If `K_(sp)` of `Ag_(2)CO_(3)` is `8 xx 10^(-12)`, the molar solubility of `Ag_(2)CO_(3)` in 0.1 M `AgNO_(3)` is `x xx 10^(-10) M`. The numerical value of x is ________.

Options

A. 8
B. 18
C. 10
D. 12

Explanation

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

### Step 1: Write the dissociation equation for \( \text{Ag}_2\text{CO}_3 \)
The dissociation of silver carbonate 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 solubility of \( \text{Ag}_2\text{CO}_3 \) in moles per liter be \( s \). From the dissociation equation, we can see that:
- For every mole of \( \text{Ag}_2\text{CO}_3 \) that dissolves, it produces \( 2s \) moles of \( \text{Ag}^+ \) and \( s \) moles of \( \text{CO}_3^{2-} \).

### Step 3: Consider the presence of \( \text{Ag}^+ \) from \( \text{AgNO}_3 \)
Since we have a 0.1 M solution of \( \text{AgNO}_3 \), it dissociates completely to give:
\[ \text{AgNO}_3 (aq) \rightleftharpoons \text{Ag}^+ (aq) + \text{NO}_3^- (aq) \]
Thus, the concentration of \( \text{Ag}^+ \) ions from \( \text{AgNO}_3 \) is 0.1 M.

### Step 4: Write the expression for the solubility product \( K_{sp} \)
The solubility product \( K_{sp} \) for \( \text{Ag}_2\text{CO}_3 \) can be expressed as:
\[ K_{sp} = [\text{Ag}^+]^2 [\text{CO}_3^{2-}] \]
Substituting the concentrations:
\[ K_{sp} = (2s + 0.1)^2 \cdot s \]
Given that \( K_{sp} = 8 \times 10^{-12} \), we can set up the equation:
\[ 8 \times 10^{-12} = (2s + 0.1)^2 \cdot s \]

### Step 5: Expand and simplify the equation
Expanding \( (2s + 0.1)^2 \):
\[ (2s + 0.1)^2 = 4s^2 + 0.04 + 0.4s \]
Thus, we have:
\[ 8 \times 10^{-12} = (4s^2 + 0.04 + 0.4s) \cdot s \]
This leads to:
\[ 8 \times 10^{-12} = 4s^3 + 0.04s + 0.4s^2 \]

### Step 6: Rearranging the equation
Rearranging gives us:
\[ 4s^3 + 0.4s^2 + 0.04s - 8 \times 10^{-12} = 0 \]

### Step 7: Solve for \( s \)
To find \( s \), we can use numerical methods or approximations. Given the small value of \( K_{sp} \), we can assume \( s \) is small compared to 0.1 M. Thus, we can neglect \( s \) in comparison to 0.1 M in the \( (2s + 0.1)^2 \) term:
\[ (0.1)^2 \cdot s \approx 8 \times 10^{-12} \]
This simplifies to:
\[ 0.01s = 8 \times 10^{-12} \implies s = \frac{8 \times 10^{-12}}{0.01} = 8 \times 10^{-10} \text{ M} \]

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