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 AgNO₃ is added drop by drop to a solution containing both chloride (Cl⁻) and iodide (I⁻) ions, we need to compare the solubility products (Ksp) of AgCl and AgI.

### Step-by-Step Solution:

1. **Identify the Solubility Products:**
- The solubility product of AgCl (Ksp₁) = 1.1 × 10^(-10)
- The solubility product of AgI (Ksp₂) = 1.6 × 10^(-16)

2. **Write the Equilibrium Expressions:**
- For AgCl:
\[ Ksp_{AgCl} = [Ag^+][Cl^-] \]
- For AgI:
\[ Ksp_{AgI} = [Ag^+][I^-] \]

3. **Calculate the Concentration of Ag⁺ Required for Precipitation:**
- For AgCl:
\[ Ksp_{AgCl} = [Ag^+]^2 \]
\[ [Ag^+] = \sqrt{Ksp_{AgCl}} = \sqrt{1.1 \times 10^{-10}} \]
\[ [Ag^+] = 1.048 \times 10^{-5} \, \text{M} \]

- For AgI:
\[ Ksp_{AgI} = [Ag^+]^2 \]
\[ [Ag^+] = \sqrt{Ksp_{AgI}} = \sqrt{1.6 \times 10^{-16}} \]
\[ [Ag^+] = 1.265 \times 10^{-8} \, \text{M} \]

4. **Compare the Concentrations:**
- The concentration of Ag⁺ required to start precipitation of AgCl is approximately \( 1.048 \times 10^{-5} \, \text{M} \).
- The concentration of Ag⁺ required to start precipitation of AgI is approximately \( 1.265 \times 10^{-8} \, \text{M} \).

5. **Determine Which Precipitates First:**
- Since AgI requires a lower concentration of Ag⁺ to precipitate than AgCl, AgI will precipitate first when AgNO₃ is added to the solution.

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

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