At `25^(@)C`, the solubility of `CaF_(2)` in water is 0.0002 moles per litre what is `K_(sp)`

Options

A. `32 xx 10^(-11)`
B. `1.6 xx 10^(-11)`
C. `8.0 xx 10^(-11)`
D. `3.2 xx 10^(-11)`

Answer

To find the solubility product constant \( K_{sp} \) for calcium fluoride \( CaF_2 \) given its solubility in water, 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 \( S = 0.0002 \) moles per liter.

### Step 3: Determine the concentrations of ions
From the dissociation equation, we can see that:
- For every 1 mole of \( CaF_2 \) that dissolves, 1 mole of \( Ca^{2+} \) is produced.
- For every 1 mole of \( CaF_2 \) that dissolves, 2 moles of \( F^{-} \) are produced.

Thus, the concentrations of the ions at equilibrium will be:
- \([Ca^{2+}] = S = 0.0002 \, \text{mol/L}\)
- \([F^{-}] = 2S = 2 \times 0.0002 = 0.0004 \, \text{mol/L}\)

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

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

### Step 5: Substitute the ion concentrations into the \( K_{sp} \) expression
Substituting the concentrations we found:

\[ K_{sp} = (0.0002)(0.0004)^2 \]

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

\[ K_{sp} = (0.0002)(0.0004 \times 0.0004) = (0.0002)(0.00000016) \]

\[ K_{sp} = 0.0002 \times 0.00000016 = 3.2 \times 10^{-11} \]

### Final Result
Thus, the solubility product constant \( K_{sp} \) for \( CaF_2 \) at \( 25^\circ C \) is:

\[ K_{sp} = 3.2 \times 10^{-11} \]

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