Sodium phosphate is added to a solution that contains 0.0099 M aluminum nitrate

Question

Sodium phosphate is added to a solution that contains 0.0099 M aluminum nitrate and 0.021 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? Hint: Note that well over 99% of the first ion has precipitated before the second ion starts to precipitate. Correct Answer: 2.613e-7 (I need an "EXPERT" to show me how to arrive at this answer that I gave above with full work). *Ok there is no sodium phosphate concentration given, but somehow still you should be able to solve this without it.. Maybe you chem experts know another way of going about this and arriving at that answer or what are your conclusions or methods you guys are planning on using to solve this problem. Anything is appreciated! (If you are not too sure on how to solve this problem, then give me a good reasoning at least on why this problem cannot be solved but it should be solvable according to my general chem 2 college professor, but all i can tell you is that there is other similar questions to this getting some kind of answer here on chegg.

Explanation / Answer

Aluminum phosphate dissolves according to
AlPO(s) Al³(aq) + PO³(aq)
=>
Ksp = [Al³][PO³]
with Ksp = 9.84×10²¹

Calcium phosphate dissolves according to
Ca(PO)(s) 3 Ca²(aq) + 2 PO³(aq)
=>
Ksp = [Ca²]³[PO³]²
with Ksp = 2.07×10³³

By addition of sodium phosphate increases while the metal ion concentrations are constant. When the product of the ion concentrations reaches the value of Ksp. When you solve the equilibrium equation of each salt for [PO³] you get the level of phosphate concentration at which the salt starts to precipitate:
- aluminum phosphate
[PO³] = Ksp / [Al³] = 9.84×10²¹ / 0.0099 = 9.93×1019 M
- calcium phosphate
[PO³] = (Ksp / [Ca²]²) = (2.07×10³³ / 0.021³) = 1.495 ×1014 M

Because [PO³] is smaller for Al³, aluminum phosphate precipitates first. After this point solution is saturated with aluminum phosphate. Further addition of sodium phosphate increases the phosphate concentration, so aluminum ion concentration decreases due precipitation such that solubility equilibrium equation is always satisfied. So at the instant when calcium phosphate starts to precipitate the aluminum ion concentration has decreased to:
[Al³] = Ksp / [PO³] = 9.84×10²¹ / 1.495 ×10¹ = 6.58×10 M

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