Question

Let's explore the efficiency of chloride ion precipitation in seperating Ag^+ from Pb^2+ in a solution. Suppose that [Pb^2+]= 0.045 mol/L and [Ag^+]= 0.084 mol/L. Calculate the chloride ion concentraion required to just start precipitation of Pb^2+ ion and then calculate the concentration of Ag^+ left in solution at that chloride ion concentratio(Assume the volume of the solution does not change). Is this an efficient method of separating the two ions?

Answer 1

Ksp = (Pb^2+)(Cl^-)^2
Substitute 0.045M for Pn^2+ and solve for Cl^-.
Then Ksp = (Ag^+)(Cl^-)
Use that Cl^- and solve for Ag^+ remaining.