Solid `AgNO_(3)` is gradually added to a solution which is `0.01M` n `Cl^(-)` and `0.01 M` in `CO_(3)^(2-) K_(sp) AgCl=1.8xx10^(-10)` and `K_(sp)Ag_(2)CO_(3)=4xx10^(-12)`
The minimum concentration of `Ag^(+)` required to start the precipation of `Ag_(2)CO_(3)` is

Options

A. `2xx10^(-5)M`
B. `4xx10^(-10)M`
C. `4xx10^(-5)M`
D. `2xx10^(-10)M`

Explanation

To find the minimum concentration of \( \text{Ag}^+ \) required to start the precipitation of \( \text{Ag}_2\text{CO}_3 \), we need to use the solubility product constant (\( K_{sp} \)) of \( \text{Ag}_2\text{CO}_3 \) and the concentration of carbonate ions (\( \text{CO}_3^{2-} \)) in the solution.

### Step-by-step Solution:

1. **Understand the Reaction**:
The precipitation of silver carbonate can be represented by the following equilibrium:
\[ \text{Ag}_2\text{CO}_3 (s) \rightleftharpoons 2 \text{Ag}^+ (aq) + \text{CO}_3^{2-} (aq) \]

2. **Write the Expression for \( K_{sp} \)**:
The solubility product constant (\( K_{sp} \)) for \( \text{Ag}_2\text{CO}_3 \) is given by:
\[ K_{sp} = [\text{Ag}^+]^2 [\text{CO}_3^{2-}] \]

3. **Substitute Known Values**:
We know that the concentration of \( \text{CO}_3^{2-} \) is \( 0.01 \, M \) (from the problem statement). The \( K_{sp} \) for \( \text{Ag}_2\text{CO}_3 \) is given as \( 4 \times 10^{-12} \).

Plugging these values into the \( K_{sp} \) expression, we have:
\[ 4 \times 10^{-12} = [\text{Ag}^+]^2 \times (0.01) \]

4. **Rearranging the Equation**:
To find \( [\text{Ag}^+] \), rearrange the equation:
\[ [\text{Ag}^+]^2 = \frac{4 \times 10^{-12}}{0.01} \]

5. **Calculate \( [\text{Ag}^+]^2 \)**:
\[ [\text{Ag}^+]^2 = 4 \times 10^{-10} \]

6. **Take the Square Root**:
Now, take the square root to find \( [\text{Ag}^+] \):
\[ [\text{Ag}^+] = \sqrt{4 \times 10^{-10}} = 2 \times 10^{-5} \, M \]

### Final Answer:
The minimum concentration of \( \text{Ag}^+ \) required to start the precipitation of \( \text{Ag}_2\text{CO}_3 \) is \( 2 \times 10^{-5} \, M \).

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