# An Introduction to Solubility Products

• • Contributed by Jim Clark
• Former Head of Chemistry and Head of Science at Truro School in Cornwall

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

## Solubility products are equilibrium constants

Barium sulfate is almost insoluble in water. It is not totally insoluble—very small amounts do dissolve. This is true of any "insoluble" ionic compound. If solid barium sulfate is shaken with water, a small amount of barium ions and sulfate ions break away from the surface of the solid and go into solution. Over time, some of these return from solution to stick onto the solid again.

An equilibrium is established when the rate at which some ions are breaking away from the solid lattice is exactly matched by the rate at which others are returning. Consider the balanced equation for the barium sulfate reaction:

$BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$

The position of this equilibrium lies very far to the left. The great majority of the barium sulfate is present as solid—​there are no visible changes to the solid. However, the equilibrium does exist, and an equilibrium constant can be written. The equilibrium constant is called the solubility product, and is given the symbol Ksp:

$K_{sp} = [Ba^{2+} (aq)] [SO_4^{2-}(aq)]$

For simplicity, solubility product expressions are often written without the state symbols:

$K_{sp} = [Ba^{2+}] [SO_4^{2-}]$

Notice that there is no solid barium sulfate term. For many simple equilibria, the equilibrium constant expression has terms for the right side of the equation divided by terms for the left side. But in this case, there is no term for the concentration of the solid barium sulfate. This is a heterogeneous equilibrium, one which contains substances in more than one state. In a heterogeneous equilibrium, concentration terms for solids are left out of the expression.

## Solubility products for more complicated solids

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

$Ca_3(PO_4)_2(s) \rightleftharpoons 3Ca^{2+}(aq) + 2PO_4^{3-}(aq)$

Below is the solubility product expression:

$K_{sp} = [Ca^{2+}]^3[PO_4^{3-}]^2$

As with any other equilibrium constant, the concentrations are raised to the power of their respective stoichiometric coefficients in the equilibrium equation. Solubility products only apply to sparingly soluble ionic compounds, not to normally soluble compounds such as sodium chloride. Interactions between the ions in the solution interfere with the simple equilibrium.

## The units for solubility products

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

Example 1: BaSO4

Below is the solubility product expression for barium sulfate:

$K_{sp} = [Ba^{2+}] [SO_4^{2-}]$

Each concentration has the unit mol dm-3. In this case, the units for the solubility product in this case are the following: (mol dm-3) x (mol dm-3) = mol2 dm-6

Example 2: CaPO4

Recall the solubility product expression for calcium phosphate:

$K_{sp} = [Ca^{2+}]^3[PO_4^{3-}]^2$

The units this time will be: (mol dm-3)3 x (mol dm-3)2 = (mol dm-3)5 = mol5 dm-15

## Solubility products apply only to saturated solutions

Recall the barium sulfate equation:

$BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$

The corresponding solubility product expression is the following:

$K_{sp}= [Ba^{2+}][SO_4^{2-}]$

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

In order for this equilibrium constant (the solubility product) to apply, solid barium sulfate must be present in a saturated solution of barium sulfate. This is indicated by the equilibrium equation. If barium ions and sulfate ions exist in solution in the presence of some solid barium sulfate at 298 K, and multiply the concentrations of the ions together, the answer will be 1.1 x 10-10 mol2 dm-6. It is possible to multiply ionic concentrations and obtain a value less than this solubility product; in such cases the solution is too dilute, and no equilibrium exists. ​If the concentrations are lowered enough, no precipitate can form.

However, it is impossible to calculate a product greater than the solubility product. If solutions containing barium ions and sulfate ions are mixed such that the product of the concentrations would exceed Ksp​, a precipitate forms. Enough solid is produced to reduce the concentrations of the barium and sulfate ions down to the value of the solubility product.

## Summary

• The value of a solubility product relates only to a saturated solution.
• If the ionic concentrations give a value less than the solubility product, the solution is not saturated. No precipitate would be formed in such a case.
• If the ionic concentrations give a value more than the solubility product, enough precipitate would be formed to reduce the concentrations to give an answer equal to the solubility product.

## Contributors

Jim Clark (Chemguide.co.uk)