Problem 35 - 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, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N} .\) 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 \( K_b \) is approximately \( 6.33 \times 10^{-5} \).

Step 1: Understand the Relationship between pKa and Ka

The given pKa value is 9.80 for \( (\mathrm{CH}_3)_3\mathrm{NH}^+ \). The relationship between pKa and Ka is \( \mathrm{Ka} = 10^{-\mathrm{pKa}} \). This equation lets us convert the pKa value into the Ka value.

Step 2: Calculate Ka from pKa

Using the pKa value of 9.80, we calculate \( \mathrm{Ka} \) as follows: \[ \mathrm{Ka} = 10^{-9.80} \approx 1.58 \times 10^{-10} \]\.

Step 3: Use Water's Ion Product at 25°C

In neutral water at 25°C, the ion product \( Kw \) is \( 1.0 \times 10^{-14} \). The relationship between \( \mathrm{Ka} \), \( \mathrm{Kb} \), and \( Kw \) is \( \mathrm{Ka} \times \mathrm{Kb} = Kw \).

Step 4: Rearrange to Find Kb

From the relation \( \mathrm{Ka} \times \mathrm{Kb} = Kw \), rearrange to solve for \( \mathrm{Kb} \):\[ \mathrm{Kb} = \frac{Kw}{\mathrm{Ka}} = \frac{1.0 \times 10^{-14}}{1.58 \times 10^{-10}} \approx 6.33 \times 10^{-5} \]

Step 5: Conclusion: Value of Kb for Trimethylamine

So, the value of \( \mathrm{K}_{\mathrm{b}} \) for \( (\mathrm{CH}_3)_3\mathrm{N} \) is approximately \( 6.33 \times 10^{-5} \).

Key Concepts

pKa and Ka relationship

The concept of pKa and Ka is fundamental in acid-base chemistry. When dealing with acids and bases, it is often more useful to work with pKa values because they are easier to handle than very small Ka values. The pKa is the negative logarithm of the acid dissociation constant, Ka, which quantifies the strength of an acid in solution. In mathematical terms, it is defined as:

• \[ \mathrm{pKa} = -\log_{10}(\mathrm{Ka}) \]

This means that a lower pKa corresponds to a stronger acid, as the Ka value, indicating the tendency to donate protons, is larger. Conversely, a higher pKa indicates a weaker acid with a smaller Ka. Being able to convert between pKa and Ka allows chemists to understand and predict the behavior of acids in different chemical environments with more ease. For example, given a pKa of 9.80, you can find Ka using:

• \[ \mathrm{Ka} = 10^{-9.80} \]

This represents the actual concentration of dissociated acid particles in the solution, making it crucial for precise calculations.

ion product of water

Water is unique because it can act as both an acid and a base, and the ion product of water, symbolized as \(K_w\), is a central value in acid-base chemistry. At 25°C, \(K_w\) is typically equal to \(1.0 \times 10^{-14}\). This constant arises from the reaction in which water self-ionizes to create hydroxide \((OH^- )\) and hydronium \((H_3O^+ )\) ions:

• \[ \text{H}_2 ext{O} \rightleftharpoons ext{H}^+ + ext{OH}^- \]

In pure water, the concentrations of \([H^+]\) and \([OH^- ]\) are equal, each at \(1 \times 10^{-7} \), leading to the \(K_w\) expression:

• \[ K_w = [ ext{H}^+][ ext{OH}^-] \]

Understanding \(K_w\) is key to calculating the dissociation constants of weak acids and bases because of the relationship:

• \[ K_a \times K_b = K_w \]

This equation illustrates how \(K_w\) serves as a bridge to find \(K_b\) when the \(K_a\) for an acid is known, demonstrating the tightly linked nature of acid-base equilibria.

conjugate acid-base pairs

In acid-base chemistry, conjugate acid-base pairs are critical concepts that describe how substances transform into each other by gaining or losing protons. A conjugate acid-base pair consists of two species that differ by a single proton \((H^+)\). When an acid donates a proton, it becomes its conjugate base, whereas a base accepting a proton becomes its conjugate acid. For instance, in the case of trimethylamine, \((CH_3)_3N\), which acts as a weak base, its conjugate acid is the trimethylammonium ion \((CH_3)_3NH^+\).
This relationship is vital for understanding chemical reactions, as it helps predict the direction of acid-base reactions and allows the calculation of equilibrium constants. The strength of conjugate acids and bases is inversely related; a strong acid will have a weak conjugate base and vice versa. Assessing conjugate pairs provides insights into the acidity or basicity of solutions and is a cornerstone for many analytical techniques used in chemistry.

Kb calculation

Calculating the base dissociation constant \( K_b \) helps predict how well a base ionizes in water. It is crucial for understanding the strength of a base in a solution, similar to \( K_a \) for acids. The calculation of \( K_b \) becomes straightforward when you relate it back to the ion product of water, \( K_w \). Given \( K_w \) and an acid's \( K_a \), you can find \( K_b \) using:

• \[ K_b = \frac{K_w}{K_a} \]

This formula arises from the equation \( K_a \times K_b = K_w \), highlighting their interdependence. For trimethylamine, using the known \( K_a \) from its conjugate acid's \( pK_a = 9.80 \), the process starts by converting to \( K_a \):

• \[ K_a = 10^{-9.80} \approx 1.58 \times 10^{-10} \]

Next, plug this into the equation with \( K_w = 1.0 \times 10^{-14} \):

• \[ K_b = \frac{1.0 \times 10^{-14}}{1.58 \times 10^{-10}} \approx 6.33 \times 10^{-5} \]

This value of \( K_b \) signifies a moderately weak base, matching the characteristics of many amine compounds in chemistry. Understanding \( K_b \) allows chemists to predict how a base will behave and interact in various chemical environments.