Question

Sodium phosphate is added to a solution that contains 0.0071 M aluminum nitrate and 0.016 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

Sodium phosphate will dissolve as

Na₃PO₄ --> 3Na⁺ + PO4^-3

Aluminum phosphate dissolves as
AlPO₄ --> Al³⁺ + PO4^-3
we can say that Ksp = Ksp = [Al³⁺]∙[PO₄^-3]
with Ksp = 9.84×10^-21 (standard value)

Calcium phosphate dissolves as
Ca₃(PO₄)₂ --> 3Ca²⁺ + 2PO4^-3

Ksp = [Ca+2]^3 [PO₄-3]^2
with Ksp = 2 X 10^-29

The solution initally contains aluminium ions and calcium ions, when we will add sodium phosphate, we are adding phsophate ions which will cause precipitation of metal ions as phosphate

The concentration of phosphate at which the two salts precipitate will be:

Aluminium phosphate
Ksp = [Al³⁺]∙[PO₄^-3]

[PO₄^-3] = Ksp / [Al³⁺] = 9.84×10^-21 / 0.0071 = 1.385 X 10^-18

Calcium phosphate

Ksp = [Ca+2]^3 [PO₄-3]^2 = 2 X 10^-29

[PO₄^-3] = √(Ksp / [Ca+2]^3 = √[2 X 10^-29 / (0.016)^3] = √ [4.88 X 10^-24] = 2.21 X 10^-12
So [PO₄^-3] is less required for aluminium to be precipitate as compared to calcium. so after this concentration of phosphate ions the solution will be saturated with aluminium phosphate and if we will add more of phosphate ions the the concentration of aluminium ions will decrease.

Hence, to precipiate calcium phosphate the concentration of aluminium ion should be

[Al³⁺] = Ksp / [PO₄^-3] = 9.84×10^-21 /2.21 X 10^-12 = 4.45 X 10^-9