Cu(en)2+2(
Michael S. answered • 08/05/26
B.S. in Chemistry, Indiana University; Organic Chem Teaching Intern
[Cu2+] = 1.76 × 10-18 M
The strategy is the one the huge Kf is hinting at. With Kf = 1.00 × 1020, complex formation is effectively complete, so do not try to solve the equilibrium directly. Instead run the reaction fully to the right, then let a tiny amount dissociate back.
Step 1: react to completion. Cu2+ is the limiting reagent, and the stoichiometry is 2 en per copper.
en consumed = 2(2.75 × 10-4) = 5.50 × 10-4 M
en left over = 1.80 × 10-3 - 5.50 × 10-4 = 1.25 × 10-3 M
Cu(en)22+ formed = 2.75 × 10-4 M
That excess ligand is the part students most often drop. The en does not all get used up, and the leftover 1.25 × 10-3 M is what suppresses the free copper in the next step.
Step 2: let the complex dissociate slightly. Rearrange the Kf expression for the free metal ion:
[Cu2+] = [Cu(en)22+] / (Kf[en]2) = (2.75 × 10-4) / [(1.00 × 1020)(1.25 × 10-3)2]
(1.25 × 10-3)2 = 1.5625 × 10-6, so the denominator is 1.5625 × 1014, giving
[Cu2+] = 1.76 × 10-18 M
Two things worth noticing. The amount that dissociates is so small that the concentrations you carried into step 2 do not need correcting, which is exactly the approximation the large Kf buys you. And square the free ligand concentration, do not just multiply by it. That exponent comes from the 2 en in the balanced equation, and dropping it changes the answer by three orders of magnitude while leaving it looking perfectly reasonable.
Step back and read the result physically. 1.76 × 10-18 M in one litre is roughly a million free copper ions, against about 1.7 × 1020 that went in. Essentially every copper ion is locked up in the complex. That is what a formation constant of 1020 means in practice, and it is why chelating agents work as antidotes for metal poisoning: they do not merely bind the metal, they reduce the free concentration to the point where it is chemically unavailable.
One footnote on the title: this is a complex-ion formation problem rather than an acid equilibrium one. The algebra is the same shape you use for weak acids, which is probably why it sits in that section.
