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Redox Reactions

Carrying out and calculating a redox titration: redox titration calculations

  • 3 things to know
  • A2 Unit 3

Overview

Sixteen recall questions on Carrying out and calculating a redox titration: redox titration calculations.

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Answer each one before you reveal it.

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Numbers

25.0 cm³ of an iron(II) solution requires 22.40 cm³ of 0.0200 mol dm⁻³ potassium manganate(VII). Calculate the concentration of the iron(II) solution.

Numbers

Calculate the volume of 0.0200 mol dm⁻³ potassium manganate(VII) needed to react with 25.0 cm³ of 0.0500 mol dm⁻³ sodium ethanedioate in acid solution.

Numbers

Calculate the volume of 0.100 mol dm⁻³ iron(II) sulfate solution that would react with 25.0 cm³ of 0.0250 mol dm⁻³ potassium dichromate(VI) in acid solution.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Explain how each piece of apparatus should be rinsed before a titration.

    Model answer

    The burette is rinsed with the solution it will hold and the pipette with the solution it will deliver, so that no residual water dilutes them. The conical flask is rinsed with distilled water only, because extra water there changes the volume but not the number of moles present.

  2. 2

    25.0 cm³ of an iron(II) solution requires 22.40 cm³ of 0.0200 mol dm⁻³ potassium manganate(VII). Calculate the concentration of the iron(II) solution.

    Model answer

    n(MnO₄⁻) = 0.02240 dm³ × 0.0200 mol dm⁻³ = 4.480 × 10⁻⁴ mol. The equation MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O gives a 1 : 5 ratio, so n(Fe²⁺) = 5 × 4.480 × 10⁻⁴ = 2.240 × 10⁻³ mol. Concentration = 2.240 × 10⁻³ / 0.0250 dm³ = 0.0896 mol dm⁻³.

  3. 3

    Five iron tablets are dissolved in dilute sulfuric acid and the solution made up to 250 cm³. A 25.0 cm³ portion requires 12.50 cm³ of 0.0100 mol dm⁻³ potassium manganate(VII). Calculate the mass of iron in one tablet. Use A(Fe) = 55.8.

    Model answer

    n(MnO₄⁻) = 0.01250 dm³ × 0.0100 mol dm⁻³ = 1.250 × 10⁻⁴ mol. n(Fe²⁺) in the portion = 5 × 1.250 × 10⁻⁴ = 6.250 × 10⁻⁴ mol. In the whole 250 cm³, n(Fe²⁺) = 10 × 6.250 × 10⁻⁴ = 6.250 × 10⁻³ mol. Mass of iron = 6.250 × 10⁻³ × 55.8 = 0.34875 g for five tablets, so each tablet contains 0.34875 / 5 = 0.0698 g, that is 69.8 mg.

  4. 4

    1.00 g of iron wire is dissolved in excess dilute sulfuric acid and the solution made up to 250 cm³. A 25.0 cm³ portion needs 17.20 cm³ of 0.0200 mol dm⁻³ potassium manganate(VII). Calculate the percentage of iron in the wire. Use A(Fe) = 55.8.

    Model answer

    n(MnO₄⁻) = 0.01720 dm³ × 0.0200 mol dm⁻³ = 3.440 × 10⁻⁴ mol. n(Fe²⁺) in the portion = 5 × 3.440 × 10⁻⁴ = 1.720 × 10⁻³ mol. In 250 cm³, n(Fe²⁺) = 1.720 × 10⁻² mol. Mass of iron = 1.720 × 10⁻² × 55.8 = 0.960 g. Percentage = (0.960 / 1.00) × 100 = 96.0%.

  5. 5

    25.0 cm³ of a diluted hydrogen peroxide solution requires 21.00 cm³ of 0.0200 mol dm⁻³ potassium manganate(VII) in acid. Calculate the concentration of the hydrogen peroxide.

    Model answer

    n(MnO₄⁻) = 0.02100 dm³ × 0.0200 mol dm⁻³ = 4.200 × 10⁻⁴ mol. The equation 2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O gives a 2 : 5 ratio, so n(H₂O₂) = (5/2) × 4.200 × 10⁻⁴ = 1.050 × 10⁻³ mol. Concentration = 1.050 × 10⁻³ / 0.0250 dm³ = 0.0420 mol dm⁻³.

  6. 6

    Calculate the volume of 0.0200 mol dm⁻³ potassium manganate(VII) needed to react with 25.0 cm³ of 0.0500 mol dm⁻³ sodium ethanedioate in acid solution.

    Model answer

    n(C₂O₄²⁻) = 0.0250 dm³ × 0.0500 mol dm⁻³ = 1.25 × 10⁻³ mol. The ratio of manganate(VII) to ethanedioate is 2 : 5, so n(MnO₄⁻) = (2/5) × 1.25 × 10⁻³ = 5.00 × 10⁻⁴ mol. Volume = 5.00 × 10⁻⁴ / 0.0200 = 0.0250 dm³, which is 25.0 cm³.

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