What is the molar solubility of `Ag_(2)CO_(3)(K_(sp)=4xx10^(-13))` in `0.1M Na_(2)CO_(3)` solution ?

Options

A. `10^(-6)`
B. `10^(-7)`
C. `2xx10^(-6)`
D. `2xx10^(-7)`

Explanation

To find the molar solubility of `Ag2CO3` in a `0.1 M Na2CO3` solution, we will follow these steps:

### Step 1: Write the dissociation equation for Ag2CO3
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: Set up the expression for Ksp
The solubility product constant (Ksp) for this reaction is given by:
\[ K_{sp} = [\text{Ag}^+]^2 [\text{CO}_3^{2-}] \]

### Step 3: Define the solubility in terms of x
Let the molar solubility of `Ag2CO3` in the solution be \( x \). From the dissociation equation, we can express the concentrations at equilibrium as:
- \([\text{Ag}^+] = 2x\)
- \([\text{CO}_3^{2-}] = x\)

### Step 4: Consider the common ion effect
Since we have `0.1 M Na2CO3`, which provides `0.1 M CO3^{2-}`, we need to account for this in our Ksp expression. Thus, the concentration of carbonate ions will be:
\[ [\text{CO}_3^{2-}] = 0.1 + x \approx 0.1 \] (since \( x \) is expected to be small)

### Step 5: Substitute into the Ksp expression
Now substituting into the Ksp expression:
\[ K_{sp} = (2x)^2 (0.1) \]
Given that \( K_{sp} = 4 \times 10^{-13} \), we have:
\[ 4 \times 10^{-13} = 4x^2 \times 0.1 \]

### Step 6: Solve for x
Rearranging the equation gives:
\[ 4 \times 10^{-13} = 0.4x^2 \]
\[ x^2 = \frac{4 \times 10^{-13}}{0.4} \]
\[ x^2 = 1 \times 10^{-12} \]
Taking the square root:
\[ x = 1 \times 10^{-6} \, \text{mol/L} \]

### Conclusion
The molar solubility of `Ag2CO3` in `0.1 M Na2CO3` solution is:
\[ \text{Molar solubility} = 1 \times 10^{-6} \, \text{mol/L} \]

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