In an acidic medium, 10 mL of 0.25 M oxalic acid is titrated with KMnO 4 solution. If the volume of KMnO 4 solution required to reach end point is 10 mL, the strength of the KMnO 4 solution isNE

Question

In an acidic medium, 10 mL of 0.25 M oxalic acid is titrated with KMnO 4 ​ solution. If the volume of KMnO 4 ​ solution required to reach end point is 10 mL, the strength of the KMnO 4 ​ solution isNEET 2026 (Re-NEET)

TestHub · Accepted Answer

In an acidic medium, the reaction of KMnO $_{4}$ with oxalic acid involves the following changes in oxidation states:

For KMnO $_{4}$ , Mn $^{7 +} \to$ Mn $^{2 +}$ , so the n-factor is $5$ .

For oxalic acid (H $_{2}$ C $_{2}$ O $_{4}$ ), C $^{3 +} \to$ C $^{4 +}$ , so the n-factor is $2 \times 1 = 2$ .

At the equivalence point, the number of equivalents of KMnO $_{4}$ equals the number of equivalents of oxalic acid:

$N_{1} V_{1} = N_{2} V_{2}$

$(M_{1} \times n_{1}) V_{1} = (M_{2} \times n_{2}) V_{2}$

Substituting the given values:

$M_{1} \times 5 \times 10 = 0.25 \times 2 \times 10$

$50 M_{1} = 5$

$M_{1} = 0.10$ M

The strength of the KMnO $_{4}$ solution is $0.10$ M.

Chemistry - Redox Reactions Question with Solution | TestHub

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