Question

3. To a solution of 0.0100 M Al(NO3)3 and 0.0200 M CaCl2 , Na3PO4 is added. The Ksp of AlPO4 is 9.8×10−22; The Ksp of Ca3(PO4)2 is 9.8×10−22.
a) Determine which of these salts is more soluble in water.
b) Calculate [PO43−] when the first cation begins to precipitate.
c) Calculate the concentration of the first ion to precipitate, when the second ion begins to precipitate.

Answer 1

a)

the salt which is most soluble (ignoring Al(NO3)3 CaCl2 and Na3PO4)

For solubility comparisson:

AlPO4:

AB <-> A+ + B-; Ksp = [A+][B-] = S*S = S^2 --> S = sqrt(Ksp)

S = sqrt(9.8*10^-22) = 3.13049*10^-11 M

Ca3(PO4)2

A3B2 <-->3A+2 + 2B-3; Ksp = [A+2]^3[B-3]^2 = (3S)^3 * (2S)^2 = 108*S^5 --> S = (Ksp/108)^(1/5)

S = ((9.8*10^-22)/(108))^(1/5)

S = 2.4635*10^-5

clearly,

Ca3(PO4)2 is much more soluble

b)

when AlPO4 starts to precipitate

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

9.8*10^-22 = (0.01)*[PO4-3]

[PO4-3] = 9.8*10^-20 M

c)

find [PO4-3] left when Ca3(PO4)2 precipitates

Ksp = [Ca+2]^3[PO4-3]^2

9.8*10^-22 = (0.02)^3 * [PO4-3]^2

sqrt((9.8*10^-22) / ((0.02)^3)) =  [PO4-3]

1.1067*10^-8 =  [PO4-3]

get [Al+3]

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

9.8*10^-22 = ([Al+3])*(1.1067*10^-8)

[Al+3] = (9.8*10^-22 ) / ((1.1067*10^-8)) = 8.855*10^-14 M