Copper(I) iodide can be prepared by heating iodine and copper in concentrated hydriodic acid, HI. In the laboratory however, copper(I) iodide is prepared by simply mixing an aqueous solution of sodium or potassium iodide and a soluble copper(II) salt such copper sulfate.
Cu2+ + 2I− → CuI2
The CuI2 immediately decomposes to iodine and insoluble copper(I) iodide, releasing I2.[7]
2 CuI2 → 2 CuI + I2
This reaction has been employed as a means of assaying copper(II) samples, since the evolved I2 can be analyzed by redox titration. The reaction in itself may look rather odd, as using the rule of thumb for a proceeding redox reaction, Eooxidator − Eoreductor > 0, this reaction fails. The quantity is below zero, so the reaction should not proceed. But the equilibrium constant[8] for the reaction is 1.38*10−13. By using fairly moderate concentrates of 0.1 mol/L for both iodide and Cu2+, the concentration of Cu+ is calculated as 3*10−7. As a consequence, the product of the concentrations is far in excess of the solubility product, so copper(I)iodide precipitates. The process of precipitation lowers the copper(I) concentration, providing an entropic driving force according to Le Chatelier's principle, and allowing the redox reaction to proceed.
CuI is poorly soluble in water (0.00042 g/L at 25 °C), but it dissolves in the presence of NaI or KI to give the linear anion [CuI2]−. Dilution of such solutions with water reprecipitates CuI. This dissolution–precipitation process is employed to purify CuI, affording colorless samples.[5]