A solution contains 0.049 M Hg22 and 0.021 M Pb2 . If you add Cl–, Hg2Cl2 and Pb

Question

A solution contains 0.049 M Hg22 and 0.021 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 Cl– concentration range in which (Hg2)2 can be completely separated from Pb2 by precipitation. E. Give the lowest Cl– concentration for the F. Give the highest Cl– concentration for the separation of (Hg2)2 from Pb2 : separation of (Hg2)2 from Pb2

Explanation / Answer

Ksp Hg2Cl2 = 1.3 x 10^-18 and Ksp PbCl2 = 1.7 x 10^-5

So, Hg2Cl2 will precipitate first from the solution

A. [Cl-] required to begin Hg2Cl2 precipitation

= sq.rt.(Ksp/[Hg2^2+])

= sq.rt.(1.3 x 10^-18/0.049)

= 5.15 x 10^-9 M

B. when Hg2Cl2 precipitation is 99.99% complete

[Hg2^2+] precipitated = 0.9999 x 0.049 M = 0.0489951 M

[Cl-] required for 99.99% Hg2Cl2 precipitation

= sq,rt(1.3 x 10^-18/0.0489951)

= 5.151 x 10^-9 M

C. [Cl-] when PbCl2 begins precipitation

= sq.rt.(Ksp/0.021)

= sq.rt.(1.7 x 10^-5/0.021)

= 0.028 M

D. when PbCl2 precipitation is 99.99% complete

[Cl-] needed = sq.rt.(1.7 x 10^-5/(0.021 x 0.9999))

                     = 0.0284 M

at this point all of Hg2Cl2 has already precipitated

So,

total [Cl-] needed = 5.15 x 10^-9 + 0.0284

                             = 0.0284 M

E. lowest [Cl-] for separation = sq.rt.(1.3 x 10^-18/0.049 x 0.99) = 5.177 x 10^-9 M

for 1% precipitation of Hg2Cl2

F. higest [Cl-] for separation = sq.rt.(1.3 x 10^-18/0.049 x 0.01) = 5.151 x 10^-8 M

for 99% precipitation of Hg2Cl2

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