Potassium permanganate and its titration
KMnO₄ is made by fusing MnO₂ with KOH, then oxidising the green manganate. In acid it takes 5 electrons (MnO₄⁻ → Mn²⁺) and is its own indicator.
Preparation of potassium permanganate
- 12MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂OFuse pyrolusite (MnO₂) with KOH and an oxidising agent such as KNO₃ or air. Dark green potassium manganate forms.
- 2★3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂OIn neutral or acidic solution, manganate disproportionates to purple permanganate and MnO₂.
- 3MnO₂ → MnO₄²⁻ (fused with KOH, oxidised with air or KNO₃) → MnO₄⁻ (electrolytic oxidation in alkaline solution)This is the commercial method.
- 42Mn²⁺ + 5S₂O₈²⁻ + 8H₂O → 2MnO₄⁻ + 10SO₄²⁻ + 16H⁺Laboratory method: Mn(II) salts are oxidised by peroxodisulphate.
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Properties and structure
- Dark purple, almost black, crystals; not very soluble in water (6.4 g per 100 g at 293 K).
- On heating to 513 K: 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂.
- Green manganate (MnO₄²⁻, Mn +6, d¹) is paramagnetic with one unpaired electron. Purple permanganate (MnO₄⁻, Mn +7, d⁰) is diamagnetic.
- How to draw: both ions are tetrahedral: Mn in the centre, four O atoms at the corners; write 2⁻ for manganate and 1⁻ for permanganate. The π bonds come from overlap of oxygen p orbitals with manganese d orbitals.
- Uses: a favourite oxidant in the laboratory and industry; bleaching wool, cotton and silk; decolourising oils.
Oxidising reactions in acid solution
Dilute sulphuric acid is used. HCl is not used, because KMnO₄ would oxidise it to chlorine.
- 110I⁻ + 2MnO₄⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O + 5I₂Iodine is set free from iodide.
- 2★5Fe²⁺ + MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺Iron(II) to iron(III).
- 3★5C₂O₄²⁻ + 2MnO₄⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O + 10CO₂Oxalate to CO₂, at about 333 K.
- 45S²⁻ + 2MnO₄⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O + 5SH₂S (as S²⁻) to sulphur.
- 55SO₃²⁻ + 2MnO₄⁻ + 6H⁺ → 2Mn²⁺ + 3H₂O + 5SO₄²⁻Sulphite (from SO₂) to sulphate.
- 65NO₂⁻ + 2MnO₄⁻ + 6H⁺ → 2Mn²⁺ + 5NO₃⁻ + 3H₂ONitrite to nitrate.
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Oxidising reactions in neutral or faintly alkaline solution
- 12MnO₄⁻ + H₂O + I⁻ → 2MnO₂ + 2OH⁻ + IO₃⁻Iodide is oxidised to iodate (not to iodine).
- 28MnO₄⁻ + 3S₂O₃²⁻ + H₂O → 8MnO₂ + 6SO₄²⁻ + 2OH⁻Thiosulphate is oxidised almost completely to sulphate.
- 32MnO₄⁻ + 3Mn²⁺ + 2H₂O → 5MnO₂ + 4H⁺Mn²⁺ is oxidised to MnO₂; zinc sulphate or zinc oxide is added as a catalyst.
Activity: molarity of KMnO₄ by titration against oxalic acid or Mohr's salt
You need a burette, pipette, conical flask, volumetric flask (250 mL), weighing bottle, dilute H₂SO₄ and a burner.
- 1Make the standard: for 250 mL of 0.050 M oxalic acid, weigh 0.050 × 0.250 × 126 = 1.575 g of H₂C₂O₄·2H₂O. For 250 mL of 0.050 M Mohr's salt, weigh 0.050 × 0.250 × 392 = 4.90 g and add a little dilute H₂SO₄ to stop hydrolysis. Dissolve and make up to the mark.
- 2Fill the burette with KMnO₄ solution. Read the upper meniscus, because the solution is dark.
- 3Pipette 10.0 mL of the standard into a conical flask and add about 10 mL of dilute H₂SO₄.
- 4For oxalic acid only, warm the flask to about 60 °C (333 K), since the reaction is slow when cold. Mohr's salt is titrated cold.
- 5★Add KMnO₄ drop by drop with swirling. Its purple colour disappears. End point: the first permanent light pink colour. KMnO₄ is its own indicator.
- 6Repeat until three readings agree (concordant). Calculate: with oxalic acid, M(KMnO₄) = (2/5) × M(oxalic) × V(oxalic) ÷ V(KMnO₄); with Mohr's salt, M(KMnO₄) = (1/5) × M(Mohr) × V(Mohr) ÷ V(KMnO₄).
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Worked example: molarity and strength of KMnO₄
(a) 10.0 mL of 0.050 M oxalic acid needs 10.0 mL of KMnO₄ solution. (b) In another run, 20.0 mL of 0.10 M Mohr's salt needs 16.0 mL of KMnO₄. Find the molarity of KMnO₄ in each case, and the strength in g L⁻¹ for (a).
- (a) Equation: 5C₂O₄²⁻ + 2MnO₄⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O + 10CO₂, so moles KMnO₄ : moles oxalic acid = 2 : 5.
- M(KMnO₄) × 10.0 = (2/5) × 0.050 × 10.0, so M(KMnO₄) = 0.020 M.
- Strength = molarity × molar mass = 0.020 × 158 = 3.16 g L⁻¹.
- (b) Equation: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O, so moles KMnO₄ : moles Fe²⁺ = 1 : 5.
- M(KMnO₄) × 16.0 = (1/5) × 0.10 × 20.0 = 0.40, so M(KMnO₄) = 0.40 ÷ 16.0 = 0.025 M.
(a) 0.020 M, 3.16 g L⁻¹; (b) 0.025 M
Dilute "pinky" KMnO₄ solution has long been used in Indian homes and villages to wash vegetables and to disinfect well water, because it is a strong oxidising agent that kills germs.
3 marks: “How is potassium permanganate prepared from pyrolusite ore? Write the equations.” Fuse MnO₂ with KOH and an oxidant: 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O. Green manganate disproportionates in neutral or acid solution: 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O (commercially, by electrolytic oxidation).
2 marks: “Write ionic equations for the reaction of acidified KMnO₄ with (i) iron(II) ions, (ii) oxalate ions.” (i) 5Fe²⁺ + MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. (ii) 5C₂O₄²⁻ + 2MnO₄⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O + 10CO₂.
3 marks: “Write the reactions of KMnO₄ with iodide ions in (i) acidic and (ii) neutral or faintly alkaline solution. Why is no indicator needed when it is titrated?” (i) 10I⁻ + 2MnO₄⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O + 5I₂. (ii) 2MnO₄⁻ + H₂O + I⁻ → 2MnO₂ + 2OH⁻ + IO₃⁻. Purple MnO₄⁻ turns colourless; one extra drop gives a pink colour, so it is its own indicator.
Permanganate in acid: 8 H⁺, 5 electrons, Mn²⁺ and 4 waters. Neutral: 3 electrons, brown MnO₂.