Question

Suppose you have a solution that contains 0.0430 M Ca2 and 0.0910 M Ag . If solid Na3PO4 is added to this mixture, which of the following phosphate species would precipitate out of solution first?

Ca3(PO4)2

When the second cation just starts to precipitate, what percentage of the first cation remains in solution?

Answer 1

Precipiates:

3Ca+2 + 2PO4-3 --> Ca3(PO4)2 Ksp = 2.07*10^-33

3Ag+ + PO4-3 --> Ag3PO4 Ksp = 8.89*10^-17

Calculate concentration of PO4-3 required for each

Ksp Ca3(PO4)2 = [Ca+2]^3[PO4-3]^2

2.07*10^-33 = (0.043^3)(PO4-3)^2

[PO4-3]^2 = (2.07*10^-33 / ( (0.043^3))

[PO4-3] = (2.6035*10^-29)^0.5 = 5.102*10^-15

for Ag3PO4

Ksp Ag3PO4 = [Ag+]^3[PO4-3]

8.89*10^-17 = (0.091^3)[PO4-3]

[PO4-3] = (8.89*10^-17)/((0.091^3)) = 1.179*10^-13

Clearly, the Ca3(PO4)2will precipitate first, since it requires less PO4-3

Now...

Find % of Ca+2

[PO4-3] when Ag precipitates --> 1.179*10^-13

2.07*10^-33 = [CA+2]^3[1.179*10^-13]^2

[Ca+2] = ((2.07*10^-33 / (1.179*10^-13)^2)^(1/3) = 0.0053004 M

% ion soluition = 0.0053004 / 0.0430 *100 = 12.326%

precipitate --> 100-12.326 = 87.674 %