How much volume of 0.40 M` Na_(2)S_(2)O_(3)` would be required to react with the `I_(2)` liberated by adding excess of KI of 50 mLof 0.20 M `M CuSO_(4)`

Options

A. 12.5 mL
B. 25 mL
C. 50 mL
D. 2.5 mL

Answer

To solve the problem, we need to determine the volume of 0.40 M Na₂S₂O₃ required to react with the iodine (I₂) liberated from the reaction of excess potassium iodide (KI) with 50 mL of 0.20 M CuSO₄.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The iodine (I₂) is produced when KI reacts with CuSO₄. The balanced chemical equation for the reaction is:
\[ 2KI + CuSO₄ \rightarrow CuI + K₂SO₄ + I₂ \]
From this equation, we see that 1 mole of CuSO₄ produces 1 mole of I₂.

2. **Calculate Moles of CuSO₄**:
Using the molarity and volume of CuSO₄, we can calculate the moles of CuSO₄ present:
\[ \text{Moles of CuSO₄} = \text{Molarity} \times \text{Volume} = 0.20 \, \text{M} \times 0.050 \, \text{L} = 0.010 \, \text{moles} \]

3. **Determine Moles of I₂ Produced**:
From the stoichiometry of the reaction, 1 mole of CuSO₄ produces 1 mole of I₂. Therefore, the moles of I₂ produced will also be 0.010 moles.

4. **Reaction Between Na₂S₂O₃ and I₂**:
The reaction between sodium thiosulfate (Na₂S₂O₃) and iodine (I₂) is given by:
\[ Na₂S₂O₃ + I₂ \rightarrow 2NaI + S \]
From this equation, we see that 1 mole of Na₂S₂O₃ reacts with 1 mole of I₂.

5. **Calculate Moles of Na₂S₂O₃ Required**:
Since 0.010 moles of I₂ are produced, we will need 0.010 moles of Na₂S₂O₃ to react with it.

6. **Calculate Volume of Na₂S₂O₃ Required**:
We can use the molarity of Na₂S₂O₃ to find the volume required:
\[ \text{Volume of Na₂S₂O₃} = \frac{\text{Moles of Na₂S₂O₃}}{\text{Molarity}} = \frac{0.010 \, \text{moles}}{0.40 \, \text{M}} = 0.025 \, \text{L} = 25 \, \text{mL} \]

### Final Answer:
The volume of 0.40 M Na₂S₂O₃ required to react with the iodine produced is **25 mL**.

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