calculate the grams per liter of solid lead(II) phosphate (Ksp = 1.00e-54) that should dissolve in .560 M lead(II) nitrate solution

Question

calculate the grams per liter of solid lead(II) phosphate (Ksp = 1.00e-54) that should dissolve in .560 M lead(II) nitrate solution

Answer 1

Pb₃(PO₄)₂(s) ==> 3Pb²⁺(aq) + 2PO₄^3-(aq)

Ksp = 1x10^-54 = [Pb²⁺]³[PO₄^3-]²

This is a common ion problem since the common ion is Pb2+ coming from Pb(NO3)2.

Ksp = 1x10^-54 = [0.560]³[PO₄^3-]²

[PO₄^3-] = 2.39x10^-27 M

2.39x10^-27 mol/L PO₄^3- x 1 mol Pb₃(PO₄)₂/2 mol PO₄^3- x 811.5 g/mole = 4.06x10^-25 grams of Pb₃(PO₄)₂