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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 molecular equation for the reaction is K2O + H2O 鈫 2KOH. The net ionic equation is O虏鈦 + 2H鈦 鈫 2OH鈦. In this reaction, the base is the oxide ion (O虏鈦), the acid is water (H2O), and the spectator ion is the potassium ion (K鈦).

Step by step solution

01

Write the molecular equation

When K2O dissolves in water, it dissociates into K+ and O虏鈦 ions. The oxide ion (O虏鈦) then reacts with water molecules to form hydroxide ions (OH鈦). The molecular equation for this reaction is: K2O + H2O 鈫 2KOH
02

Write the total ionic equation

Now, we write the total ionic equation by breaking down the molecular equation into ions: 2K鈦 + O虏鈦 + 2H鈦 + 2OH鈦 鈫 2K鈦 + 2OH鈦
03

Write the net ionic equation

To determine the net ionic equation, we remove the spectator ions from the total ionic equation. In this case, the spectator ion is K鈦: O虏鈦 + 2H鈦 鈫 2OH鈦
04

Identify the base, the acid, and the spectator ion

Based on the definitions of acid and base, the substance that donates protons (H鈦 ions) is the acid, and the substance that accepts protons (H鈦 ions) is the base. In this reaction, we can see that: - O虏鈦 is the proton acceptor or base. - H2O is the proton donor or acid. - K鈦 is the spectator ion, as it does not participate in the reaction. So, in this reaction: - The base is the oxide ion (O虏鈦) - The acid is water (H2O) - The spectator ion is the potassium ion (K鈦)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Equations
A molecular equation provides a concise way of representing a chemical reaction by showing the reactants and products in their molecular forms. It often doesn't indicate the state of matter or elucidate whether the compounds exist as molecules or ions in a solution. Taking the example given from the textbook, the molecular equation looks like this:

\[ \mathrm{K}_2\mathrm{O} + \mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{KOH} \.
\] Here, potassium oxide (\(\mathrm{K}_2\mathrm{O}\)) reacts with water (\(\mathrm{H}_2\mathrm{O}\)) to form potassium hydroxide (\(\mathrm{KOH}\)). Although this type of equation is straightforward, it's not very informative about the behavior of the compounds in water. To uncover the actual reaction dynamics, we look into the ionic substances' dissociation into their respective ions, which is detailed in the total ionic equation.
Acid-Base Reactions
Acid-base reactions are a cornerstone of chemical reactions where an acid and a base interact to form water and a salt. According to Arrhenius's concept, acids produce hydrogen ions (\(\mathrm{H}^+\)), while bases produce hydroxide ions (\(\mathrm{OH}^-\)) when dissolved in water. Br酶nsted-Lowry theory takes this a step further, defining acids as proton donors and bases as proton acceptors.

In the context of the textbook example, water acts as the acid since it donates a hydrogen ion, and the oxide ion (\(\mathrm{O}^{2-}\)) functions as the base because it accepts the hydrogen ions to form hydroxide ions (\(\mathrm{OH}^-\)).

Identifying the Base and Acid in a Reaction

By examining how molecules interact with protons, we can identify the base and acid in a given reaction. For instance, the reaction \(\mathrm{O}^{2-} + \mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{OH}^-\) reveals that water (the proton donor) is the acid, while the oxide ion (the proton acceptor) is the base.
Spectator Ions
Spectator ions are ions that appear unchanged on both the reactant and product sides of a chemical equation. They do not partake in the actual chemical reaction and are thus 'spectators' to the main chemical event. In ionic reactions, we find that some ions go through the reaction without getting involved.

The reason we talk about spectator ions is that they can be spotted in total ionic equations. When computing the net ionic equation, these spectator ions are eliminated because they effectively cancel out, leaving behind only the ions that participate in the reaction.

Distinguishing the Spectator Ions

Using the exercise solution as a framework, we know the potassium ion (\(\mathrm{K}^+\)) is the spectator ion. The net ionic equation, after removing the spectator ions, accurately depicts the chemistry at play: \[\mathrm{O}^{2-} + 2\mathrm{H}^+ \rightarrow 2\mathrm{OH}^-\]. By omitting the spectator ion, we achieve a clearer perspective of the reaction's essence.

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Most popular questions from this chapter

An \(8.65-g\) sample of an unknown group 2 A metal hydroxide is dissolved in \(85.0 \mathrm{~mL}\) of water. An acid-base indicator is added and the resulting solution is titrated with \(2.50 \mathrm{M} \mathrm{HCl}(a q)\) solution. The indicator changes color signaling that the equivalence point has been reached after \(56.9 \mathrm{~mL}\) of the hydrochloric acid solution has been added. (a) What is the molar mass of the metal hydroxide? (b) What is the identity of the metal cation: \(\mathrm{Ca}^{2+}, \mathrm{Sr}^{2+}, \mathrm{Ba}^{2+} ?\)

What is the difference between (a) a monoprotic acid and a diprotic acid, \((\mathbf{b})\) a weak acid and a strong acid, \((\mathbf{c})\) an acid and a base?

Tartaric acid, \(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\), has two acidic hydrogens. The acid is often present in wines and precipitates from solution as the wine ages. A solution containing an unknown concentration of the acid is titrated with \(\mathrm{NaOH}\). It requires \(24.65 \mathrm{~mL}\) of \(0.2500 \mathrm{M} \mathrm{NaOH}\) solution to titrate both acidic protons in \(50.00 \mathrm{~mL}\) of the tartaric acid solution. Write a balanced net ionic equation for the neutralization reaction, and calculate the molarity of the tartaric acid solution.

Name the spectator ions in any reactions that may be involved when each of the following pairs of solutions are mixed. (a) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)\) and \(\mathrm{MgSO}_{4}(a q)\) (b) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) and \(\mathrm{Na}_{2} \mathrm{~S}(a q)\) (c) \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4}(a q)\) and \(\mathrm{CaCl}_{2}(a q)\)

Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or ions in each reaction. (a) \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(a q) \longrightarrow\) (b) \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{KOH}(a q) \longrightarrow\).

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