2 mole , equimolar mixture of `Na_(2)C_(2)O_(4) and H_(2)C_(2)O_(4) "required" V_(1) L of 0.1 M KMnO_(4)` in acidic medium for complete oxidation. The same amount of the mixture required ` V_(2) L of 0.2 M NaOH` for neutralisaation. The raation of ` V_(1) and V_(2)` is:

Options

A. `1:2`
B. `2:1`
C. `4:5`
D. `5:4`

Answer

To solve the problem, we need to find the ratio of \( V_1 \) and \( V_2 \) given the conditions of the reaction involving an equimolar mixture of sodium oxalate (\( \text{Na}_2\text{C}_2\text{O}_4 \)) and oxalic acid (\( \text{H}_2\text{C}_2\text{O}_4 \)).

### Step-by-Step Solution:

1. **Identify the Moles of Each Component:**
Since we have a 2 mole equimolar mixture, we have:
- Moles of \( \text{Na}_2\text{C}_2\text{O}_4 = 1 \) mole
- Moles of \( \text{H}_2\text{C}_2\text{O}_4 = 1 \) mole

2. **Calculate the Equivalents for Oxidation:**
- For \( \text{Na}_2\text{C}_2\text{O}_4 \):
- \( n \) factor = 2 (as each oxalate ion loses 2 electrons when oxidized to \( \text{CO}_2 \)).
- Equivalents = \( 1 \, \text{mole} \times 2 = 2 \, \text{equivalents} \).
- For \( \text{H}_2\text{C}_2\text{O}_4 \):
- \( n \) factor = 2 (same reason as above).
- Equivalents = \( 1 \, \text{mole} \times 2 = 2 \, \text{equivalents} \).

Total equivalents from the mixture:
\[ \text{Total equivalents} = 2 + 2 = 4 \, \text{equivalents} \]

3. **Relate to KMnO4:**
- The \( n \) factor for \( \text{KMnO}_4 \) is 5 (since it goes from +7 to +2).
- The number of equivalents of \( \text{KMnO}_4 \) can be expressed as:
\[ \text{Equivalents of KMnO}_4 = \text{Normality} \times V_1 \]
Since Normality = Molarity × \( n \) factor:
\[ \text{Equivalents of KMnO}_4 = 0.1 \times 5 \times V_1 = 0.5 V_1 \]

Setting the equivalents equal:
\[ 0.5 V_1 = 4 \implies V_1 = \frac{4}{0.5} = 8 \, \text{L} \]

4. **Calculate the Equivalents for Neutralization:**
- Only \( \text{H}_2\text{C}_2\text{O}_4 \) reacts with \( \text{NaOH} \).
- The equivalents of \( \text{H}_2\text{C}_2\text{O}_4 \) = 2 (as calculated before).
- The equivalents of \( \text{NaOH} \) can be expressed as:
\[ \text{Equivalents of NaOH} = \text{Normality} \times V_2 \]
Since the \( n \) factor for \( \text{NaOH} \) is 1:
\[ \text{Equivalents of NaOH} = 0.2 \times V_2 \]

Setting the equivalents equal:
\[ 2 = 0.2 V_2 \implies V_2 = \frac{2}{0.2} = 10 \, \text{L} \]

5. **Calculate the Ratio \( \frac{V_1}{V_2} \):**
\[ \frac{V_1}{V_2} = \frac{8}{10} = \frac{4}{5} \]

### Final Answer:
The ratio of \( V_1 \) to \( V_2 \) is \( 4:5 \).

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