A solution of `Na_(2)S_(2)O_(3)` is standardized iodometrically against 0.167g of `KBrO_(3)`. The process requires 50mL of the `Na_(2)S_(2)O_(4)` solution. What is the normality of the `Na_(2)S_(3)O_(3)?`

Options

A. 0.2N
B. 0.12N
C. 0.72N
D. 0.02N

Answer

To find the normality of the sodium thiosulphate solution (Na₂S₂O₃), we will follow these steps:

### Step 1: Determine the number of equivalents of KBrO₃
The number of equivalents can be calculated using the formula:
\[ \text{Number of equivalents} = \frac{\text{Given weight}}{\text{Equivalent weight}} \]
The equivalent weight of KBrO₃ can be calculated as follows:
- The molecular weight of KBrO₃ = K (39) + Br (80) + 3 × O (16) = 39 + 80 + 48 = 167 g/mol.
- The n-factor for KBrO₃ is 6 (as it gains 6 electrons during the reaction).

Thus, the equivalent weight of KBrO₃ is:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} = \frac{167}{6} \approx 27.83 \text{ g/equiv} \]

Now, we can calculate the number of equivalents of KBrO₃:
\[ \text{Number of equivalents of KBrO₃} = \frac{0.167 \text{ g}}{27.83 \text{ g/equiv}} \approx 0.006 \text{ equiv} \]

### Step 2: Set the equivalents of Na₂S₂O₃ equal to KBrO₃
Since the reaction is stoichiometric, the number of equivalents of Na₂S₂O₃ will equal the number of equivalents of KBrO₃:
\[ \text{Number of equivalents of Na₂S₂O₃} = 0.006 \text{ equiv} \]

### Step 3: Calculate the normality of Na₂S₂O₃
The number of equivalents of Na₂S₂O₃ can also be calculated using the formula:
\[ \text{Number of equivalents} = \text{Volume (L)} \times \text{Normality (N)} \]
Given that the volume of Na₂S₂O₃ solution used is 50 mL (or 0.050 L), we can set up the equation:
\[ 0.006 \text{ equiv} = 0.050 \text{ L} \times N \]

Solving for N (normality):
\[ N = \frac{0.006 \text{ equiv}}{0.050 \text{ L}} = 0.12 \text{ N} \]

### Final Answer
The normality of the Na₂S₂O₃ solution is **0.12 N**.

---