Chemistry · Redox Equilibrium · NEET
n-factor is the number of electrons ONE molecule gains, not the oxidation state of manganese. In acidic medium Mn goes from +7 to +2, a change of 5 units, so KMnO4 gains 5 electrons. The number 7 is only the starting oxidation state, not the electron change. Always use 5 for acidic KMnO4 titrations.
Oxalic acid is (COOH)2. Each carbon is at oxidation state +3 and gets oxidised to +4 in CO2, a change of 1 per carbon. With two carbons the total electron loss is 2, so the n-factor is 2. Do not confuse this with its 2 acidic protons; here the 2 comes from the redox electron change, which happens to also be 2.
Both work if you are consistent. With normality use N1V1 = N2V2 directly (no n-factor needed because normality already includes it). With molarity you MUST add n-factors: n1 M1 V1 = n2 M2 V2. NEET options are usually in molarity, so the safe method is: n(KMnO4) x M x V = n(oxalic) x M x V.
In acidic medium KMnO4 is reduced all the way to Mn2+ (colourless), giving the full 5-electron change and a sharp colourless-to-pink end point. In neutral or basic medium Mn stops at MnO2 (brown, n-factor 3) or MnO4^2- (green, n-factor 1), which changes the whole calculation. We use dilute H2SO4, never HCl, because HCl would itself be oxidised and give a wrong reading.
KMnO4 is its own indicator (a self-indicator). While oxalic acid is still present, added KMnO4 loses its purple colour instantly. The moment all oxalic acid is used up, one extra drop of KMnO4 stays and the solution turns permanent light pink. That first lasting pink is the end point.
In acidic medium, 10 mL of 0.25 M oxalic acid is titrated with KMnO4 solution. If the volume of KMnO4 required to reach the end point is 10 mL, the strength of the KMnO4 solution is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
2 MnO4^- + 5 (COOH)2 + 6 H+ -> 2 Mn2+ + 10 CO2 + 8 H2O. The mole ratio is 2 KMnO4 to 5 oxalic acid, which matches the n-factor ratio 5:2 you use in the equivalence formula.
n1 x M1 x V1 = n2 x M2 x V2, where n is the n-factor (electrons transferred per molecule). For this titration n(KMnO4) = 5 and n(oxalic acid) = 2.
The reaction of oxalic acid with KMnO4 is slow at room temperature. Warming the oxalic acid to about 60-70 degrees C speeds it up. Do not boil, or oxalic acid decomposes and the reading becomes wrong.
No. HCl gets oxidised by KMnO4 to chlorine, so extra KMnO4 is used up and the strength value comes out too high. Always use dilute sulphuric acid for KMnO4 titrations.
Equivalent weight = molar mass / n-factor = 158 / 5 = 31.6 g/equiv. This is because Mn changes by 5 units (+7 to +2) in acidic medium.
