A well is dry in a bed of rock containing fluorspar `(CaF_(2))`. If the well contains `20,000 L` of `H_(2)O`, what is the amount of `F^(-)` in it
`K_(SP)` of `CaF_(2) = 4 xx 10^(-11)`

Options

A. `4.3` mole
B. `6.8` mole
C. `8.6` mole
D. `13.6` mole

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the dissociation equation for CaF₂
The dissociation of calcium fluoride (CaF₂) in water can be represented as:
\[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^- (aq) \]

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

### Step 3: Write the expression for the solubility product (K_sp)
The solubility product \( K_{sp} \) for the dissociation of CaF₂ is given by:
\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \]
Substituting the equilibrium concentrations:
\[ K_{sp} = S \cdot (2S)^2 = S \cdot 4S^2 = 4S^3 \]

### Step 4: Set K_sp equal to the given value
We know that:
\[ K_{sp} = 4 \times 10^{-11} \]
Thus, we can set up the equation:
\[ 4S^3 = 4 \times 10^{-11} \]

### Step 5: Solve for S
Dividing both sides by 4:
\[ S^3 = 10^{-11} \]
Taking the cube root:
\[ S = (10^{-11})^{1/3} = 10^{-11/3} \approx 2.15 \times 10^{-4} \text{ moles per liter} \]

### Step 6: Calculate the concentration of F⁻
The concentration of fluoride ions \( [\text{F}^-] \) is:
\[ [\text{F}^-] = 2S = 2 \times 2.15 \times 10^{-4} = 4.3 \times 10^{-4} \text{ moles per liter} \]

### Step 7: Calculate the total amount of F⁻ in the well
The total volume of water in the well is 20,000 L. The amount of fluoride ions can be calculated as:
\[ \text{Amount of } F^- = [\text{F}^-] \times \text{Volume} = 4.3 \times 10^{-4} \text{ moles/L} \times 20,000 \text{ L} \]

### Step 8: Perform the calculation
\[ \text{Amount of } F^- = 4.3 \times 10^{-4} \times 20,000 = 8.6 \text{ moles} \]

### Final Answer
The amount of \( F^- \) in the well is **8.6 moles**.

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