Question

Order the following solids (a-d) from least soluble to most soluble: 

Answer 1

At equilibrium:

SrSO4 <----> Sr2+ + SO42-

   s s

Ksp = [Sr2+][SO42-]

3.2*10^-7=(s)*(s)

3.2*10^-7= 1(s)^2

s = 5.657*10^-4 M

At equilibrium:

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

   3s 2s

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

1.3*10^-32=(3s)^3*(2s)^2

1.3*10^-32= 108(s)^5

s = 1.645*10^-7 M

At equilibrium:

PbI2 <----> Pb2+ + 2 I-

   s 2s

Ksp = [Pb2+][I-]^2

1.4*10^-8=(s)*(2s)^2

1.4*10^-8= 4(s)^3

s = 1.518*10^-3 M

At equilibrium:

MnS <----> Mn2+ + S2-

   s s

Ksp = [Mn2+][S2-]

2.3*10^-13=(s)*(s)

2.3*10^-13= 1(s)^2

s = 4.796*10^-7 M

The compounds in increasing order of solubility are:

Ca3(PO4)2 whose solubility is 1.645*10^-7 M

MnS whose solubility is 4.796*10^-7 M

SrSO4 whose solubility is 5.657*10^-4 M

PbI2 whose solubility is 1.518*10^-3 M

Answer: Ca3(PO4)2 < MnS < SrSO4 < PbI2