In permanganate titrations, potassium pemanganats is used an an oxidizing agent in acidic medium.
The medium is maintained acidic by the use of dilute sulphuric acid. Potassium permanganate acts as self indicator . The potential equation, when potassium permanganate acts as an oxidizing agent is
`2KMnO_(4) + 3H_(2)SO_(4) to K_(2)SO_(4) + 2MnSO_(4) + 3H_(2)O + 5[O]`
or `MnO_(4)^(-) + 8H^(+) + 5e^(-) to Mn^(2+) + 4H_(2)O`
Before the end point, the solution remains colourless but after teh equivalence point only one extra drop of `KMnO_(4)` solution imparts pink colour , i.e. appearance of pink colour indicates end point. These titrations are used for estimation of ferrous salts, oxalic acid, oxalates, hydrogen peroxide, `As_(2)O_(3)` etc.
In alkaline condition, `KMnO_(4)` reacts as follows :
`2KMnO_(4) + 2KOH to 2K_(2)MnO_(4) + H_(2)O + [O]`
Therefore , its equivalent mass will be

Options

A. `31.6`
B. `52.7`
C. `72.0`
D. `158.0`

Answer

To find the equivalent mass of potassium permanganate (KMnO₄) when it acts as an oxidizing agent in alkaline conditions, we can follow these steps:

### Step 1: Write down the reaction
The reaction of potassium permanganate in alkaline medium is given as:
\[ 2KMnO_4 + 2KOH \rightarrow 2K_2MnO_4 + H_2O + [O] \]

### Step 2: Determine the change in oxidation state
In KMnO₄, manganese (Mn) has an oxidation state of +7. In the product \( K_2MnO_4 \), the oxidation state of manganese is +6.

- Change in oxidation state = Final oxidation state - Initial oxidation state
- Change in oxidation state = +6 - (+7) = -1

### Step 3: Calculate the equivalent weight
The equivalent weight (E) can be calculated using the formula:
\[ E = \frac{\text{Molar Mass}}{\text{Change in Oxidation State}} \]

The molar mass of KMnO₄ is 158 g/mol.

Substituting the values:
\[ E = \frac{158 \text{ g/mol}}{1} = 158 \text{ g/equiv} \]

### Step 4: Conclusion
The equivalent mass of potassium permanganate (KMnO₄) when it acts as an oxidizing agent in alkaline conditions is 158 g/equiv.

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