A solution contains 0.047 M Ag and 0.032 M Pb2. If you add Cr, AgCl and PbCl2 wi

Question

A solution contains 0.047 M Ag and 0.032 M Pb2. If you add Cr, AgCl and PbCl2 will begin to precipitate. Wha is the concentration of CI required, in molarity, when A AgCI precipitation begins? B. AgCl precipitation is 99.99% complete? Number Number 3.77x 10 M 3.77x 10 M C. PbCl2 precipitation begins? D. PbCl2 precipitation is 99.99% complete? Number Number 0.023 Finally, give the CI concentration range in which Ag" can be completely separated from Pb2* by precipitation. E. Give the lowest Cr concentration for the F. Give the highest CI concentration for the separation of Ag from Pb2+ separation of Ag from Pb2

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^+]

                                                                                        = 1.77×10-10/0.047 M = 37.65 *10^-10 M

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

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

[Cl-] = Ksp/[Ag+] = 1.77×10-10/ 4.7*10^-6 M = 3.76*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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