🧪

Solubility Rules & Precipitation Reactions

Water treatment plants in Australia remove dissolved lead and mercury from contaminated water supplies using precisely calculated precipitation reactions — adding the exact chemical that will convert the invisible dissolved toxin into a visible, filterable solid.

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 — Before You Read

📚 Know

  • The NAGSAG solubility rules framework
  • The three NESA-specified precipitation reactions
  • How to write balanced molecular, complete ionic, and net ionic equations

🔗 Understand

  • Why the four-ion method reliably predicts precipitation
  • The difference between molecular and net ionic equations and when each is appropriate
  • How precipitation reactions remove heavy metals in water treatment

✅ Can Do

  • Predict precipitation from any combination of four ions using solubility rules
  • Write balanced molecular and net ionic equations for precipitation reactions
  • Apply precipitation chemistry to environmental and industrial contexts

Three pairs of ionic solutions are mixed. Predict whether a precipitate forms in each case and, if so, name the precipitate:

  1. Potassium chloride + silver nitrate
  2. Potassium iodide + lead nitrate
  3. Sodium sulfate + barium nitrate

Write your predictions now — you may use any prior knowledge from Module 3. At the end of this lesson you will have the systematic rules to predict any precipitation reaction.

Module 5 — Key Formulas: Lesson 16

Four-ion method: identify all 4 ions → find 2 possible products → apply NAGSAG → write equation
Full ionic equation: split all (aq) species into ions; precipitate stays as formula (s)
Net ionic equation: cancel spectator ions (identical on both sides); only reacting species remain
Spectator ions: appear unchanged (same formula, charge, state) on both sides of the full ionic equation

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.

Choose how you work — type your answers below or write in your book.

Key Terms — scan these before reading
Dynamic equilibriumA state where forward and reverse reaction rates are equal.
Le Chatelier's PrincipleA system at equilibrium shifts to minimise applied disturbances.
Equilibrium constant (Keq)The ratio of product to reactant concentrations at equilibrium.
Reaction quotient (Q)The ratio of product to reactant concentrations at any instant.
Closed systemA system where neither matter nor energy can escape to surroundings.
Reversible reactionA reaction that can proceed in both forward and reverse directions.
01

Card 1 — Solubility Rules: The NAGSAG Framework

Rather than memorising the solubility of every possible ionic compound individually, a small set of rules covers the vast majority of cases — ordered by priority to resolve any conflicts.

N
Nitrates
All soluble — no exceptions
A
Ammonium
All NH₄⁺ compounds soluble — no exceptions
G
Group 1 metals
Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺ — all soluble, no exceptions
S
Sulfates
Mostly soluble; except BaSO₄, PbSO₄ (insoluble); CaSO₄, Ag₂SO₄ (sparingly soluble)
A
Acetates
CH₃COO⁻ — all soluble, no exceptions
G
Generally insoluble
CO₃²⁻, PO₄³⁻, OH⁻, S²⁻ — insoluble unless paired with Group 1 or NH₄⁺.
Cl⁻, Br⁻, I⁻: mostly soluble except Ag⁺ and Pb²⁺
Solubility rule
All soluble
All soluble
Mostly soluble
Mostly soluble
Mostly insoluble
Mostly insoluble
Key insoluble exceptions
None
None
Ba²⁺, Pb²⁺ (insoluble); Ca²⁺, Ag⁺ (sparingly)
Ag⁺ (all three); Pb²⁺
Soluble with Group 1, NH₄⁺
Soluble with Group 1, Ba²⁺; Ca(OH)₂ sparingly
Must Know: Apply NAGSAG to BOTH possible products when two solutions are mixed. Both products must be assessed — only one may be insoluble (the precipitate). A common error is checking only one product and missing the other.
Common Error: CaSO₄ is sparingly soluble — students often assume all calcium compounds are soluble. CaCl₂ is soluble, Ca(NO₃)₂ is soluble, CaCO₃ is insoluble, Ca(OH)₂ is sparingly soluble. Know the sulfate exceptions especially: Ba²⁺, Pb²⁺, Ca²⁺.
NAGSAG — Solubility Quick Reference N Nitrates ALL soluble no exceptions A Ammonium NH₄⁺ ALL soluble no exceptions G Group 1 Li Na K Rb Cs ALL soluble S Sulfates Mostly soluble ⚠ Ba²⁺ Pb²⁺ Ca²⁺ A Acetates CH₃COO⁻ ALL soluble G Generally insoluble CO₃²⁻ PO₄³⁻ OH⁻ S²⁻ HALIDE EXCEPTIONS (Cl⁻ Br⁻ I⁻ mostly soluble) Insoluble: AgCl, AgBr, AgI (all Ag⁺) • PbCl₂, PbBr₂, PbI₂ (all Pb²⁺) PRIORITY RULE N, A, G1 override all = soluble (STAYS dissolved) = mostly soluble (check exceptions) = mostly insoluble → PRECIPITATE
Exam TipWhen explaining equilibrium shifts, always state the direction (left or right) and justify using Le Chatelier's Principle language — simply stating the direction alone will not earn full marks.
02

Card 2 — Predicting Precipitation: The Four-Ion Method

Every precipitation prediction follows the same four-step procedure — identify all ions, determine possible products, apply solubility rules, write the equation.

Step 1: Identify the four ions present when two solutions are mixed.

Step 2: Identify the two possible new ionic combinations (each cation paired with the other anion).

Step 3: Apply NAGSAG to each possible product.

Step 4: Write the molecular, full ionic, and net ionic equations.

Applied to KCl(aq) + AgNO₃(aq):

  • Ions present: K⁺, Cl⁻ (from KCl) and Ag⁺, NO₃⁻ (from AgNO₃)
  • Possible products: AgCl (Ag⁺ + Cl⁻) and KNO₃ (K⁺ + NO₃⁻)
  • NAGSAG: AgCl — Cl⁻ with Ag⁺ → INSOLUBLE. KNO₃ — Group 1 + NO₃⁻ → SOLUBLE

Molecular: $\text{KCl}(aq) + \text{AgNO}_3(aq) \rightarrow \text{AgCl}(s) + \text{KNO}_3(aq)$

Full ionic: $\text{K}^+(aq) + \text{Cl}^-(aq) + \text{Ag}^+(aq) + \text{NO}_3^-(aq) \rightarrow \text{AgCl}(s) + \text{K}^+(aq) + \text{NO}_3^-(aq)$

Net ionic: $\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)$   (K⁺ and NO₃⁻ are spectator ions)

Must Know: The net ionic equation shows only the species that actually participate in the precipitation. The net ionic equation for any silver chloride precipitation is Ag⁺(aq) + Cl⁻(aq) → AgCl(s) — regardless of what the counter-ions are. This is also the test for chloride ions.
Common Error: Writing the precipitate in the full ionic equation as separate ions: Ag⁺(aq) + Cl⁻(aq) on the right side instead of AgCl(s). The precipitate is an insoluble solid — it must be written as the formula unit with state symbol (s), never as separate aqueous ions. Only soluble species are split into ions.
03

Card 3 — The Three NESA-Specified Precipitation Reactions

NESA specifies three precipitation reactions for Module 5 — know these completely, including precipitate formula, colour, balanced molecular equation, and net ionic equation.

ReactionPrecipitateColourNet ionic equation
KCl + AgNO₃ AgCl(s) White (curdy) Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
KI + Pb(NO₃)₂ PbI₂(s) Bright yellow ★ Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)
Na₂SO₄ + Ba(NO₃)₂ BaSO₄(s) White Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

Balanced molecular equations:

$\text{KCl}(aq) + \text{AgNO}_3(aq) \rightarrow \text{AgCl}(s) + \text{KNO}_3(aq)$

$2\text{KI}(aq) + \text{Pb(NO}_3)_2(aq) \rightarrow \text{PbI}_2(s) + 2\text{KNO}_3(aq)$

$\text{Na}_2\text{SO}_4(aq) + \text{Ba(NO}_3)_2(aq) \rightarrow \text{BaSO}_4(s) + 2\text{NaNO}_3(aq)$

Must Know: Know all three NESA reactions: precipitate formula, colour, balanced molecular equation, and net ionic equation. The most commonly tested detail is the bright yellow colour of PbI₂ — distinctive and frequently used to confirm the identity of a precipitate.
Common Error: In the KI + Pb(NO₃)₂ reaction, writing PbI as the formula — wrong. Pb²⁺ requires two I⁻ to balance charge → PbI₂. Always derive the formula of the precipitate from charge balance, not by assuming a 1:1 ratio.
BaSO₄ in Medicine: BaSO₄ is so insoluble (Ksp = 1.1 × 10⁻¹⁰) it is used as a medical contrast agent (barium meal for GI imaging) — it passes through the body without being absorbed. The extreme insolubility that makes Ba²⁺ dangerous as a dissolved ion makes BaSO₄ safe as a diagnostic tool.
04

Card 4 — Writing Balanced Molecular and Net Ionic Equations

The three equations for a precipitation reaction are three levels of detail about the same event. Knowing all three shows complete understanding.

5-Step Procedure:

  1. Identify the precipitate using NAGSAG
  2. Write the balanced molecular equation: both (aq) reactants, precipitate (s), soluble product (aq)
  3. Write the full ionic equation: split all (aq) ionic compounds into ions; leave precipitate as formula (s)
  4. Identify and cancel spectator ions — ions appearing identically on both sides
  5. Write the net ionic equation; verify charge balance on both sides

Applied to Na₂SO₄ + Ba(NO₃)₂:

Molecular: $\text{Na}_2\text{SO}_4(aq) + \text{Ba(NO}_3)_2(aq) \rightarrow \text{BaSO}_4(s) + 2\text{NaNO}_3(aq)$

Full ionic: $2\text{Na}^+(aq) + \text{SO}_4^{2-}(aq) + \text{Ba}^{2+}(aq) + 2\text{NO}_3^-(aq) \rightarrow \text{BaSO}_4(s) + 2\text{Na}^+(aq) + 2\text{NO}_3^-(aq)$

Spectator ions: 2Na⁺ and 2NO₃⁻. Cancel them:

Net ionic: $\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s)$

Charge check: left = +2 + (−2) = 0; right = 0 (neutral solid) ✓

Must Know: Always verify charge balance in the net ionic equation. For Ba²⁺ + SO₄²⁻ → BaSO₄(s): +2 + (−2) = 0 = 0 ✓. For Pb²⁺ + 2I⁻ → PbI₂(s): +2 + 2(−1) = 0 = 0 ✓.
Common Error: Splitting the precipitate into ions in the full ionic equation — writing Ba²⁺(aq) + SO₄²⁻(aq) on the right side. The precipitate is a solid — it is not aqueous and must not be split. Only species labelled (aq) are split into ions.
05

Card 5 — Water Treatment: Removing Heavy Metals by Precipitation

The same precipitation chemistry is applied daily in water treatment plants to remove dissolved lead, mercury, and arsenic from drinking water — turning invisible dissolved toxins into filterable solids.

Heavy metal contamination: lead from corroded pipes, mercury from industrial discharge, arsenic from geological sources. Standard treatment: add a reagent that forms an insoluble compound with the target ion, then filter or sediment the precipitate.

Target ionPrecipitant addedPrecipitate formedNet ionic equation
Pb²⁺Na₂CO₃PbCO₃(s) — insolublePb²⁺ + CO₃²⁻ → PbCO₃(s)
Pb²⁺Na₂SO₄PbSO₄(s) — insolublePb²⁺ + SO₄²⁻ → PbSO₄(s)
Hg²⁺Na₂SHgS(s) — black, extremely insolubleHg²⁺ + S²⁻ → HgS(s)

The choice of precipitant must be specific — it must form an insoluble compound with the target metal but NOT with the other ions in the water (Na⁺, Ca²⁺, Mg²⁺, Cl⁻, etc.). This is an applied NAGSAG decision: which anion forms an insoluble compound with the target cation while remaining soluble with all other cations present?

Must Know: For questions about selecting a precipitant for water treatment: use NAGSAG to select an anion that forms an insoluble compound with the target metal cation, and verify that the same anion forms soluble compounds with all other cations present in the water.
06

Worked Examples

Example 1 — Predicting Precipitation Using the Four-Ion Method Band 4–5

Predict whether a precipitate forms when each pair of solutions is mixed. If a precipitate forms, name it, state its colour, write the balanced molecular equation, and write the net ionic equation. (a) Na₂CO₃(aq) + CaCl₂(aq). (b) KNO₃(aq) + NaCl(aq). (c) FeCl₃(aq) + NaOH(aq).

Part (a): Na₂CO₃ + CaCl₂

Ions: Na⁺, CO₃²⁻ and Ca²⁺, Cl⁻. Possible products: CaCO₃ (Ca²⁺ + CO₃²⁻) and NaCl (Na⁺ + Cl⁻).

NAGSAG: CaCO₃ — CO₃²⁻ generally insoluble; Ca²⁺ is not Group 1 → INSOLUBLE. NaCl — Na⁺ Group 1 → soluble.

Precipitate: CaCO₃ (white).

Molecular: $\text{Na}_2\text{CO}_3(aq) + \text{CaCl}_2(aq) \rightarrow \text{CaCO}_3(s) + 2\text{NaCl}(aq)$

Net ionic: $\text{Ca}^{2+}(aq) + \text{CO}_3^{2-}(aq) \rightarrow \text{CaCO}_3(s)$   Charge check: +2 + (−2) = 0 ✓

Part (b): KNO₃ + NaCl

Ions: K⁺, NO₃⁻ and Na⁺, Cl⁻. Possible products: KCl (K⁺ Group 1 → soluble) and NaNO₃ (Na⁺ Group 1, NO₃⁻ all soluble → soluble).

No precipitate forms. Both possible products are soluble — mixing these solutions produces no observable change.

Part (c): FeCl₃ + NaOH

Ions: Fe³⁺, Cl⁻ and Na⁺, OH⁻. Possible products: Fe(OH)₃ (Fe³⁺ + 3OH⁻) and NaCl (Na⁺ + Cl⁻).

NAGSAG: Fe(OH)₃ — OH⁻ generally insoluble; Fe³⁺ not Group 1 → INSOLUBLE. NaCl → soluble.

Precipitate: Fe(OH)₃ (rust-brown gelatinous precipitate).

Molecular: $\text{FeCl}_3(aq) + 3\text{NaOH}(aq) \rightarrow \text{Fe(OH)}_3(s) + 3\text{NaCl}(aq)$

Net ionic: $\text{Fe}^{3+}(aq) + 3\text{OH}^-(aq) \rightarrow \text{Fe(OH)}_3(s)$   Charge check: +3 + 3(−1) = 0 ✓

Answer: (a) CaCO₃(s) — white; Ca²⁺ + CO₃²⁻ → CaCO₃(s). (b) No precipitate — both possible products soluble. (c) Fe(OH)₃(s) — rust-brown; Fe³⁺ + 3OH⁻ → Fe(OH)₃(s). ✓

Example 2 — Selecting a Precipitant for Water Treatment Band 5–6

A water sample contains Pb²⁺ (5.0 × 10⁻³ mol/L) along with Na⁺, Ca²⁺, Cl⁻, and NO₃⁻. (a) Identify two anions that would precipitate Pb²⁺. (b) Evaluate which is preferable. (c) Write the net ionic equation for the preferred reaction.

Step 1 — Part (a): Identify Precipitants

From NAGSAG — Pb²⁺ forms insoluble compounds with: SO₄²⁻ (PbSO₄ — insoluble) and CO₃²⁻ (PbCO₃ — generally insoluble). Adding Na₂SO₄ or Na₂CO₃ would precipitate Pb²⁺.

Step 2 — Part (b): Evaluate Options

SO₄²⁻ (Na₂SO₄): Na⁺ + SO₄²⁻ → Na₂SO₄ — soluble ✓. But Ca²⁺ + SO₄²⁻ → CaSO₄ — sparingly soluble. Adding excess Na₂SO₄ might cause partial CaSO₄ precipitation. Problem: Ca²⁺ may be unintentionally removed.

CO₃²⁻ (Na₂CO₃): Na⁺ + CO₃²⁻ → Na₂CO₃ — soluble ✓. But Ca²⁺ + CO₃²⁻ → CaCO₃ — insoluble. Na₂CO₃ would co-precipitate Ca²⁺ as well.

Preferred: Na₂CO₃ efficiently precipitates Pb²⁺ (more decisively insoluble than CaSO₄). If Ca²⁺ removal is acceptable, Na₂CO₃ is the better choice. If Ca²⁺ must remain, Na₂SO₄ is preferable as CaSO₄ is only sparingly soluble and its precipitation is less complete.

Step 3 — Part (c): Net Ionic Equation

$\text{Pb}^{2+}(aq) + \text{CO}_3^{2-}(aq) \rightarrow \text{PbCO}_3(s)$

Charge check: +2 + (−2) = 0 = 0 ✓

Answer: (a) SO₄²⁻ and CO₃²⁻ both precipitate Pb²⁺. (b) Na₂CO₃ preferred for Pb²⁺ removal; Na₂SO₄ risks CaSO₄ co-precipitation. (c) Pb²⁺(aq) + CO₃²⁻(aq) → PbCO₃(s). ✓
Interactive — Solubility Classifier
Revisit Your Thinking

Precipitation prediction requires two steps: (1) use solubility rules to identify possible insoluble products, and (2) write the net ionic equation to confirm. For example, KCl + AgNO₃ forms AgCl(s) and KNO₃(aq). AgCl is insoluble, so a white precipitate forms. KNO₃ is soluble, so K⁺ and NO₃⁻ remain as spectator ions. Always name the precipitate by cation first, then anion.

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?

07

Practice Questions

Q1. Which of the following pairs of solutions will produce a precipitate when mixed?

A NaNO₃(aq) + KCl(aq)
B Ba(NO₃)₂(aq) + Na₂SO₄(aq)
C CaCl₂(aq) + KNO₃(aq)
D NH₄Cl(aq) + NaOH(aq) — dilute conditions

Q1. Select the option that pairs of solutions will produce a precipitate when mixed?

ANaNO₃(aq) + KCl(aq)
BBa(NO₃)₂(aq) + Na₂SO₄(aq)
CCaCl₂(aq) + KNO₃(aq)
DNH₄Cl(aq) + NaOH(aq) — dilute conditions

Q2. When KI(aq) is mixed with Pb(NO₃)₂(aq), a bright yellow precipitate forms. Which is the correct net ionic equation?

A K⁺(aq) + NO₃⁻(aq) → KNO₃(s)
B Pb²⁺(aq) + I⁻(aq) → PbI(s)
C Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)
D 2K⁺(aq) + 2I⁻(aq) + Pb²⁺(aq) → PbI₂(s) + 2K⁺(aq)

Q3. A student writes the full ionic equation for Na₂CO₃(aq) + CaCl₂(aq) as: 2Na⁺(aq) + CO₃²⁻(aq) + Ca²⁺(aq) + 2Cl⁻(aq) → Ca²⁺(aq) + CO₃²⁻(aq) + 2Na⁺(aq) + 2Cl⁻(aq). What error has the student made?

A The student used the wrong reactants — Na₂CO₃ and CaCl₂ do not react
B The student failed to show that CaCO₃ is a precipitate — the right side should have CaCO₃(s) instead of Ca²⁺(aq) + CO₃²⁻(aq)
C The student split too many ions — Na⁺ should remain as Na₂CO₃(s) on the left
D The student's equation is correct — all species are aqueous and appear on both sides

Lesson 16 complete — Solubility Rules & Precipitation