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 order to prepare one litre of normal solution of `KMnO_(4)`, how many grams of `KMnO_(4)` is required if solution is to be used in acid medium for oxidation ?

Options

A. `158` gm
B. `31.6` gm
C. `62` gm
D. `790` gm

Answer

To determine how many grams of potassium permanganate (KMnO₄) are required to prepare one liter of a normal solution in acidic medium, we can follow these steps:

### Step 1: Understand the Reaction
In acidic medium, the oxidation state of manganese in potassium permanganate changes from +7 to +2. The half-reaction can be represented as:
\[ \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]
This indicates that one mole of KMnO₄ can accept 5 moles of electrons.

### Step 2: Determine the n-factor
The n-factor for KMnO₄ in this reaction is 5, as it corresponds to the number of electrons transferred per mole of KMnO₄.

### Step 3: Calculate Normality and Molarity
Given that we want to prepare a 1 Normal (N) solution of KMnO₄:
- Normality (N) = Molarity (M) × n-factor
- Since the normality is 1, we can set up the equation:
\[ 1 = M \times 5 \]
From this, we can find the molarity:
\[ M = \frac{1}{5} = 0.2 \text{ M} \]

### Step 4: Calculate the Number of Moles
For a 1-liter solution, the number of moles of KMnO₄ required is:
\[ \text{Moles} = \text{Molarity} \times \text{Volume} = 0.2 \text{ moles/L} \times 1 \text{ L} = 0.2 \text{ moles} \]

### Step 5: Calculate the Mass of KMnO₄
To find the mass of KMnO₄ needed, we use the molar mass of KMnO₄, which is approximately 158 g/mol:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.2 \text{ moles} \times 158 \text{ g/mol} = 31.6 \text{ grams} \]

### Conclusion
To prepare one liter of a normal solution of KMnO₄ in acidic medium, you will need **31.6 grams** of KMnO₄.

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