Redox and Equivalent ConceptMedium

Question

In an acidic medium, 10 mL of 0.25 M oxalic acid is titrated with ${KMnO}_{4}$ solution. If the volume of ${KMnO}_{4}$ solution required to reach end point is 10 mL , the strength of the ${KMnO}_{4}$ solution is

Options

A.0.25 M
B.0.15 M
C.0.10 M
D.0.20 M

Solution

Sol. [Key Concept: Redox titration of oxalic acid with KMnO₄]

$${KMnO}_{4} + H_{2}C_{2}O_{4}\overset{\phantom{H^{+}\ }}{\rightarrow}{Mn}^{2 +} + 2{CO}_{2} + H_{2}O $$No. of equivalent of ${KMnO}_{4} = No$. of equivalent of $H_{2}C_{2}O_{4}$

$$\begin{matrix} & 10 \times 5 \times M = 10 \times 0.25 \times 2 \\ & M = \frac{5}{50} \\ & M = \frac{1}{10} = 0.10M \end{matrix}$$

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