Problem 63 - Chemistry: Principles and Reactions

Question

Write the ionization expression and the \(K_{b}\) expression for \(0.1 M\) aqueous solutions of the following bases. (a) \(\mathrm{F}^{-}\) (b) \(\mathrm{HCO}_{3}^{-}\) (c) \(\mathrm{CN}^{-}\)

Step-by-Step Solution

Answer

Question: Determine the ionization reactions and the \(K_b\) expressions for the following bases in water: F⁻, \(\mathrm{HCO}_{3}^{-}\), and \(\mathrm{CN}^{-}\). Answer: a) For F⁻, the ionization reaction is given by \(\mathrm{F}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HF} + \mathrm{OH}^{-}\), with \(K_{b} = \frac{[\mathrm{HF}][\mathrm{OH}^{-}]}{[\mathrm{F}^{-}]}\). b) For \(\mathrm{HCO}_{3}^{-}\), the ionization reaction is given by \(\mathrm{HCO}_{3}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{2}\mathrm{CO}_{3} + \mathrm{OH}^{-}\), with \(K_{b} = \frac{[\mathrm{H}_{2}\mathrm{CO}_{3}][\mathrm{OH}^{-}]}{[\mathrm{HCO}_{3}^{-}]}\). c) For \(\mathrm{CN}^{-}\), the ionization reaction is given by \(\mathrm{CN}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HCN} + \mathrm{OH}^{-}\), with \(K_{b} = \frac{[\mathrm{HCN}][\mathrm{OH}^{-}]}{[\mathrm{CN}^{-}]}\).

Step 1: a) F⁻ ionization and \(K_{b}\) expression

Step 1: Write the ionization reaction in water for F⁻ \(\mathrm{F}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HF} + \mathrm{OH}^{-}\) Step 2: Write the \(K_{b}\) expression for the ionization For the equilibrium reaction, the \(K_{b}\) expression can be written as: \(K_{b} = \frac{[\mathrm{HF}][\mathrm{OH}^{-}]}{[\mathrm{F}^{-}]}\) #a)end#

Step 2: b) \(\mathrm{HCO}_{3}^{-}\) ionization and \(K_{b}\) expression

Step 1: Write the ionization reaction in water for \(\mathrm{HCO}_{3}^{-}\) \(\mathrm{HCO}_{3}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{2}\mathrm{CO}_{3} + \mathrm{OH}^{-}\) Step 2: Write the \(K_{b}\) expression for the ionization For the equilibrium reaction, the \(K_{b}\) expression can be written as: \(K_{b} = \frac{[\mathrm{H}_{2}\mathrm{CO}_{3}][\mathrm{OH}^{-}]}{[\mathrm{HCO}_{3}^{-}]}\) #b)end#

Step 3: c) \(\mathrm{CN}^{-}\) ionization and \(K_{b}\) expression

Step 1: Write the ionization reaction in water for \(\mathrm{CN}^{-}\) \(\mathrm{CN}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HCN} + \mathrm{OH}^{-}\) Step 2: Write the \(K_{b}\) expression for the ionization For the equilibrium reaction, the \(K_{b}\) expression can be written as: \(K_{b} = \frac{[\mathrm{HCN}][\mathrm{OH}^{-}]}{[\mathrm{CN}^{-}]}\) #c)end# By following these steps, we have determined the ionization reactions and the \(K_b\) expressions for aqueous solutions of the bases F⁻, \(\mathrm{HCO}_{3}^{-}\), and \(\mathrm{CN}^⁻\).

Key Concepts

Ionization Expression

When a base dissolves in water, it undergoes a process called ionization. In this process, the base reacts with water to form products, typically a hydroxide ion (\(\mathrm{OH}^{-}\)). The ionization expression is essential as it shows how the base interacts with water. For instance, the ionization expression for \(\mathrm{F}^{-}\) can be represented as:

• \(\mathrm{F}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HF} + \mathrm{OH}^{-}\)

This represents the double-headed arrow, which indicates that the reaction can proceed in both directions, reaching a state of equilibrium. This reversible nature is common in ionization reactions.
The expression is crucial as it provides insight into how the equilibrium is established in the solution.

Equilibrium Reaction

An equilibrium reaction occurs when the rates of the forward and reverse reactions are equal. This balance results in constant concentrations of products and reactants over time. In the context of ionization, the equilibrium reaction helps clarify how bases behave in an aqueous environment.
For example, the reaction for \(\mathrm{HCO}_{3}^{-}\) looks like this:

• \(\mathrm{HCO}_{3}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{2}\mathrm{CO}_{3} + \mathrm{OH}^{-}\)

At equilibrium, the rate at which \(\mathrm{HCO}_{3}^{-}\) and water react to form \(\mathrm{H}_{2}\mathrm{CO}_{3}\) and \(\mathrm{OH}^{-}\) matches the reverse reaction's rate. Understanding this dynamic helps explain why concentrations do not change in an apparent equilibrium state.

Kb Expression

The \(K_b\) expression, or base ionization constant, quantifies the strength of a base in solution. It reflects how readily a base ionizes in water. Calculating \(K_b\) involves setting up an equation, derived from the equilibrium reaction, which incorporates the concentrations of the reactants and products.
For the base \(\mathrm{CN}^{-}\), the equilibrium expression is:

• \(K_{b} = \frac{[\mathrm{HCN}][\mathrm{OH}^{-}]}{[\mathrm{CN}^{-}]}\)

This formula conveys the proportions of reactants and products at equilibrium. A larger \(K_b\) suggests a stronger base, as it indicates a greater extent of ionization into OH⁻ ions. Conversely, a smaller \(K_b\) implies a weaker base.

Aqueous Solutions

An aqueous solution is simply a solution where water is the solvent. When talking about bases, it's crucial to understand that in aqueous solutions, water plays a vital role as the medium facilitating the ionization of the base.
This type of solution is characterized by its ability to dissolve compounds, creating an environment where chemical reactions, such as the ones described in ionization, occur. For example, in the case of \(\mathrm{CN}^{-}\), water allows the base to interact significantly, promoting its transformation into \(\mathrm{HCN}\) and \(\mathrm{OH}^{-}\).

• Aqueous solutions are essential for understanding reactions because they replicate conditions in biological and environmental systems.
• They allow acids and bases to conduct electricity, which is why they are often referenced in discussions about electrolytes.

Knowing how bases behave in aqueous solutions aids in predicting their activity in various chemical and real-world scenarios.