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