Problem 17 - Chemistry: The Central Science in SI Units, Global Edition

Question

Identify the Brønsted-Lowry acid and the Brønsted-Lowry base on the left side of each of the following equations, and also identify the conjugate acid and conjugate base of each on the right side: (a) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{NH}_{3}(a q)\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) $$ \left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ (c) \(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons\) $$ \mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q) $$

Step-by-Step Solution

Answer

For the given reactions: (a) Acid: \(\mathrm{NH}_{4}^{+}\), Base: \(\mathrm{CN}^{-}\), Conjugate Acid: \(\mathrm{HCN}\), Conjugate Base: \(\mathrm{NH}_{3}\). (b) Acid: \(\mathrm{H}_{2}\mathrm{O}\), Base: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}\), Conjugate Acid: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\), Conjugate Base: \(\mathrm{OH}^{-}\). (c) Acid: \(\mathrm{HCOOH}\), Base: \(\mathrm{PO}_{4}^{3-}\), Conjugate Acid: \(\mathrm{HPO}_{4}^{2-}\), Conjugate Base: \(\mathrm{HCOO}^{-}\).

Step 1: Identify Acid and Base on left side

In this reaction, the species on the left side are \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{CN}^{-}\). Since \(\mathrm{NH}_{4}^{+}\) is a proton donor and \(\mathrm{CN}^{-}\) is a proton acceptor, we have: - Acid: \(\mathrm{NH}_{4}^{+}\) - Base: \(\mathrm{CN}^{-}\)

Step 2: Identify Conjugate Acid and Conjugate Base on the right side

The species on the right side are \(\mathrm{HCN}\) and \(\mathrm{NH}_{3}\). After the proton transfer, we have: - Conjugate Acid: \(\mathrm{HCN}\) - Conjugate Base: \(\mathrm{NH}_{3}\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}(a q)+\mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q)\)

Step 1: Identify Acid and Base on left side

In this reaction, the species on the left side are \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}\) and \(\mathrm{H}_{2}\mathrm{O}\). Since \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}\) is a proton acceptor and \(\mathrm{H}_{2}\mathrm{O}\) is a proton donor, we have: - Acid: \(\mathrm{H}_{2}\mathrm{O}\) - Base: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}\)

Step 2: Identify Conjugate Acid and Conjugate Base on the right side

The species on the right side are \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\) and \(\mathrm{OH}^{-}\). After the proton transfer, we have: - Conjugate Acid: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\) - Conjugate Base: \(\mathrm{OH}^{-}\) (c) \(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons\mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q)\)

Step 1: Identify Acid and Base on left side

In this reaction, the species on the left side are \(\mathrm{HCOOH}\) and \(\mathrm{PO}_{4}^{3-}\). Since \(\mathrm{HCOOH}\) is a proton donor and \(\mathrm{PO}_{4}^{3-}\) is a proton acceptor, we have: - Acid: \(\mathrm{HCOOH}\) - Base: \(\mathrm{PO}_{4}^{3-}\)

Step 2: Identify Conjugate Acid and Conjugate Base on the right side

The species on the right side are \(\mathrm{HCOO}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\). After the proton transfer, we have: - Conjugate Acid: \(\mathrm{HPO}_{4}^{2-}\) - Conjugate Base: \(\mathrm{HCOO}^{-}\)

Key Concepts

Conjugate Acid

In a Brønsted-Lowry acid-base reaction, when the acid donates a proton, it transforms into a conjugate base. The species that receives the proton becomes the conjugate acid. Understanding conjugate acids is crucial because it helps determine the behavior of molecules in various chemical reactions.

• For example, when \(\text{NH}_4^+\) (ammonium) donates a proton, it becomes \(\text{NH}_3\) (ammonia), which acts as its conjugate base.
• A proton is transferred to \(\text{CN}^-\) (cyanide) forming \(\text{HCN}\) (hydrocyanic acid), which acts as the conjugate acid.

By understanding this transformation, students can better predict how molecules will react in different chemical equations.

Conjugate Base

The conjugate base is what remains of an acid after it loses a proton. Brønsted-Lowry theory beautifully illustrates this by pairing acids and bases with their conjugates, showing that every acid-base reaction is a reversible process.

• Take the reaction of the ammonium ion \(\text{NH}_4^+\): once it donates a proton, it is converted into \(\text{NH}_3\), representing its conjugate base.
• Consider the formic acid \(\text{HCOOH}\): after donating a proton, it becomes \(\text{HCOO}^-\), its conjugate base.

Conjugate bases play a vital role in buffering solutions, which helps maintain pH levels in environments such as the human body.

Proton Transfer

Proton transfer is a fundamental concept in Brønsted-Lowry acid-base theory. It involves the movement of a proton (hydrogen ion, \(\text{H}^+\)) from the acid to the base. This transfer is what leads to the formation of conjugate acid-base pairs.

• In the reaction \(\text{NH}_4^+ + \text{CN}^- \rightleftharpoons \text{HCN} + \text{NH}_3\), the ammonium ion \(\text{NH}_4^+\) transfers a proton to the cyanide ion \(\text{CN}^-\).
• This process results in the formation of \(\text{HCN}\) and \(\text{NH}_3\) as products, illustrating how proton transfer leads to new chemical species.

Understanding proton transfer helps explain many acid-base reactions, enabling us to predict the direction and outcome of chemical processes.