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 need to calculate the concentration of chloride ions \([Cl^-]\) in a solution containing both silver carbonate \((Ag_2CO_3)\) and silver chloride \((AgCl)\) in equilibrium with a carbonate ion concentration of \(1.00 \, M\).

### Step-by-Step Solution:

1. **Dissociation of \(Ag_2CO_3\)**:
The dissociation of silver carbonate can be represented as:
\[ Ag_2CO_3 (s) \rightleftharpoons 2Ag^+ (aq) + CO_3^{2-} (aq) \]
The solubility product constant \(K_{sp}\) for \(Ag_2CO_3\) is given by:
\[ K_{sp} = [Ag^+]^2 [CO_3^{2-}] \]
Given \(K_{sp} = 6.2 \times 10^{-12}\) and \([CO_3^{2-}] = 1.00 \, M\), we can substitute into the equation:
\[ 6.2 \times 10^{-12} = [Ag^+]^2 \times 1.00 \]
Thus,
\[ [Ag^+]^2 = 6.2 \times 10^{-12} \]
Taking the square root:
\[ [Ag^+] = \sqrt{6.2 \times 10^{-12}} \approx 2.49 \times 10^{-6} \, M \]

2. **Dissociation of \(AgCl\)**:
The dissociation of silver chloride can be represented as:
\[ AgCl (s) \rightleftharpoons Ag^+ (aq) + Cl^- (aq) \]
The solubility product constant \(K_{sp}\) for \(AgCl\) is given by:
\[ K_{sp} = [Ag^+][Cl^-] \]
Given \(K_{sp} = 2.8 \times 10^{-10}\), we can substitute \([Ag^+]\) from the previous step:
\[ 2.8 \times 10^{-10} = (2.49 \times 10^{-6})[Cl^-] \]
Solving for \([Cl^-]\):
\[ [Cl^-] = \frac{2.8 \times 10^{-10}}{2.49 \times 10^{-6}} \approx 1.12 \times 10^{-4} \, M \]

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