`50 mL` of `0.1 M` solution of a salt reacted with `25 mL` of `0.1 M` solution of sodium sulphite. The half reaction for the oxidation of sulphite ion is:
`SO_(3)^(2-)(aq)+H_(2)O(l) rarr (aq)+2H^(+)(aq)+2e^(-)`
If the oxidation number of metal in the salt was `3`, what would be the new oxidation number of metal:

Options

A. 0
B. 1
C. 2
D. 4

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction
We have a salt with a metal that has an oxidation number of +3. This salt reacts with sodium sulfite (Na2SO3). The half-reaction for the oxidation of the sulfite ion (SO3^2-) is given, which shows that it gets oxidized to sulfate (SO4^2-).

### Step 2: Determine the Oxidation Numbers
In the sulfite ion (SO3^2-), the oxidation number of sulfur (S) is +4, and in the sulfate ion (SO4^2-), the oxidation number of sulfur is +6. The change in oxidation number indicates that sulfur is oxidized from +4 to +6.

### Step 3: Calculate the n-factor for Sulfite
The n-factor for the oxidation of sulfite to sulfate can be calculated as the change in oxidation number:
- Change in oxidation number = +6 - (+4) = 2
Thus, the n-factor for the sulfite ion is 2.

### Step 4: Calculate the Equivalent of Sodium Sulfite
Now we calculate the equivalents of sodium sulfite used in the reaction:
- Molarity (M1) = 0.1 M
- Volume (V1) = 25 mL = 0.025 L
- n-factor = 2

Using the formula for equivalents:
\[ \text{Equivalents} = M1 \times V1 \times n-factor = 0.1 \, \text{mol/L} \times 0.025 \, \text{L} \times 2 = 0.005 \, \text{equivalents} \]

### Step 5: Calculate the Equivalents of the Salt
Let’s denote the molarity of the salt as M2 and its volume as V2:
- Molarity (M2) = 0.1 M
- Volume (V2) = 50 mL = 0.050 L
- Let the n-factor for the salt be (3 - x), where x is the new oxidation number after the reaction.

Using the equivalents formula:
\[ \text{Equivalents} = M2 \times V2 \times n-factor = 0.1 \, \text{mol/L} \times 0.050 \, \text{L} \times (3 - x) \]

### Step 6: Set Up the Equation
Equate the equivalents of sodium sulfite and the salt:
\[ 0.005 = 0.1 \times 0.050 \times (3 - x) \]
\[ 0.005 = 0.005 \times (3 - x) \]

### Step 7: Solve for x
Dividing both sides by 0.005:
\[ 1 = 3 - x \]
Rearranging gives:
\[ x = 3 - 1 = 2 \]

### Conclusion
The new oxidation number of the metal after the reaction is +2.

---