Question

In neutral or faintly alkaline medium, $MnO_4^-$ oxidizes $I^-$ to iodate. What is the volume (in L) of 0.02M $KMnO_4$ required to completely convert 1 L of 0.5 M KI solution to iodate in neutral or faintly alkaline medium?

• A. $5$
• B. $50$
• C. $20$
• D. $30$

Hint:

For redox titrations, always balance the chemical equation to determine the correct mole ratio between the reactants.

Answer 1

The Correct Option is B

Step 1: Write the balanced redox reaction in neutral/alkaline medium.
$2MnO_4^- + I^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^-$ Step 2: Determine the stoichiometry.
2 moles of $MnO_4^-$ react with 1 mole of $I^-$.
Step 3: Calculate moles of KI.
Moles of $KI = \text{Molarity} \times \text{Volume} = 0.5 \, M \times 1 \, L = 0.5 \, mol$. This means moles of $I^- = 0.5 \, mol$.
Step 4: Calculate moles of $KMnO_4$ required.
From stoichiometry, moles of $KMnO_4 = 2 \times \text{moles of } I^- = 2 \times 0.5 \, mol = 1 \, mol$.
Step 5: Calculate the volume of $KMnO_4$ solution.
Volume of $KMnO_4 = \frac{\text{moles of } KMnO_4}{\text{Molarity of } KMnO_4} = \frac{1 \, mol}{0.02 \, M} = 50 \, L$.
Thus, the volume of 0.02M $KMnO_4$ required is $ \boxed{50} \, L $.