Question

Which cobalt complex was favored when the solution was taken off the heat plate and put in an ice bath? The solution contained deionzied water, ethanol, and cobalt II chloride. The solution went from blue to pink. If you could explain this using Le Chatelier's principle, thank you!

Answer 1

The element cobalt can form compounds in two different oxidation states, +2 and +3. The +2 state is more common. The ion Co²⁺ (aq) is pink. Other compounds of cobalt(II), which include both anhydrous Co²⁺ and complex ions, are commonly blue.

If an aqueous solution contains both cobalt(II) and chloride ions, the blue ion CoCl₄^2- forms, in equilibrium with the pink Co²⁺ (aq) ion.

		CoCl42- (aq) <===========> Co2+ (aq) + 4Cl1-(aq)
			Blue		Pink

At relatively low concentrations of chloride, the equilibrium lies far to the right, and the solution is pink. If there is a large concentration of excess chloride, the equilibrium tends to the left, and the solution tends to be blue.

The equilibrium is sensitive to temperature as well as to concentration of solutes. At lower temperatures, the equilibrium tends to lie to the right, that is, to be more pink; at higher temperatures, it lies to the left and appears more blue.