AN INTRODUCTION TO SOLUBILITYPRODUCTS

This page looks at how solubility products are defined, togetherwith their units. It also explores the relationship between thesolubility product of an ionic compound and its solubility.

What are solubility products, Ksp?

Solubility products are equilibrium constants

Barium sulphate is almost insoluble in water. It isn't totally insoluble- very, very small amounts do dissolve. That's true of any so-called"insoluble" ionic compound.

if you shook some solid barium sulphate with water, a tinyproportion of the barium ions and sulphate ions would break awayfrom the surface of the solid and go into solution. Over time, someof these will return from solution to stick onto the solid again.

You get an equilibrium set up when the rate at which some ions arebreaking away is exactly matched by the rate at which others arereturning.

The position of this equilibrium lies very far to the left. The greatmajority of the barium sulphate is present as solid. In fact, if youshook solid barium sulphate with water you wouldn't be aware justby looking at it that any had dissolved at all.

But it is an equilibrium, and so you can write an equilibriumconstant for it which will be constant at a given temperature - likeall equilibrium constants.

The equilibrium constant is called the solubility product, and isgiven the symbol Ksp.

To avoid confusing clutter, solubility product expressions are oftenwritten without the state symbols. Even if you don't write them, youmust be aware that the symbols for the ions that you write are forthose in solution in water.

Why doesn't the solid barium sulphate appear in the equilibriumexpression?

For many simple equilibria, the equilibrium constant expressionhas terms for the right-hand side of the equation divided by termsfor the left-hand side. But in this case, there is no term for theconcentration of the solid barium sulphate. Why not?

This is a heterogeneous equilibrium - one which containssubstances in more than one state. In a heterogeneousequilibrium, concentration terms for solids are left out of theexpression.

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Note: The simplest explanation for this is that theconcentration of a solid can be thought of as a constant.Rather than have an expression with two constants in it (theequilibrium constant and the concentration of the solid), theconstants are merged to give a single value - the solubilityproduct.

Solubility products for more complicated solids

Here is the corresponding equilibrium for calcium phosphate,Ca3(PO4)2:

And this is the solubility product expression:

Just as with any other equilibrium constant, you raise theconcentrations to the power of the number in front of them in theequilibrium equation. There's nothing new here.

Solubility products only apply to sparingly soluble ioniccompounds

You can't use solubility products for normally soluble compoundslike sodium chloride, for example. Interactions between the ions inthe solution interfere with the simple equilibrium we are talkingabout.

The units for solubility products

The units for solubility products differ depending on the solubilityproduct expression, and you need to be able to work them outeach time.

Working out the units in the barium sulphate case

Here is the solubility product expression for barium sulphate again:

Each concentration has the unit mol dm-3. So the units for thesolubility product in this case will be:

(mol dm-3) x (mol dm-3)

= mol2 dm-6

Working out the units in the calcium phosphate case

Here is the solubility product expression for calcium phosphateagain:

The units this time will be:

(mol dm-3)3 x (mol dm-3)2

= (mol dm-3)5

= mol5 dm-15

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If you are asked to calculate a solubility product in an exam, therewill almost certainly be a mark for the correct units. It isn't very hard- just take care!

Solubility products apply only to saturated solutions

Let's look again at the barium sulphate case. Here is theequilibrium expression again:

. . . and here is the solubility product expression:

Ksp for barium sulphate at 298 K is 1.1 x 10-10 mol2 dm-6.

In order for this equilibrium constant (the solubility product) toapply, you have to have solid barium sulphate present in asaturated solution of barium sulphate. That's what the equilibriumequation is telling you.

If you have barium ions and sulphate ions in solution in thepresence of some solid barium sulphate at 298 K, and multiply theconcentrations of the ions together, your answer will be 1.1 x 10-10

mol2 dm-6.

What if you mixed incredibly dilute solutions containing barium ionsand sulphate ions so that the product of the ionic concentrationswas less than the solubility product?

All this means is that you haven't got an equilibrium. The reason forthat is that there won't be any solid present. If you lower theconcentrations of the ions enough, you won't get a precipitate - justa very, very dilute solution of barium sulphate.

So it is possible to get an answer less than the solubility productwhen you multiply the ionic concentrations together if the solutionisn't saturated.

Can you get an answer greater than the solubility product if youmultiply the ionic concentrations together (allowing for any powersin the solubility product expression, of course)? No!

The solubility product is a value which you get when the solution issaturated. If there is any solid present, you can't dissolve any moresolid than there is in a saturated solution.

Note: In the absence of any solid, a few substances produceunstable supersaturated solutions. As soon as you add anysolid, or perhaps just scratch the glass to give a rough bitthat crystals can form on, all the excess solid precipitatesout to leave a normal saturated solution.

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If you mix together two solutions containing barium ions andsulphate ions and the product of the concentrations would exceedthe solubility product, you get a precipitate formed. Enough solid isproduced to reduce the concentrations of the barium and sulphateions down to a value which the solubility product allows.

Summary

The value of a solubility product relates to a saturatedsolution.

If the ionic concentrations give a value less than the solubilityproduct, the solution isn't saturated. No precipitate would beformed.

If the ionic concentrations give a value more than thesolubility product, enough precipitate would be formed toreduce the concentrations to give an answer equal to thesolubility product.

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.

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© Jim Clark 2011 (modified November 2013)

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