As Kl is added, the equilibrium is shifted towards right giving more Agl precipitate, because

Options

A. Both AgCl and Agl are sparingly soluble
B. The `K_(sp)` of Agl is lower than `K_(sp)` of AgCl
C. The `K_(sp) ` of Ag I is higher than `K_(sp) ` of AgCl
D. Both AgCl and Agl have same soBubility product

Answer

To solve the question regarding the addition of KI (potassium iodide) and its effect on the equilibrium involving silver chloride (AgCl) and silver iodide (AgI), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**:
The equilibrium reaction can be represented as follows:
\[ \text{AgCl (s)} + \text{KI (aq)} \rightleftharpoons \text{KCl (aq)} + \text{AgI (s)} \]
Here, AgCl is a solid precipitate, and KI is a soluble salt that dissociates into K⁺ and I⁻ ions in solution.

2. **Identify the Effect of Adding KI**:
When KI is added to the system, it increases the concentration of I⁻ ions in the solution. According to Le Chatelier's principle, if a change is made to a system at equilibrium, the system will adjust to counteract that change.

3. **Shift in Equilibrium**:
The increase in I⁻ ions will shift the equilibrium to the right to reduce the concentration of I⁻ ions. This shift favors the formation of AgI, which is less soluble than AgCl.

4. **Formation of AgI Precipitate**:
As the equilibrium shifts to the right, more AgI precipitate is formed. This occurs because AgI has a lower solubility product constant (Ksp) compared to AgCl, meaning that it is less soluble in water.

5. **Conclusion**:
Therefore, the addition of KI results in the formation of more AgI precipitate due to the shift in equilibrium towards the right, driven by the increased concentration of I⁻ ions.

### Final Answer:
As KI is added, the equilibrium shifts towards the right, resulting in the formation of more AgI precipitate because AgI is less soluble than AgCl, leading to a decrease in the concentration of I⁻ ions.

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