Further Applications of Ksp 

So far we have used the solubility product, Ksp, to determine the concentration of salts at equilibrium and also worked backward to determine the Ksp value, given a salts solubility.

In this section we will further explore the application of the solubility product, Ksp, to:

  • Predicting the formation of precipitates
  • Common ion effect

Predicting the Formation of Precipitates

The use of Ksp for the dissolution of insoluble or sparingly soluble salts, only applies when the system is at equilibrium. We have seen in a previous section that the ionic product, Qsp, is correctly used when the system is not at equilibrium. Comparing the Ksp and Qsp values provides a convenient way of determining whether a precipitate will form:

  • When Qsp < Ksp, the forward reaction will be favoured to reach equilibrium. Therefore, it is an unsaturated solution and no precipitate has formed
  • When Qsp = Ksp, the reaction has reached dynamic equilibrium. Therefore, it is a saturated solution.
  • When Qsp > Ksp, the reverse reaction will be favoured to reach equilibrium. Therefore, it is a supersaturated solution and a precipitate will form.


1g of BaSO4 was placed into 200 mL of water at 25°C and stirred. Determine whether a precipitate forms. The Ksp value for BaSO4 is 1.08 × 10-10


1g × 1L/0.2L = 5g/L (convert mass of solute into g/L)

Then determine the moles of BaSO4 in a 5g sample:

5g/233.37 = 0.0214 moles

[BaSO4] = 0.0214 mol/L

[Ba2+] = [SO42-] = 0.0214 mol/L

Qsp = [Ba2+][SO42-]

Qsp = [0.0214][0.0214]

Qsp = 4.6 × 10-4

Qsp > Ksp, therefore a precipitate will form

Common Ion Effect

The common ion effect considers problems where the solute is being dissolved in a solution that already contains one of the ions present in the sample. For example, attempting to dissolve a sample of BaSO4 in a solution of 0.1M Na2SO4. As the solution already has SO42- ions present, we can assume less BaSOwill dissolve as a result. This can be analysed quantitatively and the concentration of BaSOthat will dissolve can be determined.

With these problems, it is worth noting that the solution in which the insoluble salt is being added is usually a soluble salt. Therefore, we can assume the concentration of the solution will equal the concentration of the common ion. In the example above, in a 0.1M Na2SO4, the [SO42-] = 0.1M. This may not always be the case and in more difficult problems you may have to determine the concentration of the common ion from its Ksp value.


Determine the concentration of Ba2+ (from BaSO4) that will dissolve in a solution of 0.1M Na2SO4. Ksp for BaSO4 = 1.08 × 10-10


Initially, we would assume:

Ksp = 1.08 × 10-10 = [Ba2+][SO42-] = s2

However, the sample is being placed in a solution with a [SO42-] = 0.1M, so we state:

[Ba2+] = s

[SO42-] = 0.1 + s ≈ 0.1 (remember that s is very small and will have no significant impact on the final [SO42-])


Ksp = 1.08 × 10-10 = [Ba2+][SO42-]

1.08 × 10-10 = s × 0.1

s = \cfrac { { 1.08\times 10 }^{ -10 } }{ 0.1 }

s = 1.08 × 10-9

Therefore, the concentration of BaSOand Ba2+ = 1.08 × 10-9 M