Question

1. In step1, aqueous ammonia is added to a solution of the three ions. The equilibrium for the ammonia is

EQ1: NH­_3(aq + H­₂O_(l) ßà NH₄⁺_(aq) Kb=1.8x10^-5

The bismuth(III) ion is precipitated out of the solution as Bi(OH)₃. The equilibrium involved is related to the one provided below:

EQ2: Bi(OH)_3(s) ßà Bi³⁺_(aq) +3OH_(aq)^- K_sp =3.2x10^-40

Manipulate EQ 1 and 2 to solve for the net equation, EQ3:

EQ3 (net): 3H₂O+ Bi³⁺_(aq) +3NH_3(aq) ßàBi(OH)­^-_(s) +3NH­­₄⁺_(aq)

Write the mass law expression for the net reaction and calculate the K. Does the reaction form mostly products, or mostly reactants? EXPLAIN.

2. Addition of excess aqueous ammonia to equilibrium 3(net) causes the equilibrium to shift to the (a)­_­­_______, resulting in an increase in the amount of (b)_______ and a decrease in the amount of (c)_______.

3. In step 1, excess aqueous ammonia reacts with the Cu²⁺ ion to form a complex ion:

EQ4: Cu²⁺_(aq)+ 4NH_3(aq)ßà[Cu(NH₃)₄]²⁺_(aq) K­­_f = 1.1x10¹³

A blue precipitate initially forms and then dissolves in the excess NH­₃. The precipitate that forms initially is Cu(OH)­­₂.

EQ5: Cu²⁺_(aq) + 2OH^-_(aq) ßà Cu(OH)_2(s) K=2.08x10¹⁹

The needed OH­^-_­­to dissolve the Cu(OH)_2(s)­­ comes from the ionization of ammonia in water:

EQ 6: NH_3(aq) + H­₂O_(l) ßà NH₄⁺_(aq)+ OH^-_(aq) Kb =1.8x10^-5

Manipulate EQ 5 and 6 to find the net reaction (EQ 7) for the precipitation of the copper(II) hydroxide. Write the mass law expression and calculate the K­_net.

EQ 7: Cu²⁺_(aq) +2NH_3(aq) + 2 H­₂O_(l)ßàCu(OH)­­_2(s) +2 NH₄⁺_(aq)

4. The deep blue color of the solution of [Cu(NH₃)₄]²⁺_(aq) comes from dissolving the Cu(OH)­­_2(s) in excess NH_3(aq). The net equation for the reaction is

EQ 8: Cu(OH)­­_2(s) + 2OH^-_(aq) + 4 NH₄⁺_(aq) ßà [Cu(NH₃)₄]²⁺_(aq) + 4 H­₂O_{(l) ­}

Manipulate EQ 4, 5, and 6 to solve for the net reaction, EQ 8. Write the mass law expression and calculate K_net. Compare the K for equation 8 and equation 7. Which lies further to the right? How does excess ammonia helps dissolve the Cu(OH)₂?

5. Acetic acid is used to release the Cu²⁺ from the [Cu(NH₃)₄]²⁺_(aq)

EQ 9: [Cu(NH₃)₄]²⁺_(aq) + 4CH­­₃CO₂H­_(aq)ßà Cu²⁺ + 4 NH₄⁺_(aq) + 4CH­­₃CO₂^-_(aq)

Explain in terms of leChatelier’s principle and EQ9

6. Calculate the equilibrium [Cu²⁺] and the [CN^-] if the initial concentration of [Cu(CN)₄]² =0.10M. Cu²⁺_(aq) +4CN^-_(aq) ßà[Cu(CN)₄]²Kf =2.0x10³⁰

7. Calculate the equilibrium [Cd²⁺] if the initial concentration of [Cd(CN)_4­­]^2- =0.10M

Cd^2+­_(aq)­­­ + 4CN^-_(aq)­ßà [Cd(CN)_4­­]² Kf =7.7x10¹⁶

Compare your answer to the answer from question 6. Which complex ion is more stable explain.

Please Show Work!

Answer 1

According to LeChatelier principle, if an equilibrium is put under stress then equilibrium will shift in that direction such that the stress is minimised.

The value of equilibrium constant of EQ8 is greater than that of EQ7. Hence EQ8 will lies further to the right. The excess of ammonia can dissolve Cu(OH)₂ , because Copper can form a stable complex with ligand(NH₃).

Acetic acid is used to release Cu²⁺ from the complex because acetic acid neutralize the liquid ammonia. And concentration of ammonia decrease and equilibrium will shift backward i.e. dissociation of complex.