Question

What reagents would you employ to separate these pairs of ions in solution: (a) \(\mathrm{Na}^{+}\) and \(\mathrm{Ba}^{2+},\) (b) \(\mathrm{K}^{+}\) and \(\mathrm{Pb}^{2+},\) (c) \(\mathrm{Zn}^{2+}\) and \(\mathrm{Hg}^{2+}\) ?

Short answer

The reagents that can be used to separate the ions are: \(\mathrm{Na}^{+}\) and \(\mathrm{Ba}^{2+}\) by H2SO4, \(\mathrm{K}^{+}\) and \(\mathrm{Pb}^{2+}\) by K2CrO4, \(\mathrm{Zn}^{2+}\) and \(\mathrm{Hg}^{2+}\) by KI.

Step-by-step solution

Na+ and Ba2+

Considering the pair of ions \(\mathrm{Na}^{+}\) and \(\mathrm{Ba}^{2+},\) sulfuric acid (H2SO4) can be used. Sulfate (SO4^{2-}) forms a precipitate with \(\mathrm{Ba}^{2+}\) (BaSO4), but not with sodium ion \(\mathrm{Na}^{+}\). Therefore, adding sulfuric acid will result in the precipitation of BaSO4, leaving Na+ in solution.

K+ and Pb2+

For the pair of ions \(\mathrm{K}^{+}\) and \(\mathrm{Pb}^{2+},\) potassium chromate (K2CrO4) can be used. The chromate anion (CrO4^{2-}) forms a yellow precipitate with lead ion \(\mathrm{Pb}^{2+}\) (PbCrO4), but not with \(\mathrm{K}^{+}\). In this case, adding potassium chromate will precipitate PbCrO4, leaving K+ in solution.

Zn2+ and Hg2+

The last pair consists of \(\mathrm{Zn}^{2+}\) and \(\mathrm{Hg}^{2+}\). Potassium iodide (KI) can be used. The iodide anion (I-) forms a precipitate with \(\mathrm{Hg}^{2+}\) (HgI2), but not with \(\mathrm{Zn}^{2+}\). Hence, the addition of potassium iodide will precipitate HgI2, leaving \(\mathrm{Zn}^{2+}\) in solution.

Key concepts

Precipitation Reactions

Precipitation reactions are an important process in chemistry where two solutions react to form an insoluble solid known as a precipitate. This occurs when the ions in the solution come together to form a compound that does not dissolve in water. Precipitation reactions are particularly useful in separating ions from solutions.

The key to a successful precipitation reaction is choosing the right pairs of ions and finding a reagent that will form a precipitate with one, but not the other. For example, in the case of sodium ion (\(\mathrm{Na}^{+}\)) and barium ion (\(\mathrm{Ba}^{2+}\)), when sulfuric acid (\(\mathrm{H}_2\mathrm{SO}_4\)) is added, it reacts with barium ions to form barium sulfate (\(\mathrm{BaSO}_4\)), a white precipitate. Sodium ions remain in the solution as they do not form a precipitate with sulfate.

Precipitation not only helps in ion separation but also finds applications in various industries, including wastewater treatment and chemical analysis. It's a simple yet powerful method for isolating specific components from a mixture.

Reagent Selection

Reagent selection is crucial when performing precipitation reactions to separate ions. The choice of reagent depends on the characteristics of the ions in the solution and their potential reactions.

Key factors to consider in reagent selection include:

• Specificity: Choose a reagent that reacts specifically with the desired ion to form an insoluble compound.
• Solubility Rules: Familiarity with solubility rules helps identify which ions will form a precipitate with a certain reagent.
• Reactivity: The reagent should not react with other ions in the solution in a way that might cause unwanted results or secondary reactions.

In our examples, select a reagent that forms a distinct and separable precipitate. For instance, to separate potassium (\(\mathrm{K}^{+}\)) and lead (\(\mathrm{Pb}^{2+}\)), potassium chromate (\(\mathrm{K}_2\mathrm{CrO}_4\)) can be used. The chromate ion reacts with lead ions to form lead chromate (\(\mathrm{PbCrO}_4\)), a yellow precipitate.

Reagent selection requires careful consideration and knowledge of chemical properties to achieve efficient ion separation.

Chemical Solutions

Chemical solutions consist of a solute dissolved in a solvent, forming a homogenous mixture at the molecular level. Understanding the nature of chemical solutions is fundamental when discussing ion separation.

Each ion in a solution carries both charge and specific reactivity. To separate them, one must understand their chemical behavior, which is determined by interactions within the solution. The process involves:

• Dissolution: The initial step of mixing a solute into a solvent, resulting in individual ions distributed evenly in the liquid.
• Reaction: Adding another compound (reagent) that leads to the formation of a precipitate with specific ions.
• Separation: Once the precipitate forms, it can be removed by filtration, leaving the other ions in solution.

In the instance of separating zinc ions (\(\mathrm{Zn}^{2+}\)) from mercury ions (\(\mathrm{Hg}^{2+}\)), when potassium iodide (\(\mathrm{KI}\)) is introduced, iodide ions form a precipitate with mercury ions, resulting in mercuric iodide (\(\mathrm{HgI}_2\)), allowing zinc ions to remain in the solution.

By understanding and manipulating the interactions within chemical solutions, we can effectively separate and isolate desired ions.