The solubility of `CaF_(2)` is `2 xx 10^(-4) "mole"//"litre"`. Its solubility product is

Options

A. `2 xx 10^(-4)`
B. `4 xx 10^(-3)`
C. `8 xx 10^(-12)`
D. `3.2 xx 10^(-11)`

Answer

To find the solubility product (Ksp) of calcium fluoride (CaF₂), we can follow these steps:

### Step 1: Write the dissociation equation
Calcium fluoride dissociates in water according to the following equation:
\[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^{-} (aq) \]

### Step 2: Define solubility
Let the solubility of CaF₂ be \( S \) (in moles per litre). 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 2 moles of fluoride ions are produced for every mole of CaF₂ that dissolves).

### Step 3: Write the expression for Ksp
The solubility product (Ksp) is given by the product of the concentrations of the ions, each raised to the power of their coefficients in the balanced equation:
\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^{-}]^2 \]
Substituting the concentrations we defined:
\[ K_{sp} = [S][2S]^2 \]

### Step 4: Simplify the expression
Now, substituting \( [\text{F}^{-}] = 2S \):
\[ K_{sp} = S \cdot (2S)^2 = S \cdot 4S^2 = 4S^3 \]

### Step 5: Substitute the value of S
We know from the problem that the solubility \( S = 2 \times 10^{-4} \) moles/litre. Now we can substitute this value into the Ksp expression:
\[ K_{sp} = 4(2 \times 10^{-4})^3 \]

### Step 6: Calculate \( Ksp \)
Calculating \( (2 \times 10^{-4})^3 \):
\[ (2 \times 10^{-4})^3 = 8 \times 10^{-12} \]
Now substituting this back into the Ksp equation:
\[ K_{sp} = 4 \times 8 \times 10^{-12} = 32 \times 10^{-12} \]

### Step 7: Express in scientific notation
To express \( 32 \times 10^{-12} \) in scientific notation:
\[ K_{sp} = 3.2 \times 10^{-11} \]

### Final Answer
The solubility product \( K_{sp} \) of \( \text{CaF}_2 \) is:
\[ K_{sp} = 3.2 \times 10^{-11} \]

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