a solution contains both Ag + and Pb 2+ ions, the concentration of Ag + ions is 1.5 X 10^-2 M and Pb 2+?

Question

0.5x10^-2 M. Selectively precipitate the Ag + and Pb 2+ ions by adding an aqueous solution of Chloride ions.Given the Ksp for Silver Chloride as 1.8x10^-10 and that for Pb 2 Chloride as 1.6x10^-5, determine the concentration of Chloride ions required to bring about precipitation of each ion.also determine which ion (which salt) will precititate first


plz,plz help with this home work question, i don't even know where to begin. Thanks

Answer 1

Write down the reactions and the respective equations. Then solve for the chloride concentration. In detail:

AgCl <=> Ag+ + Cl-

Ksp1= [Ag+][Cl-] => [Cl-]=Ksp1/[Ag+] = (1.8*10^-10)/(1.5*10^-2) => [Cl-]= 1.2*10^-8 M.
So AgCl will precipitate if [Cl-]> or =1.20*10^-8 M

PbCl2 <=> Pb+2 +2Cl-

Ksp2= [Pb+2][Cl-]^2 =>
[Cl-]= Squareroot (Ksp2/[Pb+2]) =SQRT((1.6*10^-5)/(0.5*10^-2))
=> [Cl-]= 0.057 M
So PbCl2 will precipitate if [Cl-]> or = 0.057 M

As you can see AgCl requires extremely less Cl- to precipitate so it will be the first to precipitate