The solubility of `PbF_(2)` (formula mass =245) is 0.46 g/L. What is the solubility product?

Options

A. `1.1xx10^(-10)`
B. `2.6xx10^(-8)`
C. `1.1xx10^(-7)`
D. `6.8xx10^(9)`

Answer

To find the solubility product (Ksp) of lead(II) fluoride (PbF2) given its solubility, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the solubility in moles per liter (mol/L)**:
- Given solubility of PbF2 = 0.46 g/L
- Molar mass of PbF2 = 245 g/mol
- To convert grams per liter to moles per liter, use the formula:
\[ \text{Solubility (mol/L)} = \frac{\text{Solubility (g/L)}}{\text{Molar mass (g/mol)}} \]
- Calculation:
\[ \text{Solubility (mol/L)} = \frac{0.46 \text{ g/L}}{245 \text{ g/mol}} = 0.00187 \text{ mol/L} \]

2. **Write the dissociation equation for PbF2**:
- The dissociation of PbF2 in water can be represented as:
\[ \text{PbF}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2\text{F}^- (aq) \]
- From the equation, we see that 1 mole of PbF2 produces 1 mole of Pb²⁺ and 2 moles of F⁻.

3. **Define the solubility (s)**:
- Let the solubility of PbF2 be \( s \) mol/L. From the dissociation:
- [Pb²⁺] = \( s \)
- [F⁻] = \( 2s \)

4. **Express the solubility product (Ksp)**:
- The solubility product expression for PbF2 is given by:
\[ K_{sp} = [\text{Pb}^{2+}][\text{F}^-]^2 \]
- Substituting the concentrations:
\[ K_{sp} = (s)(2s)^2 = s \cdot 4s^2 = 4s^3 \]

5. **Substitute the value of s**:
- We have calculated \( s = 0.00187 \) mol/L.
- Therefore:
\[ K_{sp} = 4(0.00187)^3 \]

6. **Calculate Ksp**:
- First, calculate \( (0.00187)^3 \):
\[ (0.00187)^3 = 6.55 \times 10^{-9} \]
- Now, multiply by 4:
\[ K_{sp} = 4 \times 6.55 \times 10^{-9} = 2.62 \times 10^{-8} \]

### Final Answer:
The solubility product \( K_{sp} \) of PbF2 is approximately \( 2.62 \times 10^{-8} \).

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