Calculate the solubility of the solid Pb3 (PO4) 2 Ksp=1×10^-54 in a 0.10M Pb (NO3) 2

Question

Calculate the solubility of the solid Pb3 (PO4) 2 Ksp=1×10^-54 in a 0.10M Pb (NO3) 2

Answer 1

Ksp = 1x10^-54 = [Pb²⁺]³[SO₄^2-]²

This is a common ion problem so the [ Pb2+] can be taken as 0.1 M

1x10^-54 = (0.1)³(x)²

x = 3.2x10^-26 M