Question

Dissolving \(\mathrm{K}_{2} \mathrm{CO}_{3}\) in water gives a basic solution. Write a balanced equation showing how this can occur.

Short answer

The equation \(\mathrm{K}_{2}\mathrm{CO}_{3} + \mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{K}^+ + \mathrm{HCO}_3^{-} + \mathrm{OH}^-\) shows the basic nature.

Step-by-step solution

Identify the starting compound

Start with the compound \(\mathrm{K}_{2}\mathrm{CO}_{3}\). This is potassium carbonate, which will dissolve in water.

Dissolution in water

When \(\mathrm{K}_{2}\mathrm{CO}_{3}\) is added to water, it dissociates into its ions: - \(\mathrm{K}^+\) ions - \(\mathrm{CO}_3^{2-}\) ionsThe reaction can be written as:\[\mathrm{K}_{2}\mathrm{CO}_{3}(s) \rightarrow 2\mathrm{K}^+(aq) + \mathrm{CO}_3^{2-}(aq)\]

Determine the reaction of carbonate ion with water

The carbonate ion \(\mathrm{CO}_3^{2-}\) can interact with water to form bicarbonate and hydroxide ions. The reaction is:\[\mathrm{CO}_3^{2-}(aq) + \mathrm{H}_2\mathrm{O}(l) \rightleftharpoons \mathrm{HCO}_3^{-}(aq) + \mathrm{OH}^-(aq)\]This reaction will cause the solution to become basic due to the formation of \(\mathrm{OH}^-\) ions.

Combine the overall reaction

Initially, the dissolution of \(\mathrm{K}_2\mathrm{CO}_3\) in water is given by:\[\mathrm{K}_{2}\mathrm{CO}_{3}(s) \xrightarrow{\text{water}} 2\mathrm{K}^+(aq) + \mathrm{CO}_3^{2-}(aq)\]Then the carbonates react with water as:\[\mathrm{CO}_3^{2-} + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{HCO}_3^{-} + \mathrm{OH}^-\]Together, these show how the solution becomes basic.

Key concepts

Balanced chemical equations

A balanced chemical equation accurately represents the conservation of mass in a chemical reaction. It means having equal numbers of each type of atom on both sides of the reaction. When we write chemical equations, we ensure that the number of atoms for each element is the same before and after the reaction.

For example, in the dissolution of potassium carbonate (\(\mathrm{K}_{2}\mathrm{CO}_{3}\)), the balanced equation is represented as:

• \(\mathrm{K}_{2}\mathrm{CO}_{3}(s) \rightarrow 2\mathrm{K}^{+}(aq) + \mathrm{CO}_3^{2-}(aq)\)

Here, the equation shows that two potassium ions (\(\mathrm{K}^+\)) and one carbonate ion (\(\mathrm{CO}_3^{2-}\)) are produced. This equation is balanced because the number of potassium and carbonate atoms is consistent on both sides.

Acid-base reactions

Acid-base reactions involve the transfer of protons between reactants. When potassium carbonate dissolves in water, the carbonate ion (\(\mathrm{CO}_3^{2-}\)) behaves as a base. It reacts with water (\(\mathrm{H}_2\mathrm{O}\)) to form bicarbonate (\(\mathrm{HCO}_3^{-}\)) and hydroxide ions (\(\mathrm{OH}^-\)). This reaction:

• \(\mathrm{CO}_3^{2-}(aq) + \mathrm{H}_2\mathrm{O}(l) \rightleftharpoons \mathrm{HCO}_3^{-}(aq) + \mathrm{OH}^-(aq)\)

shows that carbonate is acting as a base because it accepts protons from water.

The creation of \(\mathrm{OH}^-\) ions is what makes the solution basic. Hydroxide ions contribute to a higher pH, indicating a basic or alkaline solution.

Ion dissociation

Ion dissociation is a process where ionic compounds separate into individual ions when dissolved in water. When potassium carbonate (\(\mathrm{K}_2\mathrm{CO}_3\)) is submerged in water, it dissociates into two types of ions:

• \(\mathrm{K}^{+}\): Potassium ions
• \(\mathrm{CO}_3^{2-}\): Carbonate ions

Understanding ion dissociation is essential, as it explains how a compound can influence the properties of a solution, such as its electrical conductivity and pH level. The dissociated ions are free to move in the water, which allows them to participate in further reactions, like the carbonate ion's reaction with water to form hydroxide ions.

Chemical equilibrium

Chemical equilibrium occurs when the rate of the forward reaction is equal to the rate of the reverse reaction, resulting in stable concentrations of reactants and products. In the dissolution process of potassium carbonate, the equilibrium can be seen in the reaction of carbonate ions (\(\mathrm{CO}_3^{2-}\)) with water to form bicarbonate and hydroxide ions:

• \(\mathrm{CO}_3^{2-}(aq) + \mathrm{H}_2\mathrm{O}(l) \rightleftharpoons \mathrm{HCO}_3^{-}(aq) + \mathrm{OH}^-(aq)\)

This reaction achieves equilibrium where both reactants and products exist in a constant ratio, while the solution remains basic. The equilibrium concept explains why the basicity stays consistent and doesn’t keep increasing indefinitely. Such equilibrium is crucial in determining the final characteristics of the solution.