Problem 71 - A Complete Resource Book in Chemistry for JEE Main

Question

The ion that cannot be precipitated by both \(\mathrm{HCl}\) and \(\mathrm{H}_{2} \mathrm{~S}\) is (a) \(\mathrm{Ag}^{+}\) (b) \(\mathrm{Cu}^{+}\) (c) \(\mathrm{Sn}^{2+}\) (d) \(\mathrm{Pb}^{2+}\)

Step-by-Step Solution

Answer

(c) \(\mathrm{Sn}^{2+}\) cannot be precipitated by \(\mathrm{HCl}\) and \(\mathrm{H}_{2}\mathrm{~S}\) simultaneously.

Step 1: Understanding precipitation with HCl

When hydrochloric acid (HCl) is added to a solution, it can precipitate ions that form insoluble chlorides. This includes ions like \( \mathrm{Ag}^{+} \) and \( \mathrm{Pb}^{2+} \), which form \( \mathrm{AgCl} \) and \( \mathrm{PbCl}_{2} \) respectively, both of which are insoluble in water.

Step 2: Understanding precipitation with H2S

Hydrogen sulfide (\( \mathrm{H}_{2}\mathrm{S} \)) is used to precipitate metal ions as sulfides. \( \mathrm{Pb}^{2+} \) and \( \mathrm{Cu}^{+} \) can form insoluble sulfides, \( \mathrm{PbS} \) and \( \mathrm{Cu}_{2} \mathrm{S} \), respectively, which are not soluble in acidic conditions.

Step 3: Analyzing \( \mathrm{Sn}^{2+} \)

\( \mathrm{Sn}^{2+} \) ions, or tin(II) ions, do not form a precipitate with hydrochloric acid because tin(II) chloride is soluble. However, \( \mathrm{Sn}^{2+} \) can react with \( \mathrm{H}_{2}\mathrm{S} \) to form \( \mathrm{SnS} \), which is an insoluble sulfide.

Step 4: Evaluate all options

All ion options except \( \mathrm{Sn}^{2+} \) form insoluble chlorides or sulfides with either \( \mathrm{HCl} \) or \( \mathrm{H}_{2}\mathrm{S} \). Thus, \( \mathrm{Sn}^{2+} \) doesn't precipitate with \( \mathrm{HCl} \) but does precipitate with \( \mathrm{H}_{2}\mathrm{S} \).

Key Concepts

Insoluble Compounds

Insoluble compounds are those that do not dissolve well in water or other solvents. They often form precipitates or solids when mixed with their respective solutes. This happens because their constituent ions do not separate easily in water. For example, when hydrochloric acid (HCl) is added to silver ions (Ag^+), silver chloride (AgCl) forms, which is insoluble. This means that it does not dissolve in water and appears as a white solid precipitate.
Insolubility can be due to the strong ionic bonds within the compound or because of specific interactions with the solvent that do not favor dispersion. The formation of insoluble compounds is at the heart of precipitation reactions.
Experimenting with various ions in solutions can often lead to these visible reactions, a common occurrence in chemistry labs.

Hydrochloric Acid (HCl)

Hydrochloric acid (HCl) is a strong acid commonly used in various chemical reactions. In aqueous solutions, it dissociates completely into hydrogen ions (H^+) and chloride ions (Cl^-). This complete dissociation is what makes HCl so potent as an acid.
HCl is especially useful in precipitation reactions because it can react with metal ions to form insoluble chloride salts. For example, when silver ions (Ag^+) in a solution come into contact with HCl, silver chloride (AgCl) is formed as a white precipitate.
This precipitate formation is a classic demonstration of how HCl is used to isolate certain ions in a solution due to its ability to form insoluble compounds under specific conditions.

Hydrogen Sulfide (H2S)

Hydrogen sulfide (H_2S) is a colorless gas known for its pungent odor. In aqueous solutions, it acts as a weak acid and can release hydrogen sulfide ions (HS^-) and sulfide ions (S^{2-}). These ions can create a series of reactions that are useful for precipitating metal ions as sulfides. The use of H_2S in chemistry is significant because it helps in identifying and separating metal ions through the formation of insoluble metal sulfides.
For instance, when lead ions (Pb^{2+}) interact with H_2S, lead sulfide (PbS) forms, which is an insoluble black precipitate. This ability to create specific precipitates is crucial in qualitative analysis and helps chemists identify the presence of various metal ions in mixtures.

Metal Ions

Metal ions are atoms of metals that have lost electrons, which gives them a positive charge. The charge indicates the number of electrons the ion has lost. For example, Cu^+ denotes a copper ion that has lost one electron. Metal ions play a key role in precipitation reactions because their interactions with anions can lead to the formation of insoluble compounds.
In the presence of certain reagents like HCl and H_2S, metal ions can form specific insoluble salts. For instance, Pb^{2+} in solution will precipitate as PbCl_2 with HCl, and as PbS with H_2S.
Recognizing which metal ions form which types of insoluble compounds helps in the identification and separation processes in chemistry.

Solubility Rules

Solubility rules are guidelines that help predict the solubility of compounds in water. These rules are essential in determining whether a compound will dissolve or form a precipitate in a solution. For example, most salts containing alkali metal ions and ammonium are soluble. However, chlorides, bromides, and iodides of lead, silver, and mercury are generally insoluble.
These rules are crucial in understanding and conducting precipitation reactions. By using solubility rules, one can predict the result of mixing different solutions and determine the products formed. Understanding these rules is vital for students and chemists alike as they work to anticipate and explain the outcomes of various chemical reactions.