Problem 103 - Chemistry An Atoms First Approach

Question

The \(\mathrm{pH}\) of a \(0.016-M\) aqueous solution of \(p\) -toluidine \(\left(\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{NH}_{2}\right)\) is \(8.60 .\) Calculate \(K_{\mathrm{b}}\).

Step-by-Step Solution

Answer

The base dissociation constant (Kb) of the p-toluidine solution is approximately \(9.96 \times 10^{-10}\).

Step 1: Calculate the hydroxide ion concentration

Since we are given the pH value of the solution (8.60), we can calculate the corresponding pOH value as: pOH = 14 - pH pOH = 14 - 8.60 = 5.4 Now we can calculate the hydroxide ion concentration (OH⁻) using the pOH value: \[\text{OH}^- = 10^{-\text{pOH}}\] \[ [\text{OH}^-] = 10^{-5.4}\] From this calculation, we find that the hydroxide ion concentration is approximately: \[ [\text{OH}^-] \approx 3.98 \times 10^{-6} \text{ M} \]

Step 2: Write the dissociation reaction for p-toluidine

The dissociation reaction for p-toluidine can be represented as: \[\text{CH}_{3}\text{C}_{6}\text{H}_{4}\text{NH}_{2} + \text{H}_{2}\text{O} \rightleftarrows \text{CH}_{3}\text{C}_{6}\text{H}_{4}\text{NH}_{3}^{+} + \text{OH}^-\]

Step 3: Calculate the change in concentration of reactants and products during dissociation

Let x be the concentration of dissociated p-toluidine, which is also equal to the concentration of hydroxide ions (OH-) at equilibrium. As a result, we have: [CH3C6H4NH3+ ] = [OH-] = x Since the initial concentration of p-toluidine (0.016 M) is much larger than the equilibrium concentration of hydroxide ions (about 3.98 × 10⁻⁶ M), we can assume that the change in p-toluidine concentration is negligible, and the equilibrium concentration is still approximately equal to 0.016 M.

Step 4: Set up the Kb expression and find its value

The Kb expression for the dissociation reaction of p-toluidine is given by: \[\text{Kb} = \frac{[\text{CH}_{3}\text{C}_{6}\text{H}_{4}\text{NH}_{3}^{+}] \times [\text{OH}^-]}{[\text{CH}_{3}\text{C}_{6}\text{H}_{4}\text{NH}_{2}]}\] Substituting the values obtained in the previous steps, we have: \[\text{Kb} = \frac{(3.98 \times 10^{-6})^{2}}{0.016}\] Now, we can calculate the Kb value: \[\text{Kb} \approx 9.96 \times 10^{-10}\] So, the base dissociation constant (Kb) of the p-toluidine solution is approximately 9.96 × 10⁻¹⁰.

Key Concepts

Base Dissociation Constant

Understanding the base dissociation constant, or \( K_b \), is key when working with weak bases such as p-toluidine. \( K_b \) measures how completely a base dissociates in water. It provides valuable information about the strength of the base and its ability to produce hydroxide ions \( (\text{OH}^-) \).

For the dissociation of p-toluidine in water, the reaction can be written as: \[\text{CH}_3\text{C}_6\text{H}_4\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{C}_6\text{H}_4\text{NH}_3^+ + \text{OH}^-\] This expression indicates that the base accepts a proton from water, creating \( \text{OH}^- \) ions as a product of the reaction.

The expression for \( K_b \) is formulated as the concentration of products divided by the concentration of reactants. The lower the \( K_b \) value, the weaker the base, as it indicates fewer \( \text{OH}^- \) ions in solution. In our case, the calculated \( K_b \) for p-toluidine is approximately \( 9.96 \times 10^{-10} \), highlighting its nature as a weak base.

Hydroxide Ion Concentration

Hydroxide ion concentration \( ([\text{OH}^-]) \) is crucial in determining the basicity of a solution. The concentration of \( \text{OH}^- \) ions helps ascertain how strongly a base has dissociated in water. For p-toluidine, calculating the hydroxide ion concentration starts with knowing the pH of the solution.

First, convert pH to pOH using the formula: \[p\text{OH} = 14 - p\text{H}\] With a pH of 8.60, the pOH is 5.4. This implies: \[[\text{OH}^-] = 10^{-p\text{OH}}\] Using these calculations, the hydroxide ion concentration is approximately \( 3.98 \times 10^{-6} \text{ M} \). This tells us about the extent of dissociation of the base in the solution, shedding light on its basic nature.

Dissociation Reaction

The dissociation reaction is at the heart of understanding how bases like p-toluidine operate in solution. A dissociation reaction involves a base reacting with water to form hydroxide ions \( (\text{OH}^-) \) and its conjugate acid. For p-toluidine, this equation looks like:

\[\text{CH}_3\text{C}_6\text{H}_4\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{C}_6\text{H}_4\text{NH}_3^+ + \text{OH}^-\] It demonstrates the reversible nature of acid-base equilibrium, where the base gains a proton forming the conjugate acid and liberates \( \text{OH}^- \) ions.

Significantly, the concentration of the hydroxide ions \( ([\text{OH}^-]) \) equals the concentration of the conjugate acid \( ([\text{CH}_3\text{C}_6\text{H}_4\text{NH}_3^+]) \) at equilibrium since each molecule of base that dissociates forms one hydroxide ion. This specific equilibrium condition sets the stage for calculating the \( K_b \) value, reflecting the extent of dissociation in the equilibrium.