Free iodine is titrated against standard reducing agent usually with sodium thiosulphate , i.e.
`K_(2)Cr_(2)O_(7) + 6Kl + 7H_(2)SO_(4) to Cr_(2)(SO_(4))_(3) + 4K_(2) SO_(4) + 7H_(2)O+l_(2)`
` 2CuSO_(4) + 4Kl to Cu_(2)l_(2) + 2K_(2)SO_(4) + l_(2)`
` l_(2) + Na_(2)S_(2)O_(3) to 2Nal + Na_(2)S_(4)O_(6)`
In iodometric titration, starch solution is used as an indicator. Starch solution gives blue or violet colour with free iodine .At the end point , blue or violet colour disappears when iodine is completely changed to iodide.
50 ml of an aqueous solution of `H_(2)O_(2)` was treated with excess of Kl in dil. `H_(2)SO_(4)` . The liberated iodine required 20 ml of `0.1 N Na_(2)S_(2)O_(3)` for complete reaction. The concentration of `H_(2)O_(2)` is

Options

A. `0.34`
B. `0.8`
C. `0.17`
D. `0.85`

Answer

To solve the problem step by step, we will follow the titration reactions and calculations based on the information provided in the question.

### Step 1: Write the relevant reactions
When hydrogen peroxide (H₂O₂) reacts with potassium iodide (KI) in an acidic medium (dilute H₂SO₄), the following reaction occurs:
\[ \text{H}_2\text{O}_2 + 2 \text{KI} \rightarrow \text{K}_2\text{SO}_4 + 2 \text{H}_2\text{O} + \text{I}_2 \]
This reaction shows that 1 mole of H₂O₂ produces 1 mole of iodine (I₂).

When iodine is titrated with sodium thiosulfate (Na₂S₂O₃), the following reaction occurs:
\[ \text{I}_2 + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2 \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 \]
This indicates that 1 mole of iodine reacts with 2 moles of sodium thiosulfate.

### Step 2: Calculate the milli equivalents of sodium thiosulfate
Given that 20 mL of 0.1 N sodium thiosulfate was used, we can calculate the milli equivalents of sodium thiosulfate:
\[ \text{Milli equivalents of Na}_2\text{S}_2\text{O}_3 = \text{Normality} \times \text{Volume (mL)} = 0.1 \, \text{N} \times 20 \, \text{mL} = 2 \, \text{milli equivalents} \]

### Step 3: Relate milli equivalents of iodine to H₂O₂
From the stoichiometry of the reactions:
- 1 mole of H₂O₂ produces 1 mole of I₂.
- 1 mole of I₂ reacts with 2 moles of Na₂S₂O₃.

Thus, the milli equivalents of iodine will be equal to the milli equivalents of sodium thiosulfate:
\[ \text{Milli equivalents of I}_2 = \text{Milli equivalents of Na}_2\text{S}_2\text{O}_3 = 2 \, \text{milli equivalents} \]

### Step 4: Calculate the equivalent weight of H₂O₂
The equivalent weight of H₂O₂ is given by:
\[ \text{Equivalent weight of H}_2\text{O}_2 = \frac{\text{Molar mass}}{n} \]
Where \( n \) is the number of moles of electrons transferred. For H₂O₂, \( n = 2 \) (since it can give up 2 electrons):
- Molar mass of H₂O₂ = 34 g/mol
\[ \text{Equivalent weight of H}_2\text{O}_2 = \frac{34}{2} = 17 \, \text{g/equiv} \]

### Step 5: Calculate the mass of H₂O₂ in the solution
Using the formula:
\[ \text{Mass of H}_2\text{O}_2 = \text{Milli equivalents} \times \text{Equivalent weight} \times \frac{1}{1000} \]
Substituting the values:
\[ \text{Mass of H}_2\text{O}_2 = 2 \, \text{milli equivalents} \times 17 \, \text{g/equiv} \times \frac{1}{1000} = 0.034 \, \text{g} \]

### Step 6: Calculate the concentration of H₂O₂ in g/L
Since the volume of the H₂O₂ solution is 50 mL, we convert this to liters:
\[ \text{Concentration (g/L)} = \frac{\text{Mass (g)}}{\text{Volume (L)}} \]
\[ \text{Concentration (g/L)} = \frac{0.034 \, \text{g}}{0.050 \, \text{L}} = 0.68 \, \text{g/L} \]

### Final Answer
The concentration of H₂O₂ is **0.68 g/L**.

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