Question

Only \( 2 \, \text{mL} \) of \( \text{KMnO}_4 \) solution of unknown molarity is required to reach the end point of a titration of \( 20 \, \text{mL} \) of oxalic acid (\( 2 \, \text{M} \)) in acidic medium. The molarity of \( \text{KMnO}_4 \) solution should be ______ \( \text{M} \).

Answer 1

Correct Answer: 8

In an acidic medium, the reaction between potassium permanganate (\(\text{KMnO}_4\)) and oxalic acid (\(\text{H}_2\text{C}_2\text{O}_4\)) can be represented as:

\[ 2 \text{KMnO}_4 + 5 \text{H}_2\text{C}_2\text{O}_4 + 6 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 + 8 \text{H}_2\text{O} \]

Using the Concept of Equivalents:

According to the principle of equivalents:

\[ \text{equivalents of KMnO}_4 = \text{equivalents of H}_2\text{C}_2\text{O}_4 \]

Calculate Equivalents for Each Solution:

For oxalic acid (\(\text{H}_2\text{C}_2\text{O}_4\)):

\[ \text{Molarity} \times \text{Volume} \times \text{n-factor} = 2 \times 20 \times 2 = 80 \, \text{meq} \]

where \(\text{n-factor} = 2\) for oxalic acid.

For \(\text{KMnO}_4\):

\[ M \times 2 \times 5 = 10M \, \text{meq} \]

where \(\text{n-factor} = 5\) for \(\text{KMnO}_4\).

Equating Equivalents:

\[ 10M = 80 \]

Solving for \(M\):

\[ M = \frac{80}{10} = 8 \, \text{M} \]

Conclusion:

The molarity of the \(\text{KMnO}_4\) solution is \(8 \, \text{M}\).