Problem 143 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(\mathrm{Pb}\left[\mathrm{CH}_{3} \mathrm{COO}\right]_{2}\) and \(\mathrm{Na}_{2} \mathrm{~S}\) are mixed and dissolved in water and the solution is filtered then the filtrate will give test of (a) \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) (b) \(\mathrm{Pb}^{2+}\) (c) \(\mathrm{Na}^{+}\) (d) \(\mathrm{S}^{2-}\)

Step-by-Step Solution

Answer

The filtrate will give a test for (a) \(\text{CH}_3\text{COO}^-\) and (c) \(\text{Na}^+\).

Step 1: Write the Reaction

Write the chemical reaction between lead(II) acetate and sodium sulfide in water: \[ \text{Pb(CH}_3\text{COO)}_2 + \text{Na}_2\text{S} \rightarrow \text{PbS} \downarrow + 2\text{NaCH}_3\text{COO} \] Lead sulfide is an insoluble precipitate, while sodium acetate remains in solution.

Step 2: Examine the Filtrate Contents

The filtrate is the part of the solution that remains after removing the precipitate (in this case, PbS). The filtrate will contain the ions that do not participate in forming the precipitate.

Step 3: Identify Ions in the Filtrate

In the reaction \(\text{Na}_2\text{S} + \text{Pb(CH}_3\text{COO)}_2 \rightarrow \text{PbS} \downarrow + 2\text{NaCH}_3\text{COO} \), the soluble product in the filtrate is sodium acetate, which dissociates into ions \(\text{Na}^+\) and \(\text{CH}_3\text{COO}^-\).

Key Concepts

Precipitation reactions

A precipitation reaction occurs when two solutions containing soluble salts are mixed, resulting in the formation of an insoluble solid known as a precipitate. In the given exercise, when lead(II) acetate and sodium sulfide are combined in water, a chemical reaction occurs. Lead(II) sulfide (\(\text{PbS}\)) forms as a precipitate, which means it does not dissolve in water and settles as a solid at the bottom of the solution. The equation for the precipitation reaction is:\[\text{Pb(CH}_3\text{COO)}_2 + \text{Na}_2\text{S} \rightarrow \text{PbS} \downarrow + 2\text{NaCH}_3\text{COO}\]This reaction clearly illustrates the process of forming a solid precipitate from dissolved ionic components.During precipitation reactions:

• A visible solid forms when two solutions react.
• The solid, or precipitate, separates from the liquid, resulting in a filtrate, which is the liquid that remains after the precipitate is removed.

Recognizing the role of precipitation reactions is essential for understanding how substances like lead sulfide can be separated from a solution.

Solubility rules

Solubility rules are guidelines used to predict whether a compound will dissolve in water. These rules are particularly useful in determining the outcome of a precipitation reaction. In the context of the given reaction, the solubility rules help us establish which ions will remain in the solution and which will form a solid precipitate.Here are some basic solubility rules that apply:

• Compounds containing sodium ions (\(\text{Na}^+\)) are generally soluble in water.
• Acetates (\(\text{CH}_3\text{COO}^-\)) are also usually soluble.
• Sulfides, like lead(II) sulfide (\(\text{PbS}\)), are typically insoluble, making \(\text{PbS}\) a precipitate in this reaction.

Using these rules, we identify that while sodium and acetate ions remain dissolved in the filtrate, lead forms an insoluble compound. Understanding solubility rules is vital in predicting the behavior of ionic compounds in a reaction.

Ionic compounds dissolution

Ionic compounds dissolve in water by dissociating into their constituent ions. This process is crucial in chemical reactions, particularly in the context of ion-exchange processes and precipitation reactions. When lead(II) acetate and sodium sulfide are mixed and dissolved in water, the ionic compounds dissociate into their respective ions.For example:

• Lead(II) acetate dissolves as \(\text{Pb}^{2+}\) and \(\text{CH}_3\text{COO}^-\)
• Sodium sulfide dissolves as \(\text{Na}^+\) and \(\text{S}^{2-}\)

Through dissolution, the ions become free to react with each other. This freedom enables reactions like the formation of lead sulfide precipitate, where the lead and sulfide ions bond again.In the result of this particular exercise, the filtrate ultimately contains the ions that remain dissolved: sodium ions \(\text{Na}^+\), and acetate ions \(\text{CH}_3\text{COO}^-\). The sulfide and lead ions combine to form the insoluble lead sulfide, leaving the others in solution. Comprehending the dissolution of ionic compounds aids in understanding which ions remain active in a solution after reactions.