Question

The slightly soluble ionic compound PbF2 has a solubility in water of 0.74 g/L. What is the numerical value of Ksp for PbF2?

Answer 1

Answer:

Ksp =  1.08 × 10⁻⁷

Explanation:

Given data:

Solubility of PbF₂ = 0.74 g/L

Ksp for PbF₂ = ?

Solution:

PbF₂ → Pb⁺² + 2F⁻

Number of moles dissolved = 0.74 g/L /245.2 g/mol

Number of moles dissolved = 0.003 M

[Pb⁺²] = 0.003 M

[F⁻] = 0.003 M × 2 =  0.006 M

Ksp = [Pb⁺²] [F⁻]²

Ksp = (0.003 ) (0.006 )²

Ksp =  (0.003 ) (3.6 ×10⁻⁵)

Ksp =  1.08 × 10⁻⁷