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GCSE/Chemistry/AQA

C4.6Soluble salts: preparation by reacting acid with insoluble base and crystallisation

Notes

Preparing soluble salts (general method)

The general method is the same as the required practical (C4.5): pick an acid that gives the right anion, an insoluble base that gives the right cation, react them, filter, crystallise.

Choosing the right reactants

  • For a chloride salt, use HCl.
  • For a sulfate salt, use H₂SO₄.
  • For a nitrate salt, use HNO₃.

The base provides the metal. The base must be insoluble to allow the "excess" trick (so the unreacted excess can be filtered off).

Worked exampleExamples

  • Magnesium sulfate: H₂SO₄ + MgO (or Mg(OH)₂ or MgCO₃) → MgSO₄ + H₂O (+ CO₂ if carbonate).
  • Zinc nitrate: 2HNO₃ + ZnO → Zn(NO₃)₂ + H₂O.
  • Iron(III) chloride: 6HCl + Fe₂O₃ → 2FeCl₃ + 3H₂O.

Method (universal)

  1. Warm the acid in a beaker (gentle heat speeds the reaction).
  2. Add excess insoluble base while stirring, until no more dissolves.
  3. Filter to remove unreacted base.
  4. Pour filtrate into evaporating basin.
  5. Heat over water bath until volume reduced and saturated.
  6. Leave to cool slowly — large crystals form.
  7. Filter and pat dry.

The product is a pure, dry, crystalline soluble salt.

Why solubility decides the method

Group 1 metal (e.g. Na, K) hydroxides and carbonates are soluble — adding them in excess gives a solution, not a filterable solid. So for sodium/potassium/ammonium salts, use titration (acid + soluble alkali, with an indicator) — see C4.7.

For other metals (Cu, Zn, Fe, Mg, Pb, Ca etc.), the hydroxide/carbonate/oxide is insoluble: the excess-and-filter method (C4.5) works.

Worked exampleWorked example — making zinc chloride

Aim: prepare crystals of ZnCl₂.

  • Choose: HCl (gives chloride) + ZnO (insoluble) or ZnCO₃ (insoluble).
  • React: ZnO + 2HCl → ZnCl₂ + H₂O.
  • Filter, evaporate over water bath, cool, crystallise.

Common mistakes

  • Picking a soluble base (NaOH) for this method — won't work.
  • Forgetting to balance the equation, especially when the metal is 2+ or 3+.
  • Boiling dry the solution instead of slow crystallisation — gives a powder, not crystals.
  • Skipping the filter step — gives contaminated salt.

Links

Same procedure as C4.5. Pairs with C4.7 (titration for soluble bases). The method explains why we choose certain reagents in industrial salt production.

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Practice questions

Try each before peeking at the worked solution.

  1. Question 12 marks

    Reagent choice (F)

    (F1) Which acid and which base would you use to make iron(II) sulfate? Give names.

    [Foundation — 2 marks]

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  2. Question 21 mark

    Equation balancing (F)

    (F2) Balance: HCl + ZnO → ZnCl₂ + H₂O.

    [Foundation — 1 mark]

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  3. Question 32 marks

    Method choice (C)

    (F/H3) Why can't the excess insoluble base method be used to make sodium chloride?

    [Crossover — 2 marks]

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  4. Question 42 marks

    Identifying step purpose (F)

    (F4) Explain why the solution is heated over a water bath rather than with a Bunsen flame.

    [Foundation — 2 marks]

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  5. Question 53 marks

    Iron(III) chloride equation (H)

    (H5) Write a balanced equation for the preparation of iron(III) chloride from iron(III) oxide.

    [Higher — 3 marks]

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  6. Question 63 marks

    Yield (H)

    (H6) A student starts with 4.0 g of MgO. Calculate the theoretical mass of MgSO₄ produced. (Mr MgO = 40, Mr MgSO₄ = 120).

    [Higher — 3 marks]

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  7. Question 72 marks

    Salt identification (H)

    (H7) A pale-blue crystalline solid is made by adding a green powder to dilute sulfuric acid until no more dissolves, then filtering and crystallising. Identify the green powder and the salt.

    [Higher — 2 marks]

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Flashcards

C4.6 — Soluble salts: choice of method

10-card deck on selecting reagents and using the excess-and-filter method.

10 cards · spaced repetition (SM-2)