Question

A solution contains 0.040 M Hg2^2+ and 0.015 M Pb^2+ . 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 separation of Hg22 and Pb2 .

E. Concentration of Cl– at the start:

F. Concentration of Cl– once complete:

Answer 1

Sol :-

K_sp of Hg₂Cl₂ = 1.10 x 10^-18  

K_sp of PbCl₂ = 1.7 x 10^-5

(a). Partial dissociation of Hg₂Cl₂ is :

Hg₂Cl₂ <--------------> Hg₂²⁺ + 2Cl^-

Expression of K_sp is :

K_sp = [Hg₂⁺²]. [2Cl^-]² = 1.10 x 10^-18

So,

4 [Cl^-]² = 1.10 x 10^-18 / 0.040

[Cl^-]² = 6.875 x 10^-18

[Cl^-] = 2.62 x 10^-9 M

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(b). Precipitation is 99.99% completes means the % of Hg₂⁺² left = 0.01 %

The 0.01% of Hg₂⁺² = 0.01 x 0.04 / 100 = 4 x 10^-6

K_sp = [Hg⁺²] [2 Cl^-]² = 1.10 x 10^-18

4[Cl^-]² = 1.10 x 10^-18 / 4 x 10^-6

[Cl^-]² = 0.06875 x 10^-12

[Cl^-] = 2.62 x 10^-7 M

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(c). Partial dissociation of PbCl₂ is :

PbCl₂ <---------------------> Pb²⁺ + 2 Cl^-

Expression of K_sp is :

K_sp = [Pb⁺²] [2Cl^-]²

K_sp = 4[Pb⁺²] [Cl^-]²

1.7 x 10^-5 / 4 X 0.015 = [Cl^-]²

[Cl^-] = 1.68 X 10^-2 M

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(d). PbCl₂ ppt . 99.99%

Which mean Pb⁺² left = 0.01 % of 0.015 = 1.5 X 10^-6 M

1.7 x 10^-5 = K_sp =[Pb⁺²] [2Cl^-]²

[Pb⁺²] [2Cl^-]² = 1.7 x 10^-5

[Cl^-] = 1.7 x 10^-5 / 4 x 1.5 x 10^-6

[Cl^-] = 1.68 M