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4.1.7 Preparing soluble salts

4.1.7 Preparing soluble salts

The method depends on whether the other reactant dissolves

Definition

Soluble

Able to dissolve in a particular solvent.

Definition

Insoluble

Unable to dissolve in a particular solvent.

  1. A soluble salt stays in solution until the water is removed, so everything else has to be got rid of first.
  2. When the other reactant is insoluble, it can be added in excess and filtered off afterwards.
  3. When the other reactant is soluble, excess cannot be filtered off, so the volumes are matched by titration.
  4. Choosing between the two routes starts with asking whether that reactant dissolves.
  5. Either route finishes with crystallisation to obtain the dry salt.
Key Idea
  • Insoluble reactant: add it in excess, then filter the excess off.
  • Soluble reactant: match the volumes by titration, so nothing is left over.
Practical
  • Method: warm dilute sulfuric acid gently, then add black copper(II) oxide a little at a time, stirring, until some solid stays undissolved.
  • The excess is what shows all the acid has reacted: CuO(s)+H2SO4(aq)→CuSO4(aq)+H2O(l)\text{CuO}(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{CuSO}_4(aq) + \text{H}_2\text{O}(l)CuO(s)+H2​SO4​(aq)→CuSO4​(aq)+H2​O(l)
  • Filter off the excess copper oxide, leaving blue copper sulfate solution as the filtrate.
  • Crystallise: heat the filtrate over a water bath, stop well before dryness, and leave it to cool so that crystals form.
  • Not to dryness, because the hydrated crystals would lose their water of crystallisation.

Metals and carbonates reach the same salt by the same route

  1. A metal added in excess gives the salt and hydrogen, and the leftover metal is filtered off.
  2. A metal carbonate added in excess gives the salt, water and carbon dioxide, and the leftover solid is filtered off.
  3. Zinc with sulfuric acid: Zn(s)+H2SO4(aq)→ZnSO4(aq)+H2(g)\text{Zn}(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{H}_2(g)Zn(s)+H2​SO4​(aq)→ZnSO4​(aq)+H2​(g)
  4. Copper carbonate with hydrochloric acid: CuCO3(s)+2HCl(aq)→CuCl2(aq)+H2O(l)+CO2(g)\text{CuCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CuCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)CuCO3​(s)+2HCl(aq)→CuCl2​(aq)+H2​O(l)+CO2​(g)
  5. Fizzing that stops is the sign that the carbonate route has finished.
  6. Insoluble oxides, hydroxides, metals and carbonates can all be used this way.
Example
  • The signal to stop: solid remains undissolved, or the fizzing dies away.
  • The separation: filtration removes the excess solid and leaves the salt in solution.

Titration is used when both reactants dissolve

Definition

Titration

A method that finds the exact volume of one solution that reacts with a measured volume of another.

Definition

Indicator

A substance that changes colour to show whether a solution is acidic, neutral or alkaline.

  1. Rinse the pipette with the alkali and the burette with the acid, then discard both rinses.
  2. Pipette a measured volume of the alkali into a conical flask.
  3. Add a few drops of a single indicator such as methyl orange or phenolphthalein.
  4. Fill the burette with acid, remove the funnel, and read the level at eye level to 0.05 cm30.05\ \text{cm}^30.05 cm3.
  5. Run acid in while swirling the flask, adding it dropwise as the colour begins to change.
  6. Stop at the first permanent colour change and record the final reading.
  7. The titre is the final reading minus the initial one.
  8. Repeat until the titres agree within 0.10 cm30.10\ \text{cm}^30.10 cm3, and average only those concordant results.
Note
  • The first run is a rough one, and it is left out of the mean.
  • A single indicator is used, because universal indicator passes through too many colours to show one endpoint.

Getting a pure, dry salt from a titration

Definition

Crystallisation

A method that obtains a dissolved solid from its solution by evaporating some of the solvent and letting crystals form as the solution cools.

  1. Repeat the titration with the mean volume of acid and no indicator in the flask.
  2. The solution then holds only the salt and water, with no dye in it.
  3. Transfer it to an evaporating basin and heat gently to concentrate it.
  4. Leave the concentrated solution to cool so that crystals form.
  5. Filter off the crystals and dry them between filter papers.
  6. The salt is pure because the reactants were matched exactly and no indicator was present.
Exam technique
  • Leaving the indicator out matters because it would otherwise end up in the crystals, and that reason is the part worth writing.
  • Excess and titration answer different situations, so naming which reactant dissolves comes first.
  • A method answer is judged on its order, so heating, cooling, filtering and drying belong in sequence.
Self review
  • Why is an insoluble reactant added in excess?
  • How is that excess removed from the mixture?
  • Why is titration used when both reactants are soluble?
  • How do you know when the endpoint of a titration has been reached?
  • Why is the final preparation carried out without an indicator?
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To prepare a pure soluble salt, any excess reactant must be removed or the reactants must be added in exactly the correct amounts. The solution is then concentrated by evaporating some of the water and left to cool so that crystals form.

Choose the preparation method from the type of reactant used with the acid.

If the reactant is an insoluble base, carbonate or suitable metal, add it in excess and filter off the unreacted solid. If the reactant is a soluble alkali, use titration to find the exact volumes needed for neutralisation, then repeat without indicator before crystallising the salt.

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What property determines whether excess reactant or titration is used to prepare a soluble salt?

4.1.7 Preparing soluble salts Revision Guide

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Revision notes for Edexcel GCSE Chemistry 4.1.7 Preparing soluble salts: explanations and worked examples.

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