Problem 72 - Introductory Chemistry Atoms First

Question

Suppose you have an aqueous solution of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Ba}^{2+}\) ions. (a) How could you remove just the \(\mathrm{Pb}^{2+}\) ions from the solution? (b) How could you remove both the \(\mathrm{Pb}^{2+}\) and the \(\mathrm{Ba}^{2+}\) ions from the solution at the same time?

Step-by-Step Solution

Answer

(a) To remove just the Pb²⁺ ions, carefully add a sulfate salt solution (e.g. Na₂SO₄) dropwise to the solution, forming a white precipitate of PbSO₄, and then filter or centrifuge the solution to separate the solid. Make sure the concentration of sulfate ions is low enough so that Ba²⁺ ions remain in solution. (b) To remove both Pb²⁺ and Ba²⁺ ions simultaneously, increase the concentration of sulfate ions in the solution, leading to the precipitation of both PbSO₄ and BaSO₄. Filter or centrifuge the solution to separate the respective precipitates.

Step 1: Precipitate Formation for Pb²⁺

In order to remove Pb²⁺ ions from the solution, there needs to be a reaction that selectively forms a solid with it. A common precipitating agent for Pb²⁺ is sulfate (SO₄²⁻) ions. When a soluble sulfate salt is added to the solution containing Pb²⁺ ions, lead sulfate (PbSO₄) will form as a white precipitate. The reaction is as follows: \[Pb^{2+} (aq) + SO_4^{2-} (aq) \rightarrow PbSO_4(s)\] However, we need to make sure that the sulfate ions do not also precipitate the Ba²⁺ ions, as it would defeat the purpose of selectively removing only the Pb²⁺ ions.

Step 2: Precipitate Formation for Ba²⁺

Although BaSO₄ is also a insoluble salt, it is important to carefully control the concentration of sulfate ions in the solution. If the concentration of sulfate ions is kept low enough, the solubility product constant (Ksp) for BaSO₄ will not be exceeded, and Ba²⁺ ions will remain in solution. The solubility product constant values for both salts are: \[K_{sp}(PbSO_4) = 1.6 \times 10^{-8}\] \[K_{sp}(BaSO_4) = 1.1 \times 10^{-10}\]

Step 3: Removing Only Pb²⁺ Ions

To remove just the Pb²⁺ ions from the solution, carefully add a sulfate salt solution (like sodium sulfate, Na₂SO₄) dropwise to the solution and stir gently. Monitor the formation of white precipitate, which is PbSO₄. Continue adding the sulfate salt solution until no more precipitate is formed. Filter or centrifuge the solution to separate the solid PbSO₄ from the liquid. This will leave Ba²⁺ ions in the solution.

Step 4: Removing Both Pb²⁺ and Ba²⁺ Ions

As both Pb²⁺ and Ba²⁺ ions form insoluble salts with sulfate anions, increasing the concentration of sulfate ions in the solution would lead to the precipitation of both PbSO₄ and BaSO₄. To remove both ions from the solution simultaneously, keep adding the sulfate salt solution until both Pb²⁺ and Ba²⁺ ions precipitate as their respective insoluble sulfates. Filter or centrifuge the solution, and both PbSO₄ and BaSO₄ precipitates will be separated from the liquid.

Key Concepts

Precipitate Formation

When a solid material, known as a precipitate, forms from a solution, it is called precipitate formation. This process usually occurs when the product of the concentration of the ions in solution exceeds the solubility product constant (\(K_{sp}\)) for the compound. In the context of our exercise, when \(Pb^{2+}\) ions in an aqueous solution come in contact with sulfate \(SO_4^{2-}\) ions, they combine to form lead sulfate (PbSO₄), a white solid that separates from the solution.

To ensure only lead ions are removed, the concentration of sulfate ions is adjusted just enough to precipitate out the lead ions while keeping the barium ions in solution. The difference in solubility of these salts is exploited to achieve this selective precipitation.

Solubility Product Constant

The solubility product constant, or \(K_{sp}\), is a unique number for each soluble compound that indicates the point at which a solution becomes saturated with the ions that make up the compound. Once this point is exceeded, the excess ions form a solid. In our example, the \(K_{sp}\) values for \(PbSO_4\) and \(BaSO_4\) are \(1.6 \times 10^{-8}\) and \(1.1 \times 10^{-10}\) respectively.

A lower \(K_{sp}\) value indicates a less soluble compound. Thus, by comparing these values, we can anticipate that \(PbSO_4\) will precipitate before \(BaSO_4\) as you carefully add a sulfate salt to the solution, because its \(K_{sp}\) is reached at a higher concentration of sulfate ions.

Selective Precipitation

Selective precipitation is a technique used to separate ions in a mixture by forming a precipitate with only one of the ions while leaving the other ions in solution. It relies on the different solubilities of compounds. For instance, if you wish to remove \(Pb^{2+}\) ions from a solution that also contains \(Ba^{2+}\) ions, you can selectively precipitate out the \(Pb^{2+}\) ions by adding \(SO_4^{2-}\) ions in just the right amount.

Careful control of the reagent's concentration and the solubility differences between lead and barium sulfates allow us to selectively precipitate out lead sulfate, leaving the barium ions in solution. This exemplifies the importance of understanding the solubility product constant and the principles of selective precipitation in separating ions.