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

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 $8 \, \text{M}$.