Free iodine is titrated against standard reducing agent usually with sodium thiosulphate , i.e.
`K_(2)Cr_(2)O_(7) + 6Kl + 7H_(2)SO_(4) to Cr_(2)(SO_(4))_(3) + 4K_(2) SO_(4) + 7H_(2)O+l_(2)`
` 2CuSO_(4) + 4Kl to Cu_(2)l_(2) + 2K_(2)SO_(4) + l_(2)`
` l_(2) + Na_(2)S_(2)O_(3) to 2Nal + Na_(2)S_(4)O_(6)`
In iodometric titration, starch solution is used as an indicator. Starch solution gives blue or violet colour with free iodine .At the end point , blue or violet colour disappears when iodine is completely changed to iodide.
What volume of `0.40 M Na_(2)S_(2)O_(3)` would be required to react with `l_(2)` liberated by adding `0.04` mole of Kl to 50 ml of `0.20 M CuSO_(4)` solution ?

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 sodium thiosulfate (Na₂S₂O₃) required to react with the iodine (I₂) liberated from the reaction of potassium iodide (KI) with copper(II) sulfate (CuSO₄).

### Step-by-Step Solution:

1. **Identify the reactions involved**:
- The first reaction is:
\[ 2 \text{CuSO}_4 + 4 \text{KI} \rightarrow \text{Cu}_2\text{I}_2 + 2 \text{K}_2\text{SO}_4 + \text{I}_2 \]
- The second reaction is:
\[ \text{I}_2 + \text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2 \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 \]

2. **Calculate the moles of CuSO₄**:
- The concentration of CuSO₄ is given as 0.20 M and the volume is 50 mL.
- Convert the volume from mL to L:
\[ 50 \text{ mL} = 0.050 \text{ L} \]
- Calculate the moles of CuSO₄:
\[ \text{Moles of CuSO}_4 = \text{Molarity} \times \text{Volume} = 0.20 \, \text{mol/L} \times 0.050 \, \text{L} = 0.010 \, \text{mol} \]

3. **Determine the moles of I₂ produced**:
- From the stoichiometry of the first reaction, 2 moles of CuSO₄ produce 1 mole of I₂.
- Therefore, the moles of I₂ produced from 0.010 moles of CuSO₄:
\[ \text{Moles of I}_2 = \frac{0.010 \, \text{mol CuSO}_4}{2} = 0.005 \, \text{mol I}_2 \]

4. **Relate the moles of I₂ to Na₂S₂O₃**:
- From the second reaction, 1 mole of I₂ reacts with 1 mole of Na₂S₂O₃.
- Therefore, the moles of Na₂S₂O₃ required is equal to the moles of I₂:
\[ \text{Moles of Na}_2\text{S}_2\text{O}_3 = 0.005 \, \text{mol} \]

5. **Calculate the volume of Na₂S₂O₃ required**:
- We know the molarity of Na₂S₂O₃ is 0.40 M. Using the formula:
\[ \text{Moles} = \text{Molarity} \times \text{Volume (L)} \]
- Rearranging gives us:
\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.005 \, \text{mol}}{0.40 \, \text{mol/L}} = 0.0125 \, \text{L} \]
- Convert the volume from L to mL:
\[ 0.0125 \, \text{L} = 12.5 \, \text{mL} \]

### Final Answer:
The volume of 0.40 M Na₂S₂O₃ required is **12.5 mL**.