When `HgI_(2)` is added in KI solution. The freezing point of solution:-

Options

A. increases
B. decreases
C. Remains unchanged
D. Can't predict

Answer

To solve the problem of how the addition of HgI₂ affects the freezing point of a KI solution, we can follow these steps:

### Step 1: Understand the Reaction
When HgI₂ is added to KI solution, a reaction occurs. The reaction can be represented as:
\[ 2 \text{KI} + \text{HgI}_2 \rightarrow \text{K}_2\text{HgI}_4 \]
This shows that HgI₂ reacts with KI to form a complex, K₂HgI₄.

**Hint:** Consider how the formation of a complex affects the number of particles in solution.

### Step 2: Analyze Ion Concentration
In the KI solution, KI dissociates into K⁺ and I⁻ ions. When HgI₂ is added, it reacts with KI to form K₂HgI₄, which has fewer ions in solution compared to the original KI. The complex formation reduces the total number of free ions in the solution.

**Hint:** Think about how the number of ions affects colligative properties like freezing point depression.

### Step 3: Understand Freezing Point Depression
The freezing point of a solution is affected by the number of solute particles present. The more particles there are, the lower the freezing point (freezing point depression). Conversely, if the number of particles decreases, the freezing point will increase.

**Hint:** Recall the colligative properties and how they relate to the number of solute particles.

### Step 4: Conclusion
Since the addition of HgI₂ to KI results in the formation of a complex with fewer ions, the freezing point of the solution will increase.

**Final Answer:** The freezing point of the solution will increase.

### Summary of Steps:
1. Identify the reaction between HgI₂ and KI.
2. Analyze how this reaction affects the number of ions in solution.
3. Relate the number of ions to the concept of freezing point depression.
4. Conclude that the freezing point will increase due to the formation of a complex.