Further Applications of K_sp 

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

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

• Predicting the formation of precipitates
• Common ion effect

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Predicting the Formation of Precipitates

The use of K_sp 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, Q_sp, is correctly used when the system is not at equilibrium. Comparing the K_sp and Q_sp values provides a convenient way of determining whether a precipitate will form:

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

Example:

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

Answer:

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⁺] = [SO₄^2-] = 0.0214 mol/L

Q_sp = [Ba²⁺][SO₄^2-]

Q_sp = [0.0214][0.0214]

Q_sp = 4.6 × 10^-4

Q_sp > K_sp, therefore a precipitate will form

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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 SO₄^2- ions present, we can assume less BaSO₄ will dissolve as a result. This can be analysed quantitatively and the concentration of BaSO₄ that 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 [SO₄^2-] = 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 K_sp value.

Example: 

Determine the concentration of Ba²⁺ (from BaSO₄) that will dissolve in a solution of 0.1M Na₂SO_4. Ksp for BaSO₄ = 1.08 × 10^-10

Answer:

Initially, we would assume:

K_sp = 1.08 × 10^-10 = [Ba²⁺][SO₄^2-] = s²

However, the sample is being placed in a solution with a [SO₄^2-] = 0.1M, so we state:

[Ba²⁺] = s

[SO₄^2-] = 0.1 + s ≈ 0.1 (remember that s is very small and will have no significant impact on the final [SO₄^2-])

Now:

K_sp = 1.08 × 10^-10 = [Ba²⁺][SO₄^2-]

1.08 × 10^-10 = s × 0.1

s =

s = 1.08 × 10^-9

Therefore, the concentration of BaSO₄ and Ba²⁺ = 1.08 × 10^-9 M