Question

The complex ion Cu(NH3)42 is formed in a solution made of 0.0200 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

Given,

[Cu(NO₃)₂] = [Cu²⁺] = 0.0200 M

[NH₃] = 0.400 M

K_f value for [Cu(NH₃)₄]²⁺ = 1.70 x 10¹³

Assume the volume of the solution is 1L.

Calculating the number of moles of Cu²⁺ and NH₃

= 0.0200 M x 1 L = 0.0200 mol Cu²⁺

Similarly,

= 0.400 M x 1 L = 0.400 mol NH₃

Now, the reaction of Cu²⁺ and NH₃ is,

Cu²⁺(aq) + 4NH₃(aq) [Cu(NH₃)₄]²⁺(aq)

Drawing an ICE chart,

	Cu2+(aq)	4NH3(aq)	[Cu(NH3)4]2+(aq)
I(moles)	0.0200	0.400	0
C(moles)	-0.0200	0.0800	+0.0200
E(moles)	0	0.32	0.0200

Now, the new concentrations of NH₃ and [Cu(NH₃)₄]²⁺ are,

[NH₃] = 0.32 mol / 1L = 0.32 M

[[Cu(NH₃)₄]²⁺] = 0.0200 mol / 1L = 0.0200 M

Now, the equilibrium reaction for  [Cu(NH₃)₄]²⁺ is,

[Cu(NH₃)₄]²⁺(aq) Cu²⁺(aq) + 4NH₃(aq)

Drawing an ICE chart,

	[Cu(NH3)4]2+(aq)	Cu2+(aq)	4NH3(aq)
I(M)	0.0200	0	0.32
C(M)	-x	+x	+4x
E(M)	0.0200-x	x	0.32+4x

Now, the K_d expression is,

K_d = [Cu²⁺][NH₃]⁴ / [ [Cu(NH₃)₄]²⁺]

K_d = [x][0.32+4x]⁴ / [ [0.0200-x]

Now, calculating K_d from the given K_f value,

K_d = 1 /K_f

K_d = 1 /1.70 x 10¹³

K_d = 5.88 x 10^-14

Now,

5.88 x 10^-14 = [x][0.32+4x]⁴ / [ [0.0200-x]

5.88 x 10^-14 = [x][0.32]⁴ / [ [0.0200] ---Here, [0.32+4x]0.32 and [ 0.0200-x] 0.0200, since, x <<0.02 and 0.32

x = 1.12 x 10^-13

Now, from the ICE chart,

[Cu²⁺]_eq = x = 1.12 x 10^-13 M

[NH₃]_eq = (0.32+4x) = 0.320 M

[[Cu(NH₃)₄]²⁺]_eq = (0.0200-x) = 0.0200 M