Question

A solution that may contain \(\mathrm{Cu}^{2+}, \mathrm{Bi}^{3+}, \mathrm{Sn}^{4+},\) or \(\mathrm{Sb}^{3+}\) ions is treated with thioacetamide in an acid medium. The black precipitate that forms is partly soluble in strongly alkaline solution. The precipitate that remains is soluble in \(6 \mathrm\quad{M}\) \(HNO\), and gives only a white precipitate with excess \(\mathrm{NH}_{3}\). The alkaline solution, when acidified, produces an orange precipitate. On the basis of this information, which ions are present, which are absent, and which are still in doubt? Present __________ Absent __________ In doubt __________

Short answer

Present: \(\mathrm{Bi}^{3+}\), \(\mathrm{Sn}^{4+}\) Absent: \(\mathrm{Cu}^{2+}\), \(\mathrm{Sb}^{3+}\) In doubt: None

Step-by-step solution

Formation of a black precipitate with thioacetamide in an acidic medium

All the given ions form precipitates in an acidic medium, so the formation of a black precipitate with thioacetamide does not give conclusive information about the ions present. At this point, all ions are still in doubt.

Partial solubility in strong alkali

We know that \(\mathrm{Bi}^{3+}\) and \(\mathrm{Cu}^{2+}\) precipitates are soluble in a strong alkali, while \(\mathrm{Sn}^{4+}\) and \(\mathrm{Sb}^{3+}\) precipitates are insoluble. Based on this information, we can say that either \(\mathrm{Bi}^{3+}\) or \(\mathrm{Cu}^{2+}\) ions are present, and either \(\mathrm{Sn}^{4+}\) or \(\mathrm{Sb}^{3+}\) ions are also present.

Solubility in \(6\,\mathrm{M}\) \(HNO\)

The precipitate that remains (partly insoluble in strong alkali) is soluble in \(6\,\mathrm{M}\) \(HNO\). Since \(\mathrm{Sn}^{4+}\) precipitate is soluble in this solution, and \(\mathrm{Sb}^{3+}\) precipitate is not, we can conclude that \(\mathrm{Sn}^{4+}\) ions are present, and \(\mathrm{Sb}^{3+}\) ions are absent.

Formation of a white precipitate with excess \(\mathrm{NH}_{3}\)

We are now left with \(\mathrm{Bi}^{3+}\) and \(\mathrm{Cu}^{2+}\) ions. The precipitate soluble in \(6\,\mathrm{M}\) \(HNO\) gives only a white precipitate with excess \(\mathrm{NH}_{3}\). Since the precipitate of \(\mathrm{Bi}^{3+}\) is white with \(\mathrm{NH}_{3}\), and \(\mathrm{Cu}^{2+}\) would create a deep blue precipitate, we can conclude that \(\mathrm{Bi}^{3+}\) ions are present and \(\mathrm{Cu}^{2+}\) ions are absent.

Formation of an orange precipitate after acidification

We now need to confirm whether the alkaline solution contains ions other than bismuth. When the solution is acidified, it forms an orange precipitate. Since none of the ions mentioned above form orange precipitates, we can conclude that no other ions are present, and there are no ions in doubt. Based on this information, we can classify the ions as follows: Present: \(\mathrm{Bi}^{3+}\), \(\mathrm{Sn}^{4+}\) Absent: \(\mathrm{Cu}^{2+}\), \(\mathrm{Sb}^{3+}\) In doubt: None

Key concepts

Ion Identification

Ion identification is a crucial process in chemistry to determine which ions are present in a given solution. This involves performing various tests based on the unique chemical properties of each ion. For example, different ions may react with specific reagents to produce visible changes, such as color changes or precipitates. These reactions help us identify the ions by comparing the observed results against known chemical behavior.

In this exercise, ions like

• \( \text{\( \mathrm{Cu}^{2+} \)} \)
• \( \mathrm{Bi}^{3+} \)
• \( \mathrm{Sn}^{4+} \)
• \( \mathrm{Sb}^{3+} \)

are tested specifically through their interaction with thioacetamide. Identifying whether each ion is present, absent, or still in doubt helps chemists systematically categorize their presence in complex mixtures.

Precipitation Reactions

Precipitation reactions occur when two soluble salts react in a solution to form an insoluble product known as a precipitate. This can be a fast and visible way to analyze the components of a solution.

In these reactions, the formation of a solid precipitate indicates that specific ions have reacted together.

• With thioacetamide in an acid medium, a black precipitate may form, signaling the presence of metal sulfides.
• Further tests, like those using strong alkalis or nitric acid, can help determine the solubility of these precipitates and thereby the ions they are composed of.
• In this case, testing solubility in strong alkali helps differentiate between ions like \(\mathrm{Bi}^{3+}\) and \(\mathrm{Cu}^{2+}\) (soluble) and \(\mathrm{Sn}^{4+}\) and \(\mathrm{Sb}^{3+}\) (insoluble).

Understanding these solubility rules and reactions can help identify and isolate ions in a mixture.

Thioacetamide Test

The thioacetamide test is a method used to identify metal ions by adding thioacetamide to a solution to produce hydrogen sulfide gas in acidic conditions. This hydrogen sulfide can precipitate metal ions as their respective sulfides.

Here's how thioacetamide works:

• Under acidic conditions, thioacetamide hydrolyzes to produce hydrogen sulfide \((\mathrm{H}_2\mathrm{S})\) gas.
• This gas reacts with metal ions to form characteristic sulfide precipitates, which can be black or colored depending on the metal ion present.
• These precipitates can be further tested in different conditions (like strong alkali or nitric acid) to confirm the presence of specific ions.

For instance, in the given problem, thioacetamide is used to generate a black precipitate, which can then be dissolved or reacted further to explore which ions, like \(\mathrm{Bi}^{3+}\) or \(\mathrm{Sn}^{4+}\), are part of the mixture.