Problem 96 - General Chemistry: Principles and Modern Applications

Question

\(\mathrm{Cu}^{2+}\) and \(\mathrm{Pb}^{2+}\) are both present in an aqueous solution. To precipitate one of the ions and leave the other in solution, add (a) \(\mathrm{H}_{2} \mathrm{S}(\mathrm{aq}) ;\) (b) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\); (c) \(\mathrm{HNO}_{3}(\mathrm{aq}) ;\) (d) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{aq})\).

Step-by-Step Solution

Answer

The solution that will precipitate one of the ions and leave the other in solution is \(H_2 SO_4\).

Step 1: Consider the solubility rules

The first step is to consider the solubility rules in order to determine whether a particular salt will be soluble or not. According to these rules, sulfates (\(SO_4^{2-}\)) are generally soluble except with \(Pb^{2+}\), \(Ba^{2+}\), \(Hg^{2+}\), and \(Ag^{+}\). Similarly, most nitrate (\(NO_3^-\)) salts are soluble. Sulfides (\(S^{2-}\)) are typically insoluble except with group 1 elements, \(NH_4^+\), \(Ca^{2+}\), \(Sr^{2+}\), and \(Ba^{2+}\). Therefore, we can exclude (d) \(NH_4 NO_3\) as it will not precipitate either of the ions.

Step 2: Test the remaining solutions

Now, we test the remaining solutions (a), (b) and (c) for either \(Cu^{2+}\) or \(Pb^{2+}\) to precipitate. (a) \(H_2 S\) will form \(CuS\) and \(PbS\). Both these sulfides are insoluble, meaning both \(Cu^{2+}\) and \(Pb^{2+}\) will precipitate. Therefore, (a) \(H_2 S\) does not serve our purpose. (b) \(H_2 SO_4\) will form \(CuSO_4\) and \(PbSO_4\). While \(CuSO_4\) is soluble, \(PbSO_4\) is insoluble, so it will precipitate \(Pb^{2+}\) leaving \(Cu^{2+}\) in solution. So solution (b) \(H_2 SO_4\) is correct. As a verification step, we can look at (c) \(HNO_3\), which will form \(Cu(NO_3)_2\) and \(Pb(NO_3)_2\). Both these nitrates are soluble, hence \(HNO_3\) will not precipitate either of the ions.

Step 3: Summarize the Solution

From the analysis, \(H_2 SO_4\) is the solution which will precipitate \(Pb^{2+}\) and leave \(Cu^{2+}\) in solution whilst all other solutions do not meet the requirements. Thus, the correct solution is (b) \(H_2 SO_4\).

Key Concepts

Precipitation Reactions

Precipitation reactions occur when two aqueous solutions combine, and an insoluble solid called a precipitate forms. The key to understanding these reactions lies in knowing which combinations of ions will result in an insoluble product. This is often predicted using solubility rules.
When two ion solutions are mixed, the ions swap partners and form new compounds. If any of the new compounds are insoluble in water, they will form a solid precipitate and fall out of the solution.
For example, in the given exercise, adding \(H_2SO_4\) to an aqueous solution containing \(Pb^{2+}\) and \(Cu^{2+}\) leads to the formation of \(PbSO_4\), which is insoluble and precipitates, leaving \(Cu^{2+}\) in solution. This selectivity is crucial when we want to separate specific ions from a mixture. To predict such reactions, a solid understanding of solubility rules is indispensable.

Insoluble Compounds

Insoluble compounds are substances that do not dissolve significantly in water. Whether a substance is soluble or insoluble depends on the interactions between its ions and water molecules.
Certain ions form insoluble compounds when combined, following specific rules of thumb known as solubility rules. For instance:

• Sulfates are generally soluble, except when combined with ions like \(Pb^{2+}\), \(Ba^{2+}\), \(Hg^{2+}\), and \(Ag^{+}\).
• Sulfides are typically insoluble, except with group 1 elements, \(NH_4^+\), \(Ca^{2+}\), \(Sr^{2+}\), and \(Ba^{2+}\).
• Nitrates and acetates are soluble with virtually no exceptions.

In the original exercise, understanding that \(PbSO_4\) is insoluble aids in selectively precipitating \(Pb^{2+}\) ions from a mixture containing \(Cu^{2+}\) and \(Pb^{2+}\) using \(H_2SO_4 \). These rules help chemists manipulate reactions to obtain desired outcomes.

Aqueous Solutions

Aqueous solutions refer to a system where a substance is dissolved in water. In such solutions, compounds separate into ions, enabling them to interact with other ions in the mixture.
The concept of aqueous solutions is fundamental in reactions like precipitation. Water acts as a medium that allows ions to be free and mobile. This freedom supports the formation of new compounds when different aqueous solutions are mixed.
In the exercise, both \(Cu^{2+}\) and \(Pb^{2+}\) are present in an aqueous medium, meaning they are free ions ready to react with added reagents. When \(H_2SO_4\) is added, \(Pb^{2+}\) ions form an insoluble \(PbSO_4\) precipitate, demonstrating the power of aqueous solutions to demonstrate selective precipitation through ionic interactions. Understanding these interactions helps in designing solutions to complex chemical problems.