Question

A solution contains 0.041 M Ag and 0.047 M Pb2 . If you add Cl–, AgCl and PbCl2 will begin to precipitate. What is the concentration of Cl– required, in molarity, when A. AgCl precipitation begins? B. AgCl precipitation is 99.99% complete? C. PbCl2 precipitation begins? D. PbCl2 precipitation is 99.99% complete? Finally, give the Cl– concentration range in which Ag can be completely separated from Pb2 by precipitation. E. Give the lowest Cl– concentration for the F. Give the highest Cl–concentration for the separation of Ag from Pb2 : separation of Ag from Pb2 :

Answer 1

ans)

Ksp of AgCl is 1.6*10^-10 and that of PbCl2 is 2.4*10^-4.

The salts will start precipitating from the solution when their ionic product will be equal to Ksp value.

AgCl ==> Ag^+ + Cl^-

ionic product = Q = [Ag^+][Cl^-]

[Ag^+][Cl^-] = Ksp

or 0.041 M *[Cl^-] = 1.6*10^-10

or, [Cl^-] = 1.6*10^-10/0.041 = 3.90*10^-9 M

(a) Ag Cl will start precipitating when Cl- concentration is 3.90*10^-9 M.

---------------------------------------------------

When AgCl is 99.99% that means the Ag^+ is 0.01% left in the solution.

0.01% Ag^+ = 0.041 *0.01% = 4.1 *10^-6 M

[Cl-] = ksp/[Ag^+] = 1.6*10^-10/ 4.1*10^-6 = 0.39*10^-4 M

Concnetration of Cl- when 99.99% AgCl is precipitated is 0.39*10^-4 M.

-----------------------------------------------------

(c) PbCl2 ==> Pb^2+ + 2Cl^-

Ksp = [Pb^2+][Cl^-]^2 = 2.4*10^-4

[Cl-] = (2.4*10^-4/0.047 ) = 0.071 M

PbCl2 will start precipitate when Cl- concentration is 0.071 M.

---------------------------------------------------------

(d)

When PbCl2 is 99.99% that means the Pb^2+ is 0.01% left in the solution.

0.01% Ag^+ = 0.047 *0.01% = 4.7 *10^-6 M

[Cl-] = ksp/[Pb2+] = 2.4*10^-4/ 4.7*10^-6 = 7.14 M

Concnetration of Cl- when 99.99% PbCl2 is precipitated is 7.14 M.