A solution contains 0.028 M Ag and 0.032 M Pb2 . If you add Cl–, AgCl and PbCl2

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!

Explanation / Answer

Ksp of AgCl = 1.77×10-10

AgCl <==> Ag+ + Cl^-

A. precipitation of Ag+ will begin when concentration of Cl- is = Ksp/[Ag^+]

                                                                                        = 1.77×10-10/0.028 M = 63.21 *10^-10 M

B. when 99.99% Ag+ is precipitated, 0.01% Ag+ is left in the solution.

[Ag+] = 0.01 % * 0.028 M = 2.8*10^-6 M

[Cl-] = Ksp/[Ag+] = 1.77×10-10/ 2.8*10^-6 M = 6.32*10^-5 M

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C. Ksp of PbCl2 = 1.70*10^-5 M

PbCl2 <==> Pb^2+ + 2Cl-

Ksp = [Pb^2+][Cl^-] ^2

or, 1.70*10^-5 = 0.032 * [Cl^-]^2

or, [Cl^-] = 0.023 M

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D.

0.01% of 0.032 M Pb^2+ = 2.3*10^-6 M

Ksp = (2.3*10^-6) * (Cl^-)^2

or, 1.7*10^-5 = (2.3*10^-6) * (Cl^-)^2

or, [Cl-] = 2.72 M

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E. PbCl2 will precipitate first. [Cl-] = 2.72 M.

F. After the prcipitation of AgCl is complete [Cl-]= 6.32*10^-5 M

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