Chapter 23, Problem 30QAP

Interpretation Introduction

(a)

Interpretation:

The pH at which the zwitter ion [Z] and the cationic form [C⁺] of the amino acid alanine become equal needs to be calculated.

Concept introduction:

1. Amino acids are organic compounds composed of C, H, N and O. The two main functional groups include the amino −NH₂ and carboxyl −COOH group in addition to a side chain each with a characteristic pKa value.
2. Isoelectric point (pI) is the pH at which the net charge on the amino acid is zero. For amino acid with one −COOH and one −NH₂, the pI is given as:

$\text{pI = }\frac{{\text{pK}}_{\text{a1}}{\text{+pK}}_{\text{a2}}}{\text{2}}\text{ ------(1)}$

Answer to Problem 30QAP

pH = 2.29

Explanation of Solution

The given amino acid is alanine which has the following structure:

The structures of the respective cationic form (C⁺), zwitter ion (Z) and the anionic form (A^-) are:

The acid-base equilibria for alanine involving the cation and the zwitter ion forms is:

$\begin{array}{l}{\text{C}}^{\text{+}}\text{(aq) }\rightleftharpoons {\text{H}}^{\text{+}}\text{(aq) + Z(aq)}\\ {\text{K}}_{\text{a1}}=\frac{{\text{[H}}^{\text{+}}\text{][Z]}}{{\text{[C}}^{\text{+}}\text{]}}\\ {\text{When [Z] = [C}}^{\text{+}}\text{]}\\ {\text{K}}_{\text{a1 }}{\text{= [H}}^{\text{+}}\text{]}\\ \text{i}\text{.e}{\text{. [H}}^{\text{+}}\text{]}=5.1\times {10}^{-3}\text{ M}\\ {\text{pH = -log[H}}^{+}]=-\log [5.1\times {10}^{-3}]=2.29\end{array}$

Interpretation Introduction

(b)

Interpretation:

The pH at which the zwitter ion [Z] and the anionic form [A^-] of the amino acid alanine become equal needs to be calculated

Concept introduction:

1. Amino acids are organic compounds composed of C, H, N and O. The two main functional groups include the amino −NH₂ and carboxyl −COOH group in addition to a side chain each with a characteristic pKa value.
2. Isoelectric point (pI) is the pH at which the net charge on the amino acid is zero. For amino acid with one −COOH and one −NH₂, the pI is given as:

$\text{pI = }\frac{{\text{pK}}_{\text{a1}}{\text{+pK}}_{\text{a2}}}{\text{2}}\text{ ------(1)}$

Answer to Problem 30QAP

pH = 9.74

Explanation of Solution

The given amino acid is alanine which has the following structure:

The structures of the respective cationic form (C⁺), zwitter ion (Z) and the anionic form (A^-) are:

The acid-base equilibria for alanine involving the anion and the zwitter ion forms is:

$\begin{array}{l}\text{Z(aq) }\rightleftharpoons {\text{H}}^{\text{+}}{\text{(aq) + A}}^{-}\text{(aq)}\\ {\text{K}}_{\text{a2}}=\frac{{\text{[H}}^{\text{+}}{\text{][A}}^{-}\text{]}}{\text{[Z]}}\\ {\text{When [Z] = [A}}^{-}\text{]}\\ {\text{K}}_{\text{a2 }}{\text{= [H}}^{\text{+}}\text{]}\\ \text{i}\text{.e}{\text{. [H}}^{\text{+}}\text{]}=1.8\times {10}^{-10}\text{ M}\\ {\text{pH = -log[H}}^{+}]=-\log [1.8\times {10}^{-10}]=9.74\end{array}$

Interpretation Introduction

(c)

Interpretation:

The pH at the isoelectric point needs to be calculated

Concept introduction:

1. Amino acids are organic compounds composed of C, H, N and O. The two main functional groups include the amino −NH₂ and carboxyl −COOH group in addition to a side chain each with a characteristic pKa value.
2. Isoelectric point (pI) is the pH at which the net charge on the amino acid is zero. For amino acid with one −COOH and one −NH₂, the pI is given as:

$\text{pI = }\frac{{\text{pK}}_{\text{a1}}{\text{+pK}}_{\text{a2}}}{\text{2}}\text{ ------(1)}$

Answer to Problem 30QAP

pH = 6.02

Explanation of Solution

For alanine at the isoelectric point:

$\begin{array}{l}\text{pI =pH = }\frac{{\text{pK}}_{\text{a1}}{\text{+pK}}_{\text{a2}}}{\text{2}}\\ {\text{pK}}_{\text{a1 }}{\text{= -log[K}}_{\text{a1}}\text{] =}-\log [5.1\times {10}^{-3}]=2.29\\ {\text{pK}}_{\text{a2 }}{\text{= -log[K}}_{\text{a2}}\text{] =}-\log [1.8\times {10}^{-10}]=9.74\\ \text{pH = }\frac{\text{2}\text{.29 + 9}\text{.74}}{\text{2}}=6.02\end{array}$