I am given a blue solid W and it contains a tetraamine copper(II) complex and a $\ce{SO_4^{2-}}$. I added dilute sulfuric acid until the solution is pale blue; then added potassium iodide solution. A white precipitate forms ($\ce{CuI}$) in a brown solution ($\ce{I_2}$). However, I am confused about what reaction is actually going on.


Copper(II) is a hard Lewis acid that would be octahedrally coordinated but for Jahn-Teller distortion. Instead, it is square-planar coordinated with two distant weakly bound apical ligands. The lovely deep blue color, hard Lewis base ammonia versus hard Lewis acid water coordination, illustrates the nephelauxetic effect. What would obtain with triethylamine coordination of copper? Cu(II) is substitution labile.

Adding acid protonates the ammonia, its lone pair now in a dative bond with a proton and unavailable for coordination. Tetraamminecopper(II) (note the spelling of "ammine" when coordinated) becomes tetraaquocopper(II) with fast kinetics.

Aqueous Cu(II) is a modest oxidizing agent; iodide is a modest reducing agent. Cu(I) is a soft Lewis acid, iodide is a soft Lewis base, $\ce{CuI}$ is very insoluble (look up its $\ce{K_{sp}}$ -- see here.)

Iodine reacts with excess iodide to give triiodide anion. However, isolate the "$\ce{CuI}$," wash, and titrate with thiosulfate. $\ce{CuI}$ strongly adsorbs elemental iodine. It is also modesty light sensitive. Cyanide, being a pseudohalide, follows the same path to afford exceptionally insoluble CuCN plus cyanogen.

Remember who earned the "A," or do your chair parade for better references, then understanding for the next question.

  • $\begingroup$ This was really helpful. Just one thing, when you said "Adding acid protonates the ammonia," Are you referring to this equilibrium: $$[Cu(H_2O)_6] + 4NH_3<=> [Cu(NH_3)_4(H_2O)_2] + 4H_2O $$. So, by protonating the $NH_3$, The $NH_3$ is used up and the equilibrium shifts to the left forming more hexaaquacopper(II) ions? $\endgroup$ – Eliza Mar 29 '14 at 5:03
  • $\begingroup$ Ammonia as such, plus acid, is removed as ammonium. There is nothing remaining but water to coordinate Cu(II). [Co(NH3)6](+2) is instantly disassembled by acid. [Co(NH3)6](+3) will boil in concentrated HCl for weeks with no change. [Co(en)6](+3) plus formaldehyde and nitromethane in base is exceptionally elegant re Sargeson, DOI: 10.1021/ja00451a062 Chemistry is great stuff. $\endgroup$ – Uncle Al Mar 29 '14 at 23:04

The formation of the tetraammine (II) copper sulfate ion complex is made through the addition of excess ammonia to copper (II) sulfate solution:

$$\ce{CuSO4(aq) + 4NH3(aq) -> [Cu(NH3)4]^2+(aq) + SO4^{2-}(aq)}$$

If we add sulfuric acid in solution, we push the equilibrium to the left, as there will be an increase in the concentration of sulfate ions as sulfuric acid ionizes. As the equilibrium shifts to the left, there will be an increase in concentration of ammonia and copper (II) sulfate in solution according to Le Chatlier's Principle:

$$\ce{[Cu(NH3)4]2+(aq) + SO4^{2-}(aq) + H2SO4(aq) -> CuSO4(aq) + 4NH3(aq)}$$

If we add potassium iodide, an oxidation and reduction reaction takes place between the copper (II) sulfate solution and potassium iodide. Since the reaction is taking place in basic solution, we can balance the electrochemical reaction.

$$\ce{2KI (aq) + CuSO4(aq) -> CuI2(s) + K2SO4(aq)}$$

Here we can see that solid white copper(II) iodide precipitates and potassium sulfate dissociates into its ions.

Since copper (II) iodide is very unstable, it decomposes into copper(I) iodide and iodine:

$$\ce{2CuI2 (s) -> 2CuI (s) + I2(s)}$$

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    $\begingroup$ "CuI2" does not exist. $\endgroup$ – Uncle Al Mar 28 '14 at 20:17
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    $\begingroup$ @UncleAl Thanks, I fixed that accordingly. I forgot that it is unstable and immediately decomposes after formation. $\endgroup$ – Jun-Goo Kwak Mar 28 '14 at 20:27
  • $\begingroup$ @Jun-Goo Kwak: so the acid was added just to tip the equilibrium to the left? Just asking, If I didn't add the sulphuric acid, would the tetraammine complex react with KI? $\endgroup$ – Eliza Mar 29 '14 at 3:54
  • $\begingroup$ @Eliza Hi, yes. Otherwise, you would have very little copper (II) sulfate solution. By shifting the equilibrium left, you produce more copper (II) sulfate which can then react with the potassium iodide. $\endgroup$ – Jun-Goo Kwak Mar 29 '14 at 3:58
  • $\begingroup$ @Jun-Goo Kwak: That makes so much sense. Just one thing, instead of thinking of the equilibrium like as above, can I think about it like this: $$[Cu(H_2O)_6]^{2+} + 4NH_3<=> [Cu(NH_3)_6(H_2O)_2]^{2+} + 4H_2O $$ The $H^+$ added causes the $NH_3$ to be protonated - this uses it up. This in turn causes equilibrium to shift to the left producing more hexaaquacopper(II) ions? $\endgroup$ – Eliza Mar 29 '14 at 4:27

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