Question

describe how adding a solution of koh to the original sulfide precipitate results in the seperation of SN4+ ion from the other group II cation

Answer 1

My guess for number one would be that since KOH is a strong base, it helps ionize the origional sulfide, whatever that may be. Not too sure about that though.

When you use the terms weak and strong acids or bases, it means how well the compound ionizes when in solution. Weak acids and bases do not ionize completely, where as strong acids do. Since NH3 is a weak base, it won't ionize completely in solution and KOH does.
Sn in oxidation state +4 is amphoteric, and dissolves in high concentrations of OH- as stannate(IV). This is not possible with NH3 because it's such a weak alkali.

Answer 2

According to the lab manual, "Of the four Group II Sulfides, only Tin (IV) Sulfide (SnS₂, stannic sulfide) is soluble in excess concentrated KOH. The reaction occurs because of the amphoteric nature of SnS₂, its ability to act as both an acid and a base. The result of 3SnS₂ + 6OH^-$\rightleftharpoons$ [Sn(OH)₆]^2- + 2[SnS₃]^2- is that the supernatant liquid contains the two dissolved tin species, and the other Group II Sulfides remain in the precipitate.

Don't forget to rate this answer please. :-)