Question

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

Short answer

The molecular and net ionic equations for the reaction are: Molecular equation: \[ K_{2}O_{(s)} + H_{2}O_{(l)} \rightarrow 2KOH_{(aq)} \] Net ionic equation: \[ O^{2-}_{(aq)} + H_{2}O_{(l)} \rightarrow 2OH^{-}_{(aq)} \] The base in this reaction is the oxide ion (O^2-), the acid is water (H2O), and the spectator ion is the potassium ion (K+).

Step-by-step solution

Find the reaction of K2O with water

To start, let's write the balanced equation for the reaction of K2O (potassium oxide) with water (H2O). K2O will dissociate in water, and the oxide ion (O^2-) will react with water molecules to form hydroxide ions (OH-). The balanced equation is: \[ K_{2}O_{(s)} + H_{2}O_{(l)} \rightarrow 2KOH_{(aq)} \]

Write the molecular equation

In the molecular equation, we show the complete formulas of all the species involved in the reaction. In this case, the molecular equation is the same as the balanced equation. So, the molecular equation is: \[ K_{2}O_{(s)} + H_{2}O_{(l)} \rightarrow 2KOH_{(aq)} \]

Write the total ionic equation

To write the total ionic equation, we need to represent all the soluble species as ions. Total ionic equation: \[ 2K^{+}_{(aq)} + O^{2-}_{(aq)} + H_{2}O_{(l)} \rightarrow 2K^{+}_{(aq)} + 2OH^{-}_{(aq)} \]

Write the net ionic equation

The net ionic equation is found by canceling the spectator ions on both sides of the total ionic equation. In this case, the spectator ion is K+. Net ionic equation: \[ O^{2-}_{(aq)} + H_{2}O_{(l)} \rightarrow 2OH^{-}_{(aq)} \]

Identify the base, acid, and spectator ion

According to the Bronsted-Lowry theory, an acid is a proton (H+) donor, and a base is a proton (H+) acceptor. In the net ionic equation, the oxide ion (O^2-) accepts a proton from water, forming two hydroxide ions (OH-). Thus, O^2- is acting as a base. Water (H2O) donates a proton to the oxide ion, so it acts as an acid in this reaction. As mentioned earlier, the spectator ion is \( K^{+} \) because it does not participate in the reaction and appears on both sides of the equation in the same form. In conclusion, the molecular and net ionic equations for the reaction are: Molecular equation: \[ K_{2}O_{(s)} + H_{2}O_{(l)} \rightarrow 2KOH_{(aq)} \] Net ionic equation: \[ O^{2-}_{(aq)} + H_{2}O_{(l)} \rightarrow 2OH^{-}_{(aq)} \] The base in this reaction is the oxide ion (O^2-), the acid is water (H2O), and the spectator ion is the potassium ion (K+).