Question

2. (1) Describe how adding a solution of KOH to the original sulfide precipitate results in the separa- tion of Sn4+ ion from

Why wouldnt a weak base, such as NH3, work just as well as KOH in (1)?

2.2 Is the question I have. I figured out 2.1 already.

Below is relevant info. Thanks!

I. Separating Selected Group II Cations -5 group from sepurades 1 catrons one another Figure 1 is a flowchart showing the sep

M + T.X TU VI 30 (49) » 9) 1.3 x 10-18 1.0 x 10-2M H,S + 1.0 x10-2M H2O+(aq) 2 1.0 x 10-2M H,S +0.10M H30+(aq) d 1. 1.0 x 10-

Info Hoy ANAL 365 The Chemistry binding OH-ion. Only tin(IV) sulfide (SnS2, stannic sulfide) is soluble in excess concentrate

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Answer #1

Ammonia can't be used as KOH because unlike KOH ammonia does not form precipitate with Sn4+ and group II cations . Ammonia forms complexes with these cations which are soluble and hence addition of ammonia can't separate Sn2+ from group II cations

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