Kwok hei asked in 科學及數學化學 · 9 years ago

F5. Chem問題

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)

我想問隻blue precipitate係咩?同埋點解隻野出黎之後會redissolve?

4 Answers

  • andrew
    Lv 7
    9 years ago
    Favorite Answer

    The blue precipitate is copper(II) hydroxide.

    The blue precipitate is redissolved to give a blue solution of copper(II) ions.

    Explanation :

    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
  • 9 years ago

    Result of not reading the question carefully.

  • 9 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 ?

  • 9 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.

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