Question

In each equation, label the acids, bases, and conjugate pairs: (a) \(\mathrm{HCl}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{Cl}^{-}+\mathrm{H}_{3} \mathrm{O}^{+}\) (b) \(\mathrm{HClO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightleftharpoons \mathrm{ClO}_{4}^{-}+\mathrm{H}_{3} \mathrm{SO}_{4}^{+}\) (c) \(\mathrm{HPO}_{4}^{2-}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}+\mathrm{HSO}_{4}^{-}\)

Short answer

In (a), HCl is an acid and Cl\(^{-}\) is its conjugate base; H\(_2\)O is a base and H\(_3\)O\(^+\) is its conjugate acid. In (b), HClO\(_4\) is an acid and ClO\(_4^-\) is its conjugate base; H\(_2\)SO\(_4\) is a base and H\(_3\)SO\(_4^+\) is its conjugate acid. In (c), H\(_2\)SO\(_4\) is an acid and HSO\(_4^-\) is its conjugate base; HPO\(_4^{2-}\) is a base and H\(_2\)PO\(_4^-\) is its conjugate acid.

Step-by-step solution

- Identify the acids and bases in equation (a)

In the reaction \(\text{HCl + H}_{2}\text{O} \rightleftharpoons \text{Cl}^{-}\text{ + H}_{3}\text{O}^{+}\), identify the substances that donate protons (acids) and those that accept protons (bases). HCl donates a proton to \(\text{H}_{2}\text{O}\), so HCl is an acid and \(\text{H}_{2}\text{O}\) is a base.

- Identify the conjugate pairs in equation (a)

After HCl donates a proton, it becomes \(\text{Cl}^{-}\), and after \(\text{H}_{2}\text{O}\) accepts a proton, it becomes \(\text{H}_{3}\text{O}^{+}\). Therefore, \(\text{Cl}^{-}\) is the conjugate base of HCl and \(\text{H}_{3}\text{O}^{+}\) is the conjugate acid of \(\text{H}_{2}\text{O}\).

- Identify the acids and bases in equation (b)

In the reaction \(\text{HClO}_{4}\text{ + H}_{2}\text{SO}_{4} \rightleftharpoons \text{ClO}_{4}^{-}\text{ + H}_{3}\text{SO}_{4}^{+}\), identify the substances that donate protons (acids) and those that accept protons (bases). HClO\text{4} donates a proton to H\(_2\)SO\(_4\), so HClO\(_4\) is an acid and H\(_2\)SO\(_4\) is a base.

- Identify the conjugate pairs in equation (b)

After HClO\text{4} donates a proton, it becomes ClO\(_4^{-}\), and after H\text{2}SO\text{4} accepts a proton, it becomes H\(_3\)SO\(_4^{+}\). Therefore, ClO\(_4^{-}\) is the conjugate base of HClO\text{4} and H\(_3\)SO\(_4^{+}\) is the conjugate acid of H\text{2}SO\text{4}.

- Identify the acids and bases in equation (c)

In the reaction \(\text{HPO}_{4} ^{2-}\text{ + H}_{2}\text{SO}_{4} \rightleftharpoons \text{H}_{2}\text{PO}_{4}^{-}\text{ + HSO}_{4}^{-}\), identify the substances that donate protons (acids) and those that accept protons (bases). H\text{2}SO\text{4} donates a proton to \(\text{HPO}_{4}^{2-}\), so H\text{2}SO\text{4} is an acid and \(\text{HPO}_{4}^{2-}\) is a base.

- Identify the conjugate pairs in equation (c)

After H\text{2}SO\text{4} donates a proton, it becomes HSO\(_{\text{4}}^{-}\), and after H\(\text{PO}_{4}^{2-}\) accepts a proton, it becomes H\(_2\)PO\(_4^{-}\). Therefore, HSO\(_4^{-}\) is the conjugate base of H\(_2\)SO\(_4\) and H\(_2\)PO\(_4^{-}\) is the conjugate acid of HPO\(_4^{2-}\).

Key concepts

Bronsted-Lowry acids and bases

In the realm of acid-base chemistry, the Bronsted-Lowry theory provides a detailed perspective. According to this theory, acids and bases are defined by their ability to donate or accept protons (hydrogen ions). In simple terms:

• Acids are substances that donate protons (H⁺ ions).
• Bases are substances that accept protons.

Let's explore equation (a) from our exercise: \( \text{HCl + H}_2\text{O} \rightleftharpoons \text{Cl}^{-}\text{ + H}_3\text{O}^{+} \) Here, HCl donates a proton to water (H₂O), making HCl an acid and H₂O a base.
By labeling the substances, we can understand how protons move from one molecule to another.
Did you know? The strength of an acid or base can be determined by its degree of proton donation or acceptance. Strong acids and bases completely dissociate in water, whereas weak ones do not.
In summary, the Bronsted-Lowry definition expands our understanding by focusing on proton transfer processes that occur during acid-base reactions.

Conjugate acid-base pairs

When an acid donates a proton, it becomes its conjugate base. Conversely, when a base accepts a proton, it becomes its conjugate acid. This transformation forms what we call a conjugate acid-base pair.
Using equation (b) from the exercise: \( \text{HClO}_4\text{ + H}_2\text{SO}_4 \rightleftharpoons \text{ClO}_4^{-}\text{ + H}_3\text{SO}_4^{+} \)
Here, HClO₄ donates a proton to become ClO₄^-. Therefore, HClO₄ is an acid, and ClO₄^- is its conjugate base.
Conversely, H₂SO₄ accepts a proton to become H₃SO₄⁺. Hence, H₂SO₄ is a base, and H₃SO₄⁺ is its conjugate acid. This elegant relationship simplifies complex reactions by grouping related substances.
It's fascinating to note that:

• Every acid has a conjugate base formed after it donates a proton.
• Every base has a conjugate acid formed after it accepts a proton.

Understanding these pairs helps predict the direction of acid-base reactions and determine their equilibrium states.

Proton transfer reactions

Proton transfer reactions are at the heart of acid-base chemistry. They involve the movement of protons from acids to bases during a chemical reaction. Let's examine equation (c) from our exercise: \( \text{HPO}_4^{2-}\text{ + H}_2\text{SO}_4 \rightleftharpoons \text{H}_2\text{PO}_4^{-}\text{ + HSO}_4^{-} \)
In this reaction, H₂SO₄ donates a proton to HPO₄^2- to form H₂PO₄^- and HSO₄^-. This demonstrates a proton transfer from the acid (H₂SO₄) to the base (HPO₄^2-).
Such reactions are characterized by:

• A clear proton donor (acid).
• A clear proton acceptor (base).

The direction and extent of these transfer reactions are influenced by the relative strengths of the acids and bases involved.
Always remember, in proton transfer reactionions:

• Proton donors transform into their corresponding conjugate bases.
• Proton acceptors transform into their corresponding conjugate acids.

These reactions are essential for many biological processes and industrial applications, illustrating the foundational role of acids and bases in chemistry.