Question

A solution contains 0.028 M Ag and 0.032 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 concentration range of Cl– for the complete precipitation of Ag and Pb2 .

E. Concentration of Cl– at the start of precipitation:

F. Concentration of Cl– once precipitation complete:

So confused. Please help! Already not doing so well on this quiz. Greatly appreciate the help!

Answer 1

Ksp (AgCl) = 1.0 x 10-10

Ksp (PbCl2) = 1.7 x 10-5

(a)

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

AgCl <==> Ag+ + Cl-

Q = [Ag+][Cl-]

[Ag+][Cl-] = Ksp

0.028 M x [Cl-] = 1.0 x 10-10

[Cl-] = 3.57 x 10-9 M

AgCl will start precipitating when Cl- concentration is 3.57 x 10-9 M

(b)

When AgCl is 99.99 % ====> Ag+ is 0.01 % left in the solution.

0.01% Ag+ = 0.028 x 0.01% = 2.8 x 10-6 M

[Cl-] = Ksp/[Ag+] = 1.0 x 10-10 / 2.8 x 10-6 = 3.57 x 10-5 M

Concnetration of Cl- when 99.99% AgCl is precipitated = 3.57 x 10-5 M

(c)

PbCl2 ==> Pb2+ + 2Cl-

Ksp = [Pb2+][Cl-]2

1.7 x 10-5 = 0.032 M x [Cl-]2   

[Cl-] = 0.023 M

PbCl2 will start precipitate when Cl- concentration is 0.023 M.

(d)

When PbCl2 is 99.99% ====> Pb2+ is 0.01 % left in the solution.

0.01 % Ag+ = 0.032 x 0.01 % = 3.2 x 10-6 M

[Cl-] = √Ksp/[Pb2+] = √1.7 x 10-5/ 3.2 x 10-6 = 2.3 M

Concnetration of Cl- when 99.99 % PbCl2 is precipitated is 2.3 M.