Copper ion (pale blue in solution) forms a precipitate with hydroxide, as shown

Question

Copper ion (pale blue in solution) forms a precipitate with hydroxide, as shown below.

                                                                     Cu2+ (aq) + 2OH- (aq)     Cu(OH)2(s)

Copper ion also reacts with NH3 to form a complex ion, as shown in the next reaction.

Cu2+ (aq)    + 6NH3(aq) Cu(NH3)62+

                                                                     blue                                 deep blue

If all species are present at the same time, they form a set of competing equilibria. Since the equilibrium constants for the two reactions are different, one of the reactions will be more strongly favored than the other. By observing the various reactions that take place, you should be able to deduce the relative magnitude of the equilibrium constants.

Part III 4a. What happens at the molecular/ionic level as NH3 is added?

Part III 4b. How is the position of the equilibrium between copper ion and hydroxide affected by the addition of NH3? Explain your reasoning.

Part III 4c. What does this experiment prove about the relative magnitude of the equilibrium constant for the reaction of copper ion with OH- as compared with NH3? Explain.

Explanation / Answer

Part III 4a. When NH3 is added to the solution, it reacts with Cu2+ ion to form complex Cu(NH3)62+.

Part III 4b. By addition of NH3, The equilibrium will shift towards right. One of the reactants are increased, so according to Le-chatliers principle, more Cu(NH3)62+ will be formed.

Part III 4c. When Cu2+ ion reacts with Cu(OH)2, the reaction is exothermic, so the reaction is spontaneous. Thus K is more than 1.

When Cu2+ ion reacts with Cu(NH3)62+, the reaction is endothermic, so the reaction is not spontaneous. Thus K is less than 1.

Thus magnitude of equilibrium constant for reaction of copper ion with OH- is more than as compared with NH3.

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