Common Ion effect: Calculate solubility

Question

The molar solubility of silver phosphate in a 0.259 M silver nitrate solution is _____ M.



Answer 1

Common ion effect:

Silver phosphate = Ag₃PO₄

Ag₃PO₄(s) <==> 3Ag⁺⁽aq) + PO₄^3-(aq)

Because there is the presence of a common ion (Ag⁺) from AgNO₃, the [Ag+] is taken as 0.259 M

Ksp =[Ag⁺]³[PO₄^3-]

Looking up the Ksp for Ag₃PO₄, I find it to be 1.8x10^-18

1.8x10^-18 = (0.259)3(x) where x = [PO₄^3-]

1.8x10^-18 = 0.0174 [PO₄^3-]

[PO₄^3-] = 1.03x10^-16 M = solubility of silver phosphate in 0.259 M silver nitrate solution.

[PO43-] = 1.1x10-16