← Unit 2: Molecular interactions and reactions
Topic 2: Aqueous solutions and acidity
Apply solubility rules to predict whether ionic compounds are soluble in water, predict precipitation reactions between aqueous solutions, and write balanced full and net ionic equations including spectator ions
A focused answer to the QCE Chemistry Unit 2 dot point on solubility and precipitation. Lists the QCAA solubility rules for common ionic compounds, walks through predicting whether a precipitation reaction occurs when two aqueous solutions are mixed, and shows how to write balanced molecular, complete ionic and net ionic equations.
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What this dot point is asking
QCAA wants you to predict whether two aqueous solutions will give a precipitate when mixed (using a memorised set of solubility rules), and represent the reaction in three forms: molecular (formula equation), complete ionic equation and net ionic equation. Spectator-ion identification is the test of understanding what is actually happening.
The answer
When two aqueous solutions are mixed, the cations and anions are free to swap partners. A precipitation reaction occurs if one of the possible new combinations is insoluble; that combination crashes out as a solid. The remaining ions stay in solution as spectator ions. Solubility rules are an empirical lookup table that tells you which ionic combinations are soluble.
QCAA solubility rules (memorise this set)
| Class | Soluble | Important exceptions |
|---|---|---|
| Group 1 cations (Li+, Na+, K+, ...) | All soluble | None |
| Ammonium NH_4+ | All soluble | None |
| Nitrates NO_3- | All soluble | None |
| Acetates CH_3COO- (ethanoates) | All soluble | Ag+ (slightly soluble) |
| Chlorides Cl-, bromides Br-, iodides I- | Most soluble | Ag+, Pb^2+, Hg2^2+ insoluble |
| Sulfates SO_4^2- | Most soluble | Ba^2+, Pb^2+, Sr^2+ insoluble; Ca^2+, Ag+ slightly soluble |
| Sulfides S^2- | Most insoluble | Group 1, group 2 (except Be) and NH_4+ soluble |
| Carbonates CO_3^2-, phosphates PO_4^3- | Most insoluble | Group 1 and NH_4+ soluble |
| Hydroxides OH- | Most insoluble | Group 1 and NH_4+ soluble; group 2 increasingly soluble down the group (Ca(OH)_2 slightly soluble, Ba(OH)_2 soluble) |
Rules of thumb:
- Anything sodium, potassium, ammonium or nitrate is soluble. Always.
- Most halides are soluble; remember "silver, lead, mercury(I)" for the insoluble cases.
- Sulfates are mostly soluble; remember "barium, lead, calcium (slightly), strontium" for the insoluble ones.
- Carbonates, phosphates, sulfides and hydroxides are mostly insoluble; remember "group 1 and ammonium" as the soluble exception class.
QCAA data booklets sometimes include a solubility table; verify which version applies to your cohort. The rules above are the most common 2026 set.
Predicting whether a precipitate forms
Procedure:
- Identify the cations and anions in each starting solution.
- List the two possible new ionic combinations (the "swap partners" products).
- Look up each product in the solubility rules.
- If at least one product is insoluble, a precipitate forms. If both new products are soluble, no reaction.
Example: BaCl_2(aq) + Na_2SO_4(aq).
Cations: Ba^2+, Na+. Anions: Cl-, SO_4^2-.
Possible products: BaSO_4 (insoluble per rules) and NaCl (soluble).
Precipitate: BaSO_4(s), white.
Writing the three equation forms
Once you know a precipitate forms, you can write the reaction in three equivalent ways. Each makes different information explicit.
Molecular equation. Shows the formulas of the reactants and products. Always include state symbols.
Complete ionic equation. All aqueous strong electrolytes (soluble ionic compounds and strong acids/bases) are written as separated ions. Insoluble solids, weak electrolytes, gases and molecular liquids stay together.
Net ionic equation. Cancel the spectator ions (those that appear unchanged on both sides).
The net ionic equation captures the chemistry of the precipitation. Whether the original counter-ions were Cl- and Na+, or NO_3- and K+, or anything else soluble, the net change is the same.
Check both balance (atoms on each side equal) and charge balance (total charge on left equals total charge on right). In the example: left 2+ + 2- = 0; right 0. Balanced.
Common precipitate colours
Worth knowing for IA stimulus and EA short response:
| Precipitate | Colour |
|---|---|
| AgCl | White (darkens to violet/grey in light) |
| AgBr | Cream |
| AgI | Yellow |
| PbI_2 | Bright yellow |
| BaSO_4 | White |
| CaCO_3 | White |
| Cu(OH)_2 | Pale blue |
| Fe(OH)_2 | Pale green |
| Fe(OH)_3 | Red-brown |
| CuS | Black |
Stimulus questions often give a colour change and ask which precipitate formed.
Worked example: identifying an unknown by selective precipitation
A solution contains either Na_2SO_4 or NaCl. Adding a few drops of AgNO_3 solution produces no precipitate from one sample and a white precipitate from the other. Identify each.
Reasoning. Ag+ with SO_4^2- forms Ag_2SO_4, which is slightly soluble (not a strong precipitate at low concentrations). Ag+ with Cl- forms AgCl, which is essentially insoluble (strong white precipitate immediately).
The sample that gave the white precipitate is NaCl(aq); the sample that gave no precipitate is Na_2SO_4(aq).
A confirmatory test: add BaCl_2(aq). The Na_2SO_4 sample gives a white precipitate of BaSO_4; the NaCl sample gives no precipitate (BaCl_2 stays in solution).
Selective precipitation underpins qualitative analysis schemes used to identify unknown cations and anions.
Limitations of solubility rules
"Insoluble" in the rules really means "very slightly soluble". Even AgCl has a tiny solubility (about 1.3 x 10^-5 mol/L at 25 degrees C). The solubility product Ksp formalises this at Unit 3 / 4 level.
Some ions form coloured complexes rather than precipitates (e.g. Ag+ with NH_3 dissolves AgCl as [Ag(NH_3)_2]+). These complex equilibria sit outside Unit 2.
Common traps
Forgetting state symbols. A "precipitation reaction" requires (s) for the precipitate and (aq) for the dissolved ions. Equations without state symbols typically lose marks.
Writing precipitates as separated ions. AgCl(s) is a solid lattice; do not split into Ag+ + Cl- on the product side.
Forgetting to cancel spectators in the net ionic equation. The whole point of the net ionic equation is to show only the chemistry that happens. Leaving spectators in loses the conceptual mark.
Mis-balancing for ion charges. When NaCl reacts with Pb(NO_3)_2, the equation is 2NaCl + Pb(NO_3)_2 -> PbCl_2 + 2NaNO_3 (two chlorides per lead). Always balance both atoms and charges.
Confusing "no reaction" with "no precipitate I expected". If the rules say both possible products are soluble, the correct answer is "no reaction"; do not invent a precipitate.
In one sentence
A precipitation reaction occurs when mixing two aqueous solutions produces an insoluble ionic combination as predicted by the solubility rules (Group 1 cations, NH_4+, NO_3- and most chlorides/sulfates soluble; carbonates, phosphates, sulfides and hydroxides mostly insoluble); the molecular equation shows formulas, the complete ionic equation shows all dissolved strong electrolytes as separated ions, and the net ionic equation (after cancelling spectator ions) shows the actual chemical change.
Past exam questions, worked
Real questions from past QCAA papers on this dot point, with our answer explainer.
2024 QCAA-style4 marksSolutions of silver nitrate and sodium chloride are mixed. (a) Predict whether a precipitate forms and identify it. (b) Write the balanced molecular equation, the complete ionic equation and the net ionic equation. (c) Identify the spectator ions.Show worked answer →
A 4-mark answer needs the prediction, all three equations, and the spectator ions named.
(a) Prediction. AgNO_3(aq) and NaCl(aq) potentially exchange to give AgCl and NaNO_3. By the solubility rules: silver salts are insoluble except AgNO_3 and AgF, so AgCl is insoluble (it precipitates). All sodium and all nitrate salts are soluble, so NaNO_3 stays in solution.
Precipitate: silver chloride, AgCl(s), white solid.
(b) Equations.
Molecular: AgNO_3(aq) + NaCl(aq) -> AgCl(s) + NaNO_3(aq).
Complete ionic: Ag+(aq) + NO_3-(aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq) + NO_3-(aq).
Net ionic: Ag+(aq) + Cl-(aq) -> AgCl(s).
(c) Spectator ions. Na+ and NO_3- are the spectators. They appear unchanged on both sides of the complete ionic equation.
Markers reward the rule citation, all three equations with state symbols, and the explicit spectator identification.
2023 QCAA-style3 marksPredict and explain whether a precipitation reaction occurs when (a) Pb(NO_3)_2(aq) is added to KI(aq) and (b) NaCl(aq) is added to KNO_3(aq). For any precipitation reaction that occurs, write the net ionic equation.Show worked answer →
A 3-mark answer needs both predictions and the one net ionic equation.
(a) Pb(NO_3)_2 + KI. Potential products: PbI_2 and KNO_3. Solubility rules: most lead salts are insoluble except Pb(NO_3)_2 and Pb(CH_3COO)_2; lead iodide PbI_2 is insoluble. KNO_3 is soluble (all K+ and all NO_3- salts are soluble).
Precipitate forms: PbI_2 (bright yellow solid).
Net ionic equation: Pb^2+(aq) + 2I-(aq) -> PbI_2(s).
(b) NaCl + KNO_3. Potential products: NaNO_3 and KCl. Both are soluble (all Na+, K+, and NO_3- salts are soluble; Cl- is soluble except with Ag+, Pb^2+, Hg2^2+, none of which are present).
No precipitate; no net reaction. The solution contains Na+, Cl-, K+ and NO_3- as spectators.
Markers reward correct application of the rules, the net ionic equation with charge balance and state symbols, and explicit acknowledgement that no reaction occurs in (b).
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