Q.57 Chemistry ⭐ 360° Option Elimination Active

NEET 2026 Chemistry Question 57 Solution

Topic: Redox Reactions · Official NTA Booklet Code 50

Question

In an acidic medium, 10 mL of 0.25 M oxalic acid is titrated with KMnO₄ solution. If the volume of KMnO₄ solution required to reach end point is 10 mL, the strength of the KMnO₄ solution is

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Verified Against NTA Final Answer Key

Option (1) : 0.10 M

📚 Redox Reactions Medium

📋 Given Parameters & Problem Setup

Medium Acidic medium (H₂SO₄)

Oxalic Acid Volume (V₂) 10 mL

Oxalic Acid Molarity (M₂) 0.25 M

KMnO₄ Volume (V₁) 10 mL

Target Molarity (strength) of KMnO₄ solution (M₁)

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The 30-Second Exam Hall Shortcut: ⚡ The 5:2 n-factor Redox Ratio

⏱️ Save ~60 seconds

Permanganate transfers 5 electrons; Oxalate transfers 2 electrons!

1 Equate equivalents: N₁V₁ = N₂V₂, which means M₁ × n₁ × V₁ = M₂ × n₂ × V₂.

2 In acidic medium: KMnO₄ has n₁ = 5 (Mn⁺⁷ → Mn²⁺).

3 Oxalic acid has n₂ = 2 (C₂O₄²⁻ → 2CO₂ + 2e⁻).

4 Since volumes are equal (V₁ = V₂ = 10 mL): M₁ × 5 = 0.25 × 2.

5 5 M₁ = 0.50 -> M₁ = 0.10 M!

6 Select Option (1) in 15 seconds!

🛡️

Why You Can 100% Trust This Shortcut: Balanced redox equation: 2 MnO₄⁻ + 5 C₂O₄²⁻ + 16 H⁺ → 2 Mn²⁺ + 10 CO₂ + 8 H₂O.

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Determine the n-factors of the Reactants in Acidic Medium

1. Potassium permanganate (KMnO₄): Manganese is reduced from +7 in MnO₄⁻ to +2 in Mn²⁺. The change in oxidation state per mole is |+7 - (+2)| = 5. 2. Oxalic acid (H₂C₂O₄): Carbon is oxidized from +3 in C₂O₄²⁻ to +4 in CO₂. For two carbon atoms per molecule, total electron loss = 2 × 1 = 2.

n_1(\text{KMnO}_4) = 5, \quad n_2(\text{H}_2\text{C}_2\text{O}_4) = 2

• n-factor of KMnO₄ in acidic medium = 5.
• n-factor of oxalic acid = 2.

Step 2

Apply the Law of Chemical Equivalence

At the equivalence (end) point of a titration, the number of gram equivalents of oxidizing agent equals the number of gram equivalents of reducing agent: Number of equivalents = Molarity × n-factor × Volume

M_1 \times n_1 \times V_1 = M_2 \times n_2 \times V_2

• V₁ = 10 mL.
• V₂ = 10 mL.
• M₂ = 0.25 M.

Step 3

Calculate Molarity of KMnO₄

Substitute the given values into the equivalence equation.

M_1 \times 5 \times 10 = 0.25 \times 2 \times 10 \implies 50 M_1 = 5.0 \implies M_1 = \frac{5.0}{50} = 0.10\text{ M}

• Strength = 0.10 M.
• Matches Option (1).

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 A

✓ 100% Correct

Verdict & Scientific Reason:

Exact stoichiometric result derived from the law of equivalence and standard n-factors.

💡

Why this is airtight: None; rigorous application of redox stoichiometry.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Arises from using n = 1 for oxalic acid.

⚠️

Student Fallacy (Why candidates pick this trap): Forgetting that oxalic acid (C₂O₄²⁻) possesses two carbon atoms and loses 2 electrons.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Assuming a 1:1 molar ratio (M₁V₁ = M₂V₂), which is completely invalid for redox reactions.

⚠️

Student Fallacy (Why candidates pick this trap): Using simple molarity equation without n-factor multipliers.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Arithmetic miscalculation.

⚠️

Student Fallacy (Why candidates pick this trap): Using incorrect n-factor ratio.

⚠️

NTA Trap Alert: Using M₁V₁ = M₂V₂ Instead of N₁V₁ = N₂V₂

The most widespread error in quantitative titrations is using acid-base molarity equality for a redox reaction with unequal electron transfers (5 electrons vs 2 electrons).

🛡️ How to avoid the -1 mark: Always use equivalents: M₁ × n₁ × V₁ = M₂ × n₂ × V₂.

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Core NEET Syllabus Standard

NEET Chemistry Core Syllabus · Redox Reactions Standard Benchmark

Core Scientific Principle Tested:

"In acidic solution, MnO₄⁻ acts as a 5-electron acceptor while C₂O₄²⁻ acts as a 2-electron donor."

Syllabus Focus: Redox titrations, equivalents, and oxidation number changes

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

KMnO₄ acid = 5; Oxalate = 2: Five times M₁ equals Two times M₂!

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NEET 2026 Chemistry Question 57 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ The 5:2 n-factor Redox Ratio

Step-by-Step Rigorous Workflow & Mathematical Proof

Determine the n-factors of the Reactants in Acidic Medium

Apply the Law of Chemical Equivalence

Calculate Molarity of KMnO₄

360° Option Elimination Engine Key Feature

NTA Trap Alert: Using M₁V₁ = M₂V₂ Instead of N₁V₁ = N₂V₂

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — Redox Reactions

Oxidation States of Underlined Elements in KO₂, H₂O₂ and H₂SO₄

Identification of Non-Redox (Double Displacement) Reactions

Balancing Redox Reactions by Ion-Electron Method in Acidic Medium

Classification of Redox Reactions: Metal Displacement

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