A solution of `CaF_(2)` is found to contain `4 xx 10^(-4)M` of `F^(Theta)`, `K_(sp)` of `CaF_(2)` is

Options

A. `3.2 xx 10^(-11)`
B. `0.8 xx 10^(-11)`
C. `6.4 xx 10^(-11)`
D. `32 xx 10^(-11)`

Explanation

To find the solubility product (Ksp) of calcium fluoride (CaF₂) given the concentration of fluoride ions (F⁻), we can follow these steps:

### Step 1: Write the Dissociation Equation
Calcium fluoride dissociates in water as follows:
\[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2 \text{F}^- (aq) \]

### Step 2: Define the Solubility
Let \( S \) be the solubility of CaF₂ in moles per liter. At equilibrium:
- The concentration of \( \text{Ca}^{2+} \) will be \( S \) M.
- The concentration of \( \text{F}^- \) will be \( 2S \) M.

### Step 3: Use the Given Information
We are given that the concentration of fluoride ions is:
\[ [\text{F}^-] = 4 \times 10^{-4} \, M \]
Since \( [\text{F}^-] = 2S \), we can set up the equation:
\[ 2S = 4 \times 10^{-4} \]

### Step 4: Solve for S
Now, solve for \( S \):
\[ S = \frac{4 \times 10^{-4}}{2} = 2 \times 10^{-4} \, M \]

### Step 5: Write the Expression for Ksp
The solubility product \( K_{sp} \) for CaF₂ is given by:
\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \]
Substituting the equilibrium concentrations:
\[ K_{sp} = (S)(2S)^2 \]

### Step 6: Substitute S into the Ksp Expression
Now substitute \( S = 2 \times 10^{-4} \):
\[ K_{sp} = (2 \times 10^{-4}) \times (2(2 \times 10^{-4}))^2 \]
\[ K_{sp} = (2 \times 10^{-4}) \times (4 \times 10^{-4})^2 \]
\[ K_{sp} = (2 \times 10^{-4}) \times (16 \times 10^{-8}) \]
\[ K_{sp} = 32 \times 10^{-12} \]

### Step 7: Convert to Scientific Notation
Convert \( K_{sp} \) to scientific notation:
\[ K_{sp} = 3.2 \times 10^{-11} \]

### Final Answer
Thus, the value of \( K_{sp} \) for CaF₂ is:
\[ K_{sp} = 3.2 \times 10^{-11} \]