`MnO_(4)^(-)` ions are reduced in acidic conditions to `Mn^(2+)` ions whereas they are reduced in neutral condition to `MnO_(2)`. The oxidation of 25 mL of a solution `x` containing `Fe^(2+)` ions required in acidic condition 20 mL of a solution y containing `MnO_(4)` ions. What value of solution y would be required to oxidize 25 mL of solution x containing `Fe^(2+)` ions in neutral condition ?

Options

A. 11.4 mL
B. 12.0 mL
C. 33.3 mL
D. 35.0 mL

Explanation

To solve the problem, we need to determine the volume of solution y containing MnO4^- ions required to oxidize 25 mL of solution x containing Fe^(2+) ions in neutral conditions. We already know that in acidic conditions, 20 mL of solution y is required for the same amount of Fe^(2+) ions.

### Step-by-Step Solution:

1. **Understand the Redox Reactions**:
- In acidic conditions, the half-reaction is:
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} + 5 \text{Fe}^{3+} \]
- In neutral conditions, the half-reaction is:
\[ \text{MnO}_4^- + 4 \text{H}^+ + 3 \text{Fe}^{2+} \rightarrow \text{MnO}_2 + 3 \text{Fe}^{3+} + 2 \text{H}_2\text{O} \]

2. **Determine the Stoichiometry**:
- In acidic conditions, 1 mole of MnO4^- reacts with 5 moles of Fe^(2+).
- In neutral conditions, 1 mole of MnO4^- reacts with 3 moles of Fe^(2+).

3. **Calculate the Amount of Fe^(2+) in 25 mL**:
- Given that 20 mL of MnO4^- is required to oxidize 25 mL of Fe^(2+) in acidic conditions, we can set up the relationship:
\[ 20 \text{ mL MnO}_4^- \text{ oxidizes } 25 \text{ mL Fe}^{2+} \]

4. **Calculate the Equivalent Volume of MnO4^- in Neutral Conditions**:
- Since 5 moles of Fe^(2+) are required for 1 mole of MnO4^- in acidic conditions, the amount of Fe^(2+) in 25 mL corresponds to:
\[ \frac{25 \text{ mL}}{5} = 5 \text{ mL of MnO}_4^- \]
- In neutral conditions, 3 moles of Fe^(2+) are required for 1 mole of MnO4^-:
\[ \text{Let } V \text{ be the volume of MnO}_4^- in neutral conditions. \\ V \text{ mL MnO}_4^- \text{ will oxidize } 3V \text{ mL of Fe}^{2+} \]
- Setting the amounts equal:
\[ 3V = 25 \text{ mL} \implies V = \frac{25}{3} \approx 8.33 \text{ mL} \]

5. **Calculate the Volume of MnO4^- Required**:
- Since we need to find the volume of MnO4^- that would oxidize 25 mL of Fe^(2+) in neutral conditions, we can use the stoichiometric ratio:
\[ \text{Volume of MnO}_4^- \text{ required} = \frac{20 \text{ mL (acidic)}}{5} \times 3 = 12 \text{ mL (neutral)} \]

6. **Final Calculation**:
- To convert from acidic to neutral conditions:
\[ \text{Volume of MnO}_4^- \text{ required in neutral conditions} = \frac{20 \text{ mL}}{5} \times 3 = 12 \text{ mL} \]

### Conclusion:
The volume of solution y required to oxidize 25 mL of solution x containing Fe^(2+) ions in neutral conditions is approximately **12 mL**.