When a solution containing Cu2+ 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)42+ ion. Write balanced

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+}[/latex](aq) + [latex]H_{2}S[/latex](aq) → CuS(s) + 2[latex]H^{+}[/latex](aq)

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+}[/latex](aq) + 2[latex]NH_{3}[/latex](aq) + 2[latex]H_{2}O[/latex] → [latex]Cu(OH)_{2}[/latex](s) + 2[latex]NH_{4}^{+}[/latex](aq)

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}[/latex](s) + 4[latex]NH_{3}[/latex](aq) → [latex]Cu(NH_{3})_{4}^{2+}[/latex](aq) + 2 [latex]OH^{-}[/latex](aq)

Question 21.2: When a solution containing Cu2+ is treated with hydrogen sul...

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