Question 3 - Precipitation Separation MnS is a sparingly soluble solid in aqueou

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Question 3 - Precipitation Separation MnS is a sparingly soluble solid in aqueous solution: MnS(s)--> Mn2+ +S2- Ksp = 3.00 x 10-11 FeS is also a sparingly soluble solid in aqueous solution: Fes(s)-> Fe2+ + S2- Ksp = 8.00 x 10-19 Consider a solution that has both Mn2+ and Fe2+ each at a concentration of 1.00 mM. We can separate these two varying the pH because the sulfide ion (S2-) is the dianion of the diprotic acid H2S (pK 1 = 7.01 ; pK2= 13.89). Given a total sulfide concentration of 0.100 M, find the following: a) At what pH will FeS(s) begin to precipitate? b) At what pH will MnS(s) begin to precipitate? c) What is the [Fe2+] in solution at the pH you calculate in part b? What percentage of the Fe2+ remains in solution at that pH?

Explanation / Answer

Selective precipitation

pKa = -logKa

a) H2S <==> 2H+ + S^2-

Ka1.Ka2 = [H+]^2.[S^2-]/[H2S]

10 x 10^-8 x 13 x 10^-14 = [H+]^2[S^2-]/0.1

[H+]^2 = 1.3 x 10^-21/[S^2-]

and,

FeS Ksp = [Fe2+][S^2-]

[Fe2+] = 1 x 10^-3 M

[S^2-] = Ksp/[Fe2+] = 8 x 10^-19/0.001 M = 8 x 10^-16

So,

[H+]^2 = 1.3 x 10^-21/[S^2-]

[H+] = sq.rt.(1.3 x 10^-21/8 x 10^-16) = 1.27 x 10^-3 M

pH = -log[H+] = 2.89

So, at pH 2.89, FeS would begin to precipitate.

b) H2S <==> 2H+ + S^2-

Ka1.Ka2 = [H+]^2[S^2-]/[H2S]

10 x 10^-8 x 13 x 10^-14 = [H+]^2[S^2-]/0.1

[H+]^2 = 1.3 x 10^-21/[S^2-]

and,

MnS Ksp = [Mn2+][S^2-]

[Mn2+] = 1 x 10^-3 M

[S^2-] = Ksp/[Mn2+] = 3 x 10^-11/0.001 M = 3 x 10^-8

So,

[H+]^2 = 1.3 x 10^-21/[S^2-]

[H+] = sq.rt.(1.3 x 10^-21/3 x 10^-8) = 2.1 x 10^-7 M

pH = -log[H+] = 6.68

So, at pH 6.68, MnS would begin to precipitate.

c) at pH 6.68,

pH = -log[H+]

[H+] = 2.1 x 10^-7 M

[S^2-] = 1.3 x 10^-21/[H+]^2

          = 1.3 x 10^-21/(2.1 x 10^-7)^2 = 2.95 x 10^-8 M

amount of [Fe2+] remained in solution = Ksp/[S^2-]

                                                              = 8 x 10^-19/2.95 x 10^-8 = 2.71 x 10^-11 M

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