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)`

Answer

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:
\[ Ag_2CO_3 (s) \rightleftharpoons 2Ag^+ (aq) + CO_3^{2-} (aq) \]

### Step 2: Set up the expression for Ksp
The solubility product constant (Ksp) expression for this equilibrium is given by:
\[ K_{sp} = [Ag^+]^2 [CO_3^{2-}] \]
Given that \( K_{sp} = 4 \times 10^{-13} \).

### Step 3: Define the molar solubility
Let the molar solubility of `Ag2CO3` in the solution be \( x \). Therefore, at equilibrium:
- The concentration of \( Ag^+ \) will be \( 2x \) (since 2 moles of \( Ag^+ \) are produced for every mole of \( Ag2CO3 \) that dissolves).
- The concentration of \( CO_3^{2-} \) will be \( x \) from the dissolution of \( Ag2CO3 \) plus \( 0.1 \) from the \( Na2CO3 \) solution.

Thus, the concentration of \( CO_3^{2-} \) at equilibrium is:
\[ [CO_3^{2-}] = x + 0.1 \approx 0.1 \quad (\text{since } x \text{ is small compared to } 0.1) \]

### Step 4: Substitute into the Ksp expression
Substituting the equilibrium concentrations into the Ksp expression:
\[ K_{sp} = (2x)^2 (0.1) \]
This simplifies to:
\[ K_{sp} = 4x^2 \cdot 0.1 = 0.4x^2 \]

### Step 5: Set up the equation with the known Ksp value
Now, substituting the known value of Ksp:
\[ 4 \times 10^{-13} = 0.4x^2 \]

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

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

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