A solution contains 0.040 M Hg22 and 0.032 M Pb2 . If you add Cl–, Hg2Cl2 and Pb

Question

A solution contains 0.040 M Hg22 and 0.032 M Pb2 . If you add Cl–, Hg2Cl2 and PbCl2 will begin to precipitate. What is the concentration of Cl– required, in molarity, when:

A. Hg2Cl2 precipitation begins?

B. Hg2Cl2 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 Hg22 and Pb2 .

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

F. Concentration of Cl– once precipitation complete:

Please help!!

Explanation / Answer

Ksp (Hg2Cl2) = 1.4*10^-18

Ksp (PbCl2) = 1.2*10^-5

Precipitation of Hg2Cl2 will begin when [Cl^-] :

Ksp = [Hg2^2+][Cl^-]^2

1.4*10^-18 = 0.04 * [Cl^-]^2

or, [Cl^-] = 5.92 *10^-9 M

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Precipitation of PbCl2 will occur when [Cl^-] :

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

or, 1.2*10^-5 = 0.032 * [Cl^-]^2

or, [Cl^-] = 0.0194 M

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The precipitation will be 99.99% complete when 99.99% of Hg2^2+ is precipitated.

amount of Hg2^2+ precipitated = 0.04* 99.99% = 0.0399 M

amount of Cl- added = 2* 0.0399 M =0.0799 M

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Precipitation of PbCl2 will be 99.99% complete when 0.032 * 99.99% = 0.0319 M Pb^2+ is precipitated.

amount of Cl- added = 2* 0.0319 M = 0.0639 M

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concentration of Cl- to precipitate Hg2^2+ completely = 0.08 M

concentration of Cl- to precipitate Pb^2+ completely = 0.064 M

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