Acidified `KMnO_4` oxidises oxalic acid to `CO_2`. What is the volume (in litres) of `10^(-4)MKMnO_4` required to completely oxidise 0.5 L of `10^(-2)M oxalic acid in acid medium?

Options

A. 125
B. 1250
C. 200
D. 20

Answer

To solve the problem of how much volume of \(10^{-4} M KMnO_4\) is required to completely oxidize \(0.5 L\) of \(10^{-2} M\) oxalic acid in an acid medium, we can follow these steps:

### Step 1: Write the balanced chemical equation
The oxidation of oxalic acid (\(C_2H_2O_4\)) by acidified \(KMnO_4\) can be represented as:
\[ C_2O_4^{2-} + KMnO_4 \rightarrow CO_2 + Mn^{2+} + K^+ + H_2O \]
In this reaction, the manganese is reduced from \(+7\) in \(KMnO_4\) to \(+2\) in \(Mn^{2+}\).

### Step 2: Determine the change in oxidation states
- For \(KMnO_4\): Manganese changes from \(+7\) to \(+2\), which means it gains 5 electrons (N1 = 5).
- For oxalic acid: Each carbon in oxalic acid goes from \(+3\) to \(+4\) when it is oxidized to \(CO_2\). Since there are 2 carbon atoms, the total change is \(2\) electrons per molecule of oxalic acid (N2 = 2).

### Step 3: Calculate the moles of oxalic acid
Using the molarity and volume of the oxalic acid solution:
\[ \text{Moles of } C_2O_4^{2-} = M \times V = 10^{-2} \, M \times 0.5 \, L = 5 \times 10^{-3} \, moles \]

### Step 4: Relate moles of oxalic acid to moles of \(KMnO_4\)
From the stoichiometry of the reaction:
\[ \text{Moles of } KMnO_4 = \frac{N2}{N1} \times \text{Moles of } C_2O_4^{2-} \]
Substituting the values:
\[ \text{Moles of } KMnO_4 = \frac{2}{5} \times 5 \times 10^{-3} = 2 \times 10^{-3} \, moles \]

### Step 5: Calculate the volume of \(KMnO_4\) solution required
Using the molarity of the \(KMnO_4\) solution:
\[ \text{Volume} = \frac{\text{Moles}}{\text{Molarity}} = \frac{2 \times 10^{-3} \, moles}{10^{-4} \, M} = 20 \, L \]

### Final Answer
The volume of \(10^{-4} M KMnO_4\) required to completely oxidize \(0.5 L\) of \(10^{-2} M\) oxalic acid is **20 liters**.

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