Question

Please rank the following in order of increasing solubility

Calcium sulfate Ksp=7.1x10^-5
Calcium Carbonate Ksp=5.0x10^-9
Caclium Hydroxide Ksp=4.7x10^-6
Calcium Flouride Ksp=3.9x10^-11
Calcium Phosphate Ksp=2.1x10^-33

Answer 1

1)

At equilibrium:

CaSO4 <----> Ca2+ + SO42-

   s s

Ksp = [Ca2+][SO42-]

7.1*10^-5=(s)*(s)

7.1*10^-5= 1(s)^2

s = 8.426*10^-3 M

2)

At equilibrium:

CaCO3 <----> Ca2+ + CO32-

   s s

Ksp = [Ca2+][CO32-]

5*10^-9=(s)*(s)

5*10^-9= 1(s)^2

s = 7.071*10^-5 M

3)

At equilibrium:

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

   s 2s

Ksp = [Ca2+][OH-]^2

4.7*10^-6=(s)*(2s)^2

4.7*10^-6= 4(s)^3

s = 1.055*10^-2 M

4)

At equilibrium:

CaF2 <----> Ca2+ + 2 F-

   s 2s

Ksp = [Ca2+][F-]^2

3.9*10^-11=(s)*(2s)^2

3.9*10^-11= 4(s)^3

s = 2.136*10^-4 M

5)

At equilibrium:

Ca3(PO4)2 <----> 3 Ca2+ + 2 PO43-

   3s 2s

Ksp = [Ca2+]^3[PO43-]^2

2.1*10^-33=(3s)^3*(2s)^2

2.1*10^-33= 108(s)^5

s = 1.142*10^-7 M

Answers:

The compounds in increasing order of solubility are:

Ca3(PO4)2 whose solubility is 1.142*10^-7 M

CaCO3 whose solubility is 7.071*10^-5 M

CaF2 whose solubility is 2.136*10^-4 M

CaSO4 whose solubility is 8.426*10^-3 M

Ca(OH)2 whose solubility is 1.055*10^-2 M