Problem 31 - Chemistry and Chemical Reactivity

Question

The trimethylammonium ion, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\), is the conjugate acid of the weak base trimethylamine, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N} . \mathrm{A}\) chemical handbook gives 9.80 as the \(\mathrm{p} K_{\mathrm{a}}\) value for \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+} .\) What is the value of \(K_{\mathrm{b}}\) for \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N} ?\)

Step-by-Step Solution

Answer

The value of \( K_b \) for \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{N} \) is \( 6.3 \times 10^{-5} \).

Step 1: Understand the Relationship Between Ka and Kb

When an acid donates a proton, it forms its conjugate base. The relationship between the acid dissociation constant \( K_a \), and the base dissociation constant \( K_b \) is expressed for a conjugate acid-base pair by the equation \( K_a \times K_b = K_w \), where \( K_w \) is the ion product of water, equal to \( 1.0 \times 10^{-14} \) at 25°C.

Step 2: Convert pKa to Ka

The \( pK_a \) value is related to the \( K_a \) value by the equation \( pK_a = -\log_{10}(K_a) \). So to find \( K_a \), we use the equation \( K_a = 10^{-pK_a} \). For \( pK_a = 9.80 \), this becomes: \[ K_a = 10^{-9.80} \].

Step 3: Calculate Ka

By raising 10 to the power of \(-9.80\), we get \( K_a \). Performing the calculation: \( K_a = 1.58 \times 10^{-10} \).

Step 4: Use the Ka and Kw Relationship

Using the relationship \( K_a \times K_b = K_w \), and knowing both \( K_a = 1.58 \times 10^{-10} \) and \( K_w = 1.0 \times 10^{-14} \), we can solve for \( K_b \): \[ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.58 \times 10^{-10}} \].

Step 5: Calculate Kb

Divide \( 1.0 \times 10^{-14} \) by \( 1.58 \times 10^{-10} \). The result is \( K_b = 6.3 \times 10^{-5} \).

Key Concepts

pKa and Ka relationship

The concepts of pKa and Ka are central in understanding acid-base chemistry. The pKa is a measure of the strength of an acid, representing the negative logarithm of its acid dissociation constant, Ka. In simpler terms, pKa = -log10(Ka). This relationship allows us to easily convert between these two values, providing insight into how readily an acid will donate its protons in solution.
This conversion is crucial because it helps us understand the acidity of a solution. Smaller pKa values indicate stronger acids that dissociate more completely in water. Conversely, larger pKa values signify weaker acids. Through these values, we can quantitatively compare the strength of different acids, which is essential in various chemical applications.

Kw and ion product of water

The ion product of water, represented as Kw, is a fundamental constant in acid-base equilibrium. Kw represents the product of the hydrogen ion \((H^+)\) concentration and hydroxide ion \((OH^-)\) concentration in pure water. At 25°C, this value is always \( 1.0 \times 10^{-14} \).
Kw is essential because it establishes the baseline for understanding the acidity or basicity of aqueous solutions. Whenever you're dealing with acid or base calculations, the relationship \( K_a \times K_b = K_w \) becomes a key factor. This means that once you know an acid’s Ka, you can find its conjugate base’s Kb—and vice versa. This reciprocal relationship ties into the complete picture of equilibrium between acids and bases in water.

Acid-base conjugate pairs

Understanding acid-base conjugate pairs is vital in the study of equilibrium. An acid, after donating a proton, turns into its conjugate base. Similarly, a base that accepts a proton becomes its conjugate acid. For instance, in the given exercise, trimethylammonium ion \((CH_3)_3NH^+\) is the conjugate acid of the base trimethylamine \((CH_3)_3N\).
This relationship is essential as it allows us to write reversible reactions in equilibrium, like:

• \((CH_3)_3NH^+ + OH^- \rightleftharpoons (CH_3)_3N + H_2O \)
• \((CH_3)_3N + H_2O \rightleftharpoons (CH_3)_3NH^+ + OH^- \)

These reactions underline the dynamic nature of acid-base reactions, where equilibrium can shift based on concentrations, temperature, and other factors.

Quantitative chemical calculations

Performing quantitative chemical calculations is a core skill in chemistry, allowing you to predict concentrations and behaviors in chemical reactions. Calculations involving pKa, Ka, Kb, and Kw are crucial for determining the equilibrium position of reactions.
In practice, once you are given a pKa, you can find Ka using the formula \( K_a = 10^{-pK_a} \). From there, use the relationship \( K_a \times K_b = K_w \) to solve for Kb, as demonstrated in the original exercise. This kind of problem-solving is essential, not just for "filling in the blanks" in equations, but for understanding the deeper chemical implications these numbers represent.
Mastering these calculations allows for predicting the outcome of chemical reactions, understanding the stability of ions in solutions, and designing experiments with desirable outcomes. This quantitative insight is invaluable in both academic and practical chemical work.