0.1 M `KMnO_(4)` solution completely reacts with 0.05 M `FeSO_(4)` solution under acidic conditions. The volume of `FeSO_(4)` used is 25 ml. What volume of `KMnO_(4)` was used?

Answer

To solve the question step by step, we need to find the volume of `KMnO4` solution that reacts with the given volume of `FeSO4` solution under acidic conditions. Here’s how we can do it:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction between `KMnO4` and `FeSO4` in acidic conditions is:

\[ 2 \text{KMnO}_4 + 10 \text{FeSO}_4 + 8 \text{H}_2\text{SO}_4 \rightarrow 5 \text{Fe}_2(\text{SO}_4)_3 + \text{MnSO}_4 + \text{K}_2\text{SO}_4 + 8 \text{H}_2\text{O} \]

### Step 2: Determine the stoichiometry
From the balanced equation, we can see that:
- 2 moles of `KMnO4` react with 10 moles of `FeSO4`.
- Therefore, 1 mole of `KMnO4` reacts with 5 moles of `FeSO4`.

### Step 3: Calculate the number of equivalents of `FeSO4`
To find the number of equivalents of `FeSO4`, we can use the formula:

\[ \text{Equivalents of } \text{FeSO}_4 = \text{Molarity} \times \text{Volume (in L)} \times \text{n-factor} \]

Given:
- Molarity of `FeSO4` = 0.05 M
- Volume of `FeSO4` = 25 mL = 0.025 L
- n-factor of `FeSO4` = 1 (since it provides one mole of `Fe^2+`)

Calculating the equivalents:

\[ \text{Equivalents of } \text{FeSO}_4 = 0.05 \, \text{mol/L} \times 0.025 \, \text{L} \times 1 = 0.00125 \, \text{equivalents} \]

### Step 4: Relate the equivalents of `KMnO4` to `FeSO4`
From the stoichiometry, we know that:
- 2 equivalents of `KMnO4` react with 10 equivalents of `FeSO4`.
- Therefore, 1 equivalent of `KMnO4` reacts with 5 equivalents of `FeSO4`.

Using the equivalents of `FeSO4` calculated:

\[ \text{Equivalents of } \text{KMnO}_4 = \frac{0.00125}{5} = 0.00025 \, \text{equivalents} \]

### Step 5: Calculate the volume of `KMnO4`
Now we can calculate the volume of `KMnO4` used:

Using the formula for equivalents again:

\[ \text{Equivalents of } \text{KMnO}_4 = \text{Molarity} \times \text{Volume (in L)} \times \text{n-factor} \]

Given:
- Molarity of `KMnO4` = 0.1 M
- n-factor of `KMnO4` = 5

Let the volume of `KMnO4` be \( V \) L:

\[ 0.00025 = 0.1 \times V \times 5 \]

Solving for \( V \):

\[ 0.00025 = 0.5 \times V \]
\[ V = \frac{0.00025}{0.5} = 0.0005 \, \text{L} = 0.5 \, \text{mL} \]

### Final Answer
The volume of `KMnO4` used is **0.5 mL**.

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