Given that: copper(II)ion react with ammonia in the following way:
Cu^2+ + 4(NH3) <-> Cu[(NH3)4]^2+
which is a reversible reaction
When dilute sulphuric acid is slowly added to the solution mixture, a blue precipitate is formed. When excess sulphric acid is added, the blue precipitate dissolves, forming a blue solution.
Account for the change with corresponding equation(s)
- andrewLv 79 years agoFavorite Answer
The blue precipitate is copper(II) hydroxide.
The blue precipitate is redissolved to give a blue solution of copper(II) ions.
Cu^2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4]^2+(aq)...... (*)
Reaction (*) attains a state of equilibrium in the presence of an excess of NH3. When "a small amount" of sulphuricacid is added, "a small amount" of ammonia is removed.
NH3(aq) + H^+(aq) → NH4^+(aq)
According to Le Chatelier's principle, the equilibrium position of reaction (*)shifts to the left and thus a part of the [Cu(NH3)4]^2+(aq)complex ions are converted to Cu^2+(aq) and 4NH3(aq). This makes the concentration of copper(II)ions increase.
Because only "a small amount" of ammonia is removed, there is still acertain concentration of ammonia in the reaction mixture and thus the reactionmixture is still alkaline.
NH3(aq) + H2O(l) ⇌ NH4^+(aq) +OH^-(aq)
As mentioned above, the concentration of copper(II) ions increases due to theshift of equilibrium position of reaction (*). The copper(II) ions will react with the hydroxide ions to give a blue precipitateof copper(II) hydroxide.
Cu^2+(aq) + 2OH^-(aq) ⇌ Cu(OH)2(s) ...... (#)
As more and more sulphuric acid added, the hydrogen ions in the sulphuric acid removethe hydroxide ions in the reaction mixture. According to Le Chatelier's principle, this causes the equilibriumposition of reaction (#) to shift to the left, and thus the copper(II)hydroxide precipitate redissolves.
To make it simple, one can say that the copper(II) hydroxide precipitate isredissolved in excess sulphuric acid to form a blue solution of copper(II) ions.
Cu(OH)2(s) + 2H^+(aq) → Cu^2+(aq) + 2H2O(l)
2012-06-18 15:46:14 補充：
The precipitate must NOT be CuSO4, because it is soluble in water.
2012-06-18 15:54:14 補充：
Remember that the experiment of adding ammonia solution dropwise to copper(II) sulphate solution until excess. A blue ppt of copper(II) hydroxide is firstly formed.
Cu^2+(aq) + 2OH^-(aq) → Cu(OH)2(aq)
2012-06-18 15:55:56 補充：
2012-06-18 18:12:57 補充：
Adding an excess of ammonia solution will redissolve the copper(II) hydroxide ppt to give a deep blue solution of [Cu(NH3)4]^2+(aq) ions.
Cu(OH)2(s) + 4NH3(aq) → [Cu(NH3)4]^2+(aq) + 2OH^-(aq)
Adding sulphuric acid dropwise to such reaction mixture, the series of reactions is reversed.
2012-06-18 18:15:51 補充：
It should be Cu(OH)2(s).Source(s): andrew, andrew, andrew, andrew
- 己式庚辛Lv 79 years ago
Result of not reading the question carefully.
- 冷眼旁觀Lv 79 years ago
(1) Copper(II) sulphate is soluble in water. How can it form a blue precipitate in aqueous solution ?
(2) [Cu(NH3)4]^2+ is the short form of [Cu(NH3)4(H2O)]^2+, which is deep blue but not only blue. Why can it change to CuSO4 and then change back to itself ?
- 馬王肺話Lv 69 years ago
Copper (II) sulphate
2012-06-18 14:02:14 補充：
When excess sulphric acid is added, the copper (II) sulphate react with amonia and formed tetraaminediaqua copper II complex ions ((Cu(NH3)4(H20))2+) which is soluble in water.