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(NH3)4^2+ ion. Write

Question

When a solution containing [latex]Cu^{2+}[/latex] 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 [latex]{Cu(NH_{3})_{4}}^{2+}[/latex] ion. Write balanced net ionic equations to explain these observations.

Accepted Answer

1. The black precipitate is CuS.
[latex]Cu^{2+}(aq) + H_{2}S(aq) \longrightarrow CuS(s) + 2H^{+}(aq)[/latex]
2. An aqueous solution of ammonia consists of [latex]NH_{3}[/latex] and [latex]H_{2}O[/latex]. Since the blue precipitate must be [latex]Cu(OH)_{2}[/latex], water has to be a reactant.
[latex]Cu^{2+}(aq) + 2NH_{3}(aq) + 2H_{2}O \longrightarrow Cu(OH)_{2}(s) + 2{NH_{4}}^{+}(aq)[/latex]
3. The precipitate, [latex]Cu(OH)_{2}[/latex], reacts with excess ammonia to form [latex]{Cu(NH_{3})_{4}}^{2+}[/latex].
[latex]Cu(OH)_{2}(s) + 4NH_{3}(aq) \longrightarrow {Cu(NH_{3})_{4}}^{2+}(aq) + 2OH^-(aq)[/latex]

Question 21.2: When a solution containing Cu^2+ is treated with hydrogen su...

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