Question

Given equations:

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

(2) Cu2+ (aq) + 6 NH3 (aq) ⇌ Cu(NH3)62+ (aq)

In a test tube of~1mL of Cu2+ solution was added to it (CuSO4).
Several drops of 6 M NaOH were added to the copper solution until the precipitate Cu(OH)2 was observed. Are all the Cu2+ ions part of the precipitate? Explain.

Then drops of 6M NH3 until a change was observed. You may need to stir your test tube.

a. What is happening at the molecular level when the NH3 is added? How are the positions of equilibrium affected for both equations (1) and (2)? Explain.

Drops of 6M HCl was added to the system until a change was observed.

b. What is happening at the molecular level when the HCl is added? How are the positions of equilibrium affected for both equations (1) and (2)? Explain in terms of LeChatelier’s Principle.

Answer 1

No, all of the Cu2+ ions aren't Precipitated as Cu(OH)2. Since, this is an Equilibrium Reaction, some amount of Cu2+ remains in the aqueous solution.

a. When the NH3 is added to the Solution, the Equilibrium of Equation 2 is shifted towards the Forward Direction, so that more amount of Cu(NH3)62+ (aq.) Forms in the Solution. This, in turn, Shifts the Equilibrium of Equation 1, towards the Reverse Direction. Thus, the Bluish Green Precipitate of Cu(OH)2 gradually Dissolves in the Solution, as Cu(NH3)62+ , on Addition of NH3.

b. When HCl is added, the Equilibrium of Equation 2 is shifted towards the Reverse Direction, because HCl reacts with Ammonia, to form NH4Cl. This results an increase in the Concentration of Cu2+(aq.), Which in turn, Shifts the Equilibrium of Equation 1, towards the Forward Direction. Thus, more amount of Cu(OH)2 Precipitate will form, on Addition of HCl to the Solution.