Answer:
A 1.00 L solution containing 3.00x10^-4 M Cu(NO3)2 and 2.40x10^-3 M ethylenediamine (en).
contains
0.000300 moles of Cu(NO3)2 and 0.00240 moles of ethylenediamine
by the formula Cu(en)2^2+
0.000300 moles of Cu(NO3)2 reacts with twice as many moles of en = 0.000600 mol of en
so, 0.00240 moles of ethylenediamine - 0.000600 mol of en reacted = 0.00180 mol en remains
by the formula Cu(en)2^2+
0.000300 moles of Cu(NO3)2 reacts to form an equal 0.000300 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. = [0.000300] / [Cu+2] [0.00180 ]^2
[Cu+2] = [0.000300] / (1x10^20) (3.24 e-6)
Cu+2 = 9.26 e-19 Molar
since your Kf has only 1 sig fig, you might be expected to round that off to 9 X 10^-19 Molar Cu+2
