Problem 75 - Chemistry: The Central Science in SI Units, Global Edition

Question

Phenol, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH},\) has a \(K_{a}\) of \(1.3 \times 10^{-10}\) (a) Write out the \(K_{a}\) reaction for phenol. (b) Calculate \(K_{b}\) for phenol's conjugate base. (c) Is phenol a stronger or weaker acid than water?

Step-by-Step Solution

Answer

(a) The acid dissociation reaction of phenol is: \(C_6H_5OH \rightleftharpoons C_6H_5O^- + H^+\), and its equilibrium constant expression is: \(K_a = \frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}\). (b) The \(K_b\) value for the conjugate base of phenol is \(7.69 \times 10^{-5}\), calculated using the relationship between \(K_a\) and \(K_b\), which is \(K_a \times K_b = K_w\). (c) Phenol is a stronger acid than water because its \(K_a\) value (1.3 x 10^{-10}) is greater than that of water (1.0 x 10^{-14}).

Step 1: Acid dissociation of phenol

The acid dissociation reaction of phenol can be represented as follows: \[C_6H_5OH \rightleftharpoons C_6H_5O^- + H^+\]

Step 2: Equilibrium constant (\(K_a\)) expression

The equilibrium constant expression for the above reaction is given by: \[K_a = \frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}\] Now we have the \(K_a\) reaction for phenol with its equilibrium constant expression. #b) Calculating \(K_b\) for phenol's conjugate base#

Step 3: Understanding the relationship between \(K_a\) and \(K_b\)

For a conjugate acid-base pair, the relationship between their respective equilibrium constants (\(K_a\) and \(K_b\)) is given by: \[K_a \times K_b = K_w\] \(K_w\) is the ion product of water, which is equal to \(1.0 \times 10^{-14} \) at \(25^{\circ}C\). We are given \(K_a = 1.3 \times 10^{-10}\) for the phenol, so we can calculate \(K_b\) for its conjugate base (\(C_6H_5O^-\)) using this relationship.

Step 4: Calculating \(K_b\)

Using the given \(K_a\) value, we can calculate the \(K_b\) value for phenol's conjugate base as follows: \[K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.3 \times 10^{-10}}\] \[K_b = 7.69 \times 10^{-5}\] The \(K_b\) value for the conjugate base of phenol is \(7.69 \times 10^{-5}\). #c) Comparing the acidity of phenol with water#

Step 5: Comparing \(K_a\) values

To determine if phenol is a stronger or weaker acid than water, we need to compare their respective \(K_a\) values. The \(K_a\) value of water is \(K_w = 1.0 \times 10^{-14}\) (since water is amphoteric, its \(K_a\) value is equal to its \(K_b\) value). Given that the \(K_a\) value of phenol is \(1.3 \times 10^{-10}\), we can compare the two values to determine the acidity.

Step 6: Determining the stronger acid

Based on the \(K_a\) values, we can see that: \[K_a(C_6H_5OH) = 1.3 \times 10^{-10} > 1.0 \times 10^{-14} = K_a(H_2O)\] Since the \(K_a\) value for phenol is greater than that of water, phenol is a stronger acid than water.

Key Concepts

Acid Dissociation Constant (Ka)

The acid dissociation constant, represented as \(K_a\), is a vital concept in understanding the strength of an acid. It measures how well an acid dissociates into its ions in a solution. For any acid HA dissociating in water, the general reaction is:\[ \text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{A}^- + \text{H}_3\text{O}^+ \] The \(K_a\) expression is then given by:\[ K_a = \frac{[\text{A}^-][\text{H}_3\text{O}^+]}{[\text{HA}]} \] In the context of phenol, \(\mathrm{C}_6\mathrm{H}_5\mathrm{OH}\), the reaction indicates how phenol reacts in water to form phenoxide ions \(\mathrm{C}_6\mathrm{H}_5\mathrm{O}^-\) and hydrogen ions \(\mathrm{H}^+\). The higher the \(K_a\), the stronger the acid, as it suggests a greater tendency to donate protons. Conversely, a small \(K_a\) value, like that of phenol at \(1.3 \times 10^{-10}\), suggests a weak acid.
By calculating \(K_a\), we can compare phenol with other acids like water to determine relative strengths.

Conjugate Base

The conjugate base is the species formed when an acid donates a proton. It's integral to the acid-base equilibrium as it can accept a proton back, reverting to the original acid. In the case of phenol, \(\mathrm{C}_6\mathrm{H}_5\mathrm{OH}\), the conjugate base is \(\mathrm{C}_6\mathrm{H}_5\mathrm{O}^-\) which results from phenol losing an \(\mathrm{H}^+\) ion.
The relationship between an acid and its conjugate base is essential to understanding buffer solutions and pH stability.

Acid-Base Equilibrium

- When phenol donates a proton, it transforms into its conjugate base.- The conjugate base can recombine with \(\mathrm{H}^+\) to reform the acid, highlighting the reversible nature of acid-base reactions.

Conjugate Base Strength

- The strength of a conjugate base is inversely related to the strength of the acid.
The weaker the acid, the stronger its conjugate base tends to be. This is why understanding the relationship between \(K_a\) and \(K_b\) (the base dissociation constant) is crucial in calculating equilibrium constants like the ion product of water.

Ion Product of Water (Kw)

The ion product of water, \(K_w\), plays a significant role in acid-base chemistry. It describes the equilibrium constant for the self-ionization of water:\[ \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \]At 25°C, \(K_w\) is always \(1.0 \times 10^{-14}\), meaning that - The concentrations of \([\text{H}^+]\) and \([\text{OH}^-]\) are each \(1.0 \times 10^{-7}\) \(\text{mol/L}\) in pure water.- \(K_w\) remains constant in dilute solutions of weak acids or bases.

The Link with \(K_a\) and \(K_b\)

For any acid-base conjugate pair, the following relationship holds:\[ K_a \times K_b = K_w \] This equation helps us compute either the \(K_a\) of the acid or the \(K_b\) of the conjugate base when the other value is known. It's a quick way to assess the acid-base character of substances and understand their dissociation behavior in water. Knowing \(K_w\) simplifies calculations needed to compare the strengths of acids and their conjugates, as seen in determining if phenol is a stronger acid than water.