Question

The formation constant, Kf, for Ni(NH3)6 (charge -1) is 5.5e8. Calculate the concentration of free nickel ion in a solution that contains .012M Ni2+ and 1.0M NH3.

Answer 1

Kf = [Ni(NH3)6]/ [Ni2+][NH3]^6

Kd = [Ni2+][NH3]^6 / [Ni(NH3)6]

Kd = 1/Kf = 1/(5.5*10^8) = 1.8181*10^-9

then

0.012 M of Ni2+ reacts to form 0.012 M of complex

then

initially

[Ni(NH3)6] = 0.012

[Ni2+]= 0

[NH3] = 0

in equilbirium

[Ni(NH3)6] = 0.012 - x

[Ni2+]= 0 + x

[NH3] = 0 + 6x

substitute

Kd = [Ni2+][NH3]^6 / [Ni(NH3)6]

1.8181*10^-9 = (x)(6x)^6 / (0.012 - x)

(1.8181*10^-9)(0.012) = 46656*x^7

2.1817*10^-11 = 46656*x^7

x^7 = (2.1817*10^-11 )/(46656)

x = (4.676*10^-16)^(1/7) =0.006456

[Ni2+] = 0.006456 M