How many mg of PbI2 (Ksp = 1.4x10-8 ) will dissolve in 100. mL of 0.062 M AlI3?

Answer 1

PbI₂(s) ==> Pb²⁺(aq) + 2I^-(aq)

Ksp = [Pb²⁺][I^-]²

[I^-] = 3x0.062 M = 0.186 M

1.4x10^-8 = [Pb²⁺][0.186]²

[Pb²⁺] = 4.047x10^-7 M

4.047x10^-7 mol/L x 0.100 L = 4.047x10^-6 mole PbI₂

4.0x10^-6 mol PbI² x 461 g/mol = 1.8x10^-3 g = 1.8 mg