The `K_(sp)` of `CaF_2` at 298 K is `1.7xx10^(-10)` .Calculate its solubility in mol `L^(-1)`

Explanation

To calculate the solubility of \( \text{CaF}_2 \) in mol \( L^{-1} \) given its \( K_{sp} \) value, we can follow these steps:

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

### Step 2: Define Solubility
Let the solubility of \( \text{CaF}_2 \) be \( s \) mol \( L^{-1} \). When \( \text{CaF}_2 \) dissolves:
- The concentration of \( \text{Ca}^{2+} \) ions will be \( s \).
- The concentration of \( \text{F}^- \) ions will be \( 2s \) (since two fluoride ions are produced for each formula unit of \( \text{CaF}_2 \)).

### Step 3: Write the Expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for \( \text{CaF}_2 \) can be expressed as:
\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \]
Substituting the concentrations in terms of \( s \):
\[ K_{sp} = (s)(2s)^2 = s(4s^2) = 4s^3 \]

### Step 4: Substitute the Given \( K_{sp} \) Value
We know that \( K_{sp} \) for \( \text{CaF}_2 \) is \( 1.7 \times 10^{-10} \):
\[ 4s^3 = 1.7 \times 10^{-10} \]

### Step 5: Solve for \( s \)
To find \( s \), we rearrange the equation:
\[ s^3 = \frac{1.7 \times 10^{-10}}{4} \]
Calculating the right-hand side:
\[ s^3 = 4.25 \times 10^{-11} \]
Now, take the cube root:
\[ s = \sqrt[3]{4.25 \times 10^{-11}} \]

### Step 6: Calculate \( s \)
Using a calculator:
\[ s \approx 3.49 \times 10^{-4} \, \text{mol L}^{-1} \]

### Final Answer
The solubility of \( \text{CaF}_2 \) at 298 K is approximately:
\[ \boxed{3.49 \times 10^{-4} \, \text{mol L}^{-1}} \]