The solubility product of AgCI & AgI are `1.1xx10^(-10)` and `1.6xx10^(-16)` respectively. If `AgNO_(3)` is added drop by drop to the solution containing both choride and iodide ions. The salt which will precipitate first ?

Options

A. AgI
B. `AgNO_(3)`
C. AgCI
D. both AgCI and AgI simultanneously

Answer

To determine which salt will precipitate first when silver nitrate (AgNO₃) is added to a solution containing both chloride (Cl⁻) and iodide (I⁻) ions, we will compare the solubility products (Ksp) of AgCl and AgI.

### Step-by-Step Solution:

1. **Identify the Solubility Products**:
- The solubility product of AgCl (Ksp) is given as \( 1.1 \times 10^{-10} \).
- The solubility product of AgI (Ksp) is given as \( 1.6 \times 10^{-16} \).

2. **Write the Dissociation Equations**:
- For AgCl:
\[ \text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \]
- For AgI:
\[ \text{AgI} \rightleftharpoons \text{Ag}^+ + \text{I}^- \]

3. **Express the Ksp in Terms of Solubility (S)**:
- For AgCl:
\[ Ksp = [\text{Ag}^+][\text{Cl}^-] = S \cdot S = S^2 \]
\[ S^2 = 1.1 \times 10^{-10} \implies S = \sqrt{1.1 \times 10^{-10}} \approx 1.048 \times 10^{-5} \, \text{mol/L} \]

- For AgI:
\[ Ksp = [\text{Ag}^+][\text{I}^-] = S \cdot S = S^2 \]
\[ S^2 = 1.6 \times 10^{-16} \implies S = \sqrt{1.6 \times 10^{-16}} \approx 4.0 \times 10^{-8} \, \text{mol/L} \]

4. **Compare the Solubility Values**:
- From the calculations:
- For AgCl, the solubility (S) is approximately \( 1.048 \times 10^{-5} \, \text{mol/L} \).
- For AgI, the solubility (S) is approximately \( 4.0 \times 10^{-8} \, \text{mol/L} \).

5. **Determine Which Salt Precipitates First**:
- Since AgI has a lower solubility product, it will require a lower concentration of Ag⁺ ions to start precipitating compared to AgCl.
- Therefore, AgI will precipitate first when AgNO₃ is added drop by drop.

### Final Answer:
The salt that will precipitate first is **AgI**.