For a solution of Ni2+ and ethylenediamine, the following equilibrium constants

Question

For a solution of Ni2+ and ethylenediamine, the following equilibrium constants apply at 20° C: Ni2+ + H2NCH2CH2NH2 <-> Ni(en)2+ log K1 = 7.52 Ni(en)2+ + en <-> Ni(en)22+ log K2 = 6.32 Ni(en)22+ + en <-> Ni(en)32+ log K2 = 4.49 Calculate the concentration of free Ni2+ in a solution prepared by mixing 0.100 mol of en plus 1.00 mL of 0.010 M Ni2+ and diluting to 1.00 L with dilute base (which keeps all the en in its unprotonated form. [Hint: Assume that nearly all the Ni is in the form Ni(en)32+ (i.e., [Ni(en)32+] = 1.00 x 10-5 M).] You will need to calculate the concentrations of Ni(en)2+ and Ni(en)22+ to verify the assumption does not lead to a contradiction.

Explanation / Answer

Ni2+ +en <---->Ni(en)2+

Ni(en)2+ +en <--->Ni(en)22+

Ni(en)22+ + en<----> Ni(en)32+

given: log K3=4.49,logk2=6.32, log k1=7.52

log K3+ log k2+log k1=4.49+6.32+7.52=18.33=log (k1*k2*k3)

So if k1k2k3=Kf

log Kf=18.33

K=10^18.33=2.138*10^18

Kf=overall formation constant for complex Ni(en)32+

As the value of kf is very large ,so all Ni is in the form Ni(en)32+

The equilibrium of dissociation of Ni(en)32+:

Ni(en)32+ <-->Ni2+ +3 en

keq=1/kf=1/2.138*10^18=4.677*10^-19=[Ni2+] [ en]^3/ [Ni(en)32+]

Let the equilibrium concentration of Ni2+ =x

equilibrium concentration of en =3x

equilibrium concentration of [Ni(en)32+]=[Ni2+](initial) - x

4.677*10^-19=[Ni2+] [ en]^3/ [Ni(en)32+]=x*(3x)/([Ni2+](initial) - x)

[Ni2+](initial) =0.01 mol/L *0.001L/1L=0.00001 mol/L=1*10^-5M

4.677*10^-19=x*(3x)/(1*10^-5M - x)

x<<<1*10^-5M as very less dissociation takes place

4.677*10^-19=x*(3x)/(1*10^-5M)

x=1.248e-7

[Ni2+]free=[Ni]o - x=(1*10^-5 M)-(1.248e-7)=(1*10^-5 M)- 0.0125 *10^-5M=0.987*10^-5M=9.870*10^-6M

[Ni2+]free=9.870*10^-6M

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