Problem 29 - Chemistry and Chemical Reactivity

Question

Chloroacetic acid \(\left(\mathrm{ClCH}_{2} \mathrm{CO}_{2} \mathrm{H}\right)\) has \(K_{\mathrm{a}}=1.41 \times 10^{-3}\) What is the value of \(K_{\mathrm{b}}\) for the chloroacetate ion \(\left(\mathrm{ClCH}_{2} \mathrm{CO}_{2}^{-}\right) ?\)

Step-by-Step Solution

Answer

The value of \(K_b\) for the chloroacetate ion is \(7.09 \times 10^{-12}\).

Step 1: Identify the Relationship Between Ka and Kb

In order to find the value of the base dissociation constant \(K_b\) for a conjugate base (chloroacetate ion), we need to use the relationship between \(K_a\) and \(K_b\). This is given by the equation: \[K_a \times K_b = K_w\]where \(K_w\) is the ion-product constant for water, known to be approximately \(1.0 \times 10^{-14}\) at 25°C.

Step 2: Rearrange the Formula to Solve for Kb

We need to rearrange the relationship \(K_a \times K_b = K_w\) to solve for \(K_b\). This gives us:\[K_b = \frac{K_w}{K_a}\]

Step 3: Substitute Known Values Into the Equation

Substitute the known values: \(K_w = 1.0 \times 10^{-14}\) and \(K_a = 1.41 \times 10^{-3}\) into the equation from Step 2:\[K_b = \frac{1.0 \times 10^{-14}}{1.41 \times 10^{-3}}\]

Step 4: Perform the Calculation

Divide \(1.0 \times 10^{-14}\) by \(1.41 \times 10^{-3}\) to find \(K_b\):\[K_b = 7.09 \times 10^{-12}\]

Key Concepts

Conjugate Acid-Base Pair

A conjugate acid-base pair consists of two species that transform into each other by gain or loss of a proton (H⁺). The concept is central in understanding how acids and bases react in solution.
Chloroacetic acid (\(\mathrm{ClCH}_{2} \mathrm{CO}_{2} \mathrm{H}\)) and its conjugate base, the chloroacetate ion (\(\mathrm{ClCH}_{2} \mathrm{CO}_{2}^{-}\)), make up such a pair. The acid forms by losing a proton,and the base forms by gaining a proton.

• When chloroacetic acid donates a proton, it becomes the chloroacetate ion.
• Conversely, when the chloroacetate ion accepts a proton, it reverts back to the acid.

The strength of an acid is inversely related to the strength of its conjugate base.
This means that a strong acid will have a weak conjugate base, and vice versa. Understanding conjugate pairs helps in calculating the equilibrium concentrations and understanding the behavior of acids and bases in solutions.

Dissociation Constant

The dissociation constant is a measure of the strength of an acid in solution. It specifically quantifies the tendency of the acid to donate a proton.
In the case of chloroacetic acid, the dissociation constant (\(K_a\)) is given as \(1.41 \times 10^{-3}\). Here’s how it applies:

• \(K_a\) represents the equilibrium constant for the dissociation of the acid in water:\[ \mathrm{ClCH}_2\mathrm{CO}_2\mathrm{H} \rightleftharpoons \mathrm{ClCH}_2\mathrm{CO}_2^- + \mathrm{H}^+ \]
• A larger \(K_a\) value indicates a stronger acid, as it implies a greater degree of ionization in the solution.
• Conversely, a smaller \(K_a\) value indicates a weaker acid.

Calculating \(K_b\) (base dissociation constant) for its conjugate base from \(K_a\) provides insight into how readily the base accepts protons, thereby completing our understanding of the acid-base behavior.

Ion-Product Constant

The ion-product constant of water, denoted \(K_w\), is fundamental in acid-base chemistry. It is the product of the molar concentrations of hydrogen ions and hydroxide ions in water. At 25°C, \(K_w\) is approximately \(1.0 \times 10^{-14}\). This constant is crucial because it links the \(K_a\) and \(K_b\) of conjugate acid-base pairs.
For any water-based equilibrium:

• \(K_a \times K_b = K_w\) reflects the balance between acidic and basic forms.
• This relationship helps us determine missing values if one constant is unknown. For example, knowing \(K_a\) allows us to compute \(K_b\), and vice versa.

Understanding the \(K_w\) of water enables us to predict and control the pH of various solutions, crucial for chemical reactions, biological systems, and industrial processes.