Let's start by writing down the reactions of both solids solubilizing:
[tex]PbCrO_4\rightleftarrows Pb^{2+}+CrO_4^{2-}\text{ Ksp = 1.8*10-14}[/tex][tex]PbS\rightleftarrows Pb^{2+}+S^{2-}\text{ Ksp = 3.2*10-28}[/tex]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
[tex]x=\sqrt{1.8\times10^{-14}}[/tex]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
[tex]x=\sqrt{3.2\times10^{-28}}[/tex]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
