Question

Calculate the maximum mass (in grams) of the following metal sulfides that can be dissolved in 1.0 L of a solution that is 4.90×10−5 M in Na2S.

A. PbS

B. ZnS

Answer 1

The Ksp of ZnS is 2 x 10^-25 , Ksp = [Zn⁺²][S^-2]

The Ksp of PbS is 3 x 10^{-28 ,} Ksp = [Pb⁺²][S^-2]

Now as we have Na2S solution

a) ZnS

            ZnS ---> Zn⁺² + S^-2

Initial                   0         4.9 X 10^-5

change              +x         +x

Equilibrium         x           4.9 X 10^-5 + x

Ksp = [Zn⁺²][S^-2] =  2 x 10^-25 = [x] [ 4.9 X 10^-5 + x]

we may ignore x in addition as it is solubility of ZnS which is very low

2 x 10^-25 = [x] [ 4.9 X 10^-5]

x= solubility = 4.082 X 10^-21 M

This much moles / L of ZnS can be dissolved

Grams that can be dissolved = Molartiy X molecular weight = 4.082 X 10^-21 M X 97.5 g / mole = 3.98 X 10^-19 g / L

{this is negligeble}

2) PbS

           PbS ---> Pb⁺² + S^-2

Initial                   0         4.9 X 10^-5

change              +x         +x

Equilibrium         x           4.9 X 10^-5 + x

Ksp = [Pb⁺²][S^-2] =   3 x 10^-28 = [x] [ 4.9 X 10^-5 + x]

we may ignore x in addition as it is solubility of ZnS which is very low

3 x 10^-28= [x] [ 4.9 X 10^-5]

x= solubility = 6.122X 10^-24 M

This much moles / L of PbS can be dissolved

Grams that can be dissolved = Molartiy X molecular weight = 6.122X 10^-24 M X 239.3 g / mole = 1.465 X 10^-21 g / L

{this is negligeble}