Solubility of `CaF_(2)` is `0.5xx10^(-4)" mol L"^(-1)`. The value of `K_(sp)` for the salt is

Options

A. `5xx10^(-12)`
B. `2.5xx10^(-16)`
C. `1xx10^(-13)`
D. `5xx10^(-13)`

Explanation

To find the value of \( K_{sp} \) for calcium fluoride \( CaF_2 \), we can follow these steps:

### Step 1: Write the Dissociation Equation
Calcium fluoride dissociates in water according to the following equation:

\[ CaF_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2F^{-} (aq) \]

### Step 2: Define the Solubility
Let the solubility of \( CaF_2 \) be \( s \). According to the problem, the solubility \( s \) is given as:

\[ s = 0.5 \times 10^{-4} \, \text{mol L}^{-1} \]

### Step 3: Determine Ion Concentrations
From the dissociation equation, we can see that for every mole of \( CaF_2 \) that dissolves, 1 mole of \( Ca^{2+} \) ions and 2 moles of \( F^{-} \) ions are produced. Therefore, the concentrations of the ions at equilibrium will be:

- Concentration of \( Ca^{2+} \) = \( s = 0.5 \times 10^{-4} \, \text{mol L}^{-1} \)
- Concentration of \( F^{-} \) = \( 2s = 2 \times (0.5 \times 10^{-4}) = 1.0 \times 10^{-4} \, \text{mol L}^{-1} \)

### Step 4: Write the Expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for \( CaF_2 \) can be expressed as:

\[ K_{sp} = [Ca^{2+}][F^{-}]^2 \]

### Step 5: Substitute the Ion Concentrations into the \( K_{sp} \) Expression
Now we can substitute the concentrations we found into the \( K_{sp} \) expression:

\[ K_{sp} = (0.5 \times 10^{-4}) \times (1.0 \times 10^{-4})^2 \]

### Step 6: Calculate \( K_{sp} \)
Calculating the above expression:

\[ K_{sp} = (0.5 \times 10^{-4}) \times (1.0 \times 10^{-8}) = 0.5 \times 10^{-12} \]

### Final Answer
Thus, the value of \( K_{sp} \) for \( CaF_2 \) is:

\[ K_{sp} = 0.5 \times 10^{-12} \]

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