the solubility of PbCl2 is 0.232 g per 48.5 ml of solution at 25 degrees C. determine the value of the solubility product constant for PbCl2

Answer 1

PbCl₂(s) <==> Pb²⁺(aq) + 2Cl^-(aq)

Ksp = [Pb²⁺][Cl^-]

molar mass PbCl₂ = 278.1 g / mole

moles PbCl₂ = 0.232 g x 1 mol / 278.1 g = 8.34x10^-4 moles

[PbCl₂] = 8.34x10^-4 moles / 48.5 ml x 1000 ml / L = 0.0172 M

[Pb²⁺] = 0.0172 M

[Cl^-] = 2 x 0.0172 M = 0.0344 M

Ksp = [0.0172][0.0344]² = [0.0172][2.96x10^-4]

Ksp = 5.09x10^-6