Preparing Soluble Salts
Prepare soluble salts using excess insoluble solid or titration, then separate and crystallise the product. Explain how each step protects purity.
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A soluble salt remains dissolved after its formation. This lesson explains how to obtain its crystals without leaving excess acid, base or indicator in the product. First choose between an excess-solid reaction and titration, then concentrate and cool the salt solution.
1. Definition
Salt preparation is the selection of suitable reactants and laboratory operations to make, separate and purify a required salt.
| Target and reactants | Method | Product location |
|---|---|---|
| soluble salt from acid + insoluble metal, base or carbonate | add excess solid, filter, crystallise | salt is dissolved in the filtrate before crystallisation |
| soluble salt from acid + alkali or aqueous ammonia | titration, repeat without indicator, crystallise | salt remains in solution before crystallisation |
| insoluble salt from two soluble solutions | precipitation, filter, wash, dry | salt is the residue |
- Apply the
solubility rules
to the target salt. 2. If it is insoluble, use precipitation. 3. If it is soluble, ask whether excess reactant can be removed by filtration. 4. Use titration when both chosen reactants are solutions.
2. Key Ideas
- Decide whether the target salt is soluble before choosing a method.
- Use excess solid only when the excess reactant can be removed by filtration.
- For a soluble salt from two aqueous acid–base reactants, use titration and repeat without indicator.
- Use precipitation for an insoluble salt, then filter, wash and dry the residue.
- A soluble salt is crystallised from the filtrate; an insoluble salt is collected as the residue.
3. Detailed Explanations
A. Soluble salt from an acid and an insoluble solid
Suitable solids include metals that react with the chosen dilute acid, insoluble metal oxides, insoluble hydroxides and insoluble carbonates. The resulting salt must be soluble: lead(II) oxide with sulfuric acid would instead form insoluble lead(II) sulfate, which can coat the solid and hinder further reaction. The solid is added in excess so all the acid is used up; any unreacted solid can then be filtered off.
Procedure: excess-solid reaction
- Place the dilute acid in a beaker. For an insoluble base or carbonate, warm it gently if instructed.
- Add the solid a little at a time while stirring.
- Continue stirring and allow time for each portion to react. Add enough that some solid remains even after further stirring and, where appropriate, gentle warming. A newly added lump alone does not prove the acid has been used up.
- Filter. Discard the excess solid residue; keep the salt solution filtrate.
- Heat the filtrate gently to evaporate some water. Do not heat to dryness.
- Leave the hot concentrated solution to cool so crystals form.
- Filter the crystals. If washing is required, use a small amount of cold distilled water to remove adhering solution while limiting dissolution of the product. Dry between filter papers.
Example: copper(II) sulfate from copper(II) oxide
CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l)
- Macroscopic observation: black copper(II) oxide disappears and a blue solution forms; excess black solid eventually remains.
- Particle model: H + (aq) ions react with oxide ions in the solid. Copper(II) and sulfate ions remain in the filtrate.
- Symbolic representation: the balanced equation includes the solid, aqueous and liquid state symbols.
Other suitable reactions
Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)
For a suitable metal or carbonate, look for gas production to stop while excess solid remains after sufficient reaction time. Stopping bubbles alone could mean that the solid was all used up, the reaction slowed, or a coating formed. Once the reaction is complete, filter off the excess solid. Zinc is not an exception to this separation rule: excess zinc is a solid and can be filtered off.
Use a suitable metal above hydrogen in the reactivity series. Do not choose potassium or sodium because their reactions are dangerously vigorous, or copper and silver because they do not react with dilute hydrochloric or sulfuric acid. Keep flames away when hydrogen is produced.
B. Soluble salt from an acid and an alkali
In this method, the acid and alkali are both aqueous. Adding one in excess would leave a dissolved impurity, which filtration cannot remove. Titration finds the volumes needed for the acid–base reaction. The end point is the chosen indicator’s colour change, not a requirement that every salt solution have pH 7.
Procedure: titration
- Use a pipette to transfer a fixed volume of alkali to a conical flask.
- Add a few drops of a suitable indicator for the chosen acid and base. Universal Indicator has a broad colour change and is generally unsuitable for locating a precise titration end point.
- Add acid from a burette while swirling; add it dropwise near the end point.
- Record the volume of acid used. Repeat to obtain concordant results before choosing the reacting volume.
- Repeat using the same measured volumes but without indicator.
- Gently concentrate the salt solution, cool it, filter the crystals and dry them. Use the same solution concentrations when repeating the measured volumes.
H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)
The first titration establishes the reacting volumes. Its indicator would contaminate the crystals, so prepare the salt solution again using those volumes without adding indicator.
For detailed burette readings and concordant results, revise Titration Technique.
Use Preparing insoluble salts for the precipitation method. It keeps the solid residue, whereas the first filtration in an excess-solid preparation keeps the salt-solution filtrate.
D. Separation and purity language
| Term | Exact meaning | Salt-preparation use |
|---|---|---|
| residue | solid left on the filter paper | excess reactant or insoluble salt product |
| filtrate | liquid passing through the filter paper | soluble salt solution after excess solid is removed |
| wash | rinse a collected solid with a little distilled water | removes soluble impurities from a precipitate |
| concentrate | evaporate some solvent | prepares a soluble salt solution for crystallisation |
| crystallise | form solid crystals from a solution | recovers and purifies a soluble salt |
Heat gently to remove some water, then cool. Strong heating to dryness can cause spitting, product loss or decomposition of some salts.
4. Common Mistakes
- “Soluble salt” describes the target, not necessarily every reactant.
- Filtration cannot remove dissolved acid or alkali.
- In the excess-solid route, the wanted salt is in the filtrate, not the residue.
- In precipitation, the wanted salt is the residue and must be washed before drying.
- Evaporating to complete dryness can lose or decompose product.
5. Exam Tips
Excess solid uses up the acid; filtration removes excess solid; washing removes soluble impurities; concentrating and cooling form crystals.
- Apply the solubility rules before selecting reactants.
- For titration preparation, state that the reacting volumes are repeated without indicator.
- For precipitation, choose two soluble reactants that supply the required ions.
6. Worked Examples
Modelled example 1
Choose suitable reactants
Problem
Study the worked solution
Choose acid and insoluble base
Method
Use dilute hydrochloric acid and copper(II) oxide.Reason
Hydrochloric acid supplies chloride while the insoluble oxide supplies copper(II) ions and can be added in removable excess.Working
CuO(s) + 2HCl(aq) → CuCl₂(aq) + H₂O(l).Ensure the acid is fully used
Method
Add CuO(s) until some solid remains.Reason
Solid remaining after enough time, stirring and suitable gentle warming supports that the acid has been used up. A portion that has just been added could still be reacting.Working
Warm and add portions until excess CuO is visible.Remove excess solid
Method
Filter the mixture.Reason
Insoluble excess CuO stays as residue while soluble CuCl₂ passes into the filtrate.Working
Keep the copper(II) chloride filtrate.Crystallise and dry
Method
Concentrate gently, cool, filter and dry the crystals.Reason
Evaporation and cooling produce crystals without heating to dryness.Working
Pure CuCl₂ crystals are collected and dried.
Guided practice 2
Explain why titration is required
Problem
Use solubility to choose the method
Hints
Hint 1: separation constraint
Hint 2: avoid indicator contamination
View solution step by step
Explain method choice
Method
Use titration rather than an excess reagent.Reason
Both acid and alkali are aqueous, so dissolved excess cannot be filtered away.Working
Titration establishes the neutralising volumes.Prepare clean salt solution
Method
Repeat the exact volumes without indicator.Reason
This avoids both excess reactant and indicator contamination.Working
Obtain neutral KNO₃(aq).Obtain crystals
Method
Concentrate, cool, filter and dry.Reason
Potassium nitrate is soluble, so it must be crystallised from solution.Working
Pure potassium nitrate crystals are collected.
Common misconception 3
Correct a separation error
Learner method
Predict what passes through the filter
View solution step by step
Locate the separation error
Method
Reject filtration as a way to remove aqueous sodium hydroxide.Reason
Filter paper separates insoluble solids, not dissolved ions.Working
Excess NaOH(aq) passes into the filtrate.Replace the method
Method
Use titration, repeat without indicator, then crystallise.Reason
Exact neutralising volumes avoid any dissolved excess.Working
Titration → repeat exact volumes without indicator → concentrate and cool NaCl(aq) to obtain crystals.
The calcium carbonate precipitation example is now in the insoluble-salt lesson.
7. Mind Stretchers
Mind stretcher 1: Use the separation constraintExtension
Question: Why is excess copper(II) oxide suitable for making copper(II) sulfate, while excess sodium hydroxide is not suitable for making sodium sulfate?
Show answer
Excess copper(II) oxide is insoluble and can be filtered off. Excess sodium hydroxide remains dissolved, passes through filter paper and contaminates the salt solution.
The lead(II) iodide reactant question is now in the insoluble-salt lesson.
Try independently: A learner stops adding zinc when bubbles stop, but no zinc remains. They then evaporate the solution to dryness. Identify two weaknesses in this method and explain how to correct them.
Show answer and reasoning
Bubbles stopping without excess zinc does not establish that all acid was used up: the zinc may have been consumed. With suitable dilute acid and clean zinc, add portions until some zinc remains after sufficient reaction time, then filter it off. Concentrate the filtrate gently and cool it to form crystals; heating to complete dryness can lose or damage product.
8. Quiz
Use the Acid–Base Chemistry topic check to practise selecting reagents and explaining separation and purity.
Syllabus and review details
- SEC G3 Pure Chemistry 2027 · 2027
Content structure and subject content, PDF pages 9–24
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