PART II FORMATION OF A COMPLEX Data Collection This section involves two reactio
ID: 1066569 • Letter: P
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PART II FORMATION OF A COMPLEX Data Collection This section involves two reactions with the copper ion: (1) Cu (aq) 2 OH (aq) Cu (oH), (s) (2) Cu (aq) 6 NH,(aq) Cu (NH)62 (aq) Deep Blue 1. Get a clean test tube and add ~1 mL of Cu solution to it (CuSO4) 2. Add drops of 6 M NaoH to the copper solution until the precipitate Cu(oH)2 is observed. Are all the Cu ions part of the precipitate? Explain. 2+ It turn blue. 3. Add drops of 6 M NH, until a change is observed. You may need to stir your test tube. Record your observations. Does this agree with your answer from above? Explain The blue turn to dark blueExplanation / Answer
The two given reactions are
(1) Cu2+ (aq) + 2 OH- (aq) -----> Cu(OH)2 (s)
(2) Cu2+ (aq) + 6 NH3 (aq) -----> Cu(NH3)62+ (aq)
2. All the Cu2+ ions do not precipitate out as Cu(OH)2, i.e., all the Cu2+ ions are not part of the precipitate. This is because Cu(OH)2 is sparingly soluble in water and some of the precipitated Cu(OH)2 re-enters the solution as Cu2+ and OH- as per the below dissociation reaction:
Cu(OH)2 (s) <====> Cu2+ (aq) + 2 OH- (aq)
Due to the above dissociation, all the Cu2+ ions do not end up in the precipitate.
3. When NH3 solution is added dropwise, the color of the solution turns to intense blue. The reason is Cu(OH)2 is soluble in aqueous NH3 and the precipitated Cu(OH)2 from step (2) above re-enters the solution as Cu(NH3)62+ as per the equation:
Cu(OH)2 (s) + 6 NH3 (aq) ------> Cu(NH3)62+ (aq) + 2 OH- (aq)
4. When NH3 is added to solid Cu(OH)2, the Cu-OH bonds rupture and new Cu-NH3 bonds are formed. The energy required to split the Cu-OH bonds is supplied by the release of energy due to the formation of Cu-NH3 bonds.
Consider the equilibrium constants for reactions (1) and (2) above:
K = [Cu(OH)2]/[Cu2+][OH]2
K’ = [Cu(NH3)62+]/[Cu2+][NH3]6
When NH3 is added, [NH3] increases and hence [Cu(NH3)62+] must increase to counter the increased [NH3]. This is because the temperature is held constant and hence K’ must remain constant.
This means that [Cu2+] must decrease, i.e., more complexation must occur. As [Cu2+] decreases, [Cu(OH)2] must decrease. This is because K must remain constant at the particular temperature. Therefore, dissolution of solid Cu(OH)2 occur to produce Cu2+ and hence more Cu(NH3)62+.
5. When HCl is added to the solution, the deep blue color of Cu(NH3)62+ fades to a lighter shade due to the protonation of NH3 to ammonium ion, NH4+. NH4+ cannot complex to Cu2+ and thus the color becomes pale.
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