`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 the acidic medium is

Options

A. 0.2 M
B. 0.02M
C. 0.4 M
D. 0.04 M

Explanation

To find the molarity of the `KMnO4` solution in acidic medium, we can follow these steps:

### Step 1: Understand the Reaction
In acidic medium, `KMnO4` acts as an oxidizing agent and gets reduced from `MnO4^-` (Mn in +7 oxidation state) to `Mn^2+` (Mn in +2 oxidation state). The reaction with sodium thiosulfate (`Na2S2O3`) involves the reduction of thiosulfate ions to tetrathionate ions (`S4O6^2-`).

### Step 2: Write the Balanced Reaction
The half-reaction for the reduction of `KMnO4` in acidic medium can be written as:
\[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

The half-reaction for the reduction of thiosulfate is:
\[ 2\text{S}_2\text{O}_3^{2-} \rightarrow \text{S}_4\text{O}_6^{2-} + 2\text{e}^- \]

### Step 3: Determine the Number of Equivalents
From the half-reactions, we can see that:
- The n-factor for `KMnO4` in acidic medium is 5 (it gains 5 electrons).
- The n-factor for `Na2S2O3` is 2 (it loses 2 electrons).

### Step 4: Set Up the Equation
According to the equivalence principle:
\[ N_1V_1 = N_2V_2 \]
Where:
- \( N_1 \) = Normality of `KMnO4`
- \( V_1 \) = Volume of `KMnO4` (100 mL)
- \( N_2 \) = Normality of `Na2S2O3` (0.1 M, which is also 0.1 N since its n-factor is 1)
- \( V_2 \) = Volume of `Na2S2O3` (100 mL)

### Step 5: Calculate the Normality of `KMnO4`
Since Normality (N) is related to Molarity (M) by the equation:
\[ N = M \times n \]
We can express the normality of `KMnO4` as:
\[ N_1 = M_1 \times 5 \]

Substituting into the equivalence equation:
\[ (M_1 \times 5) \times 100 = 0.1 \times 100 \]

### Step 6: Solve for Molarity of `KMnO4`
Now we can simplify:
\[ 5M_1 \times 100 = 0.1 \times 100 \]
\[ 5M_1 = 0.1 \]
\[ M_1 = \frac{0.1}{5} = 0.02 \, \text{M} \]

### Final Answer
The molarity of the `KMnO4` solution in acidic medium is **0.02 M**.

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