Which of the following solutions will exactly oxidize `25mL` of an acid solution of `0.1 M Fe` (`II`) oxalate?

Options

A. 25 mL of 0.1 M `KMnO_(4)`
B. 25 mL of 0.2 M `KMnO_(4)`
C. 25 mL of 0.6 M `KMnO_(4)`
D. 15 mL of 0.1 M `KMnO_(4)`

Answer

To solve the question of which solution will exactly oxidize 25 mL of an acid solution of 0.1 M Fe(II) oxalate, we can follow these steps:

### Step 1: Write the oxidation reactions
1. **Oxidation of Fe(II)**:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]
This indicates that 1 mole of Fe(II) loses 1 electron to become Fe(III).

2. **Oxidation of oxalate**:
\[ \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{CO}_2 + 2e^- \]
This indicates that 1 mole of oxalate loses 2 electrons.

### Step 2: Determine the stoichiometry of the reaction
- When Fe(II) and oxalate react with permanganate (\(\text{MnO}_4^{-}\)), the overall reaction can be represented as:
\[ 5 \text{Fe}^{2+} + 3 \text{MnO}_4^{-} \rightarrow 5 \text{Fe}^{3+} + 3 \text{Mn}^{2+} \]
This means that 5 moles of Fe(II) react with 3 moles of permanganate.

### Step 3: Calculate the moles of Fe(II) in the solution
- The concentration of Fe(II) is given as 0.1 M and the volume is 25 mL:
\[ \text{Moles of Fe}^{2+} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.025 \, \text{L} = 0.0025 \, \text{mol} \]

### Step 4: Determine the moles of permanganate required
- From the stoichiometry of the reaction, 5 moles of Fe(II) require 3 moles of permanganate. Therefore, the moles of permanganate required can be calculated as follows:
\[ \text{Moles of } \text{MnO}_4^{-} = \left(\frac{3}{5}\right) \times \text{Moles of Fe}^{2+} = \left(\frac{3}{5}\right) \times 0.0025 = 0.0015 \, \text{mol} \]

### Step 5: Calculate the volume of permanganate solution required
- To find the volume of the permanganate solution needed, we will use the formula:
\[ \text{Volume} = \frac{\text{Moles}}{\text{Concentration}} \]
- We will check each option provided to see which one gives us the required moles of permanganate.

### Step 6: Check each option
1. **Option A**: 25 mL of 0.1 M KMnO4
\[ \text{Moles} = 0.1 \times 0.025 = 0.0025 \, \text{mol} \quad \Rightarrow \quad \frac{0.0025}{3} = 0.833 \quad \text{(not equal to 0.0015)} \]

2. **Option B**: 25 mL of 0.2 M KMnO4
\[ \text{Moles} = 0.2 \times 0.025 = 0.005 \, \text{mol} \quad \Rightarrow \quad \frac{0.005}{3} = 1.666 \quad \text{(not equal to 0.0015)} \]

3. **Option C**: 25 mL of 0.6 M KMnO4
\[ \text{Moles} = 0.6 \times 0.025 = 0.015 \, \text{mol} \quad \Rightarrow \quad \frac{0.015}{3} = 5 \quad \text{(not equal to 0.0015)} \]

4. **Option D**: 15 mL of 0.1 M KMnO4
\[ \text{Moles} = 0.1 \times 0.015 = 0.0015 \, \text{mol} \quad \Rightarrow \quad \frac{0.0015}{3} = 0.5 \quad \text{(equal to 0.0015)} \]

### Conclusion
The solution that will exactly oxidize 25 mL of an acid solution of 0.1 M Fe(II) oxalate is **Option D: 15 mL of 0.1 M KMnO4**.

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