For lead fluoride Ksp = s³
For Aluminum hydroxide Ksp = s⁴
Step-by-step explanation:
The expression for the solubility product can be written as the product of the concentration of the ions present in the solution.
PbF₂ → Pb²⁺ + 2 F⁻
Then the Ksp can be written as,
Ksp = [ Pb²⁺] [F⁻]²
If the concentration can be considered as s, then the Ksp can be written as,
Ksp = s× s² = s³
For Aluminum hydroxide on dissociation it gives 4 ions as,
Al(OH)₃ → Al³⁺ + 3 OH⁻
Ksp = [Al³⁺] [OH⁻]³
= s × s³ = s⁴
Thus,For lead fluoride Ksp = s³
For Aluminum hydroxide Ksp = s⁴