Question

In a redox titration, 25 mL of 0.1 M KMnO$_4$ is required to completely react with oxalic acid in acidic medium. The volume of 0.2 M oxalic acid used is:
(Balanced reaction: \( 2\text{MnO}_4^- + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O} \))

• A. 25 mL
• B. 31.25 mL
• C. 12.5 mL
• D. 62.5 mL

Hint:

In redox titrations, use the stoichiometry of the balanced equation to find the moles of reactants, then calculate the volume using molarity.

Answer 1

The Correct Option is B

- From the balanced equation: \( 2\text{MnO}_4^- + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O} \), 2 moles of \( \text{KMnO}_4 \) react with 5 moles of oxalic acid (\( \text{H}_2\text{C}_2\text{O}_4 \)).
- Moles of \( \text{KMnO}_4 \): \[ \text{Volume} = 25 \, \text{mL} = 0.025 \, \text{L}, \quad \text{Molarity} = 0.1 \, \text{M} \] \[ \text{Moles of } \text{KMnO}_4 = 0.1 \times 0.025 = 0.0025 \, \text{mol} \]
- From the stoichiometry, 2 moles of \( \text{KMnO}_4 \) react with 5 moles of oxalic acid: \[ \text{Moles of oxalic acid} = \frac{5}{2} \times 0.0025 = 0.00625 \, \text{mol} \]
- Molarity of oxalic acid = 0.2 M. Volume of oxalic acid: \[ \text{Volume} = \frac{\text{moles}}{\text{molarity}} = \frac{0.00625}{0.2} = 0.03125 \, \text{L} = 31.25 \, \text{mL} \]
- This matches option (B).