Direct titration of `I_(2)` with a reducing agent is called iodimetry. If `I_(2)` is leberated by the oxidation of `I_(ɵ)` ion by a strong oxidising agent in neutral or acidic medium, the liberated `I_(2)` is then titrated with reducing agent. Iodometry is used to estimate the strngth of the oxidising agent. For example, in the estimation of `Cu^(2+)` with `S_(2)O_(3)^(2-)`
`Cu^(2+)+I^(ɵ)toCuI_(2)+I_(2)` (iodometry)
`I_(2)+S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+I^(ɵ)` (iodimetry)
Strach is used as an indicator at the end point, which forms bluecoloured complex with `I_(3)^(ɵ)` Disappearance of blue colourindicates the end point whe free `I_(2)` in not present.
Q. The volume of KI solution used for `CuSO_(4)` is:

Options

A. 1L
B. 2L
C. 4L
D. 5L

Explanation

To solve the question regarding the volume of KI solution used for the titration of CuSO₄, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The first reaction we need to consider is the oxidation of iodide ions (I⁻) by copper(II) ions (Cu²⁺):
\[ \text{Cu}^{2+} + 2 \text{I}^- \rightarrow \text{CuI}_2 + \text{I}_2 \]
This reaction shows that 1 mole of Cu²⁺ reacts with 2 moles of I⁻ to produce 1 mole of I₂.

2. **Determine Moles of CuSO₄**:
Let's assume we have 1 mole of CuSO₄. According to the stoichiometry of the reaction:
\[ 1 \text{ mole of CuSO}_4 \text{ reacts with } 2 \text{ moles of I}^- \]

3. **Relate Iodine to Potassium Iodide**:
From the reaction, we know that 2 moles of I⁻ are provided by 2 moles of KI (since KI dissociates into K⁺ and I⁻). Therefore:
\[ 2 \text{ moles of I}^- \text{ from } 2 \text{ moles of KI} \]

4. **Calculate Moles of KI Required**:
Since 1 mole of Cu²⁺ requires 2 moles of I⁻, we can conclude that:
\[ 1 \text{ mole of CuSO}_4 \text{ requires } 2 \text{ moles of KI} \]

5. **Determine the Concentration of KI Solution**:
If we are given a concentration of KI solution, for example, 0.2 M (molar), we can calculate the volume needed to provide the required moles of KI.

6. **Calculate Volume of KI Solution**:
Using the formula:
\[ \text{Volume (L)} = \frac{\text{Moles of KI}}{\text{Concentration (mol/L)}} \]
If we need 2 moles of KI and the concentration of KI is 0.2 M:
\[ \text{Volume} = \frac{2 \text{ moles}}{0.2 \text{ mol/L}} = 10 \text{ L} \]

7. **Final Result**:
Thus, the volume of KI solution used for 1 mole of CuSO₄ is 10 liters.