Question

The complex ion Cu(NH3)42 is formed in a solution made of 0.0300 M Cu(NO3)2 and 0.400 M NH3. What are the concentrations of Cu2 , NH3, and Cu(NH3)42 at equilibrium? The formation constant*, Kf, of Cu(NH3)42 is 1.70 × 10¹³.

Answer 1

solution made of 0.0300M Cu(NO3)2 and 0.400 M NH3.

Cu(NO3)2 = Cu2+ + 2NO3(-)
[NO3(-)] = 2*0.03 = 0.06 M

Initial concentrations:

[Cu(2+)] = 0.03 M
[NH3] = 0.4 M
[Cu(NH3)4]2+ = 0

Complex Equilibria:
[Cu2+] + 4[NH3] = [Cu(NH3)4]2+

Change:
[Cu(2+)] = -x
[NH3] = -4*x
[Cu(NH3)4]2+ = x

Equilibrium concentration:
[Cu(2+)] = 0.03-x
[NH3] = 0.4-4*x
[Cu(NH3)4]2+ = x

The formation constant of [Cu(NH3)4]2+, Kf = 1.70E13.
Kf = [ML4]/([M]*[L]) = ([Cu(NH3)4]2+)/([Cu2+]*[NH3]) = 1.70E13

(x/([0.03-x]*[0.4-4*x]) = 1.70E13
x = 1.70E13*4*(x^2-0.1*x-0.03*x +0.003)
x = 6.8E13*(x^2-0.13*x+0.003)

x^2-0.13*x+0.003 = 0
x = 0.03 or 0.1, 0.1 not possible
so x = 0.03 M

Final equillibrium concentration:
[Cu(2+)] = 0 M
[NH3] = 0.4-4*0.03 = 0.28 M
[Cu(NH3)4]2+ = 0.03 M