Problem 16 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) Give the conjugate base of the following BrønstedLowry acids: (i) \(\mathrm{H}_{2} \mathrm{SO}_{3},\) (ii) \(\mathrm{HSO}_{3}^{-}\) (b) Give the conjugate acid of the following Bronsted-Lowry bases: (i) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\), (ii) \(\mathrm{CH}_{3} \mathrm{COO}^{-}\).

Step-by-Step Solution

Answer

The conjugate bases of the given Brønsted-Lowry acids are: (a) (i) \( \mathrm{HSO}_{3}^{-} \), (ii) \( \mathrm{SO}_{3}^{2-} \) And the conjugate acids of the given Brønsted-Lowry bases are: (b) (i) \( \mathrm{CH}_{3} \mathrm{NH}_{3}^{+} \), (ii) \( \mathrm{CH}_{3} \mathrm{COOH} \)

Step 1: Conjugate bases of Brønsted-Lowry acids

To find the conjugate base of a Brønsted-Lowry acid, we must remove one hydrogen ion (H⁺) from the acid species. (a) (i) The first acid is H₂SO₃, so we remove one H⁺ ion, which results in the conjugate base of the acid: \[ \mathrm{H}_{2} \mathrm{SO}_{3} \rightarrow \mathrm{HSO}_{3}^{-} \] (ii) The second acid is HSO₃⁻, so we remove one H⁺ ion, which results in the conjugate base of the acid: \[ \mathrm{HSO}_{3}^{-} \rightarrow \mathrm{SO}_{3}^{2-} \]

Step 2: Conjugate acids of Brønsted-Lowry bases

To find the conjugate acid of a Brønsted-Lowry base, we must add one hydrogen ion (H⁺) to the base species. (b) (i) The first base is CH₃NH₂, so we add one H⁺ ion, which results in the conjugate acid of the base: \[ \mathrm{CH}_{3}\mathrm{NH}_{2} + \mathrm{H}^{+} \rightarrow \mathrm{CH}_{3} \mathrm{NH}_{3}^{+} \] (ii) The second base is CH₃COO⁻, so we add one H⁺ ion, which results in the conjugate acid of the base: \[ \mathrm{CH}_{3} \mathrm{COO}^{-} + \mathrm{H}^{+} \rightarrow \mathrm{CH}_{3} \mathrm{COOH} \] To recap, the conjugate bases of the given Brønsted-Lowry acids are: (a) (i) HSO₃⁻, (ii) SO₃²⁻ And the conjugate acids of the given Brønsted-Lowry bases are: (b) (i) CH₃NH₃⁺, (ii) CH₃COOH

Key Concepts

Conjugate Acid

In the Brønsted-Lowry acid-base theory, a conjugate acid is formed when a base gains a hydrogen ion (H⁺). When a compound acts as a base, it "accepts" the hydrogen ion to become the conjugate acid of that base. This concept illustrates the reversible nature of chemical reactions involving acids and bases.

Let's dive into the examples provided. Consider the base \(\mathrm{CH}_{3} \mathrm{NH}_{2} \) (methylamine). When it accepts a hydrogen ion (H⁺), it transforms into its conjugate acid:

\[ \\mathrm{CH}_{3}\mathrm{NH}_{2} + \mathrm{H}^{+} \rightarrow \mathrm{CH}_{3} \mathrm{NH}_{3}^{+} \\]
The structure of methylamine changes with the addition of H⁺, and it becomes positively charged due to the extra proton. This is the essence of becoming a conjugate acid. The addition of an H⁺ ion signifies the switch from being a base to becoming its conjugate acid.

Another example from the exercise is the base \(\mathrm{CH}_{3} \mathrm{COO}^{-} \) (acetate ion). It becomes its conjugate acid, acetic acid (\mathrm{CH}_{3}\mathrm{COOH}), by accepting an H⁺ ion:

• Gain of H⁺ by \mathrm{CH}_{3}\mathrm{COO}^{-}
• Formation of \mathrm{CH}_{3}\mathrm{COOH}

These transformations are vital to understanding the dynamic response of substances in different environments, showcasing the versatility and reactivity of chemical species.

Conjugate Base

The concept of a conjugate base is crucial in understanding acid-base reactions in the Brønsted-Lowry theory. A conjugate base results when an acid donates a hydrogen ion (H⁺). This means that the acid has lost a proton and converted into its conjugate base counterpart.

For instance, take the acid \(\mathrm{H}_{2} \mathrm{SO}_{3} \) (sulfurous acid). Upon losing an H⁺, it forms the conjugate base \(\mathrm{HSO}_{3}^{-} \). The transformation can be represented by the equation:

\[ \\mathrm{H}_{2} \mathrm{SO}_{3} \rightarrow \mathrm{HSO}_{3}^{-} + \mathrm{H}^{+} \\]
By donating an H⁺, sulfurous acid decreases its positive charge, effectively turning into the conjugate base \(\mathrm{HSO}_{3}^{-} \). As a result, this conjugate base is ready to accept an H⁺ ion in another reaction, demonstrating its potential to revert back to its acid form.

Another example is with \(\mathrm{HSO}_{3}^{-} \). Upon losing another H⁺, it becomes \(\mathrm{SO}_{3}^{2-} \) (sulfite ion), following:

• \mathrm{HSO}_{3}^{-} loses H⁺
• Forms \mathrm{SO}_{3}^{2-}

The remarkable ability for acids and bases to switch between donating and accepting protons is a key highlight of the dynamic equilibrium between compounds, allowing for compensation and adjustment to pH changes.

Hydrogen Ion (H⁺) Exchange

The exchange of hydrogen ions (H⁺) is at the heart of the Brønsted-Lowry acid-base theory. The movement of these protons helps define the nature of acidic and basic reactions.

When a hydrogen ion is transferred from one compound to another, it determines whether a substance acts as an acid or a base. An acid donates an H⁺ ion, while a base accepts one. This exchange is what leads to the formation of conjugate pairs.

Consider the process of H⁺ exchange with \(\mathrm{H}_{2} \mathrm{SO}_{3} \):

• As an acid, \(\mathrm{H}_{2} \mathrm{SO}_{3} \) donates an H⁺.
• The remaining species, \(\mathrm{HSO}_{3}^{-} \), becomes the conjugate base.

This balance maintains the dynamic nature and equilibrium seen in many acid-base systems, allowing substances to adapt and perform various biochemical functions.

Similarly, when \(\mathrm{CH}_{3} \mathrm{NH}_{2} \) receives an H⁺, it transitions from a base to its conjugate acid, \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+} \):

• H⁺ acceptance changes its nature from base to acid.

The constant exchange of H⁺ is fundamental, facilitating many of the vital chemical processes found in nature and industrial applications. By comprehending this exchange, we gain insight into how substances can switch roles, maintaining balance in chemical reactions.