Question

A 1.00 L solution contains 2.50×10^-4 M Cu(NO3)2 and 2.50×10^-3 M ethylenediamine (en). The Kf for Cu(en)22+ is 1.00×10^20. What is the concentration of Cu2+(aq ) in the solution?

Answer 1

A 1.00 L solution containing 2.50x10^-4 M Cu(NO3)2 and 2.50x10^-3 M ethylenediamine (en).
contains
0.000250 moles of Cu(NO3)2 and 0.00250 moles of ethylenediamine

by the formula Cu(en)2^2+
0.000250 moles of Cu(NO3)2 reacts with twice as many moles of en = 0.000500 mol of en
so, 0.00250 moles of ethylenediamine - 0.000500 mol of en reacted = 0.0020 mol en remains

by the formula Cu(en)2^2+
0.000250 moles of Cu(NO3)2 reacts to form an equal 0.000250 moles of Cu(en)2^2+


Kf for Cu(en)2^2+ is 1x10^20.
so
1 Cu+2 & 2 en --> Cu(en)2^2+

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

1x10^20. = [0.000250] / [Cu+2] [0.002 ]^2

[Cu+2] = [0.00025] / (1x10^20) (4 e-6)
Cu+2 = 6.25 e-19 Molar

Hence the required concentration of Cu2+ in the solution is 6.25 e-19 or 6.25 x 10^-19 M