Please help me with this

Question

Calculate the molar solubility of PbI2 (ksp= 1.0x10^-8) in .50 M Pb(NO3)2

Answer 1

Wouldn’t it turn to x^3???

Answer 2

This is a common ion effect problem. The common ion in this problem is Pb²⁺

PbI₂(s) <==> Pb²⁺(aq) + 2I^-(aq)

Ksp = [Pb²⁺][I^-]²

1.0x10^-8 = (0.5)(x)²

x² = 5.0x10^-9

x = 7.1x10^-5 M = [I^-]

Thus, the molar solubility of PbI₂ would be 3.5x10^-5 M