Problem 82 - A Complete Resource Book in Chemistry for JEE Main

Question

Three separate samples of a solution of single salt gave these results. One formed a white precipitate with excess ammonia solution, one formed a white precipitate with dil. \(\mathrm{NaCl}\) and one formed a black precipitate with \(\mathrm{H}_{2} \mathrm{~S} .\) The salt could be (a) \(\mathrm{AgNO}_{3}\) (b) \(\mathrm{MnSO}_{4}\) (c) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) (d) \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\)

Step-by-Step Solution

Answer

The salt is \( \text{Pb(NO}_3\text{)}_2 \).

Step 1: Analyze the First Reaction

The salt forms a white precipitate with excess ammonia solution. This suggests the presence of silver ions (Ag^+), as silver chloride (AgCl) is known to form a white precipitate in the presence of ammonia. However, since we are given options, further validation is needed.

Step 2: Analyze the Second Reaction

The salt forms a white precipitate with dilute NaCl. This confirms the presence of silver ions (Ag^+), as silver nitrate (AgNO_3) reacts with NaCl to form silver chloride (AgCl), a white precipitate.

Step 3: Analyze the Third Reaction

The salt forms a black precipitate with H_2S. Lead ions (Pb^2+) and mercury ions (Hg^2+) can react with H_2S to form black precipitates—lead sulfide (PbS) and mercury(II) sulfide (HgS), respectively. Silver ions do not give a black precipitate with H_2S.

Step 4: Match All Conditions

The black precipitate with H_2S indicates that lead or mercury ions are present. Only option (c) Pb(NO_3)_2 can fulfill all conditions: white precipitate with ammonia and dilute NaCl, and both ions may cause a black precipitate with H_2S.

Key Concepts

Silver Ion Reactions

Silver ions (\(\text{Ag}^+\)) play a pivotal role in many inorganic chemical reactions. A noteworthy characteristic of these ions is their ability to form a white precipitate. This occurs when they react with certain substances such as ammonia and sodium chloride.
When silver ions react with excess ammonia, they form silver chloride (\(\text{AgCl}\)), which is a white precipitate. The reaction is as follows: \[\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl(s)} \]This compound is only slightly soluble in water, hence the appearance of the precipitate.
Furthermore, silver ions react with sodium chloride to again yield silver chloride. The persistence of this white precipitate is a hallmark of the presence of silver ions in a solution.
Silver ions, though not reactive with hydrogen sulfide (\(\text{H}_2\text{S}\)) to form black precipitates, are indispensable in various analytical techniques to identify chloride ion presence in analytical samples.

Lead Ion Reactions

The chemistry of lead ions (\(\text{Pb}^{2+}\)) is particularly interesting due to their ability to form characteristic precipitates with sulfide ions. Lead ions can react with \(\text{H}_2\text{S}\) to produce lead sulfide (\(\text{PbS}\)), a compound known for its distinctive black color: \[\text{Pb}^{2+} + \text{S}^{2-} \rightarrow \text{PbS(s)} \]This black precipitate is useful in chemical assays to confirm the presence of lead ions.
Additionally, lead ions display versatility by also forming white precipitates with chloride ions, exemplified by the formation of lead chloride (\(\text{PbCl}_2\)). This dual precipitate formation, with both sulfide and chloride, underlines the significance of lead ions in analytical chemistry.
Their reactivity contributes to the determination of lead presence in various mixtures, by allowing multi-step reactions which give lead-specific precipitates to facilitate easy identification.

Precipitate Formation

In chemical analysis, precipitate formation is a critical process. A precipitate is an insoluble solid that emerges from a liquid solution as a result of a chemical reaction. The formation of precipitates helps in identifying ions present in a complex mixture.
In our analysis, silver ions form a white precipitate with both ammonia and sodium chloride. This suggests a straightforward scenario where silver chloride is produced. Conversely, lead ions form black precipitates with hydrogen sulfide.
The ability of these ions to form precipitates lies in the solubility rules of ionic compounds. Insoluble compounds, like the ones formed in these reactions, tend to form when specific ionic pairs interact, emphasizing the importance of understanding solubility in predicting and explaining precipitate formation in inorganic chemistry.

Chemical Analysis Techniques

Chemical analysis techniques frequently employ the concept of selective precipitation. This technique enables chemists to separate and identify ions based on their differing solubilities.
By introducing reagents that react with specific ions to form precipitates, chemists can discern which ions are present in a solution. For example, the introduction of \(\text{NaCl}\) leads to the formation of a white precipitate of silver chloride if silver ions are present.
Furthermore, adding \(\text{H}_2\text{S}\) to a solution can identify the presence of lead or mercury ions through their respective black precipitates. Such chemical analysis techniques are invaluable in both qualitative and quantitative analysis, providing straightforward diagnostics for complex mixtures.
Utilizing these reactions, chemists can identify, quantify, and assess the purity of various substances with precision, making them foundational tools in laboratory and industrial applications.