25 ml of 0.17 M `HSO_3^–` in strongly acidic solution required the addition of 16.9 ml of 0.1 M `MnO_4^–` for its complete oxidation to `SO_4^(2 –)`. In neutral solution it requires 28.6 ml. Assign oxidation no. of 'Mn' in each of the products.

Answer

To solve the problem, we need to determine the oxidation states of manganese (Mn) in two different scenarios: in strongly acidic solution and in neutral solution. We will also calculate the number of moles of the reactants involved.

### Step-by-Step Solution:

1. **Identify the oxidation states of sulfur in the reactants and products:**
- For `HSO3^-` (bisulfite ion), the oxidation state of sulfur (S) can be calculated as follows:
\[ x + 3(-2) + 1 = -1 \quad \text{(where x is the oxidation state of S)} \]
\[ x - 6 + 1 = -1 \implies x - 5 = -1 \implies x = +4 \]
- For `SO4^{2-}` (sulfate ion), the oxidation state of sulfur is:
\[ y + 4(-2) = -2 \quad \text{(where y is the oxidation state of S)} \]
\[ y - 8 = -2 \implies y = +6 \]

2. **Determine the change in oxidation state:**
- The oxidation state of sulfur changes from +4 in `HSO3^-` to +6 in `SO4^{2-}`. This indicates that sulfur is oxidized, losing 2 electrons.

3. **Calculate the number of moles of `HSO3^-`:**
- Given the concentration and volume of `HSO3^-`:
\[ \text{Moles of } HSO3^- = \text{Molarity} \times \text{Volume} = 0.17 \, \text{mol/L} \times 0.025 \, \text{L} = 0.00425 \, \text{mol} \]

4. **Determine the number of moles of `MnO4^-` in acidic medium:**
- The reaction requires 16.9 mL of 0.1 M `MnO4^-`:
\[ \text{Moles of } MnO4^- = 0.1 \, \text{mol/L} \times 0.0169 \, \text{L} = 0.00169 \, \text{mol} \]

5. **Calculate the n-factor for `HSO3^-`:**
- Since 2 moles of electrons are lost per mole of `HSO3^-`, the n-factor is 2. Thus, the total number of equivalents of `HSO3^-` is:
\[ \text{Equivalents of } HSO3^- = \text{Moles} \times \text{n-factor} = 0.00425 \, \text{mol} \times 2 = 0.0085 \, \text{equivalents} \]

6. **Determine the number of equivalents of `MnO4^-`:**
- In acidic medium, 1 mole of `MnO4^-` reacts with 5 moles of electrons:
\[ \text{Equivalents of } MnO4^- = 0.00169 \, \text{mol} \times 5 = 0.00845 \, \text{equivalents} \]

7. **Calculate the number of moles of `MnO4^-` in neutral medium:**
- Given that it requires 28.6 mL of 0.1 M `MnO4^-`:
\[ \text{Moles of } MnO4^- = 0.1 \, \text{mol/L} \times 0.0286 \, \text{L} = 0.00286 \, \text{mol} \]
- The equivalents in neutral medium would be:
\[ \text{Equivalents of } MnO4^- = 0.00286 \, \text{mol} \times 5 = 0.0143 \, \text{equivalents} \]

8. **Determine the oxidation states of Mn in each medium:**
- In acidic medium, `MnO4^-` (Mn in +7 oxidation state) is reduced to `Mn^{2+}` (Mn in +2 oxidation state).
- In neutral medium, `MnO4^-` is reduced to `MnO2` (Mn in +4 oxidation state).

### Final Answer:
- The oxidation state of Mn in acidic medium is **+2**.
- The oxidation state of Mn in neutral medium is **+4**.