Question

(a) What is the \(K_{\mathrm{b}}\) of the benzoate ion, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} ?\) (b) What is the \(K_{a}\) of the 2 -hydroxycthylammonium ion, \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{3}^{+}\left(\mathrm{p} K_{\mathrm{b}}\right.\) of \(\left.\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}=4.49\right) ?\)

Short answer

(a) \(K_b = 1.54 \times 10^{-10}\). (b) \(K_a = 3.09 \times 10^{-10}\).

Step-by-step solution

Understand the Given Data

Note the provided information for each part of the exercise. For part (a), we need to find the base dissociation constant, \(K_b\), of the benzoate ion. For part (b), we have the \(pK_b\) of \(\text{HOCH}_2 \text{CH}_2 \text{NH}_2 = 4.49\) and need to find the acid dissociation constant, \(K_a\), of \(\text{HOCH}_2 \text{CH}_2 \text{NH}_3^+\).

Use Relationship Between \(K_a\) and \(K_b\) for Part (a)

Recall that \(K_a \times K_b = K_w\), where \(K_w = 1.0 \times 10^{-14}\). For the benzoate ion, \(C_6 H_5 COO^-\), the conjugate acid is benzoic acid, and its \(K_a = 6.5 \times 10^{-5}\).

Calculate \(K_b\) of Benzoate Ion

Use the formula \(K_b = \frac{K_w}{K_a}\). Thus, \(K_b = \frac{1.0 \times 10^{-14}}{6.5 \times 10^{-5}} = 1.54 \times 10^{-10}\).

Determine the Relationship for Part (b)

Recall that \(pK_a + pK_b = 14\). Given \(pK_b = 4.49\), we can find \(pK_a\) by rearranging the equation: \(pK_a = 14 - pK_b = 14 - 4.49 = 9.51\).

Calculate \(K_a\) of 2-hydroxyethylammonium Ion

Convert \(pK_a\) to \(K_a\) using the formula \(K_a = 10^{-pK_a}\). Hence, \(K_a = 10^{-9.51} = 3.09 \times 10^{-10}\).

Key concepts

Base Dissociation Constant (K_b)

The base dissociation constant, or \(K_b\), is a measure of how effectively a base dissociates into its constituent ions in a solution. Imagine a basic compound in water that dissolves and separates into a conjugate acid and hydroxide ions. The \(K_b\) value indicates the extent of this dissociation.
For example, consider the benzoate ion \(C_6 H_5 COO^-\). To calculate its \(K_b\), you should know that its conjugate acid, benzoic acid, has a dissociation constant \(K_a = 6.5 \times 10^{-5}\). Furthermore, the relationship \(K_a \times K_b = K_w\), where \(K_w = 1.0 \times 10^{-14}\), allows us to find \(K_b\).
Using the formula \(K_b = \frac{K_w}{K_a}\), we get \(K_b = \frac{1.0 \times 10^{-14}}{6.5 \times 10^{-5}} = 1.54 \times 10^{-10}\).
With these steps, we can determine the ability of the benzoate ion to accept protons.

Acid Dissociation Constant (K_a)

The acid dissociation constant, or \(K_a\), measures the extent to which an acid dissociates into its ions in water. This constant helps us understand the strength of the acid.
For instance, in part b) of the exercise, we need to find the \(K_a\) for the 2-hydroxyethylammonium ion \(\text{HOCH}_2 \text{CH}_2 \text{NH}_3^+\). Often, we are given \(pK_b\) for the conjugate base \(\text{HOCH}_2 \text{CH}_2 \text{NH}_2\) and can use it to find \(K_a\) through some calculations.
Given \(pK_b = 4.49\), we can find \(pK_a\) using the relationship: \(pK_a + pK_b = 14\). Therefore, \(pK_a = 14 - pK_b = 14 - 4.49 = 9.51\).
To convert \(pK_a\) to \(K_a\), we use the formula \(K_a = 10^{-pK_a}\). Thus, \(K_a = 10^{-9.51} = 3.09 \times 10^{-10}\).
This helps us understand how readily the 2-hydroxyethylammonium ion donates protons.

Relationship between K_a and K_b

Understanding the relationship between \(K_a\) and \(K_b\) is key to many acid-base equilibrium problems. Essentially, the product of \(K_a\) (the acid dissociation constant) and \(K_b\) (the base dissociation constant) should equal the ion-product constant of water, \(K_w\).
This relationship is mathematically represented as: \(K_a \times K_b = K_w\), where \(K_w = 1.0 \times 10^{-14}\) at 25°C.
For instance, if you know the \(K_a\) of an acid, you can easily find the \(K_b\) of its conjugate base by rearranging the formula: \(K_b = \frac{K_w}{K_a}\).
Conversely, if you have \(K_b\), you can find \(K_a\). This relationship is fundamental because it shows the inverse relationship between the strength of an acid and its conjugate base. A strong acid (large \(K_a\)) has a weak conjugate base (small \(K_b\)), and vice versa.
Thus, understanding this relationship helps in predicting behavior in chemical reactions and solutions.

pK_a and pK_b Calculations

The values \(pK_a\) and \(pK_b\) are the negative logarithms of \(K_a\) and \(K_b\) respectively. These values make it easier to work with the very small numbers typical of acid and base dissociation constants.
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