Problem 54 - General Chemistry: Principles and Modern Applications

Question

Approximately 4 metric tons of quinoline, \(\mathrm{C}_{9} \mathrm{H}_{7} \mathrm{N},\) is produced annually. The principal source of quinoline is coal tar. Quinoline is a weak base in water. A handbook gives \(K_{\mathrm{a}}=6.3 \times 10^{-10}\) for protonated quinoline, \(\mathrm{C}_{9} \mathrm{H}_{7} \mathrm{NH}^{+} .\) Write the ionization reaction for \(\mathrm{C}_{9} \mathrm{H}_{7} \mathrm{NH}^{+}\) and calculate \(\mathrm{p} \bar{K}_{\mathrm{b}}\) for quinoline.

Step-by-Step Solution

Answer

The ionization reaction for protonated quinoline is: \(C_{9}H_{7}NH^{+} (aq) \rightarrow C_{9}H_{7}N (aq) + H^{+}(aq)\). The pKb for quinoline is 4.80.

Step 1: Write the ionization reaction for protonated quinoline

Considering that quinoline acts as a weak base in water, when it gets protonated, it will act as an acid in solution.The ionization of protonated quinoline (a Bronsted acid) can be written as follows: \[C_{9}H_{7}NH^{+} (aq) \rightarrow C_{9}H_{7}N (aq) + H^{+}(aq)\]This indicates that protonated quinoline donates a hydrogen ion (or proton) to become quinoline.

Step 2: Relationship between Ka, Kb, and Kw

The relationship between the acid dissociation constant (Ka) for an acid, the base dissociation constant (Kb) for its conjugate base and the ion product of water (Kw) is given by the expression: \[Kw = Ka \cdot Kb\]At 25 degrees Celsius, the value of Kw is 1.0 x 10^-14.

Step 3: Calculate Kb for quinoline

Using the equation from step 2, we can calculate Kb for quinoline as follows: \[Kb = \frac{Kw}{Ka} = \frac{1.0 x 10^{-14}}{6.3 x 10^{-10}} = 1.59 x 10^{-5}\]

Step 4: Calculate pKb for quinoline

pKb is the negative logarithm to base 10 of the Kb value. It can be calculated as follows: \[pKb = -\log(1.59 x 10^{-5}) = 4.80\]

Key Concepts

Bronsted Acids and Bases

Bronsted acids and bases are a fundamental concept in chemistry. They are defined based on their ability to donate or accept protons (H⁺). A Bronsted acid is a substance that donates a proton, while a Bronsted base accepts a proton.
In the case of protonated quinoline, it acts as a Bronsted acid when it ionizes in water. The reaction can be represented as:

• Protonated quinoline (\[ C_{9}H_{7}NH^{+} \]) donates a proton to water.
• This forms quinoline (\[ C_{9}H_{7}N \]) and a hydrogen ion (\[ H^{+} \]).

Understanding Bronsted acids and bases is key to predicting the behavior of substances in chemical reactions, especially in acid-base equilibria.

Acid-Base Equilibria

Acid-base equilibria involve the balance between acids and bases in a solution. When an acid donates a proton, it forms a conjugate base, and when a base accepts a proton, it forms a conjugate acid.
For protonated quinoline (\[ C_{9}H_{7}NH^{+} \]), the equilibrium can be written as:

• \[ C_{9}H_{7}NH^{+} (aq) \leftrightarrow C_{9}H_{7}N (aq) + H^{+}(aq) \]
• This shows how protonated quinoline acts as an acid, reaching equilibrium with its conjugate base quinoline (\[ C_{9}H_{7}N \]).

This equilibrium is dynamic, meaning it can shift based on conditions like temperature or concentration. Understanding equilibria helps in calculating the extent of reactions and anticipating changes in pH.

Ka and Kb Constants

The constants Ka and Kb are essential for understanding the strength of acids and bases. Ka is the acid dissociation constant, expressing the strength of an acid in solution, while Kb is for the base. They give insights into how completely an acid or base ionizes in water.
The relationship between Ka and Kb for a conjugate acid-base pair is linked to the ion product of water (\[ Kw \]) with the equation:

• \[ Kw = Ka \times Kb \]
• At 25°C, \[ Kw \] is \[ 1.0 \times 10^{-14} \].

For quinoline, given the Ka for its protonated form (\[ C_{9}H_{7}NH^{+} \]) as \[ 6.3 \times 10^{-10} \], we can find its Kb using the above relationship:

• \[ Kb = \frac{Kw}{Ka} \]
• This results in \[ Kb \approx 1.59 \times 10^{-5} \].
• Calculating \[ pKb \] involves taking the negative logarithm, resulting in \[ pKb = 4.80 \].

Ka and Kb are invaluable for predicting reaction tendencies and adjusting chemical processes in lab and industry.