In an experiment, 50 mL of 0.1 M solution off a metallic salt `M(NO_(3))` reacted exactly with 25 mL of 0.1 M solution of `NaSO_(3)` in the reaction `SO_(3)^(2-)` is oxidized to `SO_(4)^(2-)`. If in the original salt oxidation number of the metal was 3, what would be the new oxidation number of metal?

Explanation

To solve the problem step by step, we will analyze the reaction and calculate the new oxidation state of the metal in the salt.

### Step 1: Understand the Reaction
We have a metallic salt \( M(NO_3)_3 \) (where the metal \( M \) has an oxidation state of +3) reacting with sodium sulfite \( Na_2SO_3 \). In this reaction, the sulfite ion \( SO_3^{2-} \) is oxidized to sulfate ion \( SO_4^{2-} \).

### Step 2: Determine the Change in Oxidation State
In the sulfite ion \( SO_3^{2-} \), the oxidation state of sulfur is +4, and in the sulfate ion \( SO_4^{2-} \), the oxidation state of sulfur is +6. Therefore, the change in oxidation state for sulfur is:
\[ +6 - (+4) = +2 \]
This means that the sulfite ion loses 2 electrons during the oxidation process.

### Step 3: Calculate the Equivalent of Reactants
We need to calculate the milliequivalents of both the metallic salt and sodium sulfite. The formula for milliequivalents is:
\[ \text{Milliequivalents} = \text{Molarity} \times \text{Volume (mL)} \times \text{n factor} \]

For the metallic salt:
- Volume = 50 mL
- Molarity = 0.1 M
- n factor = \( 3 - x \) (where \( x \) is the new oxidation state of the metal)

For sodium sulfite:
- Volume = 25 mL
- Molarity = 0.1 M
- n factor = 2 (as calculated from the oxidation of \( SO_3^{2-} \) to \( SO_4^{2-} \))

### Step 4: Set Up the Equation
Since the milliequivalents of both reactants are equal, we can set up the equation:
\[ (50 \, \text{mL} \times 0.1 \, \text{M} \times (3 - x)) = (25 \, \text{mL} \times 0.1 \, \text{M} \times 2) \]

### Step 5: Simplify the Equation
Now, we can simplify the equation:
\[ 50 \times (3 - x) = 25 \times 2 \]
\[ 50(3 - x) = 50 \]
Dividing both sides by 50:
\[ 3 - x = 1 \]

### Step 6: Solve for x
Now, we can solve for \( x \):
\[ x = 3 - 1 = 2 \]

### Conclusion
The new oxidation number of the metal \( M \) after the reaction is +2.