Solubility (s) of `CaF_(2)` in terms of its solubility product is given as

Options

A. `s = (K_(sp))^(1//3)`
B. `s = (K_(sp)//2)^(1//3)`
C. `s = (K_(sp)//4)^(1//3)`
D. `s = (K_(sp)//2)^(1//2)`

Explanation

To find the solubility (s) of \( CaF_2 \) in terms of its solubility product (\( K_{sp} \)), we can follow these steps:

### Step 1: Write the Dissociation Equation
When \( CaF_2 \) dissolves in water, it dissociates into calcium ions and fluoride ions:
\[ CaF_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2F^{-} (aq) \]

### Step 2: Define the Solubility
Let the solubility of \( CaF_2 \) be \( s \) mol/L. This means that:
- The concentration of \( Ca^{2+} \) ions will be \( s \) mol/L.
- The concentration of \( F^{-} \) ions will be \( 2s \) mol/L (since there are 2 fluoride ions for every formula unit of \( CaF_2 \)).

### Step 3: Write the Expression for \( K_{sp} \)
The solubility product \( K_{sp} \) for the dissociation of \( CaF_2 \) can be expressed as:
\[ K_{sp} = [Ca^{2+}][F^{-}]^2 \]
Substituting the concentrations in terms of \( s \):
\[ K_{sp} = (s)(2s)^2 \]

### Step 4: Simplify the Expression
Now, simplify the expression:
\[ K_{sp} = s \cdot (4s^2) = 4s^3 \]

### Step 5: Solve for \( s \)
To find \( s \) in terms of \( K_{sp} \), rearrange the equation:
\[ s^3 = \frac{K_{sp}}{4} \]
Taking the cube root of both sides gives:
\[ s = \sqrt[3]{\frac{K_{sp}}{4}} \]

### Final Expression
Thus, the solubility \( s \) of \( CaF_2 \) in terms of its solubility product \( K_{sp} \) is:
\[ s = \frac{\sqrt[3]{K_{sp}}}{\sqrt[3]{4}} \]