Problem 6 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

Methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\) is a base with \(\mathrm{pK}_{\mathrm{b}} 3.34\). Identify the conjugate acid of methylamine, and determine the \(\mathrm{pK}_{\mathrm{a}}\) for this at \(298 \mathrm{K} \text { . (Section } 7.2)\)

Step-by-Step Solution

Answer

The conjugate acid is \( \mathrm{CH}_{3}\mathrm{NH}_{3}^{+} \) with \( \mathrm{pK}_{a} = 10.66 \).

Step 1: Identify the Conjugate Acid

The conjugate acid of a base is formed when the base accepts a proton (H+). For methylamine \( \mathrm{CH}_{3}\mathrm{NH}_{2} \), its conjugate acid is formed by adding a proton to the nitrogen atom, resulting in \( \mathrm{CH}_{3}\mathrm{NH}_{3}^{+} \).

Step 2: Understand the Relationship Between pK values

For a base \( B \) with \( \mathrm{pK}_{b} \), the conjugate acid \( \mathrm{HB}^{+} \) will have a corresponding \( \mathrm{pK}_{a} \). The relationship between \( \mathrm{pK}_{a} \) and \( \mathrm{pK}_{b} \) is given by \( \mathrm{pK}_{a} + \mathrm{pK}_{b} = 14.0 \) at \( 298 \mathrm{K} \).

Step 3: Calculate the pKa of the Conjugate Acid

Using the relationship from Step 2, substitute \( \mathrm{pK}_{b} = 3.34 \) into the equation: \( \mathrm{pK}_{a} + 3.34 = 14.0 \). Solving for \( \mathrm{pK}_{a} \), \( \mathrm{pK}_{a} = 14.0 - 3.34 = 10.66 \).

Key Concepts

Conjugate Acid

In acid-base chemistry, a conjugate acid is the species formed when a base accepts a proton (Known as \( \text{H}^{+} \)). When we look at methylamine \( (\text{CH}_3\text{NH}_2) \), which acts as a base, its conjugate acid is created by the addition of a proton to its nitrogen atom. This results in the formation of \( \text{CH}_3\text{NH}_3^{+} \).
The concept of conjugate acids is important because it helps us understand how acids and bases react and transform in aqueous solutions. In general, for every base \( B \), when it gains a proton, it becomes its conjugate acid \( \text{HB}^{+} \).
Understanding these transformations is crucial in predicting the direction of acid-base reactions and their equilibrium conditions.

pKa and pKb Relationship

In the field of acid-base chemistry, the relationship between \( \text{pK}_a \) and \( \text{pK}_b \) values is central to understanding the strength of acids and bases. For any acid-base conjugate pair, the sum of \( \text{pK}_a \) and \( \text{pK}_b \) is equal to 14 at \( 298 \text{ K} \). This relationship allows for easy conversion between acidity and basicity measures.

• \( \text{pK}_a \) is a measure of the strength of an acid: The lower the \( \text{pK}_a \), the stronger the acid.
• \( \text{pK}_b \), on the other hand, is a measure of the strength of a base: A lower \( \text{pK}_b \) indicates a stronger base.

In the case of methylamine, knowing the \( \text{pK}_b \) can help us determine \( \text{pK}_a \) of its conjugate acid \( \text{CH}_3\text{NH}_3^{+} \). By using the equation \( \text{pK}_a + \text{pK}_b = 14 \), one can rearrange to solve for the unknown \( \text{pK}_a \). This calculation is particularly useful because it seamlessly transitions information from basicity to acidity.

Methylamine

Methylamine \( (\text{CH}_3\text{NH}_2) \) is a simple primary amine and acts as a base. It is commonly used in organic synthesis and has a wide range of applications from manufacturing to pharmaceuticals.
The structure of methylamine consists of a methyl group \((\text{CH}_3)\) attached to an amine group \((\text{NH}_2)\). This configuration allows it to readily accept a proton, making it a basic compound.

• Its \( \text{pK}_b \) value is 3.34, indicating that it is a relatively strong base.
• The conjugate acid, formed by adding a proton to the nitrogen, is \( \text{CH}_3\text{NH}_3^{+} \).
• This conjugate acid can then dissociate in water to reform the base and a \( \text{H}^+ \) ion.

Understanding methylamine and its conjugate acid provides insight into the behavior of organic bases and their reactions in various chemical environments.