Volumetric titrations involving `KMnO_(4)`, are carried out only in presence of dilute `H_(2)SO_(4)` but not in the presence of HCI or `HNO_(3)` This is because oxygen produced from `KMnO_(4)` + dil: `H_(2)SO_(4)`, is used only for oxidizing the reducing agent. Moreover, H2SO4 does not give any oxygen of its own to oxidize the reducing agent. In case HCI is used, the oxygen produced from `KMnO_(4) + HCI` is partly used up to oxidize HCI and in case `HNO_(3)` is used, it itself acts as oxidizing agent and partly oxidizes the reducing agent. `KMnO_(4)`, in various mediums gives following products`Mn^(7+)overset(H^(+))toMn^(2+)`
`Mn^(7+)overset(H_(2)O)toMn^(4+)`
`Mn^(7+)underset(OH^(-))toMn^(6+)`
QThe number of moles of `KMnO_(4)`that will be needed to react with one mole of sulphite in an acidic solution is

Options

A. `(2)/(5)`
B. `(3)/(5)`
C. `(4)/(5)`
D. 1

Answer

To solve the problem of how many moles of KMnO4 are needed to react with one mole of sulfite (SO3^2-) in an acidic solution, we can follow these steps:

### Step 1: Identify the oxidation states
- In KMnO4, manganese (Mn) has an oxidation state of +7.
- In sulfite (SO3^2-), sulfur (S) has an oxidation state of +4.

### Step 2: Determine the changes in oxidation states
- When KMnO4 is reduced, Mn changes from +7 to +2. This is a decrease of 5 in oxidation state (7 - 2 = 5).
- When sulfite (SO3^2-) is oxidized to sulfate (SO4^2-), sulfur changes from +4 to +6. This is an increase of 2 in oxidation state (6 - 4 = 2).

### Step 3: Calculate the n-factor for each species
- The n-factor for KMnO4 (Mn) is the change in oxidation state multiplied by the number of moles of Mn involved, which is:
\[ n_{\text{KMnO4}} = 5 \times 1 = 5 \]
- The n-factor for sulfite (SO3^2-) is:
\[ n_{\text{SO3^{2-}}} = 2 \times 1 = 2 \]

### Step 4: Use the law of equivalence
According to the law of equivalence:
\[ \text{Equivalence of KMnO4} = \text{Equivalence of SO3^{2-}} \]
Where equivalence is defined as:
\[ \text{Equivalence} = \text{moles} \times \text{n-factor} \]

### Step 5: Set up the equation
Let \( x \) be the number of moles of KMnO4 needed to react with 1 mole of sulfite:
\[ x \times n_{\text{KMnO4}} = 1 \times n_{\text{SO3^{2-}}} \]
Substituting the n-factors:
\[ x \times 5 = 1 \times 2 \]

### Step 6: Solve for \( x \)
\[ x = \frac{2}{5} \]

### Conclusion
The number of moles of KMnO4 needed to react with one mole of sulfite in an acidic solution is \( \frac{2}{5} \).

---