Problem 88 - Chemistry and Chemical Reactivity

Question

The anilinium ion, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+},\) is the conjugate acid of the weak organic base aniline. If the anilinium ion has a \(\mathrm{p} K_{\mathrm{a}}\) of \(4.60,\) what is the \(\mathrm{pH}\) of a \(0.080 \mathrm{M}\) solution of anilinium hydrochloride, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\) ?

Step-by-Step Solution

Answer

The pH of the solution is approximately 2.85.

Step 1: Write the Ionization Reaction

The anilinium ion, \(\text{C}_6\text{H}_5\text{NH}_3^+\), ionizes in water according to the equation: \[\text{C}_6\text{H}_5\text{NH}_3^+ \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + \text{H}^+\] The \(\text{H}^+\) concentration from this reaction will determine the \(\text{pH}\) of the solution.

Step 2: Use the \( pK_a \) to Calculate the \(K_a\) Value

The ionization constant, \(K_a\), can be calculated from \(pK_a\) using the relation: \[K_a = 10^{-pK_a}\] Given \(pK_a = 4.60\), we find \[K_a = 10^{-4.60} = 2.51 \times 10^{-5}\]

Step 3: Set Up the Equilibrium Expression

For the equilibrium \(\text{C}_6\text{H}_5\text{NH}_3^+ \rightleftharpoons \text{C}_6\text{H}_5\text{NH}_2 + \text{H}^+\), the expression for \(K_a\) is: \[K_a = \frac{[\text{H}^+][\text{C}_6\text{H}_5\text{NH}_2]}{[\text{C}_6\text{H}_5\text{NH}_3^+]}\] Initially, \([\text{C}_6\text{H}_5\text{NH}_3^+] = 0.080\, \text{M}\) and \([\text{H}^+] = [\text{C}_6\text{H}_5\text{NH}_2] = 0\). Let \(x\) be the change due to ionization.

Step 4: Express the Concentrations in Terms of \(x\)

At equilibrium, concentrations are: \([\text{H}^+] = x\), \([\text{C}_6\text{H}_5\text{NH}_2] = x\), and \([\text{C}_6\text{H}_5\text{NH}_3^+] = 0.080 - x\). Thus the equilibrium expression becomes: \[K_a = \frac{x^2}{0.080 - x}\]

Step 5: Assume \(x\) is Small and Simplify

Assume \(x\) is small compared to the initial concentration: \[0.080 - x \approx 0.080\] So, the equation simplifies to: \[K_a = \frac{x^2}{0.080} = 2.51 \times 10^{-5}\]

Step 6: Solve for \(x\) Using the Simplified Equation

Rearrange and solve for \(x\): \[x^2 = 2.51 \times 10^{-5} \times 0.080\] \[x^2 = 2.008 \times 10^{-6}\] \[x = \sqrt{2.008 \times 10^{-6}}\] \[x \approx 1.42 \times 10^{-3}\]

Step 7: Calculate the \(pH\)

Since \(x\) is the concentration of \(\text{H}^+\), \[\text{pH} = -\log[\text{H}^+] = -\log(1.42 \times 10^{-3}) \approx 2.85\]

Key Concepts

Anilinium Ion

Understanding the anilinium ion, \(\text{C}_6\text{H}_5\text{NH}_3^+\), is crucial in acid-base chemistry. It is formed when aniline, a weak organic base, gains a proton. This creates a positively charged ion, which can donate a hydrogen ion (\(\text{H}^+\)) in solution. The anilinium ion acts as an acid in reactions because it can donate protons, even though it is derived from a base. This proton donation affects the pH of the solution, making it more acidic than the original aniline base. Understanding these interactions helps predict how compounds will behave in various chemical environments.

pK_a Calculation

Calculating \(pK_a\) is an essential part of understanding the strength of an acid in solution. The \(pK_a\) value gives insight into how likely the acid is to donate a proton. This is crucial for determining the ionization constant \(K_a\), which quantifies the degree of ionization of an acid. For the anilinium ion, the given \(pK_a\) is 4.60. To convert \(pK_a\) to \(K_a\), use the formula:

• \( K_a = 10^{-pK_a} \)

Using this approach, the \(K_a\) of the anilinium ion is calculated:

• \(K_a = 10^{-4.60} = 2.51 \times 10^{-5}\)

This small \(K_a\) value confirms that the anilinium ion is a weak acid.

Equilibrium Expression

The equilibrium expression is key to understanding how an acid behaves in a solution. For the ionization of the anilinium ion, the equilibrium can be represented as:

• \(\text{C}_6\text{H}_5\text{NH}_3^+ \rightleftharpoons \text{C}_6\text{H}_5\text{NH}_2 + \text{H}^+\)

In this reaction, the initial concentration of the anilinium ion is important as it represents the starting point before ionization begins. Initially, the concentration of \(\text{C}_6\text{H}_5\text{NH}_2\) and \(\text{H}^+\) is zero. As the reaction progresses, these concentrations increase by \(x\), where \(x\) is the degree of ionization. The equilibrium expression for \(K_a\) is:

• \(K_a = \frac{[\text{H}^+][\text{C}_6\text{H}_5\text{NH}_2]}{[\text{C}_6\text{H}_5\text{NH}_3^+]}\)

This helps us calculate how much the anilinium ion dissociates in solution.

Ionization Constant

The ionization constant, \( K_a \), is a measurement of the extent to which an acid can dissociate in a solution. This parameter shows whether an acid is strong or weak based on its ability to donate protons.For the anilinium ion, \( K_a = 2.51 \times 10^{-5} \), indicating its weak acidic nature, with limited ionization in solution.This small \( K_a \) value tells us that in a \(0.080 \, \text{M}\) anilinium hydrochloride solution, only a small fraction of the ion will dissociate to produce \( \text{H}^+ \).This relationship describes the balance of molecules and ions at equilibrium, contributing to precise predictions related to pH and chemical behavior in the context of acid-base reactions.