Why will more Cu(IO3)2 dissolve in H20 + NH3 than in pure H20?

Question

I can't find information on this in my textbook. Below is my guess:

"

Eqn 1: Cu(IO3)2 --> Cu++ + IO3-

Eqn 2: Cu++ + NH3 --> Cu(NH3)++

The reaction of the second equation would create a shift to the right in the first equation... causing an increase in the dissolving of Cu(IO3)2."

Does this seem right? Is Cu(NH3)++ even possible??

Answer 1

Your guess is almost perfect. This means that you understand the principle, but do not have information on Complexing ions formed by Copper ions and Ammonia.  

When excess ammonia is added to a solution of copper(II) ions, ( or a dilute solution of less soluble copper salts ; as your case ) == >> a complex will form between the ammonia molecules and copper(II) ions and a deep-blue solution of [ Cu(NH3)4 ] 2+ will be obtained.

Cu++ + excess NH3 ----> [ Cu(NH3) ]++ ----> [ Cu(NH3)2 ]++

----> [ Cu(NH3)3 ]++ ----> [ Cu(NH3)4 ]++

Overall Reaction :

Cu++ + 4 NH3 ----> [ Cu(NH3)4 ]++

The ammonia molecules attach one at a time, and in between each attachment, there is a chemical equilibrium. The more ammonia is added, the more complex is formed, and the equilibrium is pushed to the product side. The blue color is the result of the complex absorbing light in the visible light spectrum, and having a concentration high enough for the eyes to detect.

First answer by ID1143022789. Last edit by Golbez 9. Contributor trust: 0 [recommend contributor]. Question popularity: 18 [recommend question]