Problem 30 - Chemistry and Chemical Reactivity

Question

A titration of \(25.0 \mathrm{mL}\) of a solution of the weak base aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2},\) requires \(25.67 \mathrm{mL}\) of \(0.175 \mathrm{M}\) HCl to reach the equivalence point. $$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)$$ (a) What was the concentration of aniline in the original solution? (b) What are the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{OH}^{-}\) and \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) at the equivalence point? (c) What is the pH of the solution at the equivalence point?

Step-by-Step Solution

Answer

Concentration of aniline is 0.1797 M, pH is 2.85 at equivalence point.

Step 1: Calculate the Moles of HCl

From the given volume and concentration of HCl, which is 0.175 M and 25.67 mL, calculate the moles of HCl used in the titration:\[\text{Moles of HCl} = 0.175 \times \frac{25.67}{1000} = 0.00449175 \text{ moles.}\]

Step 2: Determine the Moles and Concentration of Aniline

Since the stoichiometry of the reaction between aniline and HCl is 1:1, the moles of aniline originally present is equal to the moles of HCl used. Thus, \[\text{Moles of aniline} = 0.00449175 \text{ moles.}\]The original concentration of aniline is:\[\text{Concentration of aniline} = \frac{0.00449175}{\frac{25.0}{1000}} = 0.17967 \text{ M.}\]

Step 3: Calculate Concentration of \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \) at Equivalence Point

At the equivalence point, all aniline is converted to \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \). Since no apparent volume change is given, assume the addition of HCl does not change the volume significantly. Therefore, the concentration of \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \) is equal to the moles of aniline divided by the new volume of the solution:\[\text{Concentration of } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} = \frac{0.00449175}{(25.0 + 25.67) / 1000} = 0.0874 \text{ M.}\]

Step 4: Use \(K_a\) to Find \([ \mathrm{H}_3 \mathrm{O}^+ ]\) at Equivalence Point

\( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \) acts as a weak acid. Calculate \( \[ \mathrm{H}_3 \mathrm{O}^+ \ \] \) using \(K_a\) (related to \(K_b\) of aniline):- \(K_w = 1 \times 10^{-14}\)- \(K_b = 4.3 \times 10^{-10}\) (for aniline)- \(K_a = \frac{K_w}{K_b} = \frac{1 \times 10^{-14}}{4.3 \times 10^{-10}} = 2.33 \times 10^{-5}\)Using initial concentration \([ \mathrm{HA} ] = 0.0874 \text{ M}\):\[K_a = \frac{[ \mathrm{H}_3 \mathrm{O}^+ ][ \mathrm{A}^- ]}{[ \mathrm{HA} ]} \implies [ \mathrm{H}_3 \mathrm{O}^+ ] \approx \sqrt{K_a \times [ \mathrm{HA} ]} = \sqrt{2.33 \times 10^{-5} \times 0.0874} = 1.42 \times 10^{-3} \text{ M.}\]

Step 5: Calculate \([ \mathrm{OH}^- ]\) Using \(K_w\)

Calculate the concentration of \( \mathrm{OH}^- \) at the equivalence point using the hydronium ion concentration:\[[ \mathrm{OH}^- ] = \frac{K_w}{[ \mathrm{H}_3 \mathrm{O}^+ ]} = \frac{1 \times 10^{-14}}{1.42 \times 10^{-3}} = 7.04 \times 10^{-12} \text{ M.}\]

Step 6: Find the pH at the Equivalence Point

Using the hydronium ion concentration to find the pH of the solution:\[pH = -\log([ \mathrm{H}_3 \mathrm{O}^+ ]) = -\log(1.42 \times 10^{-3}) = 2.85.\]

Key Concepts

Weak Base

In chemistry, a weak base is a substance that only partially accepts protons when dissolved in water. Unlike strong bases, which fully dissociate in a solution, weak bases reach an equilibrium between the base, the conjugate acid, and the hydroxide ions produced. This equilibrium can be described through the base dissociation constant, known as \(K_b\).

For example, aniline (\( \text{C}_6\text{H}_5\text{NH}_2 \)) acts as a weak base. When it reacts with water, only a fraction of the aniline molecules ionize to form \([-\text{NH}_3^+]\) and \([-\text{OH}^- ]\) ions. This limited ionization is why weak bases typically have \( K_b \) values much less than one.

• Aniline in water: \( \text{C}_6\text{H}_5\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{C}_6\text{H}_5\text{NH}_3^+ + \text{OH}^- \)
• \(K_b\) for aniline is 4.3 x 10⁻¹⁰.

Understanding weak bases is crucial in acid-base titrations as it helps in anticipating the pH changes when acids like HCl are added.

Equivalence Point

In an acid-base titration, the equivalence point is a crucial stage where the amount of acid added equals the amount of base originally present in the solution. For a reaction involving a weak base and a strong acid, like aniline and HCl, the equivalence point marks the complete conversion of the base into its conjugate acid.

At this juncture:

• All aniline is converted into \( \text{C}_6\text{H}_5\text{NH}_3^+ \).
• The solution comprises primarily of the conjugate acid and potentially slight traces of the remaining acid.
• The characteristics of the solution at this point are governed by the properties of the conjugate acid, making it slightly acidic.

Understanding the equivalence point helps in making accurate calculations for determining concentrations and calculating pH in titration processes.

pH Calculation

Calculating pH during a titration involves understanding the relationship between the hydronium ions and the strength and concentration of the acids or bases involved. When you reach the equivalence point while titrating a weak base with a strong acid, the solution consists of the weak acid formed from the base's conjugate.
Here's a step-by-step summary of how we calculated the pH at the equivalence point:

• First, establish the concentration of the conjugate acid formed (e.g., \( [\text{C}_6\text{H}_5\text{NH}_3^+] \)).
• Utilize \(K_a\), derived from \(K_w/K_b\), to find \([\text{H}_3\text{O}^+]\). The formula \( [\text{H}_3\text{O}^+] = \sqrt{K_a \times [\text{HA}]} \) applies.
• Determine the pH using \( pH = -\log([\text{H}_3\text{O}^+]) \).

In the aniline titration, the \([\text{H}_3\text{O}^+]\) concentration corresponded to a pH of 2.85, characteristic of the weak acid nature of \( \text{C}_6\text{H}_5\text{NH}_3^+ \). This understanding assists chemists in controlling reaction environments and predicting solution behaviors during experiments.

Concentration of Solutions

Concentration is a measure of how much solute is present in a given amount of solvent or solution, often expressed in molarity (moles per liter). Accurate concentration calculations are vital for chemical reactions and processes like titrations.
During a titration, you apply this concept to determine concentrations before and after reaching the equivalence point:

• Initial Concentration: For aniline, concentration is determined using its moles and the initial volume of the solution.
• Concentration at Equivalence Point: When titration reaches this stage, the weak base converts entirely to its conjugate acid. Here, you calculate the final concentration using the total volume after the addition of the titrant.

For example, in the aniline titration, we computed the initial concentration as 0.17967 M and found the concentration of \(\text{C}_6\text{H}_5\text{NH}_3^+\) at the equivalence point to be 0.0874 M.
Grasping these concepts allows students to accurately analyze chemical solutions, which is essential in both academic and practical applications.