Question

The Acid-insoluble sulfides and Base-insoluble sulfides can be separated from each other by changing the pH of the aqueous solution that contains them. At low pHs the Acid-insoluble sulfides will precipitate out. What role does the acid play in this process?

		The H+ ion is produced by the dissolving sulfide, so the presence of an acid hinders the dissolution process.
		The sulfide reacts with the H+ ion, forming the cation and H2S.
		The sulfide reacts with any OH- ions present forming S(OH)2 and the cation.
		The sulfide forms a complex ion with the H+ provided by the acid.
		The acid does not play a role in the dissolution process of metal sulfides.

Please explain. Thank you.

Answer 1

The H+ ion is produced by the dissolving sulfide, so the presence of an acid hinders the dissolution process.

The dissolved molecular hydrogen sulfide dissociates into hydrogen ions, hydrogen sulfide ions, HS^- , and sulfide ions, S^2-. Three equilibria are involved:

H₂S(g) H₂S(aq)

H₂S(aq) H⁺+ HS^- K₁ = [ H⁺] [HS^-] / [H₂S] = 5.7x10⁻⁸

HS^- H⁺ + S^2- K₂ = [ H⁺]² [ S²⁻] / [ HS^-] =1x10 ^{−1 9}

The low value of K₂ suggests that there is very little free S^2- in aqueous solutions unless they are extremely basic.

Therefore, the precipitation of metal sulfides is best written

M²⁺ + H₂S MS(s) + 2H⁺