Question

2(a). Describe how adding a solution of KOH to the original sulfide precipitate results in the separation of Sn4+ ion from the other Group II cations.

2(b). Why wouldn’t a weak base, such as aqueous NH3 work just as well as aqueous KOH in (a)?

Answer 1

You haven't given quite enough information about the "original precipitate" for me to be totally sure, but it sounds to me that you have a precipitate of group II cations (Ca2+, Ba2+, Mg2+, etc.) along with the Sn4+.

Tin IV hydroxide, which is Sn(OH)4, is extremely insoluble, just as the sulfide is (Ksp = 1.0 *10-57 and 1.0 *10-70, respectively). However, the hydroxides of the group II cations are relatively soluble, much more soluble than the corresponding sulfides. Thus, adding KOH to the solution will dissolve these other metal ions back into solution, leaving only the Sn4+ precipitate, which you could then just filter or centrifuge out.

The reason you can't use ammonia is that it forms a complex with all of the cations mentioned. Thus, it would dissolve everything, and you would have no precipitate. It has nothing to do with the fact that ammonia is a weaker base, it has to do with the fact that ammonia forms these complexes (in other words, it is a good chelating agent and isn't very selective).

The reason you can't use other bases is that they aren't more basic than the conjugate base of hydrgen sulfide. Huh? Well, hydrogen sulfide is a very weak acid, thus the conjugate base (S2-) is a pretty strong base. Thus, you have to use a base that is more basic than the sulfide, which leaves you with very few choices, especially in aqueous solution (really only hydroxide is practical).