Consider a Solution of 0.1M ZN2+ and 0.1M Cu2+ that issaturated with H2S.The sol

Question

Consider a Solution of 0.1M ZN2+ and 0.1M Cu2+ that issaturated with H2S.The solubility product constants for ZnS and CuSare 1.1x10^-21 and 6x10^-36.
A. show that both ZnS and Cus will precipitate at a PH of8

B. Show that only CuS will precipitate when (H3O+) = 0.3M Consider a Solution of 0.1M ZN2+ and 0.1M Cu2+ that issaturated with H2S.The solubility product constants for ZnS and CuSare 1.1x10^-21 and 6x10^-36.
A. show that both ZnS and Cus will precipitate at a PH of8

B. Show that only CuS will precipitate when (H3O+) = 0.3M

Explanation / Answer

0.1 M Zn2+ if the Ksp for ZnS = 1.1 x 10^-21

K = [Zn2+][S2-]

1.1x 10^-21 = [0.1] [S2-]

S2- required to saturate is 1.1 x 10^-20 M for ZnS and 0.1 M Cu2+ if Ksp for CuS is 6x10^-36.

K = [Cu2+][S2-]

[S2-] = 6 x 10^-36/0.1 = 6 x10^-35 M for CuS.

For H2S

Ka1 = 1 x 10^-17

Ka2 = 1x 10^-19

So,

K = 1 x 10^-26

(a) now, we have to show ZnS and CuS will precipitate at pH = 8

H2S <------>2H+ + S2-

K = [H+]^2 [S2-]/[H2S]

1 x 10^-26 = [ 1 x 10^-8]^2 [S2-]/[0.1]

[S2-] = 1 x 10^-11 M which exceeds what we calculated as required to saturate either solution ( 6 x 10^-35 M for CuS and 1.1 x10^-20 M for ZnS)

So both will precipitate.

(b) We have to show CuS will precipitate when [H3O+] = 0.3 M

H2S <-------->2H+ + S2-

K = [H+]^2 [S2-]/[H2S]

1.1 x 10^-26 = [0.3]^2 [S2-]/[0.1]

[S2-] = 1 x 10^-27/[0.09]

[S2-] = 1.1 x 10^-26 M

which exceeds 6 x 10^-35 M as required to saturate CuS So CuS will precipitate.

But it is 10^6 times smaller than 1.1 x 10^-20 M So ZnS will not precipitate

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