A bit of theory to refresh your mind about the solubility product:
Ks = the given solubility constant
Ks = [Pb2+].[I-]^2
7,1*10^(-3) = [Pb2+].[I-]^2
-> The solubility of this compound is then:
S= (Ks/4)^1/3 = 1,2 mol/L
Now my question:
What is the solubility of Ag2SO4 (Ks...