Redox Titration with KMnO₄
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flagWhat you'll discover
- arrow_forwardExplain why KMnO₄ needs no separate indicator
- arrow_forwardState why the oxalic acid flask must be heated to 60-70 °C before titrating
- arrow_forwardTitrate to the correct self-indicating endpoint: the first permanent pale pink
- arrow_forwardUse the 2:5 mole ratio of MnO₄⁻ to C₂O₄²⁻ in the calculation
- arrow_forwardIdentify procedure errors that invalidate a redox trial
A titration that indicates itself
Potassium permanganate is intensely purple; its reduction product Mn²⁺ is virtually colourless at titration concentrations. So as KMnO₄ runs into the oxalic acid flask, each addition decolourises instantly while oxalate remains — the purple simply vanishes into the solution. The moment all the oxalate is consumed, the very next drop has nothing left to react with and tints the whole flask a permanent pale pink.
That first permanent pink, persisting through 30 seconds of swirling, is the endpoint — KMnO₄ is its own indicator. Adding phenolphthalein or methyl orange would be wrong twice over: they are unnecessary, and KMnO₄ being a powerful oxidiser would destroy them anyway. The full ionic equation: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O.
Why the flask must be hot (and acidic)
The permanganate-oxalate reaction is surprisingly slow at room temperature. Titrate a cold flask and each purple drop lingers unreacted for many seconds, so the solution looks pink long before the true endpoint — you stop early and your titre is badly wrong. Heating the flask to 60-70 °C before starting speeds the reaction so each drop decolourises promptly. Above 70 °C, though, oxalic acid itself starts decomposing, so the window matters: 60-70 °C, checked with a thermometer.
(The reaction is also autocatalytic — the Mn²⁺ product catalyses further reaction, which is why the first few additions of a titration decolourise sluggishly and later ones instantly.) The acid matters too: the medium is made acidic with dilute H₂SO₄, never HCl — permanganate would oxidise chloride to chlorine, consuming extra titrant and faking a high titre. And never with HNO₃, itself an oxidiser.
The calculation: mind the 2:5 ratio
Unlike the 1:1 acid-base case, this redox reaction consumes 2 mol of MnO₄⁻ per 5 mol of C₂O₄²⁻. From the balanced equation: moles KMnO₄ = (2/5) × moles oxalic acid. With 25.0 mL of 0.050 M oxalic acid in the flask and a mean titre V mL: M(KMnO₄) = (2/5) × 0.050 × 25.0 / V.
In the older normality language still used in many Nepali practical books, the arithmetic hides the ratio: equivalent weight of KMnO₄ in acidic medium is M/5 (5 electrons gained per Mn) and of oxalic acid dihydrate M/2 (2 electrons lost), so N₁V₁ = N₂V₂ applies directly. Both routes give the same answer — but state which one you are using. One more practical point: read the purple KMnO₄ meniscus at its top edge, not the bottom, because the dark solution hides the lower meniscus completely.