Solubility of `CaF_(2)` is `0.5xx10^(-4)" mol L"^(-1)`. The value of `K_(sp)` for the salt is

Options

A. a.`5xx10^(-12)`
B. b.`2.5xx10^(-16)`
C. c.`1xx10^(-13)`
D. d.`5xx10^(-13)`

Explanation

To find the solubility product constant (Ksp) for calcium fluoride (CaF₂) given its solubility, we can follow these steps:

### Step 1: Write the Dissociation Equation
When calcium fluoride (CaF₂) dissolves in water, it dissociates into calcium ions (Ca²⁺) and fluoride ions (F⁻):
\[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2 \text{F}^- (aq) \]

### Step 2: Define the Solubility
Let the solubility of CaF₂ be \( s = 0.5 \times 10^{-4} \, \text{mol/L} \). This means that for every mole of CaF₂ that dissolves, we get:
- 1 mole of Ca²⁺
- 2 moles of F⁻

### Step 3: Express Concentrations in Terms of Solubility
From the dissociation:
- The concentration of Ca²⁺ ions will be \( [\text{Ca}^{2+}] = s \)
- The concentration of F⁻ ions will be \( [\text{F}^-] = 2s \)

### Step 4: Substitute the Value of Solubility
Substituting \( s \) into the expressions:
- \( [\text{Ca}^{2+}] = 0.5 \times 10^{-4} \)
- \( [\text{F}^-] = 2 \times (0.5 \times 10^{-4}) = 1.0 \times 10^{-4} \)

### Step 5: Write the Ksp Expression
The Ksp expression for CaF₂ is given by:
\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \]
Substituting the values we found:
\[ K_{sp} = (0.5 \times 10^{-4}) \times (1.0 \times 10^{-4})^2 \]

### Step 6: Calculate Ksp
Calculating the above expression:
\[ K_{sp} = (0.5 \times 10^{-4}) \times (1.0 \times 10^{-8}) \]
\[ K_{sp} = 0.5 \times 10^{-12} \]
\[ K_{sp} = 5.0 \times 10^{-12} \]

### Final Answer
Thus, the value of \( K_{sp} \) for CaF₂ is:
\[ K_{sp} = 5.0 \times 10^{-12} \]

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