`2.68xx10^(-3)` moles of solution containing anion `A^(n+)` require `1.61xx10^(-3)` moles of `MnO_(4)^(-)` for oxidation of `A^(n+)` to `AO_(3)^(-)` in acidic medium. What is the value of `n`?

Answer

To solve the problem, we need to determine the value of \( n \) in the oxidation reaction involving the anion \( A^{n+} \) and \( MnO_4^{-} \).

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction involves the oxidation of \( A^{n+} \) to \( AO_3^{-} \) in acidic medium. The half-reaction for \( MnO_4^{-} \) in acidic medium is:
\[ MnO_4^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_2O \]
Here, the change in oxidation state of manganese is from +7 to +2, which corresponds to a change of +5. Thus, the n-factor for \( MnO_4^{-} \) is 5.

2. **Determine the n-factor for \( A^{n+} \)**:
In the product \( AO_3^{-} \), the oxidation state of \( A \) is +5. Therefore, the change in oxidation state for \( A^{n+} \) can be expressed as:
\[ n + 5 \]
This is because \( A^{n+} \) is being oxidized from \( +n \) to \( +5 \).

3. **Set Up the Equation Using the Law of Equivalents**:
According to the law of equivalents, the number of equivalents of the oxidizing agent (MnO4-) must equal the number of equivalents of the reducing agent (A^n+):
\[ \text{Moles of } A^{n+} \times (n + 5) = \text{Moles of } MnO_4^{-} \times 5 \]
Plugging in the values:
\[ (2.68 \times 10^{-3}) \times (n + 5) = (1.61 \times 10^{-3}) \times 5 \]

4. **Simplify the Equation**:
First, calculate the right side:
\[ 1.61 \times 10^{-3} \times 5 = 8.05 \times 10^{-3} \]
Now, the equation becomes:
\[ 2.68 \times 10^{-3} \times (n + 5) = 8.05 \times 10^{-3} \]

5. **Isolate \( n + 5 \)**:
Divide both sides by \( 2.68 \times 10^{-3} \):
\[ n + 5 = \frac{8.05 \times 10^{-3}}{2.68 \times 10^{-3}} \approx 3 \]

6. **Solve for \( n \)**:
Subtract 5 from both sides:
\[ n = 3 - 5 = -2 \]

7. **Final Result**:
The value of \( n \) is \( 3 \).