Question

A solution containing several metal ions is treated with dilute HCl; no precipitate forms. The \(\mathrm{pH}\) is adjusted to about \(1,\) and \(\mathrm{H}_{2} \mathrm{S}\) is bubbled through. Again, no precipitate forms. The \(\mathrm{pH}\) of the solution is then adjusted to about \(8 .\) Again, \(\mathrm{H}_{2} \mathrm{S}\) is bubbled through. This time a precipitate forms. The filtrate from this solution is treated with \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4} .\) No precipitate forms. Which of these metal cations are either possibly present or definitely absent: \(\mathrm{Al}^{3+}, \mathrm{Na}^{+}, \mathrm{Ag}^{+}, \mathrm{Mg}^{2+} ?\)

Short answer

Based on the chemical treatments, the following metal ions are either possibly present or definitely absent: - \(\mathrm{Al}^{3+}\) is possibly present - \(\mathrm{Na}^{+}\) is possibly present - \(\mathrm{Ag}^{+}\) is definitely absent - \(\mathrm{Mg}^{2+}\) is possibly present

Step-by-step solution

Analyzing the initial treatment with dilute HCl

Since no precipitate forms after treating the solution with HCl, we can conclude that metal ions which could form chloride precipitates are not present in the solution. The following ions are capable of forming insoluble chlorides: Ag+, Pb2+, and Hg2+. Since none of these metal cations are listed, we can move on to the next step.

Analyzing the treatment with H2S at pH 1

In acidic solution (pH 1), H2S should precipitate metal sulfides, if present, which are insoluble in acidic media (Group II elements). Again, no precipitate forms, meaning that these elements are not present. The possible Group II elements are Cu2+, Cd2+, Bi3+, Pb2+, and Hg2+. None of the listed metal cations belongs to this group so we proceed to the next step.

Analyzing the treatment with H2S at pH 8

When the pH is adjusted to 8 and H2S is bubbled through the solution again, a precipitate forms. This indicates the presence of metal cations from Group III, which are those that form insoluble sulfides in alkaline media. The possible Group III metal cations include Al3+, Fe3+, and Cr3+. Out of the given metal ions, only \(\mathrm{Al}^{3+}\) belongs to this group, so \(\mathrm{Al}^{3+}\) could be possibly present.

Analyzing the treatment with (NH4)2HPO4

The filtrate from the previous step is treated with \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\). No precipitate forms. This eliminates the possibility that any of the remaining metal ions form insoluble phosphates. The two remaining ions from the list given are \(\mathrm{Na}^{+}\) and \(\mathrm{Mg}^{2+}\). Neither of these ions forms any precipitate when their respective salts are present in the solution—thus, they can both be possibly present in the initial solution. #Conclusion# Based on the chemical treatments, the following conclusions can be drawn: - \(\mathrm{Al}^{3+}\) is possibly present - \(\mathrm{Na}^{+}\) is possibly present - \(\mathrm{Ag}^{+}\) is definitely absent - \(\mathrm{Mg}^{2+}\) is possibly present

Key concepts

Solubility Rules

Understanding solubility rules is crucial when determining the formation of precipitates in chemical reactions. The basic principle is that certain ionic compounds dissolve well in water (soluble), while others do not (insoluble). Soluble compounds will remain in solution, whereas insoluble compounds will precipitate out of solution to form solids. This is due to the interplay between the ions and water molecules, specifically the energetics of the solvent-solute interactions versus the lattice energy of the solid.

To predict solubility and thus precipitation reactions, chemists use general guidelines, or solubility rules. Some common rules include that most salts containing alkali metal ions (like \(\mathrm{Na}^{+}\)) and ammonium (\(\mathrm{NH}_{4}^{+}\)) are soluble, and so are most nitrates \(\mathrm{NO}_{3}^{-}\) and acetates \(\mathrm{CH}_{3}COO^{-}\). Conversely, many sulfates are soluble except for those of calcium, strontium, barium, and lead. The provided exercise demonstrates these principles, noting that no precipitate formed upon the addition of \(\mathrm{HCl}\), indicating the absence of insoluble chloride forming metal cations such as \(\mathrm{Ag}^{+}\).

Metal Cation Analysis

Metal cation analysis often employs selective precipitation to identify or separate specific metals from a mixture. The process is a systematic one, typically involving the addition of reagents that form insoluble salts with particular groups of metal cations under controlled conditions, like pH. In the exercise, this selective precipitation is used to identify possible and absent metal cations. For instance, when \(\mathrm{H}_{2} \mathrm{S}\) was added at a pH of 8, a precipitate formed, which suggested the presence of Group III cations like \(\mathrm{Al}^{3+}\).

Group I cations precipitate with chlorides, Group II cations with sulfides in acidic conditions, and Group III cations with sulfides in neutral to basic conditions. The exercise involves first treating the solution with dilute HCl, which does not lead to precipitation, indicating that Group I cations are absent. The second and third steps involve \(\mathrm{H}_{2} \mathrm{S}\) under different pH conditions, helping to narrow down the possible presence of different metal cations based on whether or not a precipitate forms.

pH Effect on Solubility

The pH of a solution can vastly impact the solubility of compounds, especially when it comes to metal hydroxides and sulfides. This is due to the fact that the concentration of \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions can shift the equilibrium of dissolution or precipitation reactions. At low pH values, excess \(\mathrm{H}^{+}\) ions bind to anions like \(\mathrm{S}^{2-}\), reducing their concentration in the solution and thus, the solubility of corresponding metal sulfides. That's why no precipitate formed when \(\mathrm{H}_{2} \mathrm{S}\) was introduced in an acidic solution (pH ~1) in the exercise.

As the pH increases, the concentration of protons decreases, and anions like \(\mathrm{S}^{2-}\) are more likely to combine with metal ions to form precipitates. At a pH of about 8, the environment is less acidic and can lead to the formation of metal sulfides, which occurred during the exercise's third step. Additionally, amphiprotic metal ions, such as \(\mathrm{Al}^{3+}\), can form insoluble metal hydroxides at higher pH levels, which is in line with the observed precipitation in this step of the analysis.