100 mL of `H_(2)O_(2)` is oxidized by 100 mL of `1M KMnO_(4)` in acidic medium (`MnO_(4)^(-)` reduced to `Mn^(+2)` ) 100 mL of same `H_(2)O_(2)` is oxidized by v mL of `1M KMnO_(4)` in basic medium (`MnO_(4)^(-)` reduced to `MnO_(2)`). Find the value of v:

Options

A. 500
B. 100
C. 100/3
D. 500/3

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction in Acidic Medium
In acidic medium, the reaction between hydrogen peroxide (H₂O₂) and potassium permanganate (KMnO₄) can be represented as follows:
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]
This indicates that 1 mole of KMnO₄ reacts with 5 moles of H₂O₂.

### Step 2: Calculate the Equivalents in Acidic Medium
Given:
- Volume of H₂O₂ = 100 mL = 0.1 L
- Concentration of KMnO₄ = 1 M
- Volume of KMnO₄ used = 100 mL = 0.1 L

First, we need to find the equivalents of KMnO₄ used:
- Moles of KMnO₄ = Molarity × Volume = 1 mol/L × 0.1 L = 0.1 moles
- Since 1 mole of KMnO₄ reacts with 5 moles of H₂O₂, the equivalents of KMnO₄ = 0.1 moles × 5 = 0.5 equivalents.

### Step 3: Set Up the Equation for H₂O₂ in Acidic Medium
Using the relationship of equivalents:
\[ \text{Equivalents of H}_2\text{O}_2 = \text{Equivalents of KMnO}_4 \]
Let the normality of H₂O₂ be \( n_1 \) and the volume be \( v_1 \):
\[ n_1 \times v_1 = 0.5 \]
Where \( v_1 = 0.1 \) L (100 mL).

### Step 4: Calculate Normality of H₂O₂
Substituting the values:
\[ n_1 \times 0.1 = 0.5 \]
\[ n_1 = \frac{0.5}{0.1} = 5 \, \text{N} \]

### Step 5: Understand the Reaction in Basic Medium
In basic medium, the reaction is different:
\[ \text{MnO}_4^- + 2 \text{e}^- + 4 \text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4 \text{OH}^- \]
Here, 1 mole of KMnO₄ reacts with 2 moles of H₂O₂.

### Step 6: Set Up the Equation for H₂O₂ in Basic Medium
Let \( v \) be the volume of KMnO₄ used in basic medium:
- Moles of KMnO₄ = 1 M × \( \frac{v}{1000} \) L = \( \frac{v}{1000} \) moles
- Equivalents of KMnO₄ = \( \frac{v}{1000} \) moles × 2 = \( \frac{2v}{1000} \)

### Step 7: Set Up the Equation for H₂O₂ in Basic Medium
Using the same relationship of equivalents:
\[ \text{Equivalents of H}_2\text{O}_2 = \text{Equivalents of KMnO}_4 \]
\[ 5 \times 0.1 = \frac{2v}{1000} \]
\[ 0.5 = \frac{2v}{1000} \]

### Step 8: Solve for \( v \)
Cross-multiplying gives:
\[ 0.5 \times 1000 = 2v \]
\[ 500 = 2v \]
\[ v = \frac{500}{2} = 250 \, \text{mL} \]

### Final Answer
The value of \( v \) is **250 mL**.

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