Question

In each of the following acid-base reactions, identify the Bronsted acid and base on the left and their conjugate partners on the right. (a) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq}) \rightleftarrows\) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(\mathrm{aq})\) (b) \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq})+\mathrm{HSO}_{4}^{-}(\mathrm{aq}) \rightleftharpoons\) \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (c) \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightleftarrows\) \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}\right]^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)\)

Short answer

(a) Acid: \( \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} \); Base: \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N} \). (b) Acid: \( \mathrm{HSO}_{4}^{-} \); Base: \( \mathrm{N}_{2} \mathrm{H}_{4} \). (c) Acid: \( \left[\mathrm{Al}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{3+} \); Base: \( \mathrm{OH}^{-} \).

Step-by-step solution

Identify Reactants and Products for Part (a)

In the reaction \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N} (\mathrm{aq})+\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq}) \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(\mathrm{aq}) \), \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N} \) is the base that accepts a proton to become \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} \). \( \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} \) is the acid, donating a proton to form \( \mathrm{CH}_{3} \mathrm{CO}_{2}^{-} \).

Identify Reactants and Products for Part (b)

In the reaction \( \mathrm{N}_{2} \mathrm{H}_{4} (\mathrm{aq})+\mathrm{HSO}_{4}^{-} (\mathrm{aq}) \rightleftharpoons \mathrm{N}_{2} \mathrm{H}_{5}^{+} (\mathrm{aq})+\mathrm{SO}_{4}^{2-} (\mathrm{aq}) \), \( \mathrm{N}_{2} \mathrm{H}_{4} \) acts as a base accepting a proton to form \( \mathrm{N}_{2} \mathrm{H}_{5}^{+} \). \( \mathrm{HSO}_{4}^{-} \) acts as the acid by donating a proton to form \( \mathrm{SO}_{4}^{2-} \).

Identify Reactants and Products for Part (c)

In the reaction \( \left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} (\mathrm{aq})+\mathrm{OH}^{-} (\mathrm{aq}) \rightleftharpoons \left[\mathrm{Al} \left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}\right]^{2+} (\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} (\ell) \), \( \left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} \) is the acid, donating an \( \mathrm{H}^{+} \) ion to \( \mathrm{OH}^{-} \), which acts as the base to form water \( \mathrm{H}_{2} \mathrm{O} \).

Key concepts

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, transform into new substances, called products. In the specific context of Bronsted acid-base reactions, we focus on the transfer of protons (hydrogen ions) between molecules.

This type of chemical reaction is characterized by one molecule, the acid, donating a proton, while the other molecule, acting as a base, accepts that proton.
In Bronsted acid-base reactions, understanding the direction of proton transfer is key to identifying which reactant is the acid and which is the base.

This involves observing what each participant in the reaction loses or gains during the reaction process. For instance, in reaction (a), the hydrocarbon molecule \( \mathrm{CH}_{3}\mathrm{CO}_{2} \mathrm{H} \) donates a proton to become \( \mathrm{CH}_{3}\mathrm{CO}_{2}^{-} \). This indicates that \( \mathrm{CH}_{3}\mathrm{CO}_{2} \mathrm{H} \) functions as the acid.

Meanwhile, the pyridine molecule \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N} \) accepts the proton, allowing it to be identified as the base in this reaction.

Conjugate Acid-Base Pairs

Conjugate acid-base pairs are central to understanding Bronsted acid-base reactions. This concept describes pairs of molecules or ions that transform into each other by the gain or loss of a proton.

Once an acid donates a proton, it becomes its conjugate base. Conversely, once a base gains a proton, it becomes its conjugate acid.
These transformations help balance the equation of a chemical reaction.

In the given reactions, for example, in part (b), \( \mathrm{HSO}_{4}^{-} \) donates a proton to become its conjugate base, \( \mathrm{SO}_{4}^{2-} \).
Similarly, \( \mathrm{N}_{2}\mathrm{H}_{4} \) accepts a proton to transform into its conjugate acid, \( \mathrm{N}_{2}\mathrm{H}_{5}^{+} \).

This dynamic relationship ensures the conservation of mass and energy, fundamental principles in chemical reactions. Each reactant couples with its conjugate counterpart, providing a clear picture of proton movement, essential for balancing reactions and predicting their outcomes.

Proton Transfer

Proton transfer is the process where a proton is moved from the acid to the base in a reaction. It's the hallmark of a Bronsted acid-base reaction and is crucial in understanding the interplay between acids and bases.

This concept helps highlight how particles interact at a molecular level, influencing the behavior of substances in solution.
During the exchange, the bond within the acid weakens, releasing the proton, while the base forms a new bond with the incoming proton, stabilizing itself as it becomes a conjugate acid.

In part (c) of the exercise, we see \( \left[\mathrm{Al}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{3+} \) donate a proton to \( \mathrm{OH}^{-} \).
This proton transfer results in \( \left[\mathrm{Al}\left(\mathrm{H}_{2}\mathrm{O}\right)_{5}\mathrm{OH}\right]^{2+} \) and \( \mathrm{H}_{2}\mathrm{O} \), showcasing the completion of proton transfer.

Ultimately, this movement defines reaction progress and outcome, offering insight into how substances react, reshape, and form new products through proton exchange.