To solve the problem of determining the volume of 10 volume H₂O₂ solution that decolorizes 200 mL of 2N KMnO₄ solution in acidic medium, we can follow these steps:
### Step 1: Understand the Reaction
In acidic medium, potassium permanganate (KMnO₄) acts as an oxidizing agent and will react with hydrogen peroxide (H₂O₂). The balanced chemical reaction is:
\[ \text{KMnO}_4 + \text{H}_2\text{O}_2 \rightarrow \text{Mn}^{2+} + \text{O}_2 + \text{H}_2\text{O} \]
### Step 2: Calculate the Equivalent of KMnO₄
Given that the normality (N) of KMnO₄ is 2N, we can calculate the equivalents in the solution:
\[
\text{Equivalents of KMnO}_4 = \text{Normality} \times \text{Volume (in L)} = 2 \, \text{N} \times 0.2 \, \text{L} = 0.4 \, \text{equivalents}
\]
### Step 3: Determine the Required Equivalents of H₂O₂
Since 1 equivalent of KMnO₄ reacts with 1 equivalent of H₂O₂, the equivalents of H₂O₂ required will also be 0.4 equivalents.
### Step 4: Calculate the Moles of H₂O₂ Required
The equivalent weight of H₂O₂ can be calculated as follows:
\[
\text{Molar mass of H}_2\text{O}_2 = 2 \times 1 + 2 \times 16 = 34 \, \text{g/mol}
\]
Since H₂O₂ has a valency of 1 in this reaction, its equivalent weight is also 34 g.
Now, we can find the moles of H₂O₂ required:
\[
\text{Moles of H}_2\text{O}_2 = \text{Equivalents} \times \text{Equivalent weight} = 0.4 \, \text{equivalents} \times 34 \, \text{g/equivalent} = 13.6 \, \text{g}
\]
### Step 5: Calculate the Volume of 10 Volume H₂O₂ Solution
10 volume H₂O₂ means that 1 liter of H₂O₂ solution produces 10 liters of O₂ at NTP. From the stoichiometry of the reaction, we know that:
\[
2 \, \text{moles of H}_2\text{O}_2 \rightarrow 1 \, \text{mole of O}_2
\]
Thus, 1 mole of H₂O₂ produces 5 liters of O₂.
To find out how many moles of H₂O₂ are needed to produce the required amount of O₂:
\[
\text{Volume of O}_2 = 10 \, \text{liters} \quad \Rightarrow \quad \text{Moles of O}_2 = \frac{10}{22.4} \, \text{moles}
\]
From the stoichiometry, the moles of H₂O₂ needed:
\[
\text{Moles of H}_2\text{O}_2 = 2 \times \text{Moles of O}_2 = 2 \times \frac{10}{22.4} = \frac{20}{22.4} \approx 0.8929 \, \text{moles}
\]
### Step 6: Calculate the Volume of 10 Volume H₂O₂ Solution
Since 10 volume H₂O₂ gives 10 liters of O₂ for every 1 liter of solution, we can find the volume of H₂O₂ solution needed:
\[
\text{Volume of H}_2\text{O}_2 = \frac{\text{Moles of H}_2\text{O}_2}{\text{Volume strength}} = \frac{0.8929 \, \text{moles}}{10 \, \text{liters}} \approx 0.08929 \, \text{liters} = 89.29 \, \text{mL}
\]
### Final Answer
The volume of 10 volume H₂O₂ solution that decolorizes 200 mL of 2N KMnO₄ solution in acidic medium is approximately **89.29 mL**.
---
