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.|
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, S2-. Three equilibria are involved:
H2S(aq) H++ HS– K1 = [ H+] [HS–] / [H2S] = 5.7×10−8
HS– H+ + S2- K2 = [ H+]2 [ S2−] / [ HS–] =1×10 −1 9
The low value of K2 suggests that there is very little free S2- in aqueous solutions unless they are extremely basic.
Therefore, the precipitation of metal sulfides is best written
M2+ + H2S MS(s) + 2H+