Question

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...

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

Solutions

Expert Solution

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.


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