KMnO4 vs Oxalic Acid: Redox Titration Calculations

By Aniket Bhardwaj · 16 October 2026 · Updated 16 October 2026 · Class 11-12 Physical Chemistry

The titration of oxalic acid or Mohr's salt against potassium permanganate is the most common redox titration in the Class 12 practical and in theory numericals. The arithmetic is short, but students lose marks because they use the wrong number of electrons, the wrong equivalent mass, or the wrong acid in the flask. This page gives the equations, the method and three fully worked examples.

A reactant grid shows 2 MnO₄⁻ units paired with 5 H₂C₂O₄ units, giving a 2:5 mole ratio. — Use the balanced equation: 2 mol permanganate reacts with 5 mol oxalic acid.

The two reactions

Permanganate with oxalic acid: 2MnO₄⁻ + 5H₂C₂O₄ + 6H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O
Permanganate with Fe²⁺ (Mohr's salt): MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

In acid solution, MnO₄⁻ gains 5 electrons (Mn goes from +7 to +2). Oxalic acid loses 2 electrons per molecule (two C atoms go from +3 to +4, one electron each). Fe²⁺ loses 1 electron. These numbers decide everything else.

Two ways to calculate

Mole ratio method. Use the balanced equation. For oxalic acid, 2 mol MnO₄⁻ reacts with 5 mol H₂C₂O₄, so mol MnO₄⁻ = (2/5) × mol H₂C₂O₄. For Mohr's salt, mol MnO₄⁻ = (1/5) × mol Fe²⁺.

Normality method. N = M × (electrons exchanged per formula unit). At the end point, N₁V₁ = N₂V₂. For KMnO₄ in acid, N = 5M. For oxalic acid, N = 2M. For Mohr's salt, N = M.

Substance	Molar mass (g/mol)	Electrons per formula unit	Equivalent mass (g/eq)
KMnO₄ (acidic medium)	158.03	5	31.61
H₂C₂O₄ (anhydrous)	90.03	2	45.02
H₂C₂O₄·2H₂O (crystals)	126.07	2	63.03
FeSO₄·(NH₄)₂SO₄·6H₂O (Mohr's salt)	392.13	1	392.13

Worked example 1: find the molarity of KMnO₄

Question. 20.0 mL of 0.0500 M oxalic acid needs 25.0 mL of KMnO₄ solution at the end point. Find the molarity of KMnO₄.

Step 1 (moles of oxalic acid). 0.0500 mol/L × 20.0 mL = 1.00 mmol.

Step 2 (moles of permanganate). From 2 : 5, mmol MnO₄⁻ = 1.00 × 2/5 = 0.400 mmol.

Step 3. M = 0.400 mmol ÷ 25.0 mL = 0.0160 M.

Check by normality. N of oxalic acid = 2 × 0.0500 = 0.100 N. Then 0.100 × 20.0 = N_KMnO4 × 25.0, so N = 0.0800 N. M = N ÷ 5 = 0.0160 M. Both methods agree.

Worked example 2: how much KMnO₄ will Mohr's salt need?

Question. 9.80 g of Mohr's salt is dissolved in dilute H₂SO₄ and made up to 250 mL. A 25.0 mL portion is titrated with 0.0200 M KMnO₄. Find the volume needed.

Step 1. Moles of Mohr's salt = 9.80 ÷ 392.13 = 0.02499 mol in 250 mL. Molarity = 0.02499 ÷ 0.250 = 0.0999 M, so about 0.100 M.

Step 2. In 25.0 mL: 0.02499 × (25.0 ÷ 250) = 2.499 × 10⁻³ mol, or 2.50 mmol Fe²⁺.

Step 3. MnO₄⁻ needed = 2.50 ÷ 5 = 0.500 mmol.

Step 4. Volume = 0.500 mmol ÷ 0.0200 mmol/mL = 25.0 mL (more exactly 24.99 mL).

Worked example 3: from weighed crystals to molarity

Question. 1.575 g of oxalic acid crystals (H₂C₂O₄·2H₂O) is dissolved in water and made up to 250 mL. A 10.0 mL portion needs 12.5 mL of KMnO₄. Find the molarity of KMnO₄.

Step 1. Moles of crystals = 1.575 ÷ 126.07 = 0.012494 mol. Molarity = 0.012494 ÷ 0.250 = 0.04997 M, so about 0.0500 M.

Step 2. In 10.0 mL: 0.04997 × 10.0 = 0.4997 mmol, so about 0.500 mmol oxalic acid.

Step 3. MnO₄⁻ = 0.4997 × 2/5 = 0.1999 mmol, about 0.200 mmol.

Step 4. M = 0.1999 ÷ 12.5 = 0.0160 M.

Check by equivalents. Equivalents of oxalic acid in 10.0 mL = (1.575 ÷ 63.03) × (10.0 ÷ 250) = 0.9995 meq. Dividing by 12.5 mL gives 0.0800 N, and 0.0800 ÷ 5 = 0.0160 M.

Practical points that decide the answer

Use dilute H₂SO₄ only. Hydrochloric acid is wrong because MnO₄⁻ oxidises Cl⁻ to chlorine and the reading is too high. Nitric acid is itself an oxidising agent and interferes.
KMnO₄ is its own indicator. The end point is the first permanent very light pink colour that stays for about 30 seconds.
Read the burette from the top of the meniscus for the dark purple solution, because the bottom of the meniscus cannot be seen clearly.
Oxalic acid titrations are done warm (the usual practical instruction is about 60–70 °C), because the reaction is slow at the start. Once a little Mn²⁺ forms it speeds the reaction up, so the colour disappears faster after the first few drops. Mohr's salt titrations are done at room temperature.
KMnO₄ is not a primary standard. Its solution is standardised against oxalic acid or Mohr's salt, which can be weighed exactly.

Common mistakes

Using 1 : 1 for oxalic acid. The ratio is 2 MnO₄⁻ : 5 H₂C₂O₄. A 1 : 1 assumption makes the answer wrong by a factor of 2.5.
Taking the equivalent mass of oxalic acid crystals as 45. That is for the anhydrous acid. Crystals include 2 H₂O, so the equivalent mass is 63.
Mixing M and N. N₁V₁ = N₂V₂ works only if both sides are in normality. Convert KMnO₄ with N = 5M.
Forgetting the aliquot fraction. If only 10 mL of a 250 mL solution is titrated, take only 10/250 of the weighed amount.
Using 6 or 7 electrons for KMnO₄. In acid, Mn changes from +7 to +2, which is 5 electrons. In neutral medium the product is MnO₂ and the figure becomes 3.

Exam summary

Titration	Mole ratio	Normality link	Temperature
KMnO₄ with oxalic acid	2 MnO₄⁻ : 5 H₂C₂O₄	N(KMnO₄) = 5M; N(oxalic) = 2M	Warm, 60–70 °C
KMnO₄ with Mohr's salt	1 MnO₄⁻ : 5 Fe²⁺	N(Mohr's salt) = M	Room temperature

For practical marking schemes and the exact procedure, follow your own board's current practical syllabus and your teacher's instructions.

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