Preparation of Soluble Salts From Metal Oxides or Carbonates (SQA National 5 Chemistry): Revision Note

Producing soluble salts from carbonates or oxides

• When an acid reacts with an insoluble base (like a metal oxide or metal carbonate), a titration cannot be used

  • This is because the solid base would block the burette

• Instead, a different method is used which takes advantage of the fact that the base is a solid we can see and easily remove

The "excess" method

• The key to this method is adding more of the insoluble base than is needed to make sure the acid has been neutralised

• Why add in excess?

  • If the acid is not all used up, the final salt will be impure, contaminated with leftover acid

  • By adding excess base, it is guaranteed that the acid is the limiting reactant and is completely used up

  • The leftover base is then easily removed by filtration

• How do we know it's in excess?

  • Unreacted solid base left is seen at the bottom of the beaker

  • It will have stopped reacting (e.g., the fizzing stops for a carbonate)

Method

1. Warm dilute acid gently in a beaker

   • Add the insoluble base, or carbonate slowly while stirring until no more reacts

   • This means that the base is in excess

2. Filter the mixture to remove the excess solid

3. Transfer the filtrate (salt solution) to an evaporating basin and heat gently until the solution is concentrated

   • Check the solution is saturated by dipping a cold, glass rod into the solution and seeing if crystals form on the end

   • Leave the basin in a warm place to crystallise

4. Decant excess liquid and dry the crystals with filter paper

• For example, preparing pure, hydrated copper(II) sulfate crystals using this method:

copper(II) oxide + sulfuric acid → copper(II) sulphate + water

CuO (s) + H₂SO₄ (aq) → CuSO₄ (aq) + H₂O (l)