Problem 63 - A Complete Resource Book in Chemistry for JEE Main

Question

The solubility of AgBr in hypo is due to the formation of (a) \(\mathrm{Ag}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) (b) \(\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)\right]\) (c) \(\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right]^{3-}\) (d) \(\mathrm{Ag}_{2} \mathrm{SO}_{3}\)

Step-by-Step Solution

Answer

(c) \(\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right]^{3-}\)

Step 1: Understanding the Problem

The problem involves understanding why the solubility of silver bromide (AgBr) increases in the presence of sodium thiosulfate (hypo). We need to understand which complex ion forms during this process.

Step 2: Evaluating Options

Examine each choice to see which chemical species forms in the reaction. Potential complex ions or compounds include: (a) \(\mathrm{Ag}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\)(b) \(\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)\right]\)(c) \(\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right]^{3-}\)(d) \(\mathrm{Ag}_{2} \mathrm{SO}_{3}\)

Key Concepts

Silver Bromide Solubility

Silver bromide (AgBr) is known to have very low solubility in water, meaning that it doesn't dissolve easily. In chemical terms, this is because it has a very low solubility product constant, or \( K_{sp} \). However, its solubility can change under certain conditions, such as in the presence of complexing agents, which change the equilibrium of the reaction. One such agent is sodium thiosulfate (often referred to as hypo).
The main reason why silver bromide can dissolve more readily in the presence of sodium thiosulfate is due to the formation of complex ions. These ions are compounds where central metal atoms or ions are bonded to one or more ligands. In this case, the silver ions (Ag⁺) react with thiosulfate ions (\( ext{S}_2 ext{O}_3^{2-} \)) to form soluble complex ions. This reaction shifts the equilibrium position, thereby increasing the solubility of AgBr.

• Low solubility in water.
• Increases in presence of complexing agents.
• Forms complex ions with ligands.

Sodium Thiosulfate Reaction

Sodium thiosulfate plays a key role as a reagent in enhancing the solubility of silver bromide by forming complex ions. When sodium thiosulfate is added to a solution containing silver bromide, it reacts with silver ions, leading to the formation of complex ions such as \([(\text{Ag} (\text{S}_2\text{O}_3)_2)]^{3-}\). This type of complex ion is more stable and has higher solubility in water compared to the original silver salt.
This reaction can be simplified as follows: Silver ions combine with the thiosulfate ions, pulling more silver ions into the solution from the solid AgBr. This results in the formation of soluble thiosulfate complexes, effectively increasing the solution's capacity to dissolve more AgBr. The ability of sodium thiosulfate to form such complexes helps increase the overall solubility.

Chemical Equilibrium in Solutions

Understanding chemical equilibrium is crucial when discussing the solubility of compounds in solution. Chemical equilibrium occurs when the forward and reverse reactions happen simultaneously at the same rate, meaning that the concentrations of reactants and products remain constant over time.
In the context of the solubility of silver bromide, the introduction of sodium thiosulfate shifts this equilibrium position. When thiosulfate is present, it reacts with silver ions to form complex ions, which reduces the number of free silver ions in solution. According to Le Chatelier's principle, this shift prompts more AgBr to dissolve to restore balance.

• Equilibrium involves a balance between reactions.
• Introduction of thiosulfate shifts balance.
• More compounds dissolve to maintain equilibrium.

Understanding these principles helps explain how the presence of thiosulfate significantly impacts the solubility dynamics of silver bromide.