Question

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

a. FeS, K_sp = 3.7×10^-19
b. AgI, K_sp = 1.5×10^-16
c. Ag₂S, K_sp = 1.6×10^-49
d. PbBr₂, K_sp = 4.6×10^-6

Solubility: < < <

Answer 1

At equilibrium:

FeS <----> Fe2+ + S2-

   s s

Ksp = [Fe2+][S2-]

3.7*10^-19=(s)*(s)

3.7*10^-19= 1(s)^2

s = 6.083*10^-10 M

At equilibrium:

AgI <----> Ag+ + I-

   s s

Ksp = [Ag+][I-]

1.5*10^-16=(s)*(s)

1.5*10^-16= 1(s)^2

s = 1.225*10^-8 M

At equilibrium:

Ag2S <----> 2 Ag+ + S2-

   2s s

Ksp = [Ag+]^2[S2-]

1.6*10^-49=(2s)^2*(s)

1.6*10^-49= 4(s)^3

s = 3.42*10^-17 M

At equilibrium:

PbBr2 <----> Pb2+ + 2 Br-

   s 2s

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

4.6*10^-6=(s)*(2s)^2

4.6*10^-6= 4(s)^3

s = 1.048*10^-2 M

The compounds in increasing order of solubility are:

Ag2S < FeS < AgI < PbBr2