Calculate the Ksp for hydroxide if the solubility of Pb(OH)2 in pure water is 1.62 × 10-3 g/L.

Answer 1

Pb(OH)₂(s) <==> Pb²⁺(aq) + 2OH^-(aq)

Ksp = [Pb²⁺][OH^-]²

molar mass Pb(OH)₂ = 241 g / mole

[Pb(OH)₂] = 1.62x10^-3 g / L x 1 mol / 241 g = 6.72x10^-6 mol / L = 6.72x10^-6 M

[Pb²⁺] = 6.72x10^-6 M

[OH^-] = 2 x 6.72x10^-6 M = 1.34x10^-5 M

Ksp = (6.72x10^-6 )(1.34x10^-5)²

Ksp = 1.21x10^-15