2 2 12 a 2 pts What is the Lewis acid and what is the Lewis

2+ 2+ 12 a) (2 pts) What is the Lewis acid and what is the Lewis base in this reaction? b) (6 pts) Suppose 200. mL of 0.10 M CuSO4(aq) and 200. mL of 1.00 M NH3(aq) are mixed. Calculate the equilibrium concentrations of Cu-t, NH3, and Cu(NH3)4 after mixing and reaction. [Assume that the volumes of mixed dilute aqueous solutions add up.] c) (4 pts) What concentration of OH should be created in the resulting solution to start precipitating Cu(OH)2 (Ksp-4.8-10-20)?

Solution

a) Lewis acid is Cu²⁺

Lewis base is NH3

b) Cu²⁺(aq) + 4NH3(aq) <--------> Cu(NH3)4²⁺(aq)

K_f = [Cu(NH3)4²⁺]/[NH3]⁴[Cu²⁺] = 5.0×10¹²

Consider the reverse reaction

Cu(NH3)4²⁺(aq) <-------> Cu²⁺(aq) + 4NH3(aq)

K_d= [Cu²⁺][NH3]⁴/[Cu(NH3)] = 1/K_f = 1/5.0×10^-12 =2.0×10^-13

Now assume the reaction is complete

0.02 moles of Cu²⁺ react with 0.08 moles of NH3

remaining moles of NH3 = 0.2 - 0.08 = 0.12

[NH3] = (0.12mol/400ml)×1000ml = 0.3M

[Cu²⁺] = (0.02mol/400ml)×1000ml =0.05M

at equillibrium

[Cu²⁺] = x

[NH3] = 0.3 + 4x

[Cu(NH3)4²⁺] = 0.05 - x

   therefore,

x(0.3 + 4x)⁴/0.05-x = 2.0×10^-13

x is small value, we can assume 0.3 + 4x = 0.3 and 0.05- x = 0.05

0.0081x /0.05 = 2.0×10^-13

0.0081x = 1.0×10^-14

   x = 1.23×10^-12

Therefore, at equillibrium

[Cu²⁺] = x = 1.23×10^-12M

[NH3] = 0.3 + 4x ~ 0.3M

[Cu(NH3)4²⁺] = 0.05 - x ~ 0.05M

c) Solubility equillibrium of Cu(OH)2 is

Cu(OH)₂ (s) <-------> Cu²⁺(aq) + 2OH^-(aq)

Ksp = [Cu²⁺][OH^-]²= 4.8×10^-20

when, [Cu²⁺] = 1.23×10^-12M

1.23×10^-12M × [OH^-]² = 4.8×10^-20M³

[OH^-]² = 3.90×10^-8

     [OH^-] = 1.97×10^-4M

Therefore,

the minimum concentration of OH^- required to start precipitating of Cu(OH)₂ is 1.97×10^-4M