10 ml of `H_(2)O_(2)` solution on treatment with Kl and titration of liberated `l_(2)` required 20 mL of 1 N hypo solution. What is the normality of `H_(2)O_(2)` solution ?

Answer

To find the normality of the \( H_2O_2 \) solution, we can follow these steps:

### Step 1: Understand the Reaction
When \( H_2O_2 \) reacts with potassium iodide (KI), it liberates iodine (\( I_2 \)). The liberated iodine is then titrated with sodium thiosulfate (hypo, \( Na_2S_2O_3 \)).

### Step 2: Write the Relation of Equivalents
According to the principle of stoichiometry, the milli-equivalents of \( H_2O_2 \) will equal the milli-equivalents of \( I_2 \), which will also equal the milli-equivalents of \( Na_2S_2O_3 \).

### Step 3: Calculate the Milli-Equivalents of Sodium Thiosulfate
The formula for calculating milli-equivalents is:
\[ \text{milli-equivalents} = \text{Normality} \times \text{Volume (in mL)} \]
For the sodium thiosulfate solution:
\[ \text{milli-equivalents of } Na_2S_2O_3 = 1 \, \text{N} \times 20 \, \text{mL} = 20 \, \text{milli-equivalents} \]

### Step 4: Set Up the Equation for \( H_2O_2 \)
Since the milli-equivalents of \( H_2O_2 \) are equal to the milli-equivalents of \( Na_2S_2O_3 \):
\[ \text{milli-equivalents of } H_2O_2 = 20 \, \text{milli-equivalents} \]

### Step 5: Calculate the Normality of \( H_2O_2 \)
Using the formula for milli-equivalents again:
\[ \text{milli-equivalents of } H_2O_2 = \text{Normality of } H_2O_2 \times \text{Volume of } H_2O_2 \text{ (in mL)} \]
Let \( N \) be the normality of \( H_2O_2 \):
\[ 20 = N \times 10 \]
Solving for \( N \):
\[ N = \frac{20}{10} = 2 \, \text{N} \]

### Final Answer
The normality of the \( H_2O_2 \) solution is \( 2 \, \text{N} \).

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