Question

(A) Calculate the solubility of Ca(IO₃)₂ in a solution containing 0.10 M CaCl₂.

ksp of Ca(IO3)2 is7.1 x 10^-7
Answer:1.3 x 10^-3 M

(B) Copper (II) phosphate has a solubility of 1.67 x 10^-8 M.

(a) Calculate the K_sp for copper (II) phosphate.

(b) Suggest something you could add to a saturated solution of copper (II) phosphate to increase its solubility.

Answer of a: 1.40 x 10^-37

Answer 1

A)

Solubility is given by

Ca(IO3)2 <-> Ca2+ + 2IO3-

the ksp expression

Ksp = [Ca+2][IO3-]^2

if [CaCl2] = 0.1 then [Ca+2] = 0.1 M

Ksp = 7.1*10^-7

then, substitute all data

Ksp = [Ca+2][IO3-]^2

Assume 1 mol of Ca(IO3)2 --> 2 mol of IO3- so 1 S --> 2 S

7.1*10^-7 = (0.1)(2S)^2

2S = ((7.1*10^-7 )/(0.1))^0.5

S = 0.0026645/2

S = 0.001332 M

S = 1.33*10^-3 M

b)

Ksp is given by

Cu3(PO4)2(s) <-> 3Ca+2 + 2PO4-2

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

Note that per each 1 mol of Cu3(PO4)2 we get, 3 mol of Ca+2, and 2 mol of PO4-3

then

[Ca+2] = 2*S = 2*1.67*10^-8 = 3.34*10^-8

[PO4-3] = 3*S = 3*1.67*10^-8 = 5.01*10^-8

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

Ksp = ( 3.34*10^-8)^2 * (5.01*10^-8)^3

Ksp = 1.402*10^-37

b)

in order to increase solubility:

- Make Ca+2 ion to precipitate

- Make PO4-3 ion to precipitate

this can be achieved by

addition of OH- ions, so Ca(OH)2(S) is formed

Addition of S-2 ions, so CaS(s) forms

Addition

Al+3 or Ba+2 ions, so AlPO4(s) precipitates or Ba3(PO3)2 (s) precipitates