Question

Explain what the \(K_{\mathrm{b}}\) for aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\) tells you \(\left(K_{\mathrm{b}}=4.3 \times 10^{-10}\right)\)

Short answer

The \(K_b\) value of 4.3 x \(10^{-10}\) for aniline \((C_6H_5NH_2)\) indicates that it is a weak base in aqueous solution, as the small \(K_b\) value signifies that the reaction with water lies mostly to the left, forming only a small amount of its conjugate acid (anilinium ion) and hydroxide ions. This means that aniline does not fully ionize in water and is a weaker base compared to strong bases like sodium hydroxide or potassium hydroxide.

Step-by-step solution

Write the chemical equation for the reaction of aniline with water.

The reaction of aniline (\(C_6H_5NH_2\)) with water can be represented as: \(C_{6}H_{5}NH_{2} + H_2O \rightleftharpoons C_{6}H_{5}NH_{3}^+ + OH^-\) Aniline acts as a Bronsted-Lowry base, accepting a proton from water to form its conjugate acid, anilinium ion (\(C_6H_5NH_3^+\)), and hydroxide ion (\(OH^-\)).

Define the base ionization constant, \(K_b\).

The base ionization constant, \(K_b\), represents the equilibrium constant for the reaction of a weak base with water. In the case of aniline, \(K_b\) is given as: \(K_b = \dfrac{[C_{6}H_{5}NH_{3}^+][OH^-]}{[C_{6}H_{5}NH_{2}]}\) where the concentrations of the species are their equilibrium concentrations. The larger the \(K_b\) value, the more the equilibrium lies to the right, and the stronger the base.

Analyze the given \(K_b\) value for aniline.

The given \(K_b\) value for aniline is: \(K_b = 4.3 \times 10^{-10}\) Since the \(K_b\) value is very small (much less than 1), this indicates that the equilibrium is shifted mostly to the left, meaning that the reaction does not proceed to a large extent and only a small amount of anilinium ions and hydroxide ions are formed. This signifies that aniline is a weak base in aqueous solution.

Conclusion.

The \(K_b\) value for aniline tells us that aniline is a weak base in aqueous solution because the equilibrium of its reaction with water lies mostly to the left, and only a small amount of its conjugate acid (anilinium ion) and hydroxide ions are formed. Its small \(K_b\) value (4.3 x \(10^{-10}\)) indicates that aniline does not fully ionize in water, being a weak base when compared to strong bases like sodium hydroxide or potassium hydroxide.

Key concepts

Aniline Reaction with Water

When discussing the reaction involving aniline, it is important to consider what happens when it interacts with water. Aniline, with the chemical formula \(C_6H_5NH_2\), acts as a base. In water, aniline undergoes a reversible chemical reaction:
\[ C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5NH_3^+ + OH^- \]
In this reaction, aniline accepts a proton from a water molecule, which results in the formation of two new species. These are the anilinium ion \(C_6H_5NH_3^+\) and the hydroxide ion \(OH^-\). This behavior marks aniline as a participant in acid-base chemistry. Understanding this kind of reaction is essential for grasping how substances interact in aqueous solutions.

Bronsted-Lowry Base

The Bronsted-Lowry theory is a fundamental concept in chemistry that helps classify substances based on their ability to donate or accept protons. According to this theory, a Bronsted-Lowry base is a substance that can accept protons (\(H^+\) ions).
In the case of aniline, it acts as a typical Bronsted-Lowry base. During its interaction with water, aniline accepts a proton from water, turning into the anilinium ion \(C_6H_5NH_3^+\). This reaction signifies its nature as a base according to the Bronsted-Lowry definition.
Understanding these foundational concepts helps in identifying how different substances behave when mixed with solvents like water.

Equilibrium Constant for Weak Bases

The equilibrium constant for weak bases, like aniline, is known as the base ionization constant, \(K_b\). This constant provides crucial insights into the extent to which a base ionizes in water.
For aniline, the base ionization constant is given by:
\[ K_b = \frac{[C_6H_5NH_3^+][OH^-]}{[C_6H_5NH_2]} \]
This formula includes the concentrations of the anilinium ion \(C_6H_5NH_3^+\), hydroxide ion \(OH^-\), and un-ionized aniline \(C_6H_5NH_2\) at equilibrium. The small size of the constant \(K_b = 4.3 \times 10^{-10}\) for aniline indicates only a minor degree of ionization, confirming aniline's weak base status. Assessing these constants helps compare different bases and predict their behaviors in solutions.

Weak Base Behavior

Weak bases like aniline exhibit distinct behaviors in water compared to strong bases. The small base ionization constant \(4.3 \times 10^{-10}\) reflects the equilibrium's position being far to the left, meaning the reaction:
\( C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5NH_3^+ + OH^- \)
produces only a small amount of products.

• This signifies that only a little aniline converts to its ionized form.
• The majority of aniline remains un-ionized in the solution.
• Weak bases partially ionize in water.

These characteristics can affect properties like pH, making weak bases less effective at raising the pH of a solution compared to stronger bases. Understanding weak bases like aniline involves recognizing their limited ability to increase hydroxide ion concentration in water.