What volume of 0.2 M `KMnO_(4)` is required to react with 1.58 g of hypo solution `(Na_(2)S_(2)O_(3))` in acidc medium?

Options

A. 20 mL
B. 10 mL
C. 16.6 mL
D. 50 mL

Explanation

To solve the problem of determining the volume of 0.2 M KMnO₄ required to react with 1.58 g of sodium thiosulfate (Na₂S₂O₃) in acidic medium, we can follow these steps:

### Step 1: Write the balanced chemical equations
In acidic medium, the reaction of KMnO₄ with sodium thiosulfate can be represented as follows:

1. **Oxidation half-reaction**:
\[ \text{S}_2\text{O}_3^{2-} \rightarrow \text{S}_4\text{O}_6^{2-} \]
Here, sulfur changes from +2 oxidation state in thiosulfate to +2.5 in tetrathionate.

2. **Reduction half-reaction**:
\[ \text{MnO}_4^{-} + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

### Step 2: Determine the n-factor for both reactants
- For KMnO₄, the change in oxidation state is from +7 (in MnO₄⁻) to +2 (in Mn²⁺), which corresponds to a change of 5 electrons. Thus, the n-factor (N₁) for KMnO₄ is 5.
- For Na₂S₂O₃, the change in oxidation state is from +2 to +2.5, which corresponds to a change of 0.5 electrons. Since there are 2 sulfur atoms, the total n-factor (N₂) for Na₂S₂O₃ is 1 (0.5 × 2).

### Step 3: Calculate moles of Na₂S₂O₃
To find the moles of sodium thiosulfate, we use the formula:
\[ \text{Moles} = \frac{\text{mass}}{\text{molar mass}} \]
Given:
- Mass of Na₂S₂O₃ = 1.58 g
- Molar mass of Na₂S₂O₃ = 158 g/mol

Calculating moles:
\[ \text{Moles of Na}_2\text{S}_2\text{O}_3 = \frac{1.58 \, \text{g}}{158 \, \text{g/mol}} = 0.01 \, \text{mol} \]

### Step 4: Use the stoichiometric relationship
Using the relationship:
\[ M_1 \cdot N_1 \cdot V_1 = \text{Moles of Na}_2\text{S}_2\text{O}_3 \cdot N_2 \]
Where:
- \(M_1\) = 0.2 M (concentration of KMnO₄)
- \(N_1\) = 5 (n-factor for KMnO₄)
- \(V_1\) = volume of KMnO₄ (in liters)
- Moles of Na₂S₂O₃ = 0.01 mol
- \(N_2\) = 1 (n-factor for Na₂S₂O₃)

Substituting the values:
\[ 0.2 \cdot 5 \cdot V_1 = 0.01 \cdot 1 \]
\[ 1.0 \cdot V_1 = 0.01 \]
\[ V_1 = \frac{0.01}{1.0} = 0.01 \, \text{L} = 10 \, \text{mL} \]

### Conclusion
The volume of 0.2 M KMnO₄ required to react with 1.58 g of sodium thiosulfate in acidic medium is **10 mL**.

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