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 (Ksp) for calcium fluoride (CaF₂) given its solubility, follow these steps:

### Step 1: Write the Dissociation Equation
Calcium fluoride (CaF₂) dissociates in water as follows:
\[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^- (aq) \]

### Step 2: Define Solubility
Let the solubility of CaF₂ be \( S \) moles per liter. According to the dissociation equation:
- The concentration of \(\text{Ca}^{2+}\) ions will be \( S \).
- The concentration of \(\text{F}^-\) ions will be \( 2S \) (since there are two fluoride ions for each formula unit of CaF₂).

### Step 3: Substitute the Given Solubility
From the problem, the solubility \( S \) is given as:
\[ S = 0.0002 \, \text{moles/L} \]

### Step 4: Calculate Ion Concentrations
Now, substituting \( S \) into the concentrations:
- \([\text{Ca}^{2+}] = S = 0.0002 \, \text{mol/L}\)
- \([\text{F}^-] = 2S = 2 \times 0.0002 = 0.0004 \, \text{mol/L}\)

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

### Step 6: Substitute the Ion Concentrations into the Ksp Expression
Substituting the values we calculated:
\[ K_{sp} = (0.0002)(0.0004)^2 \]

### Step 7: Perform the Calculation
Calculating \( (0.0004)^2 \):
\[ (0.0004)^2 = 0.00000016 \]

Now substituting this back into the Ksp expression:
\[ K_{sp} = (0.0002)(0.00000016) \]
\[ K_{sp} = 0.000000032 \]

### Step 8: Convert to Scientific Notation
Converting \( 0.000000032 \) to scientific notation:
\[ K_{sp} = 3.2 \times 10^{-11} \]

### Final Answer
Thus, the solubility product \( K_{sp} \) for \( \text{CaF}_2 \) at \( 25^\circ C \) is:
\[ K_{sp} = 3.2 \times 10^{-11} \]

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