Problem 87 - 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

The net ionic equations are: \(\text{S}^{2-}(aq) + 2\text{H}^+(aq) \rightarrow \text{H}_2\text{S}(g)\) and \(\text{Pb}^{2+}(aq) + \text{S}^{2-}(aq) \rightarrow \text{PbS}(s)\).

Step 1: Identify the Gas Evolution Reaction

When Na₂S is treated with an acid (let's assume H⁺ from HCl), the reaction is as follows: \[\text{Na}_2\text{S}(aq) + 2\text{H}^+(aq) \rightarrow 2\text{Na}^+(aq) + \text{H}_2\text{S}(g)\]. This generates H₂S gas which is colorless and has a rotten egg smell.

Step 2: Write Net Ionic Equation for Gas Evolution

From the reaction above, the spectator ions, \(\text{Na}^+\), can be removed. The net ionic equation for the gas evolution becomes: \[\text{S}^{2-}(aq) + 2\text{H}^+(aq) \rightarrow \text{H}_2\text{S}(g)\].

Step 3: Identify the Reaction of the Gas with the Second Solution

The H₂S gas is bubbled into a solution of Pb(NO₃)₂. Lead(II) ions react with sulfide ions to form a black precipitate of lead(II) sulfide, PbS.

Step 4: Write Net Ionic Equation for the Formation of Precipitate

The overall reaction involves Pb(NO₃)₂ and H₂S. Breaking it down to ionic form, you remove spectator ions (NO₃⁻), leading to: \[\text{Pb}^{2+}(aq) + \text{S}^{2-}(aq) \rightarrow \text{PbS}(s)\]. This forms a black precipitate.

Key Concepts

Gas Evolution Reactions

Gas evolution reactions are chemical reactions in which a gas is produced as a product. These reactions are often characterized by bubbling or a change in pressure within a closed system. In the given exercise, the reaction between sodium sulfide \( \text{Na}_2\text{S} \) and an acid (such as hydrochloric acid \( \text{HCl} \)) is a classic example. When \( \text{Na}_2\text{S} \) is treated with \( 2\text{H}^+ \), hydrogen sulfide gas \( \text{H}_2\text{S} \) is released. The chemical equation for this particular reaction is:\[\text{Na}_2\text{S}(aq) + 2\text{H}^+(aq) \rightarrow 2\text{Na}^+(aq) + \text{H}_2\text{S}(g)\]This reaction is notable because of the distinct rotten egg smell of \( \text{H}_2\text{S} \), making it easy to identify. It is crucial to understand that not all ions participate directly in the chemical change when forming a net ionic equation. Spectator ions, in this case, \( \text{Na}^+ \), do not participate in the formation of \( \text{H}_2\text{S} \) and are removed to simplify the equation. The net ionic equation then becomes:\[\text{S}^{2-}(aq) + 2\text{H}^+(aq) \rightarrow \text{H}_2\text{S}(g)\]The net ionic equation represents only the ions directly involved in the creation of the gas, making it an essential part of understanding any gas evolution reaction.

Precipitation Reactions

Precipitation reactions occur when two solutions containing soluble salts are mixed, resulting in the formation of an insoluble salt, known as a precipitate. In the exercise, when the hydrogen sulfide gas \( \text{H}_2\text{S} \) is bubbled into a solution of lead(II) nitrate \( \text{Pb(NO}_3\text{)}_2 \), it reacts with the lead ions \( \text{Pb}^{2+} \) to form lead(II) sulfide \( \text{PbS} \), a black precipitate.The complete chemical equation looks like this:\[\text{Pb(NO}_3\text{)}_2(aq) + \text{H}_2\text{S}(g) \rightarrow \text{2HNO}_3(aq) + \text{PbS}(s)\]However, when writing the net ionic equation, we focus only on the ions that change, excluding the nitrates \( \text{NO}_3^- \) as they are spectator ions:\[\text{Pb}^{2+}(aq) + \text{S}^{2-}(aq) \rightarrow \text{PbS}(s)\]The formation of a solid \( \text{PbS} \) that can be seen as a black precipitate signifies a successful precipitation reaction. These reactions are pivotal in various fields, including chemistry and environmental science, as they allow for the removal or recovery of specific ions from solutions.

Acid-Base Reactions

Acid-base reactions are a subset of chemical reactions where an acid reacts with a base, typically resulting in the formation of a salt and water. In the context of the original exercise, while the reaction between \( \text{Na}_2\text{S} \) and \( \text{H}^+ \) does not directly lead to salt and water, it shows the reactivity of an acid with a basic component, as \( \text{S}^{2-} \) acts as a base against the acidic \( \text{H}^+ \).Here’s how acid-base reactions typically manifest:

• Acids release \( \text{H}^+ \) ions in a solution.
• Bases release \( \text{OH}^- \) ions or accept \( \text{H}^+ \) ions.
• The interaction between \( \text{H}^+ \) and \( \text{OH}^- \) typically results in water (\( \text{H}_2\text{O} \)).

In general chemistry, understanding the neutralization process where acids and bases react is crucial. While net ionic equations for classic acid-base reactions usually show the combination of \( \text{H}^+ \) and \( \text{OH}^- \) ions forming water, they help to simplify and clarify the main reactive participants in chemical processes. This basic principle is a keystone of evaluating the interactions between acids and bases even beyond simple laboratory reactions.