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5.1.7Unit 5. Transition Metals and Organic Nitrogen Chemistry

Redox Titration Calculations — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 5 parts Lesson 13 of 43 Unit quiz

📚 IAL Edexcel Linear (Chemistry) Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Part 1 of 5Manganate(VII) Titrations

Equation
  • Equation MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺, so the mole ratio MnO₄⁻ : Fe²⁺ is 1 : 5.
Self-indicating
  • Self-indicating Purple MnO₄⁻ becomes almost colourless Mn²⁺, so the end point is the first permanent pale pink colour.
Acid used
  • Acid used Dilute sulfuric acid is added, because hydrochloric acid would be oxidised by MnO₄⁻ and nitric acid is itself an oxidising agent.
Burette
  • Burette Potassium manganate(VII) is placed in the burette and read from the top of the meniscus because it is dark.

Part 2 of 5Iodine-Thiosulfate Titrations

Reaction
  • Reaction I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻, so the ratio of I₂ to thiosulfate is 1 : 2.
Indicator
  • Indicator Starch is added near the end point when the iodine colour is pale yellow, producing a blue-black colour that disappears at the end.
Analysing copper
  • Analysing copper Copper(II) ions react with excess iodide: 2Cu²⁺ + 4I⁻ → 2CuI + I₂, and the iodine formed is titrated.
Ratio
  • Ratio Overall, Cu²⁺ : S₂O₃²⁻ is 1 : 1 because each mole of I₂ comes from 2 mol of Cu²⁺.

Part 3 of 5Calculation Plan

1
Write the balanced equation and note the mole ratio between titrant and analyte.
2
Calculate moles of titrant from concentration x volume in dm³, using the mean of concordant titres.
3
Multiply or divide by the mole ratio to find the moles of the substance being analysed.
4
Scale up if only a portion of the solution, such as 25.0 cm³ of a 250 cm³ sample, was used.
5
Convert to the required quantity: concentration, mass or percentage purity, and give the answer to suitable significant figures.

Part 4 of 5Worked Example

Question

  • A 25.0 cm³ sample of iron(II) solution needs 22.40 cm³ of 0.0200 mol dm⁻³ KMnO₄ for complete reaction in acid.
  • Calculate the concentration of Fe²⁺.

  1. Moles MnO₄⁻ = 0.0224 dm³ x 0.0200 mol dm⁻³ = 4.48 x 10⁻⁴ mol.
  2. Moles Fe²⁺ = 5 x 4.48 x 10⁻⁴ = 2.24 x 10⁻³ mol.
  3. Concentration = 2.24 x 10⁻³ / 0.0250 dm³ = 0.0896 mol dm⁻³.

Answer [Fe²⁺] = 0.0896 mol dm⁻³

Part 5 of 5Exam Checklist & Tips

5.1.7 – Key Points for the Exam

  • Titre The volume of solution delivered from the burette at the end point.
  • End point The point at which the indicator shows the reaction is just complete.
  • Concordant Titres that agree within 0.10 cm³.
  • Self-indicating A reagent whose own colour change signals the end point.

Common Mistakes

  • Forgetting to use the 1 : 5 ratio for MnO₄⁻ with Fe²⁺.
  • Using volumes in cm³ without converting to dm³.
  • Using hydrochloric acid to acidify manganate(VII) titrations.
  • Adding starch too early in an iodine titration so that the end point is blurred.

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