Soluble salts: explain formation and separation
Name a salt from its reactants, follow it through a separation sequence, and distinguish solution mass from recovered salt mass.
Before you begin
- Acids, bases and neutralisation
- Solutions, filtration and conservation of mass
Name the salt from the reactants
In the reactions considered here, an acid reacts with a suitable metal oxide base to form a salt and water. The metal supplies the first part of the salt name. Hydrochloric acid forms chlorides, sulfuric acid forms sulfates, and nitric acid forms nitrates. Copper(II) oxide with sulfuric acid gives copper(II) sulfate; magnesium oxide with hydrochloric acid gives magnesium chloride. Salt means a class of compounds, not just sodium chloride. These naming patterns do not mean that any acid will react with any substance.
Use solubility to choose what to separate
Copper(II) oxide is an insoluble base; copper(II) sulfate dissolves in water. In the example, sufficient oxide is allowed to react with the acid and some oxide remains in excess. Filtration retains that insoluble excess as the residue, while the dissolved salt and water pass through as the filtrate. The salt has already formed during the reaction: filtration does not create it. A clear filtrate can still contain dissolved impurities, so clarity alone does not prove purity.
Distinguish crystallisation from neutralisation
Removing some water concentrates the salt solution. When conditions allow, crystals form from the concentrated solution as it cools. The crystals are separated from the remaining liquid and surface liquid is removed. Dissolved salt remains in that liquid, so not all the salt necessarily becomes collected crystals. Crystallisation separates an existing dissolved substance; it is not the neutralisation reaction. Copper(II) sulfate crystals contain water within their crystal structure: removing surface liquid does not mean heating out all that water.
Evaluate a method rather than memorising steps
The excess-solid method depends on the unused base being insoluble. Excess sodium hydroxide would remain dissolved and pass through ordinary filter paper, so copying the oxide method would contaminate a sodium chloride solution. Acid and alkali amounts instead need controlled measurement so neither is in excess; that measurement method is beyond this introduction. Actual preparation requires a supervised, risk-assessed laboratory method, including suitable eye protection, controlled heating and disposal. This lesson explains the reasoning, not a recipe for an unsupervised experiment.
Worked example
A student reacts copper(II) oxide with sulfuric acid, then filters. The filter paper holds black solid and the liquid below is blue. Which part should be kept to recover copper(II) sulfate, and why is filtering alone insufficient?
Show the worked solution
- Identify copper(II) sulfate as the salt: copper comes from the oxide and sulfate from sulfuric acid.
- Keep the blue filtrate: dissolved salt passes through ordinary filter paper; the black residue is excess copper(II) oxide in this stated example.
- The filtrate still contains water. Concentration, crystallisation and separation of crystals are needed; clarity or colour alone does not prove purity.
Keep the filtrate, then recover crystals from the solution. The residue is not the desired salt.
Try it yourself
A clear salt solution passes through ordinary filter paper. Which statement explains where the dissolved salt goes?
Common mistakes
- Do not confuse the first residue (excess base) with the crystals collected later.
- Solution mass includes solvent; it is not the mass of pure salt.
- A clear solution or unchanged mass alone does not prove purity.
- Excess dissolved alkali cannot be removed by ordinary filtration.
- Written method reasoning does not demonstrate practical competence.
Follow the salt, then test the explanation
Explore the soluble-salts lesson and question mind map.
- React: In this example, copper(II) oxide reacts with sulfuric acid. After sufficient mixing and reaction time, excess unreacted oxide remains as a solid.
- Filter: Keep the filtrate containing dissolved copper(II) sulfate. The first filter retains excess insoluble oxide, not the dissolved salt.
- Concentrate and crystallise: Removing some water concentrates the solution. Suitable concentration and cooling allow crystals to form. Some salt stays dissolved in the remaining liquid.
- Separate and dry: Separate the crystals from the remaining solution and remove surface liquid. This later separation collects crystals; it is different from the first filtration.
Account for the mass
Original constructed data for the copper(II) oxide example, not experimental results or quantities to use in a practical. Assume the reaction is complete, the stated residue is only unreacted oxide, and nothing is spilled or left on apparatus. During concentration, only the stated water leaves; the salt is not lost. The table stops before crystallisation and does not establish that the solution is saturated.
| Quantity | Mass (g) |
|---|---|
| Starting acid solution | 100.0 |
| Copper(II) oxide added | 8.0 |
| Unreacted oxide removed | 3.0 |
| Water removed from filtrate | 20.0 |
Write your reasoning first. These five self-check points are marking guidance for this activity, not official marks or an automatically awarded score.
- Calculate how much oxide reacted.
- Calculate the mass of filtrate after removing the residue.
- Calculate the mass remaining after the stated water loss.
- Explain why neither the reacted oxide mass nor the remaining mixture mass is a measured mass of pure salt.
- Name a real source of loss that the idealised model excludes. Explain how you would account for it when evaluating recovery.
Compare my mass reasoning
- 8.0 − 3.0 = 5.0 g of oxide reacted.
- 100.0 + 8.0 − 3.0 = 105.0 g of filtrate. It includes dissolved salt and water.
- 105.0 − 20.0 = 85.0 g remains in the vessel. The water vapour is outside this weighed boundary; mass has not been destroyed.
- The acid also contributes material to the salt, so 5.0 g of reacted oxide is not the salt mass. The 85.0 g includes remaining water, so it is not pure salt. These data do not give a dry crystal yield.
- For example, solution can remain on filter paper or be lost during transfer. Collect and measure relevant fractions where possible, record losses and keep a consistent separation/drying method when comparing recoveries. Repetition alone cannot recover lost material or prove purity.
Challenge a different method
A student adds excess sodium hydroxide solution to hydrochloric acid and plans to filter out the unused alkali. Name the expected salt, explain the flaw, and identify what must be controlled instead. Then explain why a larger mass of wet crystals does not establish a better yield of pure salt.
Compare my method reasoning
- The salt is sodium chloride.
- Sodium hydroxide is soluble. Excess dissolved alkali passes through ordinary filter paper, unlike excess insoluble copper(II) oxide.
- The reacting amounts must be measured and controlled so neither reactant remains in excess. Simply using equal volumes is insufficient if concentrations differ.
- Wet crystals carry liquid as well as crystals. Compare equivalent starting amounts and a defined surface-drying method; do not call a larger wet mass more pure salt. Copper(II) sulfate crystals can contain water as part of their structure, so surface drying is not the same as removing every form of water.
Revisit neutralisation and excess reactants · Revisit mass and the system boundary
Try fresh questions
Open the six-question soluble-salts practice. This separate set uses new scenarios and data. No score is saved.
Grade 8 is StudyForA’s suggested placement. England KS3 science covers neutralisation and separation; the insoluble-base preparation sequence is an introductory extension towards AQA GCSE soluble-salt preparation, not a claim that this is an official Grade 8 requirement. Specialist review has not been completed. Written reasoning does not demonstrate supervised practical competence.
What can you explain now?
Magnesium oxide reacts with hydrochloric acid. After the reaction, excess solid is filtered away. Name the salt and identify where it is immediately after this filtration.
Compare with the explanation
Magnesium chloride, dissolved in the filtrate.
Magnesium supplies the metal name and hydrochloric acid supplies chloride. The unused insoluble oxide is the residue; soluble magnesium chloride passes through in solution. It has not all become crystals just because the mixture was filtered.
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