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.
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Three pairs of ionic solutions are mixed. Predict whether a precipitate forms in each case and, if so, name the precipitate:
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.
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.
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.
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):
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)
NESA specifies three precipitation reactions for Module 5 — know these completely, including precipitate formula, colour, balanced molecular equation, and net ionic equation.
| Reaction | Precipitate | Colour | Net 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:
$\text{KCl}(aq) + \text{AgNO}_3(aq) \rightarrow \text{AgCl}(s) + \text{KNO}_3(aq)$
$2\text{KI}(aq) + \text{Pb(NO}_3)_2(aq) \rightarrow \text{PbI}_2(s) + 2\text{KNO}_3(aq)$
$\text{Na}_2\text{SO}_4(aq) + \text{Ba(NO}_3)_2(aq) \rightarrow \text{BaSO}_4(s) + 2\text{NaNO}_3(aq)$
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:
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) ✓
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 ion | Precipitant added | Precipitate formed | Net ionic equation |
|---|---|---|---|
| Pb²⁺ | Na₂CO₃ | PbCO₃(s) — insoluble | Pb²⁺ + CO₃²⁻ → PbCO₃(s) |
| Pb²⁺ | Na₂SO₄ | PbSO₄(s) — insoluble | Pb²⁺ + SO₄²⁻ → PbSO₄(s) |
| Hg²⁺ | Na₂S | HgS(s) — black, extremely insoluble | Hg²⁺ + 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?
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).
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 ✓
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.
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 ✓
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.
From NAGSAG — Pb²⁺ forms insoluble compounds with: SO₄²⁻ (PbSO₄ — insoluble) and CO₃²⁻ (PbCO₃ — generally insoluble). Adding Na₂SO₄ or Na₂CO₃ would precipitate Pb²⁺.
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.
$\text{Pb}^{2+}(aq) + \text{CO}_3^{2-}(aq) \rightarrow \text{PbCO}_3(s)$
Charge check: +2 + (−2) = 0 = 0 ✓
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.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Q1. Which of the following pairs of solutions will produce a precipitate when mixed?
Q1. Select the option that pairs of solutions will produce a precipitate when mixed?
Q2. When KI(aq) is mixed with Pb(NO₃)₂(aq), a bright yellow precipitate forms. Which is the correct net ionic equation?
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?
Lesson 16 complete — Solubility Rules & Precipitation