A solution containing a mixture of `0.05M NaCI` and `0.05M` Nal is taken. `(K_(sp) of AgCI = 10^(-10)`and `K_(sp)`of `AgI = 4 xx 10^(-16))`. When `AgNO_(3)` is added to such a solution:

Options

A. The concentartion fo `Ag^(o+)` required to precipitate `CI^(Theta)` is `2 xx 10^(-9) mol L^(-1)`.
B. The concentartion of `Ag^(o+)` required to precipitate `I^(Theta)` is `8 xx 10^(-15) mol L^(-1)`.
C. `AgCI` and `AgI` will be precipitate togther.
D. First `AgI` will be precipitated.

Explanation

To solve the problem step by step, we will analyze the conditions under which silver chloride (AgCl) and silver iodide (AgI) will precipitate when silver nitrate (AgNO₃) is added to a solution containing sodium chloride (NaCl) and sodium iodide (NaI).

### Step 1: Determine the concentration of Cl⁻ ions
Given:
- The concentration of NaCl = 0.05 M

Since NaCl dissociates completely in solution:
\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]

The concentration of Cl⁻ ions is also 0.05 M.

### Step 2: Calculate the concentration of Ag⁺ required to precipitate AgCl
The solubility product constant (Ksp) for AgCl is given as:
\[ K_{sp}(\text{AgCl}) = 10^{-10} \]

The expression for Ksp is:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]

Substituting the known values:
\[ 10^{-10} = [\text{Ag}^+][0.05] \]

To find the concentration of Ag⁺ required to start precipitation:
\[ [\text{Ag}^+] = \frac{10^{-10}}{0.05} = 2 \times 10^{-9} \, \text{M} \]

### Step 3: Determine the concentration of I⁻ ions
Given:
- The concentration of NaI = 0.05 M

Since NaI also dissociates completely:
\[ \text{NaI} \rightarrow \text{Na}^+ + \text{I}^- \]

The concentration of I⁻ ions is also 0.05 M.

### Step 4: Calculate the concentration of Ag⁺ required to precipitate AgI
The solubility product constant (Ksp) for AgI is given as:
\[ K_{sp}(\text{AgI}) = 4 \times 10^{-16} \]

The expression for Ksp is:
\[ K_{sp} = [\text{Ag}^+][\text{I}^-] \]

Substituting the known values:
\[ 4 \times 10^{-16} = [\text{Ag}^+][0.05] \]

To find the concentration of Ag⁺ required to start precipitation:
\[ [\text{Ag}^+] = \frac{4 \times 10^{-16}}{0.05} = 8 \times 10^{-15} \, \text{M} \]

### Step 5: Determine which salt precipitates first
Since the concentration of Ag⁺ required to precipitate AgI (8 × 10⁻¹⁵ M) is much lower than that required to precipitate AgCl (2 × 10⁻⁹ M), AgI will precipitate first when AgNO₃ is added to the solution.

### Conclusion
1. The concentration of Ag⁺ required to precipitate Cl⁻ (AgCl) is \(2 \times 10^{-9} \, \text{M}\).
2. The concentration of Ag⁺ required to precipitate I⁻ (AgI) is \(8 \times 10^{-15} \, \text{M}\).
3. AgI will precipitate first, and AgCl will not precipitate until the concentration of Ag⁺ exceeds \(2 \times 10^{-9} \, \text{M}\).