The ksp of lead(ii) hydroxide, pb(oh)2, is 1.43 × 10-20. calculate the solubility of this compound in g/l.

When Pb(OH)₂ dissolves it dissociates as follows;
Pb(OH)₂ ---> Pb²⁺ + 2OH⁻
molar solubility is the number of moles of compound that can be dissolved in 1 L of solution.
if molar solubility of Pb(OH)₂ is x then molar solubility of Pb²⁺ is x and OH⁻ is 2x
the formula for solubility product constant is as follows;
ksp = [Pb²⁺][OH⁻]²
ksp = (x)(2x)²
ksp = 4x³
ksp = 1.43 x 10⁻²⁰
4x³ = 1.43 x 10⁻²⁰
x = 1.53 x 10⁻⁷ M
molar solubility of Pb(OH)₂ is 1.53 x 10⁻⁷ M
molar mass is 241.2 g/mol 
solubility of Pb(OH)₂ is 1.53 x 10⁻⁷ M x 241.2 g/mol = 3.69 x 10⁻⁵ g/L

D.na is the answer hope this

The answer to this is the 3rd choice!