Bleaching powder and bleach solution are produced on a large scale and used in several house-hold products. The effectiveness of bleach solution is often measured by iodometry.
`25 mL` of household bleach solution was mixed with `30 mL` of `0.50 M KI` and `10 mL` of `4N` acetic acid. In the titration of the liberated iodine, `48 mL` of `0.25 N Na_(2)S_(2)O_(3)` was used to reach the end point. The molarity of the household bleach solution is :

Options

A. `0.48`M
B. `0.96`M
C. `0.24`M
D. `0.02`M

Explanation

To find the molarity of the household bleach solution, we will follow these steps:

### Step 1: Write the reactions involved
The reactions that occur when bleach is treated with potassium iodide (KI) and acetic acid can be summarized as follows:

1. **Bleach Reaction**:
\[ \text{Ca(OCl)_2} + 2 \text{CH}_3\text{COOH} \rightarrow \text{Cl}_2 + \text{Ca(CH}_3\text{COO})_2 + \text{H}_2\text{O} \]

2. **Chlorine Reaction with KI**:
\[ \text{Cl}_2 + 2 \text{KI} \rightarrow 2 \text{KCl} + \text{I}_2 \]

3. **Iodine Reaction with Sodium Thiosulfate**:
\[ \text{I}_2 + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2 \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 \]

### Step 2: Calculate milliequivalents of Na₂S₂O₃
We need to find the milliequivalents of sodium thiosulfate (Na₂S₂O₃) used in the titration:

\[ \text{Milliequivalents of Na}_2\text{S}_2\text{O}_3 = \text{Volume (mL)} \times \text{Normality (N)} \]
\[ = 48 \, \text{mL} \times 0.25 \, \text{N} = 12 \, \text{milliequivalents} \]

### Step 3: Relate milliequivalents of I₂ to Cl₂
From the stoichiometry of the reactions, we know that:
- 1 mole of I₂ reacts with 2 moles of Na₂S₂O₃.
- Therefore, the milliequivalents of I₂ will be half of the milliequivalents of Na₂S₂O₃.

\[ \text{Milliequivalents of I}_2 = \frac{12}{2} = 6 \, \text{milliequivalents} \]

### Step 4: Relate milliequivalents of Cl₂ to bleach
Since the milliequivalents of Cl₂ are equal to the milliequivalents of I₂:
\[ \text{Milliequivalents of Cl}_2 = 6 \, \text{milliequivalents} \]

### Step 5: Calculate the milliequivalents of bleach
The volume of the bleach solution used is 25 mL. Therefore, the milliequivalents of bleach in this volume can be expressed as:
\[ \text{Milliequivalents of bleach} = 6 \, \text{milliequivalents} \]

### Step 6: Calculate the molarity of the bleach solution
Molarity (M) is defined as the number of equivalents per liter of solution. Since we have 6 milliequivalents in 25 mL, we can calculate the molarity as follows:

\[ \text{Molarity} = \frac{\text{Milliequivalents}}{\text{Volume (L)}} \]
\[ = \frac{6 \, \text{milliequivalents}}{25 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}}} = \frac{6}{0.025} = 240 \, \text{molar} \]

### Step 7: Convert to molarity
Since we are calculating normality, we need to divide by the number of equivalents:
\[ \text{Molarity} = \frac{6 \, \text{milliequivalents}}{25 \, \text{mL}} = 0.24 \, \text{M} \]

### Final Answer
The molarity of the household bleach solution is **0.24 M**.

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