Preparing Insoluble Salts
Prepare an insoluble salt by precipitation: choose soluble ion sources, collect the residue, wash away dissolved impurities and dry the product.
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An insoluble salt forms as a solid when suitable dissolved ions meet. Your product is the residue on the filter paper. Collecting the wrong fraction loses the product; skipping washing leaves soluble impurities on it.
Before starting, use the solubility rules to establish that the target salt is insoluble. For a soluble target, use the soluble-salt methods.
Choose soluble sources of the required ions
Choose solutions of two soluble compounds that supply the required cation and anion. When mixed, the required ions form an insoluble solid.
Example: barium sulfate
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)
Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
- Macroscopic observation: a white precipitate forms.
- Particle model: Ba²⁺ and SO₄²⁻ ions join to form a solid lattice; Na⁺ and Cl⁻ remain aqueous spectator ions.
- Symbolic representation: the ionic equation shows only the ions that form the precipitate and balances both atoms and charge.
Separate, wash and dry the precipitate
- Mix the two solutions and stir.
- Filter to collect the insoluble salt as the residue.
- Wash the residue with distilled water to remove soluble ions on its surface.
- Dry the product between filter papers.
Choose enough of each solution to produce the required precipitate. You do not need to say that “both reactants are in excess”: one reagent can be in excess relative to the other. Any dissolved excess is removed by filtration and washing.
Use a little cold distilled water for washing. It dissolves soluble material in the solution clinging to the precipitate. Even a salt described as insoluble has a small solubility, so very large volumes of wash water can lose some product. Filtering alone does not remove liquid held among the solid particles.
For calcium carbonate made from calcium nitrate and sodium carbonate, sodium and nitrate ions remain in solution. The collected solid also initially holds some of that solution; washing removes those soluble impurities before drying.
Explain the reaction at three levels
The white solid is the observation. The required positive and negative ions join in an ionic lattice, which is the particle explanation. The net ionic equation represents that change and omits spectator ions. Do not treat the spectator salt as the precipitate just because its formula appears in the full equation.
For the barium sulfate example above, the ions that enter the solid are Ba²⁺ and SO₄²⁻. Sodium and chloride ions stay dissolved. Both charge and atoms balance in the net ionic equation.
Worked precipitation plan
Challenge 1
Plan a precipitation
Insoluble-product transfer
Keep the correct filter fraction
Hints
Hint 1: form the insoluble pair
Hint 2: purify the solid
View solution step by step
Form the precipitate
Method
Mix calcium nitrate and sodium carbonate solutions.Reason
Mobile Ca²⁺ and CO₃²⁻ ions meet and form insoluble calcium carbonate.Working
Ca(NO₃)₂(aq) + Na₂CO₃(aq) → CaCO₃(s) + 2NaNO₃(aq).Separate the product
Method
Filter and retain the residue.Reason
The required CaCO₃ is the insoluble solid; sodium nitrate stays in the filtrate.Working
Residue: white CaCO₃(s).Purify and dry
Method
Wash with distilled water and dry between filter papers.Reason
Washing removes soluble ions and drying removes water.Working
Pure dry calcium carbonate is obtained.
Check your reasoning
Mind stretcher 1: Design suitable reactantsExtension
Question: A student wants lead(II) iodide. Why should both chosen reactants be soluble even though the product is insoluble?
Show answer
Soluble reactants release the required ions into solution, where Pb²⁺ and I⁻ can meet and form the insoluble precipitate. Soluble spectator ions remain in the filtrate and are washed from the product.
This is a planning exercise: practical work with toxic lead compounds requires the school laboratory’s controlled handling and disposal procedures.
Try independently: Choose two soluble starting salts to make lead(II) iodide. Give the net ionic equation and explain what washing removes.
Show answer and reasoning
Lead(II) nitrate and potassium iodide solutions are suitable: they supply Pb²⁺(aq) and I⁻(aq).
Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)
Filter to collect the yellow precipitate. Washing with a little cold distilled water removes soluble potassium and nitrate ions, along with any dissolved excess reagent.
Try independently: A learner filters calcium carbonate, then dries the residue without washing it. Is a dry solid necessarily pure? Explain using the ions present after precipitation. Why would dilute hydrochloric acid be unsuitable for washing this calcium carbonate?
Show answer and reasoning
No. The residue holds solution containing soluble sodium and nitrate ions. Drying removes water, but those dissolved impurities can remain as solids on the calcium carbonate. Wash the residue with a little cold distilled water first, then dry it. Hydrochloric acid would react with calcium carbonate, consuming the required solid:
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)
Use the Acid–Base Chemistry topic check to practise choosing soluble reactants and purifying insoluble salts.
Syllabus and review details
- SEC G3 Pure Chemistry 2027 · 2027
Content structure and subject content, PDF pages 9–24
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