If `K_(sp)` of `Ag_(2)CO_3` is `8xx10^(-12)`, the molar solubility of `Ag_2CO_3` in `0.1M AgNO_3` is:

Options

A. `8xx10^(-12)M`
B. `8xx10^(-11)M`
C. `8xx10^(-10)M`
D. `8xx10^(-13)M`

Explanation

To find the molar solubility of \( \text{Ag}_2\text{CO}_3 \) in a \( 0.1 \, M \) \( \text{AgNO}_3 \) solution, we will follow these steps:

### Step 1: Write the dissociation equation for \( \text{Ag}_2\text{CO}_3 \)

The dissociation of silver carbonate in water can be represented as:
\[ \text{Ag}_2\text{CO}_3 (s) \rightleftharpoons 2 \text{Ag}^+ (aq) + \text{CO}_3^{2-} (aq) \]

### Step 2: Define the solubility

Let the molar solubility of \( \text{Ag}_2\text{CO}_3 \) be \( S \). From the dissociation equation, we can see that:
- For every mole of \( \text{Ag}_2\text{CO}_3 \) that dissolves, \( 2S \) moles of \( \text{Ag}^+ \) ions and \( S \) moles of \( \text{CO}_3^{2-} \) ions are produced.

### Step 3: Consider the common ion effect

Since we are dissolving \( \text{Ag}_2\text{CO}_3 \) in a \( 0.1 \, M \) \( \text{AgNO}_3 \) solution, the concentration of \( \text{Ag}^+ \) ions from \( \text{AgNO}_3 \) will also contribute to the equilibrium. The dissociation of \( \text{AgNO}_3 \) can be represented as:
\[ \text{AgNO}_3 (s) \rightleftharpoons \text{Ag}^+ (aq) + \text{NO}_3^- (aq) \]
Thus, the concentration of \( \text{Ag}^+ \) ions in the solution will be:
\[ [\text{Ag}^+] = 0.1 + 2S \]

### Step 4: Write the expression for \( K_{sp} \)

The solubility product \( K_{sp} \) for \( \text{Ag}_2\text{CO}_3 \) is given by:
\[ K_{sp} = [\text{Ag}^+]^2 [\text{CO}_3^{2-}] \]
Substituting the values we have:
\[ K_{sp} = (0.1 + 2S)^2 \cdot S \]
Given \( K_{sp} = 8 \times 10^{-12} \).

### Step 5: Simplify the equation

Since \( S \) (the solubility of \( \text{Ag}_2\text{CO}_3 \)) is expected to be very small compared to \( 0.1 \), we can neglect \( 2S \) in the expression for \( [\text{Ag}^+] \):
\[ K_{sp} \approx (0.1)^2 \cdot S \]
Thus, we have:
\[ 8 \times 10^{-12} = (0.1)^2 \cdot S \]
\[ 8 \times 10^{-12} = 0.01 \cdot S \]

### Step 6: Solve for \( S \)

Now, we can solve for \( S \):
\[ S = \frac{8 \times 10^{-12}}{0.01} = 8 \times 10^{-10} \]

### Conclusion

The molar solubility of \( \text{Ag}_2\text{CO}_3 \) in \( 0.1 \, M \) \( \text{AgNO}_3 \) is:
\[ \boxed{8 \times 10^{-10} \, M} \]