Without performing detailed calculations, show why you would expect the concentrations of the various ammine-aqua complex ions to be negligible compared with that of \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) in a solution having a total \(\mathrm{Cu}(\mathrm{II})\) concentration of \(0.10 \mathrm{M}\) and a total concentration of \(\mathrm{NH}_{3}\) of \(1.0 \mathrm{M}\). Under what conditions would the concentrations of these ammine-aqua complex ions (such as \(\left.\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{NH}_{3}\right]^{2+}\right)\) become more significant relative to the concentration of \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} ?\) Explain.

Short Answer

Expert verified
Under given conditions, the ammine complex ions dominate due to higher ammonia concentration favoring the formation of [Cu(NH_3)_4]2+. The amount of free copper ion is negligible, so the equilibrium for complexes with water ligands doesn’t go far to the right, forming tiny amounts of mixed complexes. The conditions under which these mixed complexes would become significant would be if the concentration of ammonia was lowered, or the concentration of water was increased, or both.

Step by step solution

01

Understanding the equilibrium

In this solution, an equilibrium exists between the copper ion, water, ammonia and each species of the ammine-aqua complex copper ions. The equilibrium constant expression for the fully ammonia bound copper would be: \[ K_f = \frac{[\mathrm{Cu}(NH_3)_4]^{2+}}{[\mathrm{Cu}^{2+}][NH_3]^4} \] Now we should realize that if most of the copper is in the form of \([Cu(NH_3)_4]^2+\), it implies that the equilibrium constant for its formation is large enough, thus favoring the formation of the tetraamminecopper(II) complex. This also implies that the amount of free copper ion in the solution not in the [Cu(NH_3)_4] is negligible.
02

Considering the competition of ammonium and water

The water ligand can substitute the ammonia in [Cu(NH_3)_4]2+ complex according to the equilibrium. However, under the given conditions, the concentration of ammonia is much larger than the concentration of water. Thus, the ammonia will out-compete the water for the binding site, resulting in the negligible extent to which water binds. This leads to an extremely tiny equilibrium constant, favoring the left side in the equilibria, forming the mixed complex ions such as [Cu(H2O)_3(NH_3)]. Thus, the concentration of these ions will be quite tiny in contrast to the concentration of the [Cu(NH_3)_4]2+ ions.
03

Conditions for higher concentrations of mixed-complex ions

In order to see a significant amount of mixed complex ions, the conditions should favor the side of the equilibrium with the mixed complex. This can be achieved by either increasing the concentration of water or lowering the concentration of ammonia, or both. Lower concentrations of ammonia will allow more water to bind to the copper ion, hence increasing the concentrations of ammine-aqua complex ions. Increasing the concentration of water also favors the same.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free