`KMnO_4` reacts with `Na_2S_2O_3` in acidic, strongly basic and aqueous (neutral) media. `100mL` of `LMnO_4` reacts with 100 mL of 0.1 M `Na_2S_2O_3` in acidic, basic and neutral media.
Q. The molarity (M) of `KMnO_4` solution in basic medium is:

Options

A. 0.8 M
B. 0.08 M
C. 0.26 M
D. 0.026M

Explanation

To find the molarity of \( KMnO_4 \) in basic medium, we can follow these steps:

### Step 1: Understand the Reaction
In the reaction between \( KMnO_4 \) and \( Na_2S_2O_3 \), we need to consider the equivalents of both reactants. The number of equivalents of \( KMnO_4 \) will be equal to the number of equivalents of \( Na_2S_2O_3 \).

### Step 2: Write the Equation for the Reaction
In basic medium, the half-reaction for \( KMnO_4 \) can be written as:
\[ MnO_4^- + 8e^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O \]
However, in basic medium, it reduces to \( Mn^{6+} \) instead of \( Mn^{2+} \), so we consider the change in oxidation state.

For \( Na_2S_2O_3 \):
\[ S_2O_3^{2-} \rightarrow 2SO_4^{2-} + 8e^- \]
This indicates that the oxidation state of sulfur changes from +2 to +6, resulting in a total change of +8.

### Step 3: Calculate the n-factor
- For \( KMnO_4 \) in basic medium, the n-factor (number of electrons transferred) is 1.
- For \( Na_2S_2O_3 \), the n-factor is 8 (since 2 sulfur atoms change from +2 to +6).

### Step 4: Set Up the Equation
Using the formula:
\[ n_1 v_1 = n_2 v_2 \]
Where:
- \( n_1 \) = n-factor of \( KMnO_4 \) = 1
- \( v_1 \) = volume of \( KMnO_4 \) = 100 mL = 0.1 L
- \( n_2 \) = n-factor of \( Na_2S_2O_3 \) = 8
- \( v_2 \) = volume of \( Na_2S_2O_3 \) = 100 mL = 0.1 L
- \( M_1 \) = molarity of \( KMnO_4 \) (which we need to find)
- \( M_2 \) = molarity of \( Na_2S_2O_3 \) = 0.1 M

### Step 5: Substitute Values into the Equation
Substituting the known values into the equation:
\[ M_1 \cdot 1 \cdot 0.1 = 0.1 \cdot 8 \cdot 0.1 \]
This simplifies to:
\[ M_1 \cdot 0.1 = 0.08 \]

### Step 6: Solve for Molarity
Now, solve for \( M_1 \):
\[ M_1 = \frac{0.08}{0.1} = 0.08 \text{ M} \]

### Conclusion
The molarity of \( KMnO_4 \) solution in basic medium is \( 0.08 \, M \).

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