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