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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

Expert verified
The balanced chemical equation for the reaction between \(\mathrm{K}_{2} \mathrm{O}\) and water is: \[\mathrm{K}_{2}\mathrm{O} + \mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{KOH}\] The molecular and net ionic equation is: \[\mathrm{K}_{2}\mathrm{O} + \mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{K}^{+} + 2\mathrm{OH}^{-}\] In this reaction, the base is the oxide ion (O^2-), the acid is water (\(\mathrm{H}_{2}\mathrm{O}\)), and there is no spectator ion.

Step by step solution

01

Write the balanced chemical equation

First, we write the balanced chemical equation for the reaction between \(\mathrm{K}_{2} \mathrm{O}\) and water: \[\mathrm{K}_{2}\mathrm{O} + \mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{KOH}\] Now, let's break down the soluble ionic compounds into their ions in order to write the molecular equation.
02

Write the molecular equation

The molecular equation involves breaking down all soluble ionic compounds into their individual ions: \[\mathrm{K}_{2}\mathrm{O} + \mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{K}^{+} + 2\mathrm{OH}^{-}\]
03

Write the net ionic equation

The net ionic equation can be written by removing the spectator ions, which are the ions that aren't involved in the actual reaction. In this case, there are no spectator ions, so the net ionic equation remains as: \[\mathrm{K}_{2}\mathrm{O} + \mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{K}^{+} + 2\mathrm{OH}^{-}\]
04

Identify the acid, base, and spectator ion

Now, let's identify the acid and base in the reaction. The base is the species that accepts a hydrogen ion (H+), which is the oxide ion (O^2-) reacting with water to form hydroxide ions (OH-). Therefore, O^2- is the base. The acid is the species that donates a hydrogen ion (H+). In this reaction, water donates a hydrogen ion to the oxide ion, so H2O is the acid. Since there are no spectator ions in this reaction, we can say there is no spectator ion in this case.

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Most popular questions from this chapter

Lanthanum metal forms cations with a charge of \(3+\). Consider the following observations about the chemistry of lanthanum: When lanthanum metal is exposed to air, a white solid (compound A) is formed that contains lanthanum and one other element. When lanthanum metal is added to water, gas bubbles are observed and a different white solid (compound B) is formed. Both \(A\) and \(B\) dissolve in hydrochloric acid to give a clear solution. When either of these solutions is evaporated, a soluble white solid (compound \(\mathrm{C}\) ) remains. If compound \(\mathrm{C}\) is dissolved in water and sulfuric acid is added, a white precipitate (compound D) forms. (a) Propose identities for the substances \(\mathrm{A}, \mathrm{B}, \mathrm{C}\), and \(\mathrm{D}\). (b) Write net ionic equations for all the reactions described. (c) Based on the preceding observations, what can be said about the position of lanthanum in the activity series (Table 4.5)?

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