Question

A solution contains 0.032 M Ag and 0.015 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 separation of Ag and Pb2 .

E. Concentration of Cl– at the start:

F. Concentration of Cl– once complete:

Answer 1

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 0.032 M *[Cl^-] = 1.6*10^-10

or, [Cl^-] = 1.6*10^-10/0.032 = 5*10^-9 M

(a) Ag Cl will start precipitating when Cl- concentration is 5*10^-9 M.

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

0.01% Ag^+ = 0.032 *0.01% = 3.2 *10^-6 M

[Cl-] = ksp/[Ag^+] = 1.6*10^-10/ 3.2 *10^-6 = 0.5*10^-4 M

Concnetration of Cl- when 99.99% AgCl is precipitated is 0.5*10^-4 M.

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

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

[Cl-] = (2.4*10^-4/0.015 ) = 0.126 M

PbCl2 will start precipitate when Cl- concentration is 0.126 M.

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(d)

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

0.01% Ag^+ = 0.015 *0.01% = 1.5 *10^-6 M

[Cl-] = ksp/[Pb2+] = 2.4*10^-4/ 1.5 *10^-6 = 12.65 M

Concnetration of Cl- when 99.99% PbCl2 is precipitated is 12.65 M.