Problem 44 - Chemistry: The Central Science in SI Units, Global Edition

Question

As \(\mathrm{K}_{2} \mathrm{O}\) dissolves in water, the oxide ion reacts with water molecules to form hydroxide ions. (a) Write the molecular and net ionic equations for this reaction. (b) Based on the definitions of acid and base, what ion is the base in this reaction? (c) What is the acid in the reaction? (d) What is the spectator ion in the reaction?

Step-by-Step Solution

Answer

The balanced molecular equation is: K2O(s) + H2O(l) -> 2K^(+)(aq) + 2OH^(-)(aq). The net ionic equation is: O^(2-)(aq) + H2O(l) -> 2OH^(-)(aq). The base in the reaction is the oxide ion (O^(2-)). The acid in the reaction is water (H2O), and the spectator ion in the reaction is the potassium ion (K^+).

Step 1: Write the balanced molecular equation

To write the balanced molecular equation, we need to know the reactants and the products of the reaction. In this case, the reactants are K2O (potassium oxide) and water (H2O). When K2O dissolves in water, it forms an oxide ion (O^2-) and two potassium ions (K^+). The oxide ion reacts with water to form hydroxide ions (OH^-). The balanced molecular equation is: K2O(s) + H2O(l) -> 2K^(+)(aq) + 2OH^(-)(aq)

Step 2: Write the balanced net ionic equation

To write the net ionic equation, we need to break the molecular equation into its ionic components. Keeping in mind that K2O is a solid and H2O is a liquid, for the remaining ions, we simply write them as separate entities. Note that solid and liquid substances do not break into ions. Net ionic equation: O^(2-)(aq) + H2O(l) -> 2OH^(-)(aq)

Step 3: Identify the base in the reaction

Based on the definitions of acids and bases, an acid is a substance that donates a proton (H^+), and a base is a substance that accepts a proton (H^+). From the net ionic equation, we can see that the oxide ion (O^(2-)) is accepting a proton and forming hydroxide ions (OH^(-)). Therefore, the base in the reaction is the oxide ion (O^(2-)).

Step 4: Identify the acid in the reaction

The acid in the reaction is the substance that donates a proton (H^+) to the base. In the net ionic equation, the water molecule (H2O) donates a proton to the oxide ion (O^(2-)). Hence, the acid in the reaction is water (H2O).

Step 5: Identify the spectator ion in the reaction

Spectator ions are ions that do not participate in the chemical reaction and remain unchanged. In this case, the potassium ion (K^+) does not participate in the reaction and remains unchanged throughout the process. Therefore, the spectator ion in the reaction is the potassium ion (K^+).

Key Concepts

Molecular Equations

A molecular equation is a representation of a chemical reaction where the reactants and products are expressed as compounds rather than ions. It shows the complete chemical formulas of the substances involved in the reaction. In the case of the potassium oxide (\(\mathrm{K}_2 \mathrm{O}\)) dissolving in water, the molecular equation would be written as:

• \(\mathrm{K}_{2} \mathrm{O} (s) + \mathrm{H}_2\mathrm{O} (l) \rightarrow 2\mathrm{K}^+ (aq) + 2\mathrm{OH}^- (aq)\)

This equation indicates that solid potassium oxide reacts with liquid water, producing aqueous potassium ions and hydroxide ions. During this process, the solid becomes dissolved in water, and the products are completely dissociated, contributing to the aqueous state. By writing the equation in this form, it provides a clear overview of all substances involved, making it easier for students to visualize the chemical changes occurring.

Net Ionic Equations

Net ionic equations focus on the species that actually participate in a chemical reaction, excluding the spectator ions. They are simplified versions of the full molecular equations. For the dissolution of \(\mathrm{K}_{2} \mathrm{O}\) in water, the net ionic equation can be formulated by removing the ions that do not change during the reaction. Such ions remain in the same state before and after the reaction, which in this context is the potassium ion:

• \(\mathrm{O}^{2-} (aq) + \mathrm{H}_2\mathrm{O} (l) \rightarrow 2\mathrm{OH}^- (aq)\)

This equation highlights that the oxide ion and the water react specifically to form hydroxide ions. Net ionic equations are particularly useful when analyzing reactions, as they strip away unnecessary details and focus only on the chemical changes happening directly. This helps students understand which elements and compounds are directly involved in the reaction process.

Acid-Base Reactions

In an acid-base reaction, an acid donates a proton (H\(^+\)), while a base accepts it. These reactions are fundamental concepts in chemistry, helping to understand how different substances interact with each other. In our example reaction between oxide ions and water:

• The oxide ion (\(\mathrm{O}^{2-}\)) acts as the base because it accepts protons from water.
• Water (\(\mathrm{H}_2\mathrm{O}\)) serves as the acid because it donates protons to the oxide ion.

Through this process, the oxide ions are converted into hydroxide ions (\(\mathrm{OH}^-\)). Understanding this concept of proton transfer helps illustrate the role each component plays in the reaction, clarifying the fundamental nature of acid-base interactions and helping students comprehend the direction and outcome of these reactions.

Spectator Ions

Spectator ions are those that exist in the same form on both the reactant and product sides of a chemical equation. They do not participate directly in the reaction and do not affect the equilibrium of the reaction. For the dissolution of potassium oxide in water, the potassium ions (\(\mathrm{K}^+\)):

• Remain unchanged throughout the chemical process.
• Do not participate in forming new products.

Since they do not undergo any change, spectator ions often can be omitted from net ionic equations, simplifying the process of representing the reaction. Teaching students about spectator ions is important because it allows them to differentiate between active participants in a reaction and those that merely "spectate" the chemical changes. This knowledge aids in the simplification of reactions, making it easier to focus on the actual chemistry occurring.