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

(a) \(\mathrm{CH}_3\mathrm{COO}^-\) is in the filtrate.

Step 1: Understanding the Reaction

When \(\mathrm{Pb}\left[\mathrm{CH}_{3} \mathrm{COO}\right]_{2}\) and \(\mathrm{Na}_{2} \mathrm{~S}\) are mixed in water, a double displacement reaction occurs. \(PbS\) (lead(II) sulfide) is insoluble in water and forms a precipitate. The reaction can be written as: \[ \mathrm{Pb(CH}_3\mathrm{COO})_2 + \mathrm{Na}_2\mathrm{S} \rightarrow \mathrm{PbS} \downarrow + 2\mathrm{NaCH}_3\mathrm{COO} \]

Step 2: Identifying Soluble Products

After the reaction, lead(II) sulfide \(PbS\) precipitates out, leaving \(NaCH_3COO\) (sodium acetate) in solution. Sodium acetate is soluble in water, meaning it will not form a precipitate and will remain in the filtrate.

Step 3: Analyzing the Filtration

Once the mixture is filtered, the solid \(PbS\) is removed, and the filtrate contains the soluble compound \(NaCH}_3\mathrm{COO}\). This means we need to determine which ions are present in sodium acetate.

Step 4: Ionic Components of Sodium Acetate

Sodium acetate \(NaCH_3COO\) breaks down into \(Na^+\) and \(CH_3COO^-\) ions in water. Thus, the filtrate will contain these ions. \(Pb^{2+}\) and \(S^{2-}\) are not present in the filtrate because \(PbS\) appears in the solid precipitate.

Key Concepts

Ionic Compounds

Ionic compounds are formed when positively charged ions (cations) and negatively charged ions (anions) come together. They create a structure held together by the strong forces of attraction between oppositely charged ions in what is often called an "ionic bond."
When these compounds dissolve in water, their constituent ions separate, becoming free to move throughout the solution.
In the exercise provided, both lead acetate, \( \mathrm{Pb}\left[\mathrm{CH}_{3} \mathrm{COO}\right]_{2} \), and sodium sulfide, \( \mathrm{Na}_2 \mathrm{S} \), are ionic compounds. When mixed in water, they dissociate into their respective ions:

• \( \mathrm{Pb}^{2+} \) and \( \mathrm{CH}_{3} \mathrm{COO}^{-} \) from lead acetate
• \( \mathrm{Na}^{+} \) and \( \mathrm{S}^{2-} \) from sodium sulfide

This separation of ions is essential for the reaction to take place as it sets the stage for the interchange of ions, which is fundamental in a double displacement reaction.

Precipitation Reaction

A precipitation reaction is a type of chemical reaction where two solutions of ionic substances are combined, resulting in the formation of an insoluble solid, known as a "precipitate."
The insoluble compound separates out of the solution as a solid.In the given example, when the solutions of lead acetate and sodium sulfide are mixed, lead sulfide (\( \mathrm{PbS} \)) forms as a solid precipitate.
This is because lead sulfide is not soluble in water. The reaction equation can be represented as: \[ \mathrm{Pb(CH}_3\mathrm{COO})_2 + \mathrm{Na}_2\mathrm{S} \rightarrow \mathrm{PbS} \downarrow + 2\mathrm{NaCH}_3\mathrm{COO} \] Here, the arrow pointing down (\( \downarrow \)) indicates the formation of the precipitate.
This reaction is a classic example of a double displacement reaction, where the ions swap partners and at least one of the resulting compounds is insoluble in water, forming a precipitate.

Solubility Rules

Solubility rules are essential guidelines that predict whether an ionic compound will dissolve in water. These rules help determine if an ionic compound will remain dissolved or will precipitate when formed in solution.
A few important solubility rules include:

• Most salts containing alkali metal ions (like \( \mathrm{Na}^+ \)) and ammonium (\( \mathrm{NH}_4^+ \)) are soluble.
• Nitrates (\( \mathrm{NO}_3^- \)), acetates (\( \mathrm{CH}_3 \mathrm{COO}^- \)), and most chlorates (\( \mathrm{ClO}_3^- \)) are soluble.
• Sulfides (\( \mathrm{S}^{2-} \)), carbonates (\( \mathrm{CO}_3^{2-} \)), and phosphates (\( \mathrm{PO}_4^{3-} \)) are generally insoluble, except for those of alkali metals and ammonium.
• Lead sulfide (\( \mathrm{PbS} \)), in this specific problem, is insoluble, leading to the formation of the precipitate.

In practice, these rules help chemists predict the products of a reaction and identify if a precipitate will form, guiding the analysis of filtration outcomes and identifying ions left in solution.