In: Chemistry
Calculate the equilibrium concentrations of NH3, Cu2+, [Cu(NH3)]2+, [Cu(NH3)2]2+, [Cu(NH3)3]2+, and [Cu(NH3)4]2+ in a solution made by mixing 500.0 mL of 3.00 M NH3 with 500.0 mL of 2.00 x 10-3 M Cu(NO3)2. The sequential equilibria are
Cu2+ (aq) + NH3(aq) ⇐⇒ [Cu(NH3)]2+ (aq) K1 = 1.86 x 104
[Cu(NH3)]2+ (aq) + NH3(aq) ⇐⇒ [Cu(NH3)2]2+ (aq) K2 = 3.88 x 103
[Cu(NH3)2]2+ (aq) + NH3(aq) ⇐⇒ [Cu(NH3)3]2+ (aq) K3 = 1.00 x 103
[Cu(NH3)3]2+ (aq) + NH3(aq) ⇐⇒ [Cu(NH3)4]2+ (aq) K4 = 1.55 x 102
moles of Cu2+ = 0.002 M x 0.5 L = 0.001 mol
moles of NH3 = 3 M x 0.5 L = 1.5 mol
[Cu(NH3)2]2+ = 0.001 mol/1 L = 0.001 M
[NH3] remained = (1.5 - 2 x 0.001)mol/1 L = 1.5 M
Cu2+ + 2NH3 <==> [Cu(NH3)2]2+
I - 1.5 0.001
C +x +2x -x
E x (1.5+2x) 0.001-x
So with x being a small change,
K1 = 1.86 x 10^4 = (0.001-x)/(1.5)^2.x
4.2 x 10^4x = 0.001 - x
[Cu2+] = x = 2.4 x 10^-8 M
[Cu(NH3)2]2+ = 0.001 M
[NH3] = 1.5 M
[Cu(NH3)3]2+ = 0.001 mol/1 L = 0.001 M
[NH3] remained = (1.5 - 3 x 0.001)mol/1 L = 1.5 M
[Cu(NH3)2]2+ + 3NH3 <==> [Cu(NH3)3]2+
I 0.001 1.5 0.001
C +x +3x -x
E (0.001+x) (1.5+3x) 0.001-x
So with x being a small change,
K1 = 3.88 x 10^3 = (0.001-x)/(1.5+3x)^3. (0.001+x)
1.05 x 10^5x^3 + 1.57 x 10^5x^2 + 7.86 x 10^4x + 1.32 x 10^4 = 0
[Cu(NH3)3]2+ = x = 5,43 x 10^-12 M
So, the concentrations of the remaining two complexes would be negligibly small.