Question

# 3. To a solution of 0.0100 M Al(NO3)3 and 0.0200 M CaCl2 , Na3PO4 is added....

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.

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

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