Question

A 1.0 L solution contains 3.00×10-4 M Cu(NO3)2 and 1.63×10-3 M ethylenediamine (en). The Kf for Cu(en)22+ is 1.00×1020. What is the concentration of Cu2+(aq ) in the solution?

Answer 1

Cu+2 + 2EN --> Cu(EN)2+2

Kf = [Cu(EN)2+2] / [Cu+2][EN]^2

Kd= 1/Kf = 1/(10^20) = 10^-20

Kd = [Cu+2][EN]^2 / [Cu(EN)2+2]

initially

[Cu+2] = 0

[EN] = 0

[Cu(EN)2] = assume 100% 3*10^-4 -->  3*10^-4

in equilibrium

[Cu+2] = 0 + x

[EN] = 0 +2x

[Cu(EN)2] =  3*10^-4 - x

substitute

Kd = [Cu+2][EN]^2 / [Cu(EN)2+2]

10^-20 = (x)(2X)^2 /(3*10^-4 - x)

since x <<< 3*10^-3, thn 3*10^-4 - x = 3*10^-4

10^-20 = (x)(2X)^2 /(3*10^-4)

(10^-20) * (3*10^-4) = 4*x^3

(((10^-20) * (3*10^-4)) /4 )^(1/3) = x

9.0856*10^-9 = x

then

[Cu+2] = x = 9.0856*10^-9M