Question

A solution is made byequilibratingthe twosolids lead chromate(Ksp = 1.8x10-14) andlead sulfide (Ksp3.2x10-28) with water.What are theconcentrations of thethree ions at equilibrium,if some of each of thesolids remain?[Pb²+] =[Cro4²-][s²] ==M=MM

Answer 1

Let's start by writing down the reactions of both solids solubilizing:

`PbCrO_4\rightleftarrows Pb^(2+)+CrO_4^(2-)\text{ Ksp = 1.8*10-14}`
`PbS\rightleftarrows Pb^(2+)+S^(2-)\text{ Ksp = 3.2*10-28}`

Now let's calculate the contribution of Lead (Pb) from each solid.

Starting with lead chromate. Since the compound ratio is 1:1, we can assume both ions will have the same concentration. So we can calculate using the Ksp constant and x for the unknown concentration:

Ksp = [Pb2+][CrO4 2-]

Ksp = x * x

Ksp = x²

1.8x10-14 = x²

`x=\sqrt{1.8*10^(-14)}`

x = 1.3x10-7

[Pb2+] = 1.3x10-7 M (from lead chromate)

[CrO4 2-] = 1.3x10-7 M

Now let's do the same for lead sulfide:

Ksp = [Pb2+][S 2-]

Ksp = x * x

Ksp = x²

3.2x10-28 = x²

`x=\sqrt{3.2*10^(-28)}`

x = 1.8x10-14

[Pb2+] = 1.8x10-14 M (from lead sulfide)

[S 2-] = 1.8x10-14 M

Summarizing:

[Pb2+] = 1.3x10-7 M + 1.8x10-14 M

[Pb2+] = 1.3x10-7 M

[CrO4 2-] = 1.3x10-7 M

[S 2-] = 1.8x10-14 M