Chemistry Q57 – NEET-UG 2026

Home Chemistry Redox ReactionsQuestion #560

In an acidic medium,

10 mL

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 is

Clue 1 of 3: Redox Titration Principle

In a redox titration, the equivalence point is reached when the equivalents of the oxidizing agent equal the equivalents of the reducing agent.

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Ninja Strategy•n-factor Molarity Relationship

Recognize that for equal volumes, the molarity of the reactant with a higher n-factor must be proportionally lower. Specifically, $M_{{KMnO}_{4}} = M_{oxalic acid} \times \frac{n_{oxalic acid}}{n_{{KMnO}_{4}}} = 0.25 \times \frac{2}{5} = 0.25 \times 0.4 = 0.1 M$ .

🧠 Full Solution Path

Step 1: Identify Reactants and their n-factors✦ Active

In an acidic medium, ${KMnO}_{4}$ (permanganate ion, ${MnO}_{4}^{-}$ ) acts as an oxidizing agent and is reduced to ${Mn}^{2 +}$ . The change in oxidation state of $Mn$ is from $+ 7$ to $+ 2$ , so its $n$ -factor is $5$ (i.e., $5$ electrons gained).

Oxalic acid ( $H_{2} C_{2} O_{4}$ ) acts as a reducing agent. The carbon atoms in $C_{2} O_{4}^{2 -}$ are oxidized from $+ 3$ to $+ 4$ in ${CO}_{2}$ . For two carbon atoms, the total change in oxidation state is $2 \times (4 - 3) = 2$ , so its $n$ -factor is $2$ (i.e., $2$ electrons lost).

Step 2: Apply the Equivalence Point Formula○ Expand

At the equivalence point of a redox titration, the number of gram equivalents of the oxidizing agent equals the number of gram equivalents of the reducing agent. This can be expressed using the formula:

M_{1} V_{1} n_{1} = M_{2} V_{2} n_{2}

Where $M$ is molarity, $V$ is volume, and $n$ is the $n$ -factor. Let subscript 1 be for ${KMnO}_{4}$ and subscript 2 for oxalic acid.

💡 Teacher's Secret Hint

Remember to correctly identify the n-factor for each reactant based on the change in oxidation states in the given medium.

Step 3: Calculate the Molarity of KMnO₄○ Expand

Given values:

For oxalic acid (2): $M_{2} = 0.25 M$ , $V_{2} = 10 mL$ , $n_{2} = 2$

For ${KMnO}_{4}$ (1): $V_{1} = 10 mL$ , $n_{1} = 5$ , $M_{1} = ?$

Substitute these values into the formula:

M_{1} \times 10 mL \times 5 = 0.25 M \times 10 mL \times 2

M_{1} \times 50 = 5

M_{1} = \frac{5}{50} = 0.1 M

💡 Teacher's Secret Hint

Ensure units are consistent, although in this case, volumes cancel out.

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Chemistry Question 57 – NEET-UG 2026

Clue 1 of 3: Redox Titration Principle

🧠 Full Solution Path