Question

(a) To what equilibria do the values of \(\mathrm{p} K_{\mathrm{a}}(1)=10.71\) and \(\mathrm{p} K_{\mathrm{a}}(2)=7.56\) for the conjugate acid of \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) refer? (b) Calculate the corresponding values of \(\mathrm{p} K_{\mathrm{b}}\) and write equations to show the equilibria to which these values refer.

Short answer

The values refer to the equilibria of \(\mathrm{H}_{3}\mathrm{NCH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{3}^{2+}\) losing protons. The corresponding \(\mathrm{p}K_{\mathrm{b}}\) values are 3.29 and 6.44.

Step-by-step solution

Understand the Given Values

The given values are \(\mathrm{p} K_{\mathrm{a}}(1) = 10.71\) and \(\mathrm{p} K_{\mathrm{a}}(2) = 7.56\) for the conjugate acid of \(\mathrm{H}_{2}\mathrm{NCH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{2}\). These represent the acid dissociation constants for a species that can donate protons in two steps.

Identify the Equilibria

\(\mathrm{H}_{2}\mathrm{NCH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{2}\) is a diamine, meaning it can accept protons to form a dication \((\mathrm{H}_{3}\mathrm{NCH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{3}^{2+})\). The two values refer to the equilibria where the dication loses protons: 1. \(\mathrm{H}_{3}\mathrm{NCH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{3}^{2+} \rightarrow \mathrm{H}_{2}\mathrm{NCH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{2.5}^{+} + \mathrm{H}^{+}\)2. \(\mathrm{H}_{2}\mathrm{NCH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{2.5}^{+} \rightarrow \mathrm{H}_{2}\mathrm{NCH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{2} + \mathrm{H}^{+}\)

Calculate Corresponding \(\mathrm{p}K_{\mathrm{b}}\) Values

Use the relation \(\mathrm{p}K_{\mathrm{a}} + \mathrm{p}K_{\mathrm{b}} = 14\) to find \(\mathrm{p}K_{\mathrm{b}}\):1. For \(\mathrm{p}K_{\mathrm{a}}(1) = 10.71\), \(\mathrm{p}K_{\mathrm{b}}(1) = 14 - 10.71 = 3.29\).2. For \(\mathrm{p}K_{\mathrm{a}}(2) = 7.56\), \(\mathrm{p}K_{\mathrm{b}}(2) = 14 - 7.56 = 6.44\).

Write Equations for Base Dissociation

The base form of the compound \(\mathrm{H}_{2}\mathrm{NCH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{2}\) can accept protons as follows:1. \(\mathrm{H}_{2}\mathrm{NCH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O} \leftrightarrow \mathrm{H}_{2}\mathrm{NCH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{3}^{+} + \mathrm{OH}^{-}\)2. \(\mathrm{H}_{2}\mathrm{NCH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{3}^{+} + \mathrm{H}_{2}\mathrm{O} \leftrightarrow \mathrm{H}_{3}\mathrm{NCH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{3}^{2+} + \mathrm{OH}^{-}\)

Key concepts

pKa values

The concept of \(\text{p}K_{\text{a}}\) values is central in understanding acid-base reactions. A \(\text{p}K_{\text{a}}\) value is a quantitative measure of the strength of an acid in solution. It indicates how easily an acid can donate a proton. Lower \(\text{p}K_{\text{a}}\) values correspond to stronger acids, which more readily donate protons, while higher \(\text{p}K_{\text{a}}\) values indicate weaker acids.
The formula defining \(\text{p}K_{\text{a}}\) is \[\text{p}K_{\text{a}} = -\log_{10}(K_{\text{a}})\]\where \(K_{\text{a}}\) is the acid dissociation constant. This constant expresses the equilibrium concentration ratio of the dissolved ions compared to the undissociated acid.

• A lower \(\text{p}K_{\text{a}}\) means the equilibrium lies towards the dissociated side.
• The relationship between \(\text{p}K_{\text{a}}\) and \(\text{p}K_{\text{b}}\) (base dissociation constant) is given by the formula \(\text{p}K_{\text{a}} + \text{p}K_{\text{b}} = 14\) at 25°C.

This relationship is crucial in determining the conjugate base strength when the acid behavior is known.

diamine compounds

Diamines are a special class of compounds characterized by having two amino groups (\(-\text{NH}_2\)) in their structure. These groups are basic, meaning they can accept protons in aqueous solutions to form ammonium ions. A typical example is ethylene diamine, \(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2\).
The ability of diamine compounds to accept protons twice makes them diprotic bases. The process of accepting protons occurs in steps:

• First protonation: The first amino group accepts a proton to form \(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_3^+\).
• Second protonation: The second amino group gets protonated, resulting in a dication (\(\text{H}_3\text{NCH}_2\text{CH}_2\text{NH}_3^{2+}\)).

Diamines like ethylene diamine play significant roles in various fields such as chelation therapy and as stabilizers in polymer production. Their unique reactivity also makes them important in studying acid-base equilibria.

conjugate acid-base pairs

In acid-base chemistry, every acid has a conjugate base, and every base has a conjugate acid. These pairs differ by just one proton. When an acid donates a proton, it forms its conjugate base. Conversely, when a base accepts a proton, its conjugate acid is formed.
The concept can be illustrated using ethylene diamine and its protonated forms:

• First pair: \(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2\) (base) and \(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_3^+\) (conjugate acid).
• Second pair: \(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_3^+\) (base) and \(\text{H}_3\text{NCH}_2\text{CH}_2\text{NH}_3^{2+}\) (conjugate acid).

Understanding conjugate acid-base pairs is crucial for predicting the outcomes of acid-base reactions. It reveals how substances can transition between acid and base forms, dictating the reaction's direction and extent under given conditions. This interconversion is central to equilibria and influences how various species interact in chemical solutions.