Problem 54 - Chemistry and Chemical Reactivity

Question

Calculate the \(\mathrm{pH}\) of a \(0.12 \mathrm{M}\) aqueous solution of the base aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=4.0 \times 10^{-10}\right)\). $$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$$

Step-by-Step Solution

Answer

The pH of the solution is 8.84.

Step 1: Write the Base Ionization Reaction

The reaction of the base, aniline, in water is given by: \[ \text{C}_6\text{H}_5\text{NH}_2(aq) + \text{H}_2\text{O}(\ell) \rightleftharpoons \text{C}_6\text{H}_5\text{NH}_3^+(aq) + \text{OH}^-(aq) \] This represents the base's ionization in water to produce the conjugate acid and hydroxide ions.

Step 2: Write the Expression for Base Ionization Constant

The expression for the base ionization constant \(K_b\) is given by: \[ K_b = \frac{[\text{C}_6\text{H}_5\text{NH}_3^+][\text{OH}^-]}{[\text{C}_6\text{H}_5\text{NH}_2]} \] We know from the problem, \(K_b = 4.0 \times 10^{-10}\).

Step 3: Set Up an ICE Table

Use an ICE table to track the changes in concentrations. Initially, \([\text{C}_6\text{H}_5\text{NH}_2] = 0.12\,\text{M}\) and \([\text{OH}^-] = 0\). Let the change in concentration of \([\text{OH}^-]\) be \(x\). Then: - \[ \begin{array}{c|c|c|c} & \text{C}_6\text{H}_5\text{NH}_2 & \text{C}_6\text{H}_5\text{NH}_3^+ & \text{OH}^- \hline\text{Initial Concentration (M)} & 0.12 & 0 & 0 \\text{Change (M)} & -x & +x & +x \\text{Equilibrium Concentration (M)} & 0.12-x & x & x \end{array} \]

Step 4: Solve for 'x' Using the Kb Expression

Substitute the equilibrium concentrations into the \(K_b\) expression: \[ K_b = \frac{x^2}{0.12-x} \approx \frac{x^2}{0.12} = 4.0 \times 10^{-10} \] Assuming \(x\) is small compared to 0.12, we approximate \(0.12 - x \approx 0.12\). Solving gives: \[ x^2 = 4.8 \times 10^{-11} \rightarrow x = \sqrt{4.8 \times 10^{-11}} = 6.93 \times 10^{-6} \] Thus, \([\text{OH}^-] = 6.93 \times 10^{-6}\,M\).

Step 5: Calculate pOH

Use the concentration of \([\text{OH}^-]\) to calculate \(\text{pOH}\): \[ \text{pOH} = -\log([\text{OH}^-]) = -\log(6.93 \times 10^{-6}) \approx 5.16 \]

Step 6: Calculate pH from pOH

Use the relation \(\text{pH} + \text{pOH} = 14\) to find \(\text{pH}\): \[ \text{pH} = 14 - \text{pOH} = 14 - 5.16 = 8.84 \]

Key Concepts

base ionization

Base ionization involves the reaction of a base with water to form a conjugate acid and hydroxide ions. In this context, aniline (\(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2\)), being a weak base, partially ionizes in water. This process is represented by the chemical equation: \[ \mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2(\mathrm{aq}) + \mathrm{H}_2\mathrm{O}(\ell) \rightleftharpoons \mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+(\mathrm{aq}) + \mathrm{OH}^-(\mathrm{aq}). \]

• Aniline accepts a proton (from the water) to form its conjugate acid, \(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+\).
• As a result, hydroxide ions \(\mathrm{OH}^-\) are also produced, leading to basic properties.

Understanding this reaction is crucial to calculating properties such as the hydroxide concentration and eventually the pH of the solution. Each step in the ionization helps to track the formation of ions involved.

ice table

An ICE table is a helpful tool for keeping track of the concentrations of species involved in a chemical reaction at different stages:

• I stands for the initial concentrations of the reactants and products.
• C denotes the change in concentration as the reaction progresses.
• E represents the equilibrium concentrations when the reaction reaches a steady state.

In the case of aniline's ionization, the initial concentration of aniline is 0.12 M, while the products initially have a concentration of 0 M. As the reaction proceeds, aniline’s concentration decreases by \(x\), while the concentrations of \(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+\) and \(\mathrm{OH}^-\) both increase by \(x\). The table provides a clear visualization:

• Initial: \([\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2] = 0.12 \text{ M}\), \([\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+] = 0 \text{ M}\), \([\mathrm{OH}^-] = 0 \text{ M}\).
• Change: \( -x \), \(+x\), \(+x\) respectively for the three species.
• Equilibrium: \([\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2] = 0.12-x \text{ M}\), \([\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+] = x \text{ M}\), \([\mathrm{OH}^-] = x \text{ M}\).

aniline

Aniline is an organic compound often depicted as an aromatic amine. It is a weak base due to the presence of an amino group attached to the benzene ring. The molecular formula is \(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2\). When aniline dissolves in water, it undergoes ionization, forming its conjugate acid, \(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+\), and hydroxide ions.

• Being a weak base, aniline's ionization constant \(K_b\) is typically much lower than that of strong bases, indicating less extensive ionization in solution.
• Understanding the base ionization constant is important for predicting the extent of ionization, and subsequently, the solution’s pH.

Aniline is commonly used in the chemical industry, giving significance to its basic properties and the mechanisms of its reactions.

hydroxide concentration

Hydroxide concentration, \([\mathrm{OH}^-]\), is a key factor derived from the ionization of bases in aqueous solutions. For aniline, the equilibrium concentration of hydroxide ions is determined through the expression for the base ionization constant \(K_b\):\[ K_b = \frac{[\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+][\mathrm{OH}^-]}{[\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2]} \]Given \(K_b = 4.0 \times 10^{-10}\), and assuming \(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+\) and \(\mathrm{OH}^-\) are both \(x\) at equilibrium, the expression simplifies to:\[ x = [\mathrm{OH}^-] \]Estimation shows:

• Simplifying the equation, \(x^2 = 4.8 \times 10^{-11}\) leads to \(x = 6.93 \times 10^{-6} \text{ M}\), representing \([\mathrm{OH}^-]\).
• This concentration allows the calculation of \(\text{pOH}\), which is integral for finding the solution's \(\text{pH}\).

Hydroxide concentration is crucial in understanding the basic nature of solutions and determining pH accurately.