Question

In the reaction of KMnO\(_4\) with an oxalate in acidic medium, MnO\(_4^-\) is reduced to Mn\(^{2+}\) and C\(_2\)O\(_4^{2-}\) is oxidized to CO\(_2\). Hence, 50 mL of 0.02 M KMnO\(_4\) is equivalent to:

• A. 100 mL of 0.05 M \( \text{MH}_2 \text{C}_2 \text{O}_4 \)
• B. 50 mL of 0.05 M \( \text{MH}_2 \text{C}_2 \text{O}_4 \)
• C. 25 mL of 0.2 M \( \text{MH}_2 \text{C}_2 \text{O}_4 \)
• D. 50 mL of 0.10 M \( \text{MH}_2 \text{C}_2 \text{O}_4 \)

Hint:

In redox reactions, use stoichiometry to relate the volumes and concentrations of reactants involved.

Answer 1

The Correct Option is B

Step 1: Stoichiometry of the reaction.
From the stoichiometry of the reaction, the molar ratio between KMnO\(_4\) and oxalate is 1:2. This means that 1 mole of KMnO\(_4\) reacts with 2 moles of oxalate.

Step 2: Use the molarity and volume.
For 50 mL of 0.02 M KMnO\(_4\), the equivalent volume of oxalate solution can be calculated as follows: \[ \text{moles of KMnO}_4 = 0.02 \times 0.05 = 0.001 \, \text{moles}. \] The equivalent moles of oxalate are \( 2 \times 0.001 = 0.002 \, \text{moles} \). Thus, the equivalent volume of oxalate solution is: \[ \text{volume} = \frac{0.002}{0.05} = 0.04 \, \text{L} = 50 \, \text{mL}. \] Thus, the correct answer is (2).