Question

Sodium phosphate is added to a solution that contains 0.0051 M aluminum nitrate and 0.014 M calcium chloride. The concentration of the first ion to precipitate (either Al3+ or Ca2+) decreases as its precipitate forms. What is the concentration of this ion when the second ion begins to precipitate?

Answer 1

Ksp of AlPO4 = 9.84 x 10^-21

Ksp of Ca3(PO4)2 = 2.07 x 10^-33

AlPO4 -------------> Al3+   + PO43-

                                  S            S

Ksp = [Al3+][PO43-]

9.84 x 10^-21 = S^2

S = 9.92 x 10^-11

solubility of AlPO4 = 9.92 x 10^-11 M

Ca3(PO4)2 ---------> 3 Ca2+ +   2 PO43-

                                    3S              2 S

Ksp = (3S)^3 x (2S)^2

2.07 x 10^-33 = 108 S^5

S = 1.14 x 10^-7 M

solubility of Ca3(PO4)2 = 1.14 x 10^-7 M

so AlPO4 will precipitate first.

Ca3(PO4)2 ---------> 3 Ca2+ +   2 PO43-

Ksp = [Ca2+]^3 [PO43-]^2

2.07 x 10^-33 = (0.014)^3 [PO43-]^2

[PO43-] = 2.75 x 10^-14 M

AlPO4 -------------> Al3+   + PO43-

Ksp = [Al3+][PO43-]

9.84 x 10^-21 = [Al3+] (2.75 x 10^-14)

[Al3+] = 3.58 x 10^-7 M

concentration of this ion = 3.58 x 10^-7 M