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

2(a).

Sn⁴⁺ hydroxide, which is Sn(OH)₄, is extremely insoluble, just as the sulfide is (Ksp = 1.0 x10^-57 and 1.0 x10^-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.

(b)

Ammonia can't be used because 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 NH₃ 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.

Even can't use other bases, since they aren't more basic than the conjugate base of hydrogen sulfide. Hydrogen sulfide is a very weak acid, thus the conjugate base (S^2-) is a pretty strong base. So, you have to use a base that is more basic than the sulfide, which leaves you with very few choices, especially in aqueous solution.