Question

A solution of Na₃PO₄ is added dropwise to a solution that is 0.0219 M in Ca²⁺ and 7.03e-06 M in Al³⁺.

The K_sp of Ca₃(PO₄)₂ is 2.07e-33.
The K_sp of AlPO₄ is 9.84e-21.

(a) What concentration of PO₄^3- is necessary to begin precipitation? (Neglect volume changes.)

[PO₄^3-] = ___________ M.



(b) Which cation precipitates first?

Ca²⁺

Al³⁺    

(c) What is the concentration of PO₄^3- when the second cation begins to precipitate?

[PO₄^3-] = _____________ M.

Answer 1

Ca3(PO4)2 <--> 3Ca+2 + 2PO3-3

Ksp = [Ca+2]^3 * [PO3-3]^2

2.07*10^-33 = (0.0219^3)(2S)^2

S = sqrt((2.07*10^-33) / (0.0219^3)) = 1.403844*10^-14 is the solubility

for..

AlPO4 ---> Al+3 + PO4-3

Ksp = [Al+3][PO4-3]

9.84*10^-21 = (7.03*10^-6)(S)

S = (9.84*10^-21) /( (7.03*10^-6)) = 1.399*10^-15

therefore...

Ca3(PO4)2 precipittes first...

anything more than:

[PO4-3] = 2*S = 2*1.403844*10^-14 = 2.80710^-14 will start precipitation

b)

the cation to precipitate first was already state, it is Ca+2

c)

find PO4-3 required for second precipitation...

this must be in equilibrium:

from:

Ksp = [Ca+2]^3 * [PO3-3]^2

[PO4-3] = 2*S = 2*1.403844*10^-14 = 2.80710^-14

then..

for second cation:

we require:

[PO4-3] = 1.399*10^-15

so

[PO4-3] = 1.399*10^-15 will be the concentration when AlPO3 starts to precipitate