Problem 17 - Chemistry The Central Science

Question

Designate the Bronsted-Lowry acid and the Bronsted-Lowry base on the left side of each of the following equations, and also designate the conjugate acid and conjugate base of each on the right side: $$ \text { (a) } \mathrm{NH}_{4}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{NH}_{3}(a q) $$ (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

(a) The Bronsted-Lowry acid is \(\mathrm{NH}_{4}^{+}\), the Bronsted-Lowry base is \(\mathrm{CN}^{-}\), the conjugate acid is \(\mathrm{HCN}\), and the conjugate base is \(\mathrm{NH}_{3}\). (b) The Bronsted-Lowry acid is \((\mathrm{CH}_{3})_{3} \mathrm{~N}\), the Bronsted-Lowry base is \(\mathrm{H}_{2}\mathrm{O}\), the conjugate acid \((\mathrm{CH}_{3})_{3} \mathrm{NH}^{+}\), and the conjugate base is \(\mathrm{OH}^{-}\). (c) The Bronsted-Lowry acid is \(\mathrm{HCOOH}\), the Bronsted-Lowry base is \(\mathrm{PO}_{4}^{3-}\), the conjugate acid is \(\mathrm{HPO}_{4}^{2-}\), and the conjugate base is \(\mathrm{HCOO}^{-}\).

Step 1: (a) Identify Bronsted-Lowry acid and base on the left side

In the given equation, \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{NH}_{3}(a q),\) we can see that \(\mathrm{NH}_{4}^{+}\) donates a proton to \(\mathrm{CN}^{-}.\) So, the Bronsted-Lowry acid is \(\mathrm{NH}_{4}^{+}\), and the Bronsted-Lowry base is \(\mathrm{CN}^{-}.\)

Step 2: (a) Identify the conjugate acid and base on the right side

On the right side of the equation, after \(\mathrm{NH}_{4}^{+}\) donates a proton, it forms \(\mathrm{NH}_{3}.\) Therefore, the conjugate base is \(\mathrm{NH}_{3}.\) After receiving a proton, \(\mathrm{CN}^{-}\) becomes \(\mathrm{HCN}.\) Therefore, the conjugate acid is \(\mathrm{HCN}.\)

Step 3: (b) Identify Bronsted-Lowry acid and base on the left side

In this equation, \((\mathrm{CH}_{3})_{3} \mathrm{~N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons (\mathrm{CH}_{3})_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q),\) we can see that \((\mathrm{CH}_{3})_{3} \mathrm{~N}\) donates a proton to \(\mathrm{H}_{2}\mathrm{O}.\) So, the Bronsted-Lowry acid is \((\mathrm{CH}_{3})_{3} \mathrm{~N},\) and the Bronsted-Lowry base is \(\mathrm{H}_{2}\mathrm{O}.\)

Step 4: (b) Identify the conjugate acid and base on the right side

On the right side of the equation, after donating a proton, \((\mathrm{CH}_{3})_{3} \mathrm{~N}\) forms \((\mathrm{CH}_{3})_{3} \mathrm{NH}^{+}.\) Therefore, the conjugate acid is \((\mathrm{CH}_{3})_{3} \mathrm{NH}^{+}.\) After receiving a proton, \(\mathrm{H}_{2}\mathrm{O}\) becomes \(\mathrm{OH}^{-}.\) Therefore, the conjugate base is \(\mathrm{OH}^{-}.\)

Step 5: (c) Identify Bronsted-Lowry acid and base on the left side

In this equation, \(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons \mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q),\) we can see that \(\mathrm{HCOOH}\) donates a proton to \(\mathrm{PO}_{4}^{3-}.\) So, the Bronsted-Lowry acid is \(\mathrm{HCOOH},\) and the Bronsted-Lowry base is \(\mathrm{PO}_{4}^{3-}.\)

Step 6: (c) Identify the conjugate acid and base on the right side

On the right side of the equation, after donating a proton, \(\mathrm{HCOOH}\) forms \(\mathrm{HCOO}^{-}.\) Therefore, the conjugate base is \(\mathrm{HCOO}^{-}.\) After receiving a proton, \(\mathrm{PO}_{4}^{3-}\) becomes \(\mathrm{HPO}_{4}^{2-}.\) Therefore, the conjugate acid is \(\mathrm{HPO}_{4}^{2-}.\)

Key Concepts

acid-base reactions

Acid-base reactions in the context of the Bronsted-Lowry theory are fascinating as they involve the transfer of protons between different chemical species. Unlike older theories, which focused solely on hydrogen ions, the Bronsted-Lowry concept provides a broader perspective by defining acids and bases based on their ability to donate or accept protons.

In any Bronsted-Lowry acid-base reaction, the acid is the substance that donates a proton (H⁺ ion), while the base is the one that accepts this proton. This proton transfer is the core of the reaction. For instance, in the equation \(NH_{4}^{+}(aq) + CN^{-}(aq) \rightleftharpoons HCN(aq) + NH_{3}(aq),\) \(NH_{4}^{+}\) acts as the acid since it donates a proton, and \(CN^{-}\) serves as the base by accepting that proton.

As these reactions occur, they showcase the dynamic nature of chemical interactions, moving beyond simple molecular changes to highlight the essential role of protons in dictating chemical behavior. Understanding this concept allows us to predict and explain the outcomes of acid-base reactions more accurately.

conjugate acid-base pairs

Understanding conjugate acid-base pairs is crucial for comprehending the outcome of acid-base reactions. In the Bronsted-Lowry theory, each acid and base has a conjugate partner. When an acid donates a proton, it transforms into its conjugate base. Conversely, when a base accepts a proton, it becomes its conjugate acid.

Take, for example, the reaction \((CH_{3})_{3} N (aq) + H_{2}O(l) \rightleftharpoons (CH_{3})_{3} NH^{+}(aq) + OH^{-}(aq).\) In this case, \((CH_{3})_{3} N\) acts as the base, and after it accepts a proton, it becomes \((CH_{3})_{3} NH^{+}\), its conjugate acid. Simultaneously, \(H_{2}O\) serves as the acid and becomes \(OH^{-}\) after it donates a proton, turning into its conjugate base.

This conjugate relationship emphasizes the balanced nature of acid-base reactions, where everything occurs in pairs. Neither acids nor bases exist in isolation during these reactions. This duality ensures the conservation of charge and species, enabling predictable behavior that is fundamental to understanding chemical equilibria.

proton transfer

Proton transfer is the pivotal action in Bronsted-Lowry acid-base reactions. It signifies the movement of a hydrogen ion, or proton, from one molecule to another, initiating a chemical transformation. This transfer is the defining feature that distinguishes acid-base interactions in this theory from other types of chemical reactions.

To illustrate, consider the reaction \(HCOOH(aq) + PO_{4}^{3-}(aq) \rightleftharpoons HCOO^{-}(aq) + HPO_{4}^{2-}(aq).\) In this system, \(HCOOH\) donates a proton to \(PO_{4}^{3-}\), demonstrating the concept of proton transfer. \(HCOOH\) loses a hydrogen ion, converting into \(HCOO^{-}\), while \(PO_{4}^{3-}\) accepts the proton, turning into \(HPO_{4}^{2-}\).

Proton transfer not only facilitates the creation of new substances but also maintains a charge balance, which is crucial for the stability of the reaction products. Recognizing proton transfer is essential for understanding the reactivity and functionality of acids and bases in various chemical contexts.