50 mL of 0.1 M solution of metallic salt (A) oxidised 25 mL of 0.1 M sodium sulphite. If oxidation number of the metal in the salt (A) is 3, then new oxidation number of the metal is __________

Answer

To solve the problem step by step, let's analyze the information given and apply stoichiometry principles.

### Step 1: Understand the Reaction
We know that the metallic salt (A) is oxidizing sodium sulphite (Na2SO3). The oxidation reaction can be represented as:
\[ \text{SO}_3^{2-} + \text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 2\text{H}^+ + 2e^- \]
This indicates that sodium sulphite is losing electrons (being oxidized) and is involved in a redox reaction.

### Step 2: Calculate the Number of Equivalents of Sodium Sulphite
Given:
- Volume of sodium sulphite solution = 25 mL = 0.025 L
- Molarity of sodium sulphite = 0.1 M

The number of moles of sodium sulphite is:
\[ \text{Moles of Na}_2\text{SO}_3 = \text{Volume (L)} \times \text{Molarity (mol/L)} = 0.025 \, \text{L} \times 0.1 \, \text{mol/L} = 0.0025 \, \text{mol} \]

Since the n-factor for sodium sulphite (as it gives 2 electrons) is 2, the number of equivalents of sodium sulphite is:
\[ \text{Equivalents of Na}_2\text{SO}_3 = \text{Moles} \times \text{n-factor} = 0.0025 \, \text{mol} \times 2 = 0.005 \, \text{equivalents} \]

### Step 3: Set Up the Equation for the Metallic Salt
Let the n-factor of the metallic salt (A) be \( N \). The volume of the metallic salt solution is 50 mL = 0.050 L, and its molarity is 0.1 M.

The number of moles of the metallic salt is:
\[ \text{Moles of salt A} = 0.050 \, \text{L} \times 0.1 \, \text{mol/L} = 0.005 \, \text{mol} \]

The number of equivalents of salt A is:
\[ \text{Equivalents of salt A} = \text{Moles} \times N = 0.005 \, \text{mol} \times N \]

### Step 4: Equate the Number of Equivalents
Since the number of equivalents of salt A must equal the number of equivalents of sodium sulphite:
\[ 0.005 \, N = 0.005 \]
From this, we can solve for \( N \):
\[ N = 1 \]

### Step 5: Determine the New Oxidation Number
The initial oxidation state of the metal in salt A is given as +3. The new oxidation number after the reaction can be calculated as:
\[ \text{New oxidation number} = \text{Initial oxidation number} - N = 3 - 1 = 2 \]

Thus, the new oxidation number of the metal in salt A is **2**.

### Final Answer:
The new oxidation number of the metal is **2**.

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