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Chemistry: The Formation of Complex Ions

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  • Type: Video Tutorial
  • Length: 10:01
  • Media: Video/mp4
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  • Size: 107 MB
  • Posted: 07/14/2009

This lesson is part of the following series:

Chemistry: Full Course (303 lessons, $198.00)
Chemistry: Equilibrium in Aqueous Solution (21 lessons, $31.68)
Chemistry: Complex Ion Equilibria (2 lessons, $2.97)

This lesson was selected from a broader, comprehensive course, Chemistry, taught by Professor Harman, Professor Yee, and Professor Sammakia. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/chemistry. The full course covers atoms, molecules and ions, stoichiometry, reactions in aqueous solutions, gases, thermochemistry, Modern Atomic Theory, electron configurations, periodicity, chemical bonding, molecular geometry, bonding theory, oxidation-reduction reactions, condensed phases, solution properties, kinetics, acids and bases, organic reactions, thermodynamics, nuclear chemistry, metals, nonmetals, biochemistry, organic chemistry, and more.

Dean Harman is a professor of chemistry at the University of Virginia, where he has been honored with several teaching awards. He heads Harman Research Group, which specializes in the novel organic transformations made possible by electron-rich metal centers such as Os(II), RE(I), AND W(0). He holds a Ph.D. from Stanford University.

Gordon Yee is an associate professor of chemistry at Virginia Tech in Blacksburg, VA. He received his Ph.D. from Stanford University and completed postdoctoral work at DuPont. A widely published author, Professor Yee studies molecule-based magnetism.

Tarek Sammakia is a Professor of Chemistry at the University of Colorado at Boulder where he teaches organic chemistry to undergraduate and graduate students. He received his Ph.D. from Yale University and carried out postdoctoral research at Harvard University. He has received several national awards for his work in synthetic and mechanistic organic chemistry.

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Thinkwell
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Let's look at another example of equilibria and aqueous solution, that of the formation of complex ions. Now, complex in this usage means association, as in an apartment complex, not complex, like something that's really difficult. In fact, the idea that we're going to be looking at here are basically just equilibrious. Stuff that we've looked at before, only in a slightly different way.
Now, let's first define what a complex ion is. A complex ion is an ion that has one or more molecules or ions attached to it. So, for instance, when we look at the tetrachloral cobaltate two minus anion the chlorides are attached to the cobalt. One thing that we haven't made clear up until this point is that when we dissolve a metal salt into water, water gets around the metal cation and so, for instance, when we talk about aqueous cobalt two plus, what we really mean to say is that there are waters around the cobalt and so this is also a complex ion. So this distinction we make between complex ions versus hypothetical non-complex ions is really bogus. What we're saying is that all things that dissolve in water have one or more solvents around them and so they're all complex ions.
Now, basically all we can say that or what we're interested in is the idea that if we have something that's other than the solvent in the shell, then we could call that a complex ion. So, people might call this a complex ion, but everything that's dissolved in water, including this, ex-aqua cobaltate two plus is a complex ion. Now the reactions of complex ions--actually one more idea. This reaction is a reaction of waters reacting with the cobalt two plus is an example of a Lewis acid, Lewis base reaction. That is, water is acting as the Lewis base and the cobalt two plus is reacting as a Lewis acid. Lewis bases are things that donate electrons. The waters have loan pairs on the oxygen, so they donate electron density to the cobalt two plus. The cobalt two plus wants that electron density so it behaves like a Lewis acid.
Similarly, for this complex ion over here, the chlorides have loan pairs that they can donate to the cobalt two plus and so this is an example of a Lewis acid, Lewis base reaction. Now, we introduced the special term, "ligand." That refers to the molecular ion that's attached to the metal. So a chloride here is acting as a ligand. Water here is acting as a ligand.
Let's look at a reaction between solvated cobalt two plus in water reacting with chloride to form tetrachloricobaltate two minus. The starting material is pink and the product is blue and so you'll see a color change. What I have here is a beaker that has cobalt two plus in it and I'm going to add some hydrochloric acid and we'll see a color change. Actually I wanted to put this into a bigger beaker, because I'm going to need additional volume. That's not too good. I'll add some more acid. This is still hydrochloric acid and you can see that it's working. A little strange, there we go. So, now that's now a nice blue color and again, that's the color of the tetrachloricobaltate two minus.
Now, we can define an equilibrium constant for the formation of a complex ion. Cobalt two plus reacts with ammonia to form this tetra ammonium copperate two plus cation and the equilibrium constant for this reaction is just the concentration of the product divided by the concentration of the cobalt and then the concentration of the ammonia raised to the fourth power. Remember this four comes from the fact that stoichiometric coefficient is four in front of the ammonia. Typically, formation constants for complex ions of this sort or very large. In other words, this reaction essentially goes to completion. If we take cobalt two plus dissolve it in water, it forms a complex ion that has waters around it, but if we give it some ammonia, it prefers the ammonia. It glums on to the ammonia to form a tetra ammonium copperate two plus, with an equilibrium constant 4.8 times 10^12 or something that's very large.
Now, we can also define a second sort of special equilibrium constant called the dissociation constant case of d before the formation constant with case of f. I may have forgotten to mention that. Case of d is just the reverse reaction. So take the cobalt, the tetramine copper two plus and its associates to form copper two plus, an aqueous solution plus four ammonias being the reverse reaction we write their equilibrium expression as one over what it was before. So, clearly a sub d the dissociation constant is just one over the formation constant. Some books will list formation constants. Some books will relate dissociation constants, but you can see that they're intimately related to each other and so you don't need to know both. You only need to know one.
So, how do we use things like dissociation constants? Let's go ahead and use our 4.8 times 10^12 for the formation of the tetra ammonium copper two plus cation. Suppose we start with 800 milliliters of .8 molar ammonia in a beaker and then we start adding copper sulfate? Suppose we add a tenth of a mole of copper sulfate to this solution? What's going to happen is that the copper is going to react with the ammonia to form the complex ion and there's not going to be a whole lot of copper two plus left in solution, sort of floating around free. Most of it is going to be bound up by the ammonia to form the complex ion. Why is that? It's because the equilibrium constant for this reaction is so large because the formation constant is so gigantic.
But we can still calculate what the concentrations at equilibrium are for this species and this species and in doing that we can, knowing the formation constant, calculate what the equilibrium concentration is of copper two plus that free in solution. Now, to do that, what we have to do is, first of all, a limiting reagent problem. In other words, we put in a tenth of a mole of copper sulfate, do we have enough ammonia here in order to make all of the copper form the complex ion? The way to do that is we have to calculate the number of moles of ammonia that we're going to utilize. The number of moles is a tenth of a mole of copper two plus, that's what we put into the pot originally times four moles of ammonia divided by one mole of copper two plus. These are just stoichiometric coefficients for the balanced reaction.
It means that we would need .40 moles of ammonia to react with all the copper two plus that we've added. Well, we have .64 moles of ammonia. How do we know that? Well, remember we know the concentration of the ammonia and we know the volume of the ammonia. We started with .800 liters and we know that it's .80 molar and to get the number of moles of ammonia this represents, we multiply the concentration times the volume. So, .80 times .800 is .64. So this is the number of mole of ammonia we started with. We're going to use up .40 moles of ammonia. That's going to leave us .24 moles of ammonia left over. So, we have enough ammonia for this reaction. We're going to use up all the copper to form the complex ion.
So, an expression of that--the symbolic expression is that all of or the concentration of the complex ion in solution at equilibrium is just given by the original concentration of copper that we started with. Right? Essentially all the copper is going to be converted into the complex ion. Now, that's an assumption. What we have to do is we have to check afterwards to see whether or not in fact almost all the copper is converted to this. Remember when we make assumptions, if the concentration of free copper is less than about what percent of this, then our assumption makes sense.
So, we'll calculate the concentration of the tetra ammonium copper two plus cation and that's .10 moles, because that's the number of moles that we started with divided by the final volume and that gives us .13 molar concentration of the complex ion. The concentration of ammonia at equilibrium, so that's the equilibrium concentration and similarly with the concentration of the ammonia at equilibrium as given by the ammonia that's left over divided by the volume of the solution and that's .30 molar. Now, we can take these two pieces of information and plug them into our expression for the formation constant.
We're going to rearrange this expression for the formation constant. Put the copper two plus, that's what we're interested in, by itself and that means that we rearrange this expression to this expression. We know what the equilibrium concentration is of the complex ion. We know what the formation constant is and we know what the equilibrium concentration of ammonia is. So, we plug those three values in and we get the equilibrium concentration of copper two plus. Of course, it's really tiny. So, our assumption that all of the copper that we put into the pot, essentially get converted to the complex ion makes perfect sense. Remember this is around a tenth molar compared to the concentration of free copper, which is 3.3 times 10^-12
Now, again, this is all just an application of equilibria that you should be really getting comfortable with now. It's just formation constants of complex ions as opposed to acid base or titrations or solubility products. The only new terms that we've introduced is the idea of a ligand. That is the molecule or ion that's attached to the central metal and the idea that all metals really are complex ions when you dissolve in water. All cations are really complex ions when you dissolve them in water, but we're talking about complex ions that have ligands in their shell, in their coordination shell that is different from the solvent. But other than that, all we have here is another example of equilibria in water.
Equilibrium in Aqueous Solution
Complex Ion Equilibria
The Formation of Complex Ions Page [2 of 2]

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