`CaCO_(3)` and `BaCO_(3)` have solubility product values `1 xx 10^(-8)` and `5 xx 10^(-9)`, respectively. If water is shaken up with both solids till equilibrium is reached, the concentration of `CO_(3)^(2-)` ion is

Options

A. `1.5 xx 10^(-8)`
B. `1.225 xx 10^(-4)`
C. `2.25 xx 10^(-9)`
D. None of these

Explanation

To solve the problem, we need to find the concentration of the carbonate ion \((CO_3^{2-})\) when both calcium carbonate \((CaCO_3)\) and barium carbonate \((BaCO_3)\) are in equilibrium with water. We will use the solubility product constants (\(K_{sp}\)) provided for both salts.

### Step-by-Step Solution:

1. **Write the Dissolution Reactions:**
- For calcium carbonate:
\[ CaCO_3 (s) \rightleftharpoons Ca^{2+} (aq) + CO_3^{2-} (aq) \]
- For barium carbonate:
\[ BaCO_3 (s) \rightleftharpoons Ba^{2+} (aq) + CO_3^{2-} (aq) \]

2. **Define Solubility:**
- Let the solubility of \(CaCO_3\) be \(x\) mol/L. Then, at equilibrium:
- \([Ca^{2+}] = x\)
- \([CO_3^{2-}] = x\)
- Let the solubility of \(BaCO_3\) be \(y\) mol/L. Then, at equilibrium:
- \([Ba^{2+}] = y\)
- \([CO_3^{2-}] = y\)

3. **Write the \(K_{sp}\) Expressions:**
- For \(CaCO_3\):
\[ K_{sp} = [Ca^{2+}][CO_3^{2-}] = x \cdot x = x^2 \]
Given \(K_{sp} = 1 \times 10^{-8}\), we have:
\[ x^2 = 1 \times 10^{-8} \quad \Rightarrow \quad x = \sqrt{1 \times 10^{-8}} = 1 \times 10^{-4} \, \text{mol/L} \]

- For \(BaCO_3\):
\[ K_{sp} = [Ba^{2+}][CO_3^{2-}] = y \cdot y = y^2 \]
Given \(K_{sp} = 5 \times 10^{-9}\), we have:
\[ y^2 = 5 \times 10^{-9} \quad \Rightarrow \quad y = \sqrt{5 \times 10^{-9}} = \sqrt{5} \times 10^{-4.5} \approx 7.07 \times 10^{-5} \, \text{mol/L} \]

4. **Consider the Common Ion Effect:**
- The concentration of \(CO_3^{2-}\) ions will be affected by both \(x\) and \(y\). Thus, we have:
\[ [CO_3^{2-}] = x + y \]

5. **Substitute Values:**
- From the previous calculations:
\[ [CO_3^{2-}] = 1 \times 10^{-4} + 7.07 \times 10^{-5} \approx 1.707 \times 10^{-4} \, \text{mol/L} \]

6. **Final Calculation:**
- Thus, the concentration of \(CO_3^{2-}\) ions is approximately:
\[ [CO_3^{2-}] \approx 1.707 \times 10^{-4} \, \text{mol/L} \]

### Conclusion:
The concentration of \(CO_3^{2-}\) ion when equilibrium is reached is approximately \(1.707 \times 10^{-4} \, \text{mol/L}\).