Question

Designate the Brønsted-Lowry acid and the BrønstedLowry base on the left side of each of the following equations, and also designate the conjugate acid and conjugate base 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) \(\mathrm{HCHO}_{2}(a q)+\mathrm{PO}_{4}{ }^{3-}(a q) \rightleftharpoons\) \(\mathrm{CHO}_{2}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q)\)

Short answer

(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: \((\mathrm{CH}_{3})_{3} \mathrm{~N}\), Conjugate acid: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\), Conjugate base: \(\mathrm{OH}^{-}\) (c) Acid: \(\mathrm{HCHO}_{2}\), Base: \(\mathrm{PO}_{4}{ }^{3-}\), Conjugate acid: \(\mathrm{HPO}_{4}^{2-}\), Conjugate base: \(\mathrm{CHO}_{2}^{-}\)

Step-by-step solution

Identify the acid and base on the left side

Compare the left side of the equation with the right side and look for species that have gained or lost a proton (H+ ion). We see that \(\mathrm{NH}_{4}{ }^{+}\) has lost a proton to form \(\mathrm{NH}_{3}\), and \(\mathrm{CN}^{-}\) has gained a proton to form \(\mathrm{HCN}\). Thus, \(\mathrm{NH}_{4}{ }^{+}\) is the Brønsted-Lowry acid, and \(\mathrm{CN}^{-}\) is the Brønsted-Lowry base.

Identify the conjugate acid and base on the right side

Since \(\mathrm{CN}^{-}\) gained a proton to form \(\mathrm{HCN}\), \(\mathrm{HCN}\) is the conjugate acid. Since \(\mathrm{NH}_{4}{ }^{+}\) lost a proton to form \(\mathrm{NH}_{3}\), \(\mathrm{NH}_{3}\) is the conjugate base. (b) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons$ \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q)\)

Identify the acid and base on the left side

Compare the left side of the equation with the right side and look for species that have gained or lost a proton (H+ ion). We see that \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\) has gained a proton to form \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\), and \(\mathrm{H}_{2}\mathrm{O}\) has lost a proton to form \(\mathrm{OH}^{-}\). Thus, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\) is the Brønsted-Lowry base, and \(\mathrm{H}_{2}\mathrm{O}\) is the Brønsted-Lowry acid.

Identify the conjugate acid and base on the right side

Since \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\) gained a proton to form \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\), \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\) is the conjugate acid. Since \(\mathrm{H}_{2}\mathrm{O}\) lost a proton to form \(\mathrm{OH}^{-}\), \(\mathrm{OH}^{-}\) is the conjugate base. (c) \(\mathrm{HCHO}_{2}(a q)+\mathrm{PO}_{4}{ }^{3-}(a q) \rightleftharpoons\) \(\mathrm{CHO}_{2}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q)\)

Identify the acid and base on the left side

Compare the left side of the equation with the right side and look for species that have gained or lost a proton (H+ ion). We see that \(\mathrm{HCHO}_{2}\) has lost a proton to form \(\mathrm{CHO}_{2}^{-}\), and \(\mathrm{PO}_{4}{ }^{3-}\) has gained a proton to form \(\mathrm{HPO}_{4}^{2-}\). Thus, \(\mathrm{HCHO}_{2}\) is the Brønsted-Lowry acid, and \(\mathrm{PO}_{4}{ }^{3-}\) is the Brønsted-Lowry base.

Identify the conjugate acid and base on the right side

Since \(\mathrm{PO}_{4}{ }^{3-}\) gained a proton to form \(\mathrm{HPO}_{4}^{2-}\), \(\mathrm{HPO}_{4}^{2-}\) is the conjugate acid. Since \(\mathrm{HCHO}_{2}\) lost a proton to form \(\mathrm{CHO}_{2}^{-}\), \(\mathrm{CHO}_{2}^{-}\) is the conjugate base.

Key concepts

Conjugate Acid-Base Pairs

In the realm of the Brønsted-Lowry acid-base theory, understanding the concept of conjugate acid-base pairs is essential. When an acid donates a proton \((\text{H}^+)\), it transforms into its conjugate base, while a base that accepts a proton becomes its conjugate acid. This relationship forms a pair that undergoes reversible reactions, showcasing the dynamic nature of acid-base chemistry.
Let's explore this with an example:

• Consider \(\text{NH}_4^+\) and \(\text{NH}_3\). When \(\text{NH}_4^+\) donates a proton, it becomes \(\text{NH}_3\), making \(\text{NH}_4^+\) the conjugate acid, and \(\text{NH}_3\) the conjugate base.
• Next, consider \(\text{HCN}\) and \(\text{CN}^-\). Here, \(\text{HCN}\) is the conjugate acid of \(\text{CN}^-\) after \(\text{CN}^-\) gains a proton.

Through these pairs, chemical reactions maintain a constant balance enhancing the understanding between the exchange of compounds during chemical reactions.

Proton Transfer Reactions

Proton transfer reactions are at the heart of the Brønsted-Lowry theory. These involve the movement of a proton from the acid to the base. This process is what enables the identification of acids and bases under this framework. Consider the equation \(\mathrm{CH}_3)_3 \text{~N} + \text{H}_2 \text{O}\rightleftharpoons (\mathrm{CH}_3)_3 \text{NH}^+ + \text{OH}^-\). Here's how proton transfer occurs:

• The base, \(\mathrm{CH}_3)_3 \text{~N}\), accepts a proton to become \((\mathrm{CH}_3)_3 \text{NH}^+\), showcasing its role in the reaction as a proton acceptor.
• The acid, \(\text{H}_2\text{O}\), donates a proton to become \(\text{OH}^-\), demonstrating its function in donating protons.

This exchange of protons distinguishes the reactive species, allowing us to better understand the mechanisms and behavior of acids and bases in various chemical environments.

Acid-Base Identification

Identifying acids and bases using the Brønsted-Lowry theory involves observing proton transfers within a reaction. This observation allows chemists to determine which species acts as the acid and which as the base. For instance, in the reaction \(\mathrm{HCHO}_2 + \text{PO}_4^{3-}\rightleftharpoons \mathrm{CHO}_2^- + \text{HPO}_4^{2-}\), one can pinpoint:

• \(\mathrm{HCHO}_2\) as the Brønsted-Lowry acid since it loses a proton to become \(\mathrm{CHO}_2^-\).
• \(\text{PO}_4^{3-}\) as the Brønsted-Lowry base, gaining that proton to form \(\text{HPO}_4^{2-}\).

Identifying acids and bases based on proton exchange provides a systematic approach to classifying substances and predicting their behaviors in reactions, enhancing both experimental practices and theoretical understandings in chemistry.

Chemical Equilibrium

Chemical equilibrium describes the point in a reaction where the rates of forward and reverse reactions are equal, leading both reactions to occur simultaneously without any net change in concentration of reactants and products. In the context of acid-base reactions, equilibrium showcases how acids and bases continuously donate and accept protons, maintaining a dynamic balance. Consider the general reaction:
\[ \text{HA} + \text{B} \rightleftharpoons \text{A}^- + \text{HB}^+ \]
This equation illustrates that as the forward reaction proceeds, the reverse reaction also keeps pace, thereby establishing equilibrium.

• This equilibrium is sensitive to different conditions, such as temperature or concentration, and can shift according to Le Chatelier's principle.
• The concept of equilibrium ensures that reactions can adjust to changes while maintaining the state of balance, essential in various chemical processes.

Understanding equilibrium in acid-base reactions is fundamental for predicting how different conditions can affect nature and extent of reactions, impacting everything from industrial processes to biological systems.