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

The ion that cannot be precipitated by both HCl and H2S is Sn^{2+}.

Step 1: Understanding Precipitation with HCl

Hydrochloric acid (HCl) is commonly used to precipitate Ag^+, Pb^{2+}, and Cu^+ ions from their solutions because they form insoluble chlorides with HCl.

Step 2: Understanding Precipitation with H2S

Hydrogen sulfide (H_2S) is used to precipitate metal ions that form insoluble sulfides. Most metal ions will form sulfide precipitates under acidic conditions provided by H_2S. However, the formation of sulfides depends on the solubility product (Ksp).

Step 3: Evaluating Each Ion for Precipitation

(a) Ag^+ forms AgCl with HCl and Ag_2S with H_2S. (b) Cu^+ forms CuCl with HCl generally. (c) Sn^{2+} does not form a precipitate with HCl but forms SnS with H_2S. (d) Pb^{2+} forms PbCl_2 with HCl and PbS with H_2S.

Step 4: Identifying the Non-Precipitating Ion

The ion Sn^{2+} cannot be precipitated with HCl, as it does not form an insoluble chloride, but can be precipitated with H_2S as it forms SnS. Therefore, Sn^{2+} cannot be precipitated by both HCl and H_2S.

Key Concepts

Solubility Product

The solubility product, often abbreviated as Ksp, is a very important concept in chemistry, especially when dealing with precipitation reactions. It's a special type of equilibrium constant that applies to the dissolution of a slightly soluble ionic compound. This value indicates how much of the compound can dissolve in the solution before starting to precipitate.

When we say that a compound "precipitates," it means that it forms a solid from the solution. The solubility product helps predict this occurrence. If the ionic product (the concentrations of the dissolved ions multiplied together) exceeds the Ksp, precipitation occurs. For example, in the case of chloride ions forming a precipitate with certain metal ions, understanding the Ksp of these metal chlorides helps us anticipate whether a solid will form when these ions are mixed.

• A low Ksp means that the compound is less soluble and more likely to precipitate quickly.
• A high Ksp indicates greater solubility and little tendency to form a solid precipitate.

Metal Ion Precipitation

Precipitating metal ions is a common laboratory procedure to separate or identify different ionic species. It involves adding a reagent that reacts with certain metal ions to form an insoluble solid, or precipitate.

Hydrochloric acid (HCl) is widely used for precipitating silver (\(\mathrm{Ag^+}\), lead (\(\mathrm{Pb^{2+}}\), and copper (\(\mathrm{Cu^{+}}\) ions from a solution, forming insoluble chlorides like \(\mathrm{AgCl}\) and \(\mathrm{PbCl_2}\). This effectiveness is based on their low solubility products. However, not all metal ions form precipitates with HCl. For instance, \(\mathrm{Sn^{2+}}\) does not react with HCl in the same manner.

• \(\mathrm{HCl}\) and \(\mathrm{H_2S}\) often serve as reagents for metal ion precipitation due to their ability to form low-solubility compounds with certain metals.
• \(\mathrm{H_2S}\) leads to the formation of sulfide precipitates, essential for separating ions that do not form chloride precipitates.

Insoluble Chlorides

Insoluble chlorides play a crucial role in precipitation chemistry, acting as a tool for separating specific ions from a mixture. "Insoluble" typically means a compound does not dissolve noticeably in water—or does so only minimally.

In the context of ions like \(\mathrm{Ag^+}\), \(\mathrm{Pb^{2+}}\), and \(\mathrm{Cu^{+}}\), these ions form insoluble chlorides when combined with hydrochloric acid (\(\mathrm{HCl}\)). These solids, \(\mathrm{AgCl}\), \(\mathrm{PbCl_2}\), and \(\mathrm{CuCl}\), respectively, allow easy separation from other ions in solution.

• Not all cations form insoluble chlorides. For example, \(\mathrm{Sn^{2+}}\) does not precipitate as a chloride, illustrating varying interactions with \(\mathrm{HCl}\).
• The formation of these insoluble salts depends greatly on both the ion properties and their solubility products, dictating when and if precipitation will occur.