For `Ag_(2)CO_(3), K_(sp)=6.2xx10^(-12)`. For `AgCl, K_(sp)=2.8xx10^(-10)`. Solid `Ag_(2)CO_(3)` and solid `AgCl` are added to a beaker containing `Na_(2)CO_(3)(aq)`. Under these conditions the `[CO_(3)^(2-)]=1.00M`. Calculate the `[Cl^(-)]` in solution when equilibrium is established.

Options

A. `1.1xx10^(-4)`
B. `1.26xx10^(-8)`
C. 0.15
D. `2.8xx10^(-6)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the dissociation equations and Ksp expressions
For silver carbonate (Ag₂CO₃):
\[ \text{Ag}_2\text{CO}_3 (s) \rightleftharpoons 2\text{Ag}^+ (aq) + \text{CO}_3^{2-} (aq) \]
The solubility product constant (Ksp) expression is:
\[ K_{sp} = [\text{Ag}^+]^2 [\text{CO}_3^{2-}] \]

For silver chloride (AgCl):
\[ \text{AgCl} (s) \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]
The Ksp expression is:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]

### Step 2: Substitute known values into the Ksp expression for Ag₂CO₃
Given:
- \( K_{sp} \) for Ag₂CO₃ = \( 6.2 \times 10^{-12} \)
- \([CO_3^{2-}] = 1.00 \, M\)

Using the Ksp expression:
\[ 6.2 \times 10^{-12} = [\text{Ag}^+]^2 \cdot 1.00 \]
This simplifies to:
\[ [\text{Ag}^+]^2 = 6.2 \times 10^{-12} \]

### Step 3: Calculate \([\text{Ag}^+]\)
Taking the square root:
\[ [\text{Ag}^+] = \sqrt{6.2 \times 10^{-12}} \approx 2.49 \times 10^{-6} \, M \]

### Step 4: Substitute \([\text{Ag}^+]\) into the Ksp expression for AgCl
Given:
- \( K_{sp} \) for AgCl = \( 2.8 \times 10^{-10} \)

Using the Ksp expression:
\[ 2.8 \times 10^{-10} = [\text{Ag}^+] \cdot [\text{Cl}^-] \]
Substituting \([\text{Ag}^+]\):
\[ 2.8 \times 10^{-10} = (2.49 \times 10^{-6}) \cdot [\text{Cl}^-] \]

### Step 5: Solve for \([\text{Cl}^-]\)
Rearranging gives:
\[ [\text{Cl}^-] = \frac{2.8 \times 10^{-10}}{2.49 \times 10^{-6}} \approx 1.12 \times 10^{-4} \, M \]

### Conclusion
The concentration of chloride ions \([\text{Cl}^-]\) in solution when equilibrium is established is approximately:
\[ [\text{Cl}^-] \approx 1.12 \times 10^{-4} \, M \]