The solubility of a saturated solution of calcium fluoride is `2xx10^(-4)` mol/L. Its solubility product is

Options

A. `12xx10^(-2)`
B. `14xx10^(-4)`
C. `22xx10^(-11)`
D. `32xx10^(-12)`

Explanation

To find the solubility product (Ksp) of calcium fluoride (CaF2), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Dissociation Equation:**
The dissociation of calcium fluoride in water can be represented as:
\[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2 \text{F}^- (aq) \]

2. **Define the Solubility:**
Let the solubility of calcium fluoride be \( S \) (in mol/L). Given that the solubility of the saturated solution is \( 2 \times 10^{-4} \) mol/L, we have:
\[ S = 2 \times 10^{-4} \, \text{mol/L} \]

3. **Determine Ion Concentrations:**
From the dissociation equation:
- The concentration of \( \text{Ca}^{2+} \) ions will be \( S \).
- The concentration of \( \text{F}^- \) ions will be \( 2S \) (since 2 moles of fluoride are produced for every mole of calcium fluoride that dissolves).

4. **Substitute Values:**
Therefore, we can express the concentrations as:
- \( [\text{Ca}^{2+}] = S = 2 \times 10^{-4} \, \text{mol/L} \)
- \( [\text{F}^-] = 2S = 2 \times (2 \times 10^{-4}) = 4 \times 10^{-4} \, \text{mol/L} \)

5. **Write the Expression for Ksp:**
The solubility product \( K_{sp} \) is given by the formula:
\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \]

6. **Substitute the Ion Concentrations:**
Plugging in the values:
\[ K_{sp} = (2 \times 10^{-4}) \times (4 \times 10^{-4})^2 \]

7. **Calculate \( Ksp \):**
First, calculate \( (4 \times 10^{-4})^2 \):
\[ (4 \times 10^{-4})^2 = 16 \times 10^{-8} \]
Now, substitute this back into the equation for \( K_{sp} \):
\[ K_{sp} = (2 \times 10^{-4}) \times (16 \times 10^{-8}) = 32 \times 10^{-12} \]

8. **Final Result:**
Thus, the solubility product \( K_{sp} \) of calcium fluoride is:
\[ K_{sp} = 32 \times 10^{-12} \]