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) \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq})+\mathrm{HSO}_{4}^{-}(\mathrm{aq}) \rightleftarrows\) \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (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 the Acid and Base for Reaction (a)

In reaction (a), we need to identify the Bronsted acid and base. The chemical equation is: \[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N} + \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} \rightleftarrows \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} + \mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\]The acid is the one donating a proton (H⁺), which is \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\), since it turns into \(\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\). The base is \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\), as it accepts a proton becoming \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\).

Identify Conjugate Pairs for Reaction (a)

The conjugate acid-base pairs consist of \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) and \(\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\) as one pair, and \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\) and \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) as the other pair.

Identify the Acid and Base for Reaction (b)

In reaction (b), the chemical equation is: \[\mathrm{N}_{2} \mathrm{H}_{4} + \mathrm{HSO}_{4}^{-} \rightleftarrows \mathrm{N}_{2} \mathrm{H}_{5}^{+} + \mathrm{SO}_{4}^{2-}\]Here, \(\mathrm{HSO}_{4}^{-}\) donates a proton, making it the acid, and \(\mathrm{N}_{2} \mathrm{H}_{4}\) accepts the proton, making it the base.

Identify Conjugate Pairs for Reaction (b)

The conjugate pairs are \(\mathrm{HSO}_{4}^{-}\) and \(\mathrm{SO}_{4}^{2-}\) forming one pair, and \(\mathrm{N}_{2} \mathrm{H}_{4}\) and \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}\) forming the other.

Identify the Acid and Base for Reaction (c)

In reaction (c), the chemical equation is: \[\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} + \mathrm{OH}^{-} \rightleftarrows \left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}\right]^{2+} + \mathrm{H}_{2} \mathrm{O}\]Here, \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) acts as the acid releasing a proton to form \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}\right]^{2+}\), while \(\mathrm{OH}^{-}\) acts as the base.

Identify Conjugate Pairs for Reaction (c)

The conjugate pairs are \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) and \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}\right]^{2+}\), with \(\mathrm{OH}^{-}\) and \(\mathrm{H}_{2} \mathrm{O}\) forming the other pair.

Key concepts

Conjugate Acid-Base Pairs

Conjugate acid-base pairs are fundamental concepts in the Bronsted-Lowry acid-base theory. When an acid donates a proton (\(\mathrm{H}^{+}\)), it forms its conjugate base, while a base that accepts a proton forms its conjugate acid. This means that an acid and its conjugate base differ by one proton. Let's consider reaction (a) from our exercise: \[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}+\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} \rightleftarrows \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}+\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\]In this case, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) donates a proton to become \(\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\), making them a conjugate pair. Similarly, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\) accepts a proton to become \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\), forming another conjugate pair.

• Conjugate acid-base pairs are related by the loss or gain of a single proton.
• Knowing conjugate pairs helps predict the direction of equilibrium in acid-base reactions.

Proton Transfer Reactions

In Bronsted-Lowry theory, proton transfer reactions are central to understanding how acids and bases behave. A proton transfer occurs when a proton is moved from an acid to a base. This is why acids are often called proton donors, and bases are proton acceptors. Let's look at reaction (b) from the exercise: \[\mathrm{N}_{2} \mathrm{H}_{4} + \mathrm{HSO}_{4}^{-} \rightleftarrows \mathrm{N}_{2} \mathrm{H}_{5}^{+} + \mathrm{SO}_{4}^{2-}\]Here, \(\mathrm{HSO}_{4}^{-}\) acts as the acid by donating an \(\mathrm{H}^{+}\), transferring it to \(\mathrm{N}_{2} \mathrm{H}_{4}\), which functions as the base. This transfer results in \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}\) and \(\mathrm{SO}_{4}^{2-}\).

• Proton transfer reaction is the hallmark of acid-base interactions in Bronsted-Lowry theory.
• During the reaction, the acid becomes its conjugate base, and the base becomes its conjugate acid.

Acid-Base Identification

Recognizing which substances in a reaction act as acids and which act as bases is crucial in chemistry. As per Bronsted-Lowry theory, acids are proton donors, and bases are proton acceptors. To identify them, focus on what gains and loses protons. For example, in reaction (c): \[\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}+\mathrm{OH}^{-} \rightleftarrows \left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}\right]^{2+}+\mathrm{H}_{2} \mathrm{O}\], the complex \([\mathrm{Al}(\mathrm{H}_{2} \mathrm{O})_{6}]^{3+}\) donates a proton, so it is the acid. The \( \mathrm{OH}^{-}\) ion accepts this proton, making it the base.

• Tracking proton movement helps identify acids and bases in reactions.
• Equilibrium of reactions depends on the strength of the acids and bases involved.