Question 21.2: When a solution containing Cu2+ is treated with hydrogen sul...
When a solution containing Cu^{2+} is treated with hydrogen sulfide, a black precipitate forms. When another portion of the solution is treated with ammonia, a blue precipitate forms. This precipitate dissolves in excess ammonia to form a deep blue solution containing the Cu(NH_{3})_{4}^{2+} ion. Write balanced net ionic equations to explain these observations.
1. The black precipitate is CuS.
Cu^{2+}(aq) + H_{2}S(aq) → CuS(s) + 2H^{+}(aq)
2. An aqueous solution of ammonia consists of NH_{3} and H_{2}O. Since the blue precipitate must be Cu(OH)_{2}, water has to be a reactant.
Cu^{2+}(aq) + 2NH_{3}(aq) + 2H_{2}O → Cu(OH)_{2}(s) + 2NH_{4}^{+}(aq)
3. The precipitate, Cu(OH)_{2}, reacts with excess ammonia to form Cu(NH_{3})_{4}^{2+}.
Cu(OH)_{2}(s) + 4NH_{3}(aq) → Cu(NH_{3})_{4}^{2+}(aq) + 2 OH^{-}(aq)