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2.7.9Atoms, Elements and Compounds

Practical- Prepare Lead(II)Sulfate — Cambridge IGCSE Chemistry 0620 Revision Notes & Topical Questions

5 min read 5 parts Lesson 17 of 26 Unit quiz

📚 Cambridge IGCSE Chemistry 0620 Hub / Topical Revision
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Part 1 of 5Method

2.7.9 Practical- Prepare Lead(II)Sulfate

Lead(II) sulfate is insoluble, so it forms as a precipitate when two soluble salts are mixed.

1
Measure 25.0 cm³ of 0.50 mol/dm³ lead(II) nitrate solution into a small beaker.
2
Add 25.0 cm³ of 0.50 mol/dm³ potassium sulfate solution to the beaker and stir with a glass rod to mix.
3
Filter the mixture to collect the lead(II) sulfate precipitate on filter paper.
4
Wash the precipitate with distilled water to remove traces of the other solutions, then dry it in an oven.

Part 2 of 5Equation and Ions

Writing the equations shows that only the lead and sulfate ions form the solid.

Word equation
  • Lead(II) nitrate + potassium sulfate → lead(II) sulfate + potassium nitrate.
Symbol equation
  • Pb(NO₃)₂(aq) + K₂SO₄(aq) → PbSO₄(s) + 2KNO₃(aq).
Ionic equation
  • Pb²⁺(aq) + SO₄²⁻(aq) → PbSO₄(s), since the potassium and nitrate ions are spectators.
Lead safety
  • Lead compounds are toxic, so gloves should be worn and the solid kept out of the mouth and away from food.

Part 3 of 5Why the Method Works

Why the Method Works
Insoluble product
  • Lead(II) sulfate is insoluble, so it separates as a solid from the solution.
Soluble reactants
  • Lead(II) nitrate and potassium sulfate are soluble, so they provide the lead and sulfate ions.
Equal amounts
  • Equal volumes and concentrations of each solution give equal moles of the two ions.
Drying
  • Drying in an oven removes water from the solid precipitate before it is weighed.

Part 4 of 5Worked Example

Question

  • Calculate the maximum mass of lead(II) sulfate that can be formed from 25.0 cm³ of 0.50 mol/dm³ lead(II) nitrate and excess potassium sulfate.
  • (Ar: Pb = 207, S = 32, O = 16)

  1. Moles of Pb²⁺ = 0.50 mol/dm³ × 0.0250 dm³ = 0.0125 mol.
  2. Mole ratio Pb²⁺PbSO₄ is 1 : 1, so moles of PbSO₄ = 0.0125 mol.
  3. Mr of PbSO₄ = 207 + 32 + (4 × 16) = 303, so mass = 0.0125 mol × 303 g/mol = 3.79 g.

Answer 3.79 g (about 3.8 g) of lead(II) sulfate, assuming complete precipitation and a 100% yield.

Part 5 of 5Exam Checklist & Tips

2.7.9 – Key Points for the Exam

  • Precipitate An insoluble solid that forms when two solutions react.
  • Spectator ion An ion that does not change during a reaction and is left out of the ionic equation.
  • Theoretical yield The maximum mass of product that could be obtained from the reactants.
  • Molar mass The mass of one mole of a substance, given in g/mol.

Common Mistakes

  • Including the potassium and nitrate ions in the ionic equation; they are spectators.
  • Forgetting to use the mole ratio, or using the wrong molar mass when calculating the mass of product.
  • Failing to wash the precipitate, which leaves soluble potassium nitrate on the solid.
  • Leaving out the (s) state symbol for the lead(II) sulfate precipitate.

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