SOLUBILITY PRODUCT and THECOMMON ION EFFECT

This page looks at the common ion effect related to solubilityproducts, including a simple calculation. You need to know aboutsolubility products and calculations involving them before you readthis page.

What is the common ion effect?

I need to look again at a simple solubility product calculation,before we go on to the common ion effect.

The solubility of lead(II) chloride in water

Lead(II) chloride is sparingly soluble in water, and this equilibriumis set up between the solid and its ions in solution:

If you just shook up some solid lead(II) chloride with water, then thesolution would obviously contain twice as many chloride ions aslead(II) ions.

The expression for the solubility product and its value are given by:

Note: I have no confidence in the accuracy of this value.There are serious discrepancies between the values fromdifferent sources. But that makes no difference to thediscussion.

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For comparison purposes later, I need to work out the lead(II) ionconcentration in this saturated solution.

If the concentration of dissolved lead(II) chloride is s mol dm-3,then:

[Pb2+] = s mol dm-3

[Cl- ] = 2s mol dm-3

Put these values into the solubility product expression, and do thesum.

So the concentration of lead(II) ions in the solution is 1.62 x 10-2

mol dm-3 (or 0.0162 mol dm-3 if you prefer).

What happens if you add some sodium chloride to thissaturated solution?

Now we are ready to think about the common ion effect.

Sodium chloride shares an ion with lead(II) chloride. The chlorideion is common to both of them. This is the origin of the term"common ion effect".

Look at the original equilibrium expression again:

What would happen to that equilibrium if you added extra chlorideions?

According to Le Chatelier, the position of equilibrium would shift inorder to counter what you have just done. In this case, it would tendto remove the chloride ions by making extra solid lead(II) chloride.

Note: Actually, of course, the concentration of lead(II) ions inthe solution is so small to start with, that only a tinyproportion of the extra chloride ions can be converted intosolid lead(II) chloride.

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The lead(II) chloride will become even less soluble - and, of course,the concentration of lead(II) ions in the solution will decrease.

Something similar happens whenever you have a sparingly solublesubstance. It will be less soluble in a solution which contains anyion which it has in common. This is the common ion effect.

A simple calculation to show this

Suppose you tried to dissolve some lead(II) chloride in some0.100 mol dm-3 sodium chloride solution instead of in water. Whatwould the concentration of the lead(II) ions be this time?

As before, let's call the concentration of the lead(II) ions s.

[Pb2+] = s mol dm-3

Now the sum gets different. This time the concentration of thechloride ions is governed by the concentration of the sodiumchloride solution. The number of ions coming from the lead(II)chloride is going to be tiny compared with the 0.100 mol dm-3

coming from the sodium chloride solution.

In calculations like this, you can always assume that theconcentration of the common ion is entirely due to the othersolution. This makes the maths a lot easier. In fact if you don'tmake this assumption, the maths of this can become impossible todo at this level.

So we assume:

[Cl- ] = 0.100 mol dm-3

The rest of the sum looks like this:

Finally, compare that value with the simple saturated solution westarted with:

Original solution:

[Pb2+] = 0.0162 mol dm-3

Solution in 0.100 mol dm-3 NaCl solution:

[Pb2+] = 0.0017 mol dm-3

The concentration of the lead(II) ions has fallen by a factor of about10.

If you tried the same sum with more concentrated solutions ofsodium chloride, the solubility would fall still further. Try it yourselfwith chloride ion concentrations of 0.5 and 1.0 mol dm-3.

Note: If you can be bothered to do this, don't just swap thevalues in the last sum. Work it out for yourself from scratch -it's good to practise!

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Questions to test your understanding

If this is the first set of questions you have done, please read theintroductory page before you start. You will need to use the BACK BUTTONon your browser to come back here afterwards.

questions on the common ion effect

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© Jim Clark 2011 (modified December 2013)

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