15.0 mL of 0.12 M `KMnO_(4)` solution are required to oxidise 20.0 mL of `FeSO_(4)` solution in aicdic medium what is the concentration of `FeSO_(4)` solution ?

Explanation

To find the concentration of the `FeSO4` solution that is oxidized by `KMnO4`, we can follow these steps:

### Step 1: Determine the moles of `KMnO4` used.
We know the volume and molarity of the `KMnO4` solution.

\[ \text{Moles of } KMnO_4 = \text{Molarity} \times \text{Volume (in L)} \]

Given:
- Molarity of `KMnO4` = 0.12 M
- Volume of `KMnO4` = 15.0 mL = 0.015 L

Calculating the moles:

\[ \text{Moles of } KMnO_4 = 0.12 \, \text{mol/L} \times 0.015 \, \text{L} = 0.0018 \, \text{mol} \]

### Step 2: Determine the equivalent of `KMnO4`.
In acidic medium, `KMnO4` acts as a strong oxidizing agent and has a change in oxidation state from +7 to +2. Therefore, 1 mole of `KMnO4` can oxidize 5 moles of `Fe^{2+}` ions.

\[ \text{Equivalents of } KMnO_4 = \text{Moles} \times 5 = 0.0018 \, \text{mol} \times 5 = 0.009 \, \text{equivalents} \]

### Step 3: Relate the equivalents of `FeSO4` to those of `KMnO4`.
Since `Fe^{2+}` is oxidized to `Fe^{3+}`, 1 mole of `FeSO4` provides 1 equivalent of `Fe^{2+}`. Therefore, the equivalents of `FeSO4` will be equal to the equivalents of `KMnO4`.

\[ \text{Equivalents of } FeSO_4 = 0.009 \, \text{equivalents} \]

### Step 4: Calculate the concentration of `FeSO4`.
We know that:

\[ \text{Concentration (M)} = \frac{\text{Equivalents}}{\text{Volume (in L)}} \]

Given:
- Volume of `FeSO4` = 20.0 mL = 0.020 L

Calculating the concentration:

\[ \text{Concentration of } FeSO_4 = \frac{0.009 \, \text{equivalents}}{0.020 \, \text{L}} = 0.45 \, \text{M} \]

### Final Answer:
The concentration of the `FeSO4` solution is **0.45 M**.

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