Problem 1 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

Identify the acid, base, conjugate acid, and conjugate base in the following equilibrium. (Section 7.1). (a) \(\mathrm{HCO}_{2} \mathrm{H}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}\left(\mathrm{j} \rightleftharpoons \mathrm{HCO}_{2}^{-}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})\right.\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3}^{+}(\mathrm{aq})+\mathrm{OH}^{-(\mathrm{aq})}\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \rightleftharpoons \mathrm{HSO}_{4}^{-}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}_{2}^{+}\)

Step-by-Step Solution

Answer

In (a): Acid=\(\mathrm{HCO}_{2}\mathrm{H}\), Base=\(\mathrm{H}_{2}\mathrm{O}\); In (b): Acid=\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{NH}_{2}\), Base=\(\mathrm{H}_{2}\mathrm{O}\); In (c): Acid=\(\mathrm{H}_{2}\mathrm{SO}_{4}\), Base=\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\).

Step 1: Understand the Reaction Components (a)

For the reaction \( \mathrm{HCO}_{2} \mathrm{H} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{HCO}_{2}^{-} + \mathrm{H}_{3} \mathrm{O}^{+} \), identify the acid, base, conjugate acid, and conjugate base.- Acid: \( \mathrm{HCO}_{2} \mathrm{H} \)- Base: \( \mathrm{H}_{2} \mathrm{O} \)- Conjugate Acid: \( \mathrm{H}_{3} \mathrm{O}^{+} \)- Conjugate Base: \( \mathrm{HCO}_{2}^{-} \)

Step 2: Analyze Components (b)

For the reaction \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3}^{+} + \mathrm{OH}^{-} \), identify the acid, base, conjugate acid, and conjugate base.- Acid: \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2} \)- Base: \( \mathrm{H}_{2} \mathrm{O} \)- Conjugate Acid: \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3}^{+} \)- Conjugate Base: \( \mathrm{OH}^{-} \)

Step 3: Identify Elements in Reaction (c)

For the reaction \( \mathrm{H}_{2} \mathrm{SO}_{4} + \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \rightleftharpoons \mathrm{HSO}_{4}^{-} + \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}_{2}^{+} \), identify the acid, base, conjugate acid, and conjugate base.- Acid: \( \mathrm{H}_{2} \mathrm{SO}_{4} \)- Base: \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \)- Conjugate Acid: \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}_{2}^{+} \)- Conjugate Base: \( \mathrm{HSO}_{4}^{-} \)

Key Concepts

Acid-Base Reactions

Acid-base reactions are interactions that involve the transfer of protons between chemical species. These reactions are fundamental to many processes in chemistry and biology. In a typical acid-base reaction, we have an acid reacting with a base, resulting in the formation of a conjugate base and a conjugate acid.

An acid is a substance that donates a proton ( H^+ ). In contrast, a base is a substance that accepts this proton. When an acid donates a proton, it transforms into its conjugate base. Similarly, when a base accepts a proton, it becomes its conjugate acid. This transformation plays a key role in maintaining the balance of reactions in chemical systems.

To illustrate, consider reaction (a) from the original exercise:

\[ \mathrm{HCO}_{2} \mathrm{H} + \mathrm{H}_{2} \mathrm{O} \leftrightharpoons \mathrm{HCO}_{2}^{-} + \mathrm{H}_{3} \mathrm{O}^{+} \]
Here, \( \mathrm{HCO}_{2} \mathrm{H} \) donates a proton to \( \mathrm{H}_{2} \mathrm{O} \) , making \( \mathrm{HCO}_{2} \mathrm{H} \) the acid and \( \mathrm{H}_{2} \mathrm{O} \) the base. Post donation, \( \mathrm{HCO}_{2}^{-} \) becomes the conjugate base of \( \mathrm{HCO}_{2} \mathrm{H} \), and \( \mathrm{H}_{3} \mathrm{O}^{+} \) is the conjugate acid of \( \mathrm{H}_{2} \mathrm{O} \).

This reaction shows how substances can shift their roles between acids and bases in dynamic equilibrium, key for various chemical processes.

Conjugate Acid-Base Pairs

Conjugate acid-base pairs are essential to understanding how substances interact during acid-base reactions. A conjugate pair consists of two species that transform into each other by the gain or loss of a proton. These pairs ensure that the reaction maintains balance on both sides of the equation.

For instance, in the reaction: \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2} + \mathrm{H}_{2} \mathrm{O} \leftrightharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3}^{+} + \mathrm{OH}^{-} \), the \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2} \) acts as a base, accepting a proton from \( \mathrm{H}_{2} \mathrm{O} \), leading to \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3}^{+} \) , its conjugate acid. Meanwhile, \( \mathrm{H}_{2} \mathrm{O} \), after donating a proton, becomes its conjugate base, \( \mathrm{OH}^{-} \).

In every acid-base reaction, two pairs of conjugates are always present:

• Acid-Conjugate Base Pair
• Base-Conjugate Acid Pair

These pairs help in identifying the direction the reaction can naturally progress, as well as determining which substances can act as buffers to maintain pH equilibrium in solutions. Understanding these pairs empowers one to predict reaction products and balance equations effectively.

Bronsted-Lowry Theory

The Bronsted-Lowry theory is a paradigm that provides a comprehensive framework for understanding acid-base chemistry. Unlike earlier definitions, which only considered acids and bases in solvent contexts like water, the Bronsted-Lowry theory broadens them to generic proton exchanges. This makes it highly versatile.

According to the Bronsted-Lowry theory:

• An **acid** is a proton ( H^+ ) donor.
• A **base** is a proton ( H^+ ) acceptor.

This simple definition allows it to be applied to a wide range of substances beyond those that dissolve in water.

Consider reaction (c) in our exercise:

\[ \mathrm{H}_{2} \mathrm{SO}_{4} + \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \leftrightharpoons \mathrm{HSO}_{4}^{-} + \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}_{2}^{+} \]
Here, \( \mathrm{H}_{2} \mathrm{SO}_{4} \) donates a proton to \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \). According to the Bronsted-Lowry definition, \( \mathrm{H}_{2} \mathrm{SO}_{4} \) is the acid and \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \), which accepts the proton, is the base.

This broad approach aids in understanding the behavior of substances when predicting reaction pathways and determining strengths of acids and bases based on their tendency to donate or accept protons.