Problem 79 - Chemistry and Chemical Reactivity

Question

Gas evolution was observed when a solution of \(\mathrm{Na}_{2} \mathrm{S}\) was treated with acid. The gas was bubbled into a solution containing \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2},\) and a black precipitate formed. Write net ionic equations for the two reactions.

Step-by-Step Solution

Answer

1. \(\mathrm{S}^{2-}_{(aq)} + 2 \mathrm{H}^+_{(aq)} \rightarrow \mathrm{H}_2\mathrm{S}_{(g)}\); 2. \(\mathrm{Pb}^{2+}_{(aq)} + \mathrm{S}^{2-}_{(aq)} \rightarrow \mathrm{PbS}_{(s)} \).

Step 1: Identify the Gas Evolution Reaction

When sodium sulfide \(\mathrm{Na}_2\mathrm{S}\) is treated with an acid such as \(\mathrm{H}^+\), the gas hydrogen sulfide \(\mathrm{H}_2\mathrm{S}\) is produced. The reaction is: \[ \mathrm{Na}_2\mathrm{S}_{(aq)} + 2 \mathrm{H}^+_{(aq)} \rightarrow 2 \mathrm{Na}^+_{(aq)} + \mathrm{H}_2\mathrm{S}_{(g)} \]

Step 2: Write the Net Ionic Equation for the Gas Evolution

The net ionic equation for the gas evolution reaction considers only the ions and molecules directly involved in forming \(\mathrm{H}_2\mathrm{S}\) gas: \[ \mathrm{S}^{2-}_{(aq)} + 2 \mathrm{H}^+_{(aq)} \rightarrow \mathrm{H}_2\mathrm{S}_{(g)} \]

Step 3: Identify the Precipitation Reaction

The \(\mathrm{H}_2\mathrm{S}\) gas is bubbled into a lead nitrate solution, \(\mathrm{Pb(NO}_3)_2\), leading to the formation of a black precipitate of lead sulfide \(\mathrm{PbS}\). The reaction proceeds as follows: \[ \mathrm{Pb}^{2+}_{(aq)} + \mathrm{S}^{2-}_{(aq)} \rightarrow \mathrm{PbS}_{(s)} \]

Step 4: Write the Net Ionic Equation for the Precipitation

The net ionic equation for the formation of lead sulfide precipitation is simple and only involves ions that form \(\mathrm{PbS}\): \[ \mathrm{Pb}^{2+}_{(aq)} + \mathrm{S}^{2-}_{(aq)} \rightarrow \mathrm{PbS}_{(s)} \]

Key Concepts

Gas Evolution Reaction

A gas evolution reaction is a chemical reaction where a gas is produced as one of the products. This reaction type is characterized by bubbling and fizzing as the gas escapes the liquid solution. Such reactions are quite common in chemistry, especially in laboratory settings, where they may be used to study the properties of gases or to drive a desired reaction under specific conditions.

In the provided exercise, when a solution of sodium sulfide (\(\mathrm{Na_2S}\)) is treated with an acid, it undergoes a gas evolution reaction. The acid reacts with sulfide ions (\(\mathrm{S}^{2-}\)) to form hydrogen sulfide gas (\(\mathrm{H_2S}\)). The net ionic equation simplifies this to only include the relevant species:

• Reactants: \(\mathrm{S}^{2-}_{(aq)}\) and \(2\mathrm{H}^{+}_{(aq)}\)
• Product: \(\mathrm{H_2S}_{(g)}\)

This reaction is particularly significant as \(\mathrm{H_2S}\) gas is not only a product but also a reactant in the subsequent chemical process.

Precipitation Reaction

A precipitation reaction occurs when two aqueous solutions combine to form an insoluble solid known as a precipitate. This is a straightforward and common type of reaction where the mixing of ions leads to the formation of a compound that does not dissolve in the solution.

In our exercise, the hydrogen sulfide gas produced from the gas evolution reaction is bubbled into a lead nitrate solution. Here, the lead ions (\(\mathrm{Pb}^{2+}_{(aq)}\)) and the sulfide ions (\(\mathrm{S}^{2-}_{(aq)}\)) meet and react to form lead sulfide (\(\mathrm{PbS}_{(s)}\), a black precipitate. This visible change marks the completion of the precipitation reaction.

• Reactants: \(\mathrm{Pb}^{2+}_{(aq)}\) and \(\mathrm{S}^{2-}_{(aq)}\)
• Product: \(\mathrm{PbS}_{(s)}\)

The net ionic equation captures the essence of this interaction and illustrates how the ions form a new compound, demonstrating how precipitates are detailed indicators of chemical reactions.

Lead Sulfide Reaction

The formation of lead sulfide (\(\mathrm{PbS}\)) is a classic example often used to teach the concepts of both gas evolution and precipitation reactions. Lead sulfide is notable for its characteristic black color, making it easy to visually identify in laboratory experiments.

During the precipitation reaction, as sulfide ions from \(\mathrm{H_2S}\) gas react with lead ions from lead nitrate, lead sulfide forms as a solid precipitate. This reaction can be summarized by the equation:

• Reactants: \(\mathrm{Pb}^{2+}_{(aq)}\) and \(\mathrm{S}^{2-}_{(aq)}\)
• Product: \(\mathrm{PbS}_{(s)}\)

The importance of this reaction lies in its practical applications. For example, in the study of solubility, selective precipitation, and even certain industrial processes involving the extraction and processing of minerals. Understanding this reaction helps demonstrate how different chemical principles interact to produce tangible results in scientific and industrial fields.