Question

A solution has initial [Cu2+] = 0.010 M and initial [NH3] = 1.000 M. The complex ion Cu(NH3)42+ forms in solution. Calculate the [Cu2+] when equilibrium is achieved.

Answer 1

Cu+2 + 4NH3 --> Cu(NH3)4+2

inver it

Cu(NH3)4+2 --> Cu+2 + 4NH3

Kf = 5*10^12

Kd is the inverse of Kf so

Kd = 1/Kf = 1/(5*10^12) = 2*10^-13

Kd = [Cu+2][NH3]^4 / [Cu(NH3)4+2]

initially

[Cu+2]= 0

[NH3] = 0

[Cu(NH3)4+2] = 0.01

in equilbirium

[Cu+2]= 0 + x

[NH3] = 0 + 4x

[Cu(NH3)4+2] = 0.01 - x

substitute

Kd = [Cu+2][NH3]^4 / [Cu(NH3)4+2]

2*10^-13 = x*(4x)^4 / (0.01-x)

sole

2*10^-13 = x*(4x)^4 / (0.01)

256*x^5 = 0.01*2*10^-13

x^5 = (0.01*2*10^-13 )/256 =7.812*10^-18

x = (7.812*10^-18)^(1/5) = 0.0003789

[Cu+2]= 0 + x = 0.0003789 M