# How to identify an emerald green solution with possibly two anions?

The solution is emerald green in colour. When added with water, it gives a pale blue solution. When reacted with $$\ce{BaCl2}$$, it gives a sky blue solution and a white precipitate. When reacted with $$\ce{AgNO3}$$, it gives a sky blue solution and a white precipitate.

This does not make any sense to me. The $$\ce{BaCl2}$$ is used to test sulphates ions and $$\ce{AgNO3}$$ is used to test halides. How could you have two negative ions?

• Why can't? The solution get some color, so it's reasonable to surmise there are some cations, and the solution keep electrically neutral. Besides, $\ce {SO4^2-}$ and $\ce {Cl-}$ won't react. However, I'm not sure what ions make those color, so I give a comment instead. – Asydot Apr 30 '15 at 11:54
• Can you give the procedure of the experiment. I might be some mistakes in it. – Simon-Nail-It Apr 30 '15 at 12:21
• What's to say there isn't a mixture in solution? – Geoff Hutchison Apr 30 '15 at 15:50

You can have two different anions if your solution is a mixture of two different metal salts, a chloride and a sulfate. The color change from green to blue upon dilution with water fits well with $$\ce{Cu^2+}$$ being the corresponding cation. Concentrated solutions of $$\ce{CuCl2}$$ are green due to the presence of chlorocuprates, like $$\ce{[CuCl3]^{-}}$$ and $$\ce{[CuCl4]^{2-}}$$. When water is added, a stepwise ligand exchange takes place, and the pale blue color of $$\ce{[Cu(H2O)6]^{2+}}$$ starts to appear. The image below shows copper(II) chloride solutions of different concentrations: Green when the chloride concentration is high, and blue when it is lower (Wikipedia: copper chloride).

In summary, the observations (color change, precipitates) indicate that you have an aqueous solution of $$\ce{CuCl2}$$ and $$\ce{CuSO4}$$.

Assuming some wording modifications of the problem as stated, I would suggest as a single salt candidate solution: $$\ce{[Cu(NH3)5(H2O)]Cl2}$$. Other candidates include: $$\ce{[Cu(NH3)4(H2O)2]Cl2}$$ and $$\ce{[Cu(NH3)3(H2O)3]Cl2}$$.

The appearance of the cation Cu(NH3)x(H2O)y, where x + y = 6, also changes color with high values of x producing more intense blue coloration.

In the presence of AgNO3, the creation of insoluble white silver chloride:

$$\ce{[Cu(NH3)2(H2O)4]Cl2 + 2 AgNO3 -> [Cu(NH3)2(H2O)4](NO3)2 + 2 AgCl(s)}$$

Based on the cited reaction (see, for example, this reference):

$$\ce{[Cu(H2O)4](2+) + 2 NH3 ⇔ [Cu(NH3)2](2+) + 4 H2O}$$

The alkaline nature of the above implies on contact with aqueous barium ions (from BaCl2) the formation of the insoluble white Ba(OH)2.

This is, in my opinion, a good question for grading, given the number of possible reasonable answers.

Note, technically, I am assuming the correct wording of the problem is that 'upon dilution the coloration declines' so as to arrive at a single salt answer. Or, if one is starting with little NH3 in the presence of CuCl2, and one continues to add concentrated aqueous NH3, then coloration increases, with the answer for the starting solution being $$\ce{[Cu(NH3)(H2O)5]Cl2}$$.

• Ammonia will not precipitate barium hydroxide, which is generally considered a strong base. – Oscar Lanzi Feb 5 '20 at 2:41