions; keep precipitate intact):
Pb²⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) + 2K⁺(aq) + 2NO₃⁻(aq)
Spectator ions appear identically on both sides — K⁺(aq) and NO₃⁻(aq) here. They play no role in the reaction.
Level 3 — Net ionic equation (remove spectator ions — the most chemically meaningful):
Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)
Rule for ionic equations: Only split compounds into ions if they are aqueous (aq) AND ionic. Never split a precipitate (s), a liquid (l), or a gas (g) into ions.
Critical error — splitting the precipitate: Students frequently write Pb²⁺(aq) + 2I⁻(aq) on the right side. The precipitate is insoluble — it exists as a solid lattice, not free ions. Always keep the precipitate as PbI₂(s).
03

Water Treatment — Precipitation Removing Heavy Metals

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:

Pb²⁺

Reagent added: Na₂SO₄(aq)
Precipitate formed: PbSO₄(s) — white
Net ionic equation: Pb²⁺(aq) + SO₄²⁻(aq) → PbSO₄(s)

Pb²⁺

Reagent added: Na₂CO₃(aq)
Precipitate formed: PbCO₃(s) — white
Net ionic equation: Pb²⁺(aq) + CO₃²⁻(aq) → PbCO₃(s)

Hg²⁺

Reagent added: Na₂S(aq)
Precipitate formed: HgS(s) — black
Net ionic equation: Hg²⁺(aq) + S²⁻(aq) → HgS(s)
💧 Real-World Anchor — Water Treatment: In short answer Q3, you will be asked to select an appropriate reagent to remove a specific ion from water. Use solubility rules to choose a reagent whose anion forms an insoluble compound with the target cation — and a soluble compound with all other ions present.
Practical limitation: Adding excess reagent can cause secondary contamination — for example, too much Na₂S introduces sulfide ions into the water. Real-world applications require precise stoichiometry. In HSC questions, note this limitation if asked to evaluate the method.
Choose two solutions · watch ions mix · precipitate forms when an insoluble product is predicted by solubility rules Interactive

Worked Example 1 — Predicting Whether a Precipitate Forms

Stepwise
Predict whether a precipitate will form when aqueous barium chloride (BaCl₂) is mixed with aqueous sodium sulfate (Na₂SO₄). If a precipitate forms, name it, give its colour, and write the molecular equation with state symbols.
  1. 1
    Identify all ions present when solutions are mixed
    BaCl₂ provides: Ba²⁺ and Cl⁻   |   Na₂SO₄ provides: Na⁺ and SO₄²⁻
  2. 2
    Identify the two possible product combinations
    BaSO₄ (Ba²⁺ + SO₄²⁻)   and   NaCl (Na⁺ + Cl⁻)
  3. 3
    Apply solubility rules to each product
    NaCl: Group 1 (Na⁺) — all soluble ✓   |   BaSO₄: sulfate, with Ba²⁺ — insoluble exception → precipitate forms
  4. 4
    Write and balance the molecular equation
    BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)
    Atom check: Left — 1 Ba, 2 Cl, 2 Na, 1 S, 4 O. Right — 1 Ba, 2 Cl, 2 Na, 1 S, 4 O. ✓
✓ Answer Yes — white BaSO₄ precipitate forms. BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)

Worked Example 2 — Writing the Net Ionic Equation

Stepwise
For the reaction in Worked Example 1, write (a) the full ionic equation and (b) the net ionic equation. Identify the spectator ions.
  1. 1
    Write the full ionic equation — split soluble compounds; keep BaSO₄(s) intact
    Ba²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) + 2Na⁺(aq) + 2Cl⁻(aq)
  2. 2
    Identify spectator ions
    Na⁺(aq) and Cl⁻(aq) appear identically on both sides — they are spectator ions.
  3. 3
    Cancel spectator ions to give the net ionic equation
    Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
✓ Answer Spectator ions: Na⁺ and Cl⁻. Net ionic: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

📝 How are you completing this lesson?

How Precipitation Works Solution A Pb²⁺ NO₃⁻ soluble + Solution B K⁺ I⁻ soluble Mixed PbI₂ solid NO₃⁻ K⁺ spectator ions

🧪 Activities

📝 Activity 1 — Predict + Justify

Applying Solubility Rules to Predict Precipitates

For each pair of solutions, predict whether a precipitate forms, identify it if it does, and write the net ionic equation.

  1. 1 Aqueous silver nitrate AgNO₃(aq) is mixed with aqueous sodium chloride NaCl(aq). Does a precipitate form?

    Ions present: Ag⁺, NO₃⁻, Na⁺, Cl⁻
    Possible products: AgCl (Ag⁺ + Cl⁻) and NaNO₃ (Na⁺ + NO₃⁻)
    NaNO₃: Group 1 → soluble. AgCl: halide with Ag⁺ → insoluble exception → white precipitate forms.
    Molecular: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
    Spectator ions: Na⁺ and NO₃⁻
    Net ionic: Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
  2. 2 Aqueous potassium carbonate K₂CO₃(aq) is mixed with aqueous calcium chloride CaCl₂(aq). Does a precipitate form?

    Ions present: K⁺, CO₃²⁻, Ca²⁺, Cl⁻
    Possible products: CaCO₃ (Ca²⁺ + CO₃²⁻) and KCl (K⁺ + Cl⁻)
    KCl: Group 1 → soluble. CaCO₃: carbonate with Ca²⁺ (not Group 1, not ammonium) → insoluble → white precipitate forms.
    Molecular: K₂CO₃(aq) + CaCl₂(aq) → CaCO₃(s) + 2KCl(aq)
    Spectator ions: K⁺ and Cl⁻
    Net ionic: Ca²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s)
  3. 3 Aqueous sodium sulfate Na₂SO₄(aq) is mixed with aqueous potassium nitrate KNO₃(aq). Does a precipitate form?

    Ions present: Na⁺, SO₄²⁻, K⁺, NO₃⁻
    Possible products: Na₂SO₄ (Na⁺ + SO₄²⁻ — same compound, stays dissolved) and KNO₃ (K⁺ + NO₃⁻)
    NaNO₃ (alternative pairing): Group 1 → soluble. K₂SO₄: Group 1 → soluble.
    Both possible products are soluble. No precipitate forms — no reaction.

Type your working below before revealing answers:

Complete in your workbook.

✏️ Complete in your workbook
💧 Activity 2 — Case Study: Water Treatment

Selecting Reagents to Remove Heavy Metal Ions

A water treatment plant needs to remove dissolved heavy metal ions from contaminated water. Use solubility rules to select the correct reagent for each scenario.

Strategy: Choose a reagent whose anion forms an insoluble compound with the target cation AND a soluble compound with all other ions in solution (to avoid secondary contamination).
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
Question: For the Pb²⁺ removal scenario, explain why you could NOT use sodium chloride (NaCl) as the reagent, even though Pb²⁺ does form an insoluble precipitate with Cl⁻.

Complete the table and answer the question below:

Complete the table and answer the question in your workbook.

✏️ Complete the table and answer in your workbook
Interactive: Precipitation Predictor
Challenge: Precipitation Predictor Quiz
Revisit Your Thinking

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.

Annotate your initial response in your book
Saved
Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

Misconceptions to Fix

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.

MC

Multiple Choice

5 random questions from a replayable lesson bank — feedback shown immediately

✍️ Short Answer

04

Extended Questions

UnderstandBand 3

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

Type your answer below:

Answer in your workbook.

✏️ Answer in your workbook
ApplyBand 4

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

Type your answer below:

Answer in your workbook.

✏️ Answer in your workbook
EvaluateBand 5

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

Type your answer below:

Answer in your workbook.

✏️ Answer in your workbook

✅ Comprehensive Answers

📝 Activity 1 — Predicting Precipitates

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.

💧 Activity 2 — Water Treatment Table

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.

❓ Multiple Choice

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.

📝 Short Answer Model Answers

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].

☄️
Asteroid Blaster

Blast the Correct Answer

Defend your ship by blasting the correct answers for Precipitation & Solubility Rules. Scores count toward the Asteroid Blaster leaderboard.

Play Asteroid Blaster →

Mark lesson as complete

Tick when you've finished all activities and checked your answers.