A sample of 28 mL of `H_(2)` `O_(2)` (aq) solution required 10 mL of 0.1 M `KMnO_(4)` (aq) solution for complete reaction in acidic medium. What is the valume strength of `H_(2)O_(2)` ?

Answer

To find the volume strength of the hydrogen peroxide (H₂O₂) solution, we will follow these steps:

### Step 1: Understand the Reaction
The reaction between hydrogen peroxide (H₂O₂) and potassium permanganate (KMnO₄) in acidic medium involves the oxidation of H₂O₂ and the reduction of KMnO₄. The balanced reaction can be summarized as:
\[ \text{H}_2\text{O}_2 + \text{KMnO}_4 \rightarrow \text{O}_2 + \text{Mn}^{2+} + \text{H}_2\text{O} \]

### Step 2: Determine the Change in Oxidation States
- For H₂O₂, the oxidation state of oxygen changes from -1 (in H₂O₂) to 0 (in O₂).
- The change in oxidation state for one oxygen atom is 1, and since there are 2 oxygen atoms in H₂O₂, the total change is 2. Thus, the n-factor for H₂O₂ is 2.
- For KMnO₄, manganese changes from +7 to +2, resulting in a change of 5. Therefore, the n-factor for KMnO₄ is 5.

### Step 3: Calculate the Number of Equivalents
The number of equivalents can be calculated using the formula:
\[ \text{Number of equivalents} = \text{Normality} \times \text{Volume (in liters)} \]

Since we have molarity (M) instead of normality, we can use the relationship:
\[ \text{Normality} = \text{Molarity} \times \text{n-factor} \]

### Step 4: Calculate Equivalents for KMnO₄
Given:
- Volume of KMnO₄ = 10 mL = 0.010 L
- Molarity of KMnO₄ = 0.1 M
- n-factor of KMnO₄ = 5

Calculating the number of equivalents for KMnO₄:
\[ \text{Equivalents of KMnO}_4 = \text{Molarity} \times \text{n-factor} \times \text{Volume} \]
\[ = 0.1 \, \text{M} \times 5 \times 0.010 \, \text{L} = 0.005 \, \text{equivalents} \]

### Step 5: Set Equivalents Equal
Since the number of equivalents of KMnO₄ equals the number of equivalents of H₂O₂:
\[ \text{Equivalents of H}_2\text{O}_2 = 0.005 \, \text{equivalents} \]

### Step 6: Calculate Molarity of H₂O₂
Let the molarity of H₂O₂ be \( M \).
Given:
- Volume of H₂O₂ = 28 mL = 0.028 L
- n-factor of H₂O₂ = 2

Using the equivalents formula:
\[ \text{Equivalents of H}_2\text{O}_2 = M \times 2 \times 0.028 \]
Setting this equal to the equivalents calculated:
\[ M \times 2 \times 0.028 = 0.005 \]
\[ M = \frac{0.005}{2 \times 0.028} = \frac{0.005}{0.056} = 0.0893 \, \text{M} \]

### Step 7: Calculate Volume Strength of H₂O₂
The volume strength (V) of H₂O₂ is calculated using the formula:
\[ \text{Volume Strength} = 11.2 \times \text{Molarity} \]
Substituting the molarity:
\[ \text{Volume Strength} = 11.2 \times 0.0893 \approx 1 \, \text{V} \]

### Final Answer
The volume strength of the H₂O₂ solution is **1 V**.

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