Chemguide: Support for CIE A level Chemistry


Learning outcomes 12.5 (whole section)

These statements are about stability constants in complex ion formation and ligand exchange reactions.

This is new for the 2016 syllabus, and there isn't a sample question in the specimen paper. It is therefore difficult to be sure what CIE want exactly, and that won't really be known before 2016 at the earliest. I will update this page as more information becomes available.

Before you go on, you should find and read the statement in your copy of the syllabus.


Start by refreshing your memory about ligand exchange reactions.

Then read the first half of the page about stability constants. You don't need to know about the second half on the chelate effect.


What you need to know

What a stability constant is, and how to write an expression for it

You need to know that a stability constant for a complex ion is just the equilibrium constant for its formation.

You also need to understand that in a complex ion such as [Cu(NH3)4(H2O)2]2+ the water molecules are replaced by ammonia molecules one at a time giving a series of equilibria such as

You should be able to write an equilibrium constant for each of these stages, and also one for the overall replacement of the four water molecules. You should also be able to do this in unfamiliar cases.

For example, you should be able to work out an expression for the stability constant for the overall formation of the [Cr(NH3)6]3+ ion:

Don't forget that the water isn't included in these expressions.

Or you should be able to write down any of the equilibria on the way to the final product (like the copper equilibria above) and write expressions for the equilibrium constants for each stage.


What the size of a stability constant means

A complex ion with a large stability constant is more stable than one with a smaller one.

For example, the stability constants for two copper complexes are:

[Cu(NH3)4(H2O)2]2+1 x 1013
[CuCl4]2-4 x 105

The complex with ammonia is a lot more stable than the one with chloride ions.

This is seen in the lab. The chloride ions are easily stripped off the complex again by diluting the solution - an effect of Le Chatelier's Principle.

The olive-green solution turns pale blue. (Depending on the concentration, "olive-green" could be yellow-brown!)

The complex with ammonia can still be seen as dark blue even at very low concentrations.


Competing equilibria in ligand exchange reactions

Suppose you had a solution containing the tetrachlorocuprate(II) ion with this equilibrium:

Now suppose you added concentrated ammonia solution to it, so that you also have the possibility of this equilibrium involving the hexaaquacopper(II) ions as well:

The ammonia solution will form a very stable complex with any [Cu(H2O)6]2+ ions which may be formed by the first equilibrium involving the chloro-complex.

As they are removed from the solution, that equilibrium will tip to the left to replace them.

But the newly formed aqua-ions will react with ammonia to form the more stable complex. In the competition between the two equilibria, the one with the greater stability constant wins.

The solution will therefore turn from an olive-green colour to the typical deep blue of the complex with ammonia.


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© Jim Clark 2014