**Further Applications of K _{sp }**

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

**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 BaSO_{4} was placed into 200 mL of water at 25°C and stirred. Determine whether a precipitate forms. The K_{sp} value for BaSO_{4} is 1.08 × 10^{-10}

**Answer:**

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

Then determine the moles of BaSO_{4} in a 5g sample:

5g/233.37 = 0.0214 moles

[BaSO_{4}] = 0.0214 mol/L

[Ba^{2+}] = [SO_{4}^{2-}] = 0.0214 mol/L

Q_{sp} = [Ba^{2+}][SO_{4}^{2-}]

Q_{sp} = [0.0214][0.0214]

Q_{sp} = 4.6 × 10^{-4}

Q_{sp }> K_{sp}, 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 BaSO_{4} in a solution of 0.1M Na_{2}SO_{4}. As the solution already has SO_{4}^{2-} ions present, we can assume less BaSO_{4 }will dissolve as a result. This can be analysed quantitatively and the concentration of BaSO_{4 }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 Na_{2}SO_{4}, the [SO_{4}^{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 Ba^{2+} (from BaSO_{4}) that will dissolve in a solution of 0.1M Na_{2}SO_{4. }Ksp for BaSO_{4 }= 1.08 × 10^{-10}

**Answer:**

Initially, we would assume:

K_{sp} = 1.08 × 10^{-10 }= [Ba^{2+}][SO_{4}^{2-}] = s^{2}

However, the sample is being placed in a solution with a [SO_{4}^{2-}] = 0.1M, so we state:

[Ba^{2+}] = s

[SO_{4}^{2-}] = 0.1 + s ≈ 0.1 (remember that s is very small and will have no significant impact on the final [SO_{4}^{2-}])

Now:

K_{sp} = 1.08 × 10^{-10 }= [Ba^{2+}][SO_{4}^{2-}]

1.08 × 10^{-10 }= s × 0.1

s =

s = 1.08 × 10^{-9}

Therefore, the concentration of BaSO_{4 }and Ba^{2+} = 1.08 × 10^{-9 }M