Hi can anyone show me the solution of this please

Question

Determine the Ksp of lead (II) chloride (PbCl2), if its molar solubility is 1.95 x 10^-3 M

Answer 1

PbCl₂(s) <==> Pb²⁺(aq) + 2Cl^-(aq) ... this is the dissolution equation for PbCl₂(s)

Ksp = [Pb²⁺][Cl^-]²

If the solubility of PbCl₂ is 1.9510^-3 M, then [Pb²⁺] = 1.95x10^-3 M and [Cl-] = 2 x 1.95x10^-3 = 3.9x10^-3 M

Plugging these concentrations into the Ksp expression and solving for Ksp, we have...

Ksp = (1.95x10^-3)(3.9x10^-3)²

Ksp = 2.97x10^-8