In iodometric titrations, an oxidizing agent such as `KNnO_(4), K_(2)Cr_(2)O_(7),CuSO_(4),H_(2)O_(2)` is allowed to react in neutral medium or in acidic medium with excess of potassium iodide to liberate free iodine `Kl+` oxidizon agent `to l_(2)`
Free iodine is titrated against stanard reducing agent usually with sodium thiosulphate i.e.,
`K_(2)Cr_(2)O_(7)+6Kl+7H_(2)SO_(4)toCr_(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 titrations, 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 disappear when iodine is completely changed to iodide.
What volume of `0.40M Na_(2)S_(2)O_(3)` would be required to reach with `l_(2)` liberated by adding `0.04` mole of `Kl` to `50 mL` to `0.20 M CuSO_(4)` solution?

Options

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

Explanation

To solve the problem step by step, we will follow the stoichiometry of the reactions involved in the iodometric titration.

### Step-by-step Solution:

1. **Identify the Reaction**:
The reaction between copper sulfate (CuSO₄) and potassium iodide (KI) produces iodine (I₂). The relevant reaction is:
\[ 2 \text{CuSO}_4 + 4 \text{KI} \rightarrow \text{Cu}_2\text{I}_2 + 2 \text{K}_2\text{SO}_4 + \text{I}_2 \]

2. **Calculate Moles of CuSO₄**:
Given that the concentration of CuSO₄ is 0.20 M and the volume is 50 mL, we can calculate the moles of CuSO₄:
\[ \text{Moles of CuSO}_4 = \text{Molarity} \times \text{Volume (in L)} = 0.20 \, \text{mol/L} \times 0.050 \, \text{L} = 0.01 \, \text{mol} \]

3. **Determine the Moles of KI**:
We are given that 0.04 moles of KI are added.

4. **Calculate the Equivalent Moles of I₂ Produced**:
From the stoichiometry of the reaction, 4 moles of KI produce 1 mole of I₂. Therefore, the moles of I₂ produced from 0.04 moles of KI can be calculated as follows:
\[ \text{Moles of I}_2 = \frac{0.04 \, \text{mol KI}}{4} = 0.01 \, \text{mol I}_2 \]

5. **Identify the Limiting Reagent**:
The moles of CuSO₄ available (0.01 mol) will react with the moles of KI (0.04 mol) to produce I₂. Since both CuSO₄ and I₂ produced are in a 1:1 ratio, and we have 0.01 mol of CuSO₄, it will limit the reaction.

6. **Calculate the Moles of Na₂S₂O₃ Required**:
The reaction between I₂ and sodium thiosulfate (Na₂S₂O₃) is:
\[ \text{I}_2 + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2 \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 \]
From this reaction, 1 mole of I₂ reacts with 2 moles of Na₂S₂O₃. Therefore, for 0.01 mol of I₂, the moles of Na₂S₂O₃ required will be:
\[ \text{Moles of Na}_2\text{S}_2\text{O}_3 = 0.01 \, \text{mol I}_2 \times 2 = 0.02 \, \text{mol} \]

7. **Calculate the Volume of Na₂S₂O₃ Solution Required**:
The concentration of Na₂S₂O₃ is 0.40 M. To find the volume required to provide 0.02 moles, we use the formula:
\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.02 \, \text{mol}}{0.40 \, \text{mol/L}} = 0.05 \, \text{L} = 50 \, \text{mL} \]

### Final Answer:
The volume of 0.40 M Na₂S₂O₃ required to react with the iodine liberated is **50 mL**.