In an acidic medium, 10 mL of 0.25 M oxalic acid is titrated with $\mathrm{KMnO}_4$ solution. If the volume of $\mathrm{KMnO}_4$ solution required to reach the end point is 10 mL, the strength of the $\mathrm{KMnO}_4$ solution is

Question

Accepted Answer

B. 1. In acidic medium, $\mathrm{KMnO}_4$ oxidises oxalic acid completely, so we first write the balanced redox reaction between them.
$$
2\mathrm{MnO}_4^- + 5\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4 + 6\mathrm{H}^+ ightarrow 2\mathrm{Mn}^{2+} + 10\mathrm{CO}_2 + 8\mathrm{H}_2\mathrm{O}
$$

2. Since this is a titration, the **law of equivalence** applies — the number of milliequivalents of $\mathrm{KMnO}_4$ equals the number of milliequivalents of oxalic acid at the end point.
$$
V_1 N_1(\mathrm{KMnO}_4) = V_2 N_2(\mathrm{oxalic\ acid})
$$

3. The **n-factor** of $\mathrm{KMnO}_4$ in acidic medium is 5 (it gains 5 electrons per formula unit), and the n-factor of oxalic acid is 2 (it loses 2 electrons). Given $V(\mathrm{KMnO}_4) = V(	ext{oxalic acid}) = 10\ 	ext{mL}$ and $M(	ext{oxalic acid}) = 0.25\ 	ext{M}$, we substitute:
$$
10 	imes (M 	imes 5) = 10 	imes (0.25 	imes 2)
$$

4. Solving for $M$, the molarity of $\mathrm{KMnO}_4$:
$$
M = \frac{10 	imes 0.5}{10 	imes 5} = \frac{5}{50} = 0.10\ 	ext{M}
$$

Hence, the strength of $\mathrm{KMnO}_4$ solution is **0.10 M**, so the correct answer is **Option B**.

In an acidic medium, 10 mL of 0.25 M oxalic acid is titrated with KMnO4\mathrm{KMnO}_4KMnO4​ solution. If the volume of KMnO4\mathrm{KMnO}_4KMnO4​ solution required to reach the end point is 10 mL, the strength of the KMnO4\mathrm{KMnO}_4KMnO4​ solution is

In an acidic medium, 10 mL of 0.25 M oxalic acid is titrated with $\mathrm{KMnO}_4$ solution. If the volume of $\mathrm{KMnO}_4$ solution required to reach the end point is 10 mL, the strength of the $\mathrm{KMnO}_4$ solution is