Part 5: Manganate(VII) titrations and calculations

Build redox mole ratios and carry them through aliquots, dilution factors, purity and hydrated-salt calculations.

The end point is a small excess of manganate(VII)

In a typical titration, standardised purple KMnO₄ is in the burette and the reducing agent is in the flask with excess dilute sulfuric acid. While reductant remains, added MnO₄⁻ is reduced to very pale pink Mn²⁺, usually effectively colourless at titration concentrations. The end point is the first faint pink colour persisting after swirling: a slight excess of manganate(VII). No separate indicator is required.

Use dilute sulfuric acid to provide H⁺ without a competing redox reagent. Hydrochloric acid can introduce chloride that is oxidised by manganate(VII); nitric acid can oxidise an analyte such as Fe²⁺ before the titration. Inadequate acidity can give brown MnO₂ and invalidate the assumed five-electron reduction.

Pipette a measured aliquot into a conical flask; rinse the pipette with sample and burette with titrant. Remove the filling funnel, clear any jet bubble, read consistently at eye level and add titrant dropwise near the end point over a white background. Obtain concordant titres using the required protocol and average appropriate concordant values, excluding the rough titration. Fresh Fe²⁺ solutions minimise oxidation by air.

Derive each ratio from electrons

Fe²⁺ loses one electron, whereas one oxalate ion or one H₂O₂ molecule loses two in these reactions. Manganate(VII) accepts five electrons in sufficiently acidic solution. Therefore MnO₄⁻:Fe²⁺ is 1:5 and MnO₄⁻:C₂O₄²⁻ or H₂O₂ is 2:5. Do not assume every reducing agent uses the iron ratio.

The oxalate reaction is initially slow, so the acidified solution is typically warmed to around 60 °C according to the method. It becomes faster as Mn²⁺ accumulates, an example of autocatalysis developed in Part 6. Avoid boiling or overheating the sample.

Worked example: iron in an alloy

Constructed problem: 1.50 g of an alloy is treated so that all its iron becomes Fe²⁺, with no other species consuming the titrant. The solution is made up to 250.0 cm³. A 25.00 cm³ aliquot requires 18.40 cm³ of 0.0200 mol dm⁻³ KMnO₄. Use Aᵣ(Fe) = 55.8.

n(MnO₄⁻) = 0.0200 × 18.40/1000 = 3.680 × 10⁻⁴ mol. The aliquot contains five times this amount of Fe²⁺: 1.840 × 10⁻³ mol. The full flask contains ten aliquots, so n(Fe) = 0.01840 mol. Mass of iron = 0.01840 × 55.8 = 1.02672 g. Percentage iron = 1.02672/1.50 × 100 = 68.4%.

The same sequence applies to an iron tablet after quantitative dissolution. State the analyte assumption: if other reducing agents also react, the titre cannot be assigned entirely to iron. A factor of ten is applied once for the 250/25 sampling ratio, not again simply because cm³ were converted to dm³.

Worked example: hydrated ammonium iron(II) sulfate

Constructed problem: 3.920 g of (NH₄)₂Fe(SO₄)₂·xH₂O is made up to 250.0 cm³. A 25.00 cm³ aliquot requires 20.00 cm³ of 0.0100 mol dm⁻³ KMnO₄. The anhydrous formula mass is 284.0; Mᵣ(H₂O) = 18.0. Assume purity and one Fe²⁺ per formula unit.

n(MnO₄⁻) = 2.000 × 10⁻⁴ mol. Aliquot Fe²⁺ = 1.000 × 10⁻³ mol, giving 0.01000 mol salt in the full flask. Mᵣ(hydrate) = 3.920/0.01000 = 392.0. Hence x = (392.0 − 284.0)/18.0 = 6. Report the hydration number as an integer after assessing agreement with experimental precision.

For hydrated ethanedioic acid the reducing unit instead transfers two electrons. Determine the moles of acid from the 2:5 ratio, then use the sample mass to find its formula mass and subtract the anhydrous formula mass. Do not use the 1:5 iron ratio for a colourless acid just because the same burette reagent is used.

Worked example: hydrogen peroxide after dilution

Constructed problem: 10.00 cm³ of a peroxide product is diluted to 250.0 cm³. A 25.00 cm³ aliquot of the diluted solution requires 16.00 cm³ of 0.0200 mol dm⁻³ KMnO₄ in acid. Assume peroxide is the only reductant.

n(MnO₄⁻) = 3.200 × 10⁻⁴ mol. n(H₂O₂) in the aliquot = (5/2) × 3.200 × 10⁻⁴ = 8.000 × 10⁻⁴ mol. Diluted concentration = 8.000 × 10⁻⁴/0.02500 = 0.0320 mol dm⁻³. The original product was diluted by 250/10 = 25, so its concentration was 0.800 mol dm⁻³. With Mᵣ(H₂O₂) = 34.0, that is 27.2 g dm⁻³.

The aliquot volume determines the diluted concentration; the preparation ratio restores the original concentration. Do not automatically turn g dm⁻³ into mass percentage without a density. June 2022 Paper 1 Q03.1–03.2 provides an assessment context for balancing the peroxide equation, reversing dilution and explaining the self-indicating end point.

When two parts of the same compound are oxidised

For FeC₂O₄, Fe²⁺ loses one electron and C₂O₄²⁻ loses two: three electrons per formula unit. Since MnO₄⁻ accepts five, five FeC₂O₄ units require three MnO₄⁻ ions. For FeC₂O₄·2H₂O the water of crystallisation changes the formula mass but not this electron ratio.

Original example: a pure FeC₂O₄·2H₂O sample consumes 0.000600 mol MnO₄⁻. It contains (5/3) × 0.000600 = 0.00100 mol hydrate. Using Mᵣ = 179.8, its mass is 0.180 g to three significant figures. Ignoring oxidation of oxalate would overestimate the amount by using the wrong ratio.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.

Sources