In a redox reaction `Na_(2)S_(2)O_(3)` changes to `Na_(2)S_(4)O_(6)` what is the equivalent weight of `Na_(2)S_(2)O_(3)`?

Options

A. same as its molecular weight
B. `1//2xx`(its molecular weight)
C. `1//2xx`(its molecular weight)
D. `1//3xx`(its molecular weight )

Explanation

To find the equivalent weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) in the redox reaction where it changes to \( \text{Na}_2\text{S}_4\text{O}_6 \), we can follow these steps:

### Step 1: Determine the oxidation states of sulfur in both compounds.

1. **For \( \text{Na}_2\text{S}_2\text{O}_3 \)**:
- Let the oxidation state of sulfur be \( x \).
- The equation for the oxidation state is:
\[ 2x + 3(-2) + 2(+1) = 0 \]
- Simplifying this gives:
\[ 2x - 6 + 2 = 0 \implies 2x - 4 = 0 \implies 2x = 4 \implies x = +2 \]

2. **For \( \text{Na}_2\text{S}_4\text{O}_6 \)**:
- Let the oxidation state of sulfur be \( y \).
- The equation for the oxidation state is:
\[ 4y + 6(-2) + 2(+1) = 0 \]
- Simplifying this gives:
\[ 4y - 12 + 2 = 0 \implies 4y - 10 = 0 \implies 4y = 10 \implies y = +2.5 \]

### Step 2: Calculate the change in oxidation state.

- The change in oxidation state for sulfur from \( +2 \) to \( +2.5 \) is:
\[ \Delta = 2.5 - 2 = 0.5 \]

### Step 3: Determine the total change in oxidation number.

- Since there are 2 sulfur atoms in \( \text{Na}_2\text{S}_2\text{O}_3 \), the total change in oxidation state for the two sulfur atoms is:
\[ \text{Total change} = 2 \times 0.5 = 1 \]

### Step 4: Calculate the molecular mass of \( \text{Na}_2\text{S}_2\text{O}_3 \).

- The molecular mass can be calculated as follows:
- Sodium (Na): \( 23 \, \text{g/mol} \)
- Sulfur (S): \( 32 \, \text{g/mol} \)
- Oxygen (O): \( 16 \, \text{g/mol} \)

\[ \text{Molecular mass} = 2(23) + 2(32) + 3(16) = 46 + 64 + 48 = 158 \, \text{g/mol} \]

### Step 5: Calculate the equivalent weight.

- The equivalent weight is given by the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular mass}}{\text{Total change in oxidation state}} \]
- Substituting the values:
\[ \text{Equivalent weight} = \frac{158}{1} = 158 \, \text{g/equiv} \]

### Final Answer:
The equivalent weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) is **158 g/equiv**.

---