When `KMnO_(4)` is titrated against `FeSO_(4).(NH_(4))_(2)SO_(4).6H_(2)O` in acidic medium the equivalent mass of `KMnO_(4)` is -

Options

A. Molecular mass `//10`
B. Molecular mass `//5`
C. Molecular mass `//2`
D. Molecular mass

Answer

To find the equivalent mass of potassium permanganate (KMnO₄) when titrated against Mohr's salt (FeSO₄·(NH₄)₂SO₄·6H₂O) in an acidic medium, we can follow these steps:

### Step 1: Identify the oxidation and reduction reactions
In this titration, KMnO₄ acts as an oxidizing agent, while Mohr's salt acts as a reducing agent. The oxidation half-reaction involves the oxidation of Fe²⁺ ions to Fe³⁺ ions.

**Oxidation Half-Reaction:**
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]

### Step 2: Write the reduction half-reaction
The reduction half-reaction involves the reduction of MnO₄⁻ ions to Mn²⁺ ions in an acidic medium.

**Reduction Half-Reaction:**
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 e^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

### Step 3: Balance the overall reaction
To balance the overall reaction, we need to ensure that the number of electrons lost in oxidation equals the number of electrons gained in reduction. Since the oxidation of one Fe²⁺ ion releases 1 electron and the reduction of one MnO₄⁻ ion requires 5 electrons, we multiply the oxidation half-reaction by 5.

**Overall Balanced Reaction:**
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5 \text{Fe}^{3+} + 4 \text{H}_2\text{O} \]

### Step 4: Determine the n-factor for KMnO₄
The n-factor is defined as the number of electrons gained or lost per formula unit in a redox reaction. In this case, KMnO₄ gains 5 electrons during the reduction process.

### Step 5: Calculate the equivalent mass of KMnO₄
The equivalent mass of a substance is calculated using the formula:
\[ \text{Equivalent Mass} = \frac{\text{Molecular Weight}}{\text{n-factor}} \]

1. **Molecular Weight of KMnO₄:**
- K = 39 g/mol
- Mn = 55 g/mol
- O = 16 g/mol × 4 = 64 g/mol
- Total = 39 + 55 + 64 = 158 g/mol

2. **n-factor of KMnO₄:** 5 (as determined in Step 4)

3. **Calculating Equivalent Mass:**
\[ \text{Equivalent Mass of KMnO}_4 = \frac{158 \text{ g/mol}}{5} = 31.6 \text{ g/equiv} \]

### Final Answer:
The equivalent mass of KMnO₄ is **31.6 g/equiv**.

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