To find the molarity of the hydrogen peroxide (H₂O₂) solution, we can follow these steps:
### Step 1: Write the balanced chemical equation
The reaction between hydrogen peroxide (H₂O₂) and potassium iodide (KI) in acidic medium can be represented as follows:
\[
2 \, \text{H}_2\text{O}_2 + 2 \, \text{KI} + 2 \, \text{H}_2\text{SO}_4 \rightarrow 2 \, \text{K}_2\text{SO}_4 + 2 \, \text{H}_2\text{O} + \text{I}_2
\]
### Step 2: Determine the amount of iodine (I₂) produced
From the problem, we know that the liberated iodine (I₂) required 20 ml of a 0.04 M hypo (sodium thiosulfate, Na₂S₂O₃) solution for titration.
First, calculate the moles of hypo used:
\[
\text{Moles of Na}_2\text{S}_2\text{O}_3 = \text{Molarity} \times \text{Volume (L)} = 0.04 \, \text{mol/L} \times 0.020 \, \text{L} = 0.0008 \, \text{mol}
\]
### Step 3: Relate moles of iodine to moles of hydrogen peroxide
The stoichiometry of the reaction shows that 1 mole of I₂ reacts with 2 moles of H₂O₂. Therefore, the moles of I₂ produced can be equated to the moles of Na₂S₂O₃ used:
\[
\text{Moles of I}_2 = \text{Moles of Na}_2\text{S}_2\text{O}_3 = 0.0008 \, \text{mol}
\]
Thus, the moles of H₂O₂ can be calculated as:
\[
\text{Moles of H}_2\text{O}_2 = 2 \times \text{Moles of I}_2 = 2 \times 0.0008 \, \text{mol} = 0.0016 \, \text{mol}
\]
### Step 4: Calculate the molarity of the H₂O₂ solution
The molarity (M) of the H₂O₂ solution can be calculated using the formula:
\[
\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}}
\]
Given that the volume of the H₂O₂ solution is 50 ml (or 0.050 L):
\[
\text{Molarity of H}_2\text{O}_2 = \frac{0.0016 \, \text{mol}}{0.050 \, \text{L}} = 0.032 \, \text{M}
\]
### Step 5: Convert to grams per liter (if needed)
If we want to express this in grams per liter, we can use the molar mass of H₂O₂ (approximately 34 g/mol):
\[
\text{Mass of H}_2\text{O}_2 = \text{Moles} \times \text{Molar mass} = 0.0016 \, \text{mol} \times 34 \, \text{g/mol} = 0.0544 \, \text{g}
\]
To find grams per liter:
\[
\text{Concentration in g/L} = \frac{0.0544 \, \text{g}}{0.050 \, \text{L}} = 1.088 \, \text{g/L}
\]
### Final Result
The molarity of the H₂O₂ solution is approximately **0.032 M**.
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