What is the molarity of a saturated solution of `CaCO_(3)`? `(K_(sp)=2.8xx10^(-9))`

Options

A. `2.6xx10^(-5)`
B. `2.8xx10^(-9)`
C. `5.2xx10^(-5)`
D. `5.6xx10^(-9)`

Answer

To find the molarity of a saturated solution of calcium carbonate (CaCO₃) given its solubility product constant (Ksp = 2.8 x 10⁻⁹), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Dissociation Equation:**
Calcium carbonate dissociates in water as follows:
\[ \text{CaCO}_3 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + \text{CO}_3^{2-} (aq) \]

2. **Define Solubility:**
Let the solubility of calcium carbonate be \( S \) moles per liter. At equilibrium, the concentration of \(\text{Ca}^{2+}\) ions will be \( S \) and the concentration of \(\text{CO}_3^{2-}\) ions will also be \( S \).

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

4. **Substitute the Given Ksp Value:**
We know that \( K_{sp} = 2.8 \times 10^{-9} \). Therefore, we can set up the equation:
\[ S^2 = 2.8 \times 10^{-9} \]

5. **Calculate the Solubility (S):**
To find \( S \), take the square root of both sides:
\[ S = \sqrt{2.8 \times 10^{-9}} = \sqrt{28 \times 10^{-10}} = \sqrt{28} \times 10^{-5} \]
Calculating \( \sqrt{28} \) gives approximately \( 5.29 \). Thus:
\[ S \approx 5.29 \times 10^{-5} \text{ moles per liter} \]

6. **Determine Molarity:**
For a saturated solution, the molarity is equal to the solubility. Therefore, the molarity of the saturated solution of \( \text{CaCO}_3 \) is:
\[ \text{Molarity} = S \approx 5.29 \times 10^{-5} \text{ M} \]

### Final Answer:
The molarity of a saturated solution of \( \text{CaCO}_3 \) is approximately \( 5.29 \times 10^{-5} \) M.

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