Question

Identify the conjugate acid-base pairs in the following reactions. (a) \(\mathrm{HNO}_{3}+\mathrm{N}_{2} \mathrm{H}_{4} \rightleftharpoons \mathrm{NO}_{3}^{-}+\mathrm{N}_{2} \mathrm{H}_{5}^{+}\) (b) \(\mathrm{NH}_{3}+\mathrm{N}_{2} \mathrm{H}_{5}{ }^{+} \rightleftharpoons \mathrm{NH}_{4}^{+}+\mathrm{N}_{2} \mathrm{H}_{4}\) (c) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}+\mathrm{CO}_{3}{ }^{2-} \rightleftharpoons \mathrm{HPO}_{4}^{2-}+\mathrm{HCO}_{3}^{-}\) (d) \(\mathrm{HIO}_{3}+\mathrm{HC}_{2} \mathrm{O}_{4}^{-} \rightleftharpoons \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\)

Short answer

Conjugate acid-base pairs for each reaction are: (a) \(\mathrm{HNO}_{3}/\mathrm{NO}_{3}^{-}\) and \(\mathrm{N}_{2} \mathrm{H}_{4}/\mathrm{N}_{2} \mathrm{H}_{5}^{+}\), (b) \(\mathrm{NH}_{3}/\mathrm{NH}_{4}^{+}\) and \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}/\mathrm{N}_{2} \mathrm{H}_{4}\), (c) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}/\mathrm{HPO}_{4}^{2-}\) and \(\mathrm{CO}_{3}^{2-}/\mathrm{HCO}_{3}^{-}\), and (d) \(\mathrm{HIO}_{3}/\mathrm{IO}_{3}^{-}\) and \(\mathrm{HC}_{2} \mathrm{O}_{4}^{-}/\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\).

Step-by-step solution

Reaction (a) Analysis

In reaction (a), identify the acid-base pairs by looking at what each species loses or gains. In this reaction, \begin{itemize} item \(\mathrm{HNO}_{3}\) donates a proton (\(H^+\)) and becomes \(\mathrm{NO}_{3}^{-}\), thus \(\mathrm{HNO}_{3}/\mathrm{NO}_{3}^{-}\) form a conjugate acid-base pair. item \(\mathrm{N}_{2} \mathrm{H}_{4}\) accepts a proton (\(H^+\)) and becomes \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}\), so \(\mathrm{N}_{2} \mathrm{H}_{4}/\mathrm{N}_{2} \mathrm{H}_{5}^{+}\) form a conjugate acid-base pair. end{itemize}

Reaction (b) Analysis

For reaction (b), again identify the substances that lose or gain a proton. In this reaction, \begin{itemize} item \(\mathrm{NH}_{3}\) gains a proton and becomes \(\mathrm{NH}_{4}^{+}\), so \(\mathrm{NH}_{3}/\mathrm{NH}_{4}^{+}\) form a conjugate acid-base pair. item Conversely, \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}\) loses a proton to form \(\mathrm{N}_{2} \mathrm{H}_{4}\), which makes \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}/\mathrm{N}_{2} \mathrm{H}_{4}\) a conjugate acid-base pair. end{itemize}

Reaction (c) Analysis

For reaction (c), examine the changes in protonation of the species involved. In this reaction, \begin{itemize} item \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) donates a proton to become \(\mathrm{HPO}_{4}^{2-}\), forming the conjugate acid-base pair \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}/\mathrm{HPO}_{4}^{2-}\). item \(\mathrm{CO}_{3}^{2-}\) accepts a proton to become \(\mathrm{HCO}_{3}^{-}\), so \(\mathrm{CO}_{3}^{2-}/\mathrm{HCO}_{3}^{-}\) is the conjugate acid-base pair. end{itemize}

Reaction (d) Analysis

Lastly, for reaction (d), identify which species undergo proton transfer. In this case: \begin{itemize} item \(\mathrm{HIO}_{3}\) acts as an acid by donating a proton to become \(\mathrm{IO}_{3}^{-}\), forming the conjugate acid-base pair \(\mathrm{HIO}_{3}/\mathrm{IO}_{3}^{-}\). item \(\mathrm{HC}_{2} \mathrm{O}_{4}^{-}\) acts as a base by accepting a proton, thereby becoming \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), and establishes the conjugate acid-base pair \(\mathrm{HC}_{2} \mathrm{O}_{4}^{-}/\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\). end{itemize}

Key concepts

Acid-Base Reaction

An acid-base reaction is a chemical process where an acid and a base interact with each other. The core idea behind this interaction is the exchange of a proton (hydrogen ion, H+). When an acid releases a proton, a base is ready to accept it. This exchange leads to the creation of new substances named conjugate acid-base pairs.

Using the textbook solution as an example, the reaction between \textbf{HNO3} and \textbf{N2H4} (part a) illustrates an acid-base reaction where \(\textbf{HNO3}\) acts as the acid by donating a proton, and \(\textbf{N2H4}\) behaves as a base by accepting that proton. As a result, \(\textbf{NO3}^-\) and \(\textbf{N2H5}^+\) are formed which are the conjugate base and conjugate acid, respectively. It's essential to identify what each reactant donates or accepts to understand the reaction outcome.

Proton Transfer

Proton transfer is the hallmark of acid-base reactions in the Brønsted-Lowry theory. In this perspective, acids are proton donors and bases are proton acceptors. The movement of the proton from the acid to the base is what characterizes the acid-base process.

In the step-by-step solution provided for the textbook exercises, proton transfer can be observed in each reaction. For instance, in reaction (c), the \(\textbf{H2PO4}^-\) donates a proton to \(\textbf{CO3}^{2-}\), effectuating the proton transfer. This results in the formation of \(\textbf{HPO4}^{2-}\) and \(\textbf{HCO3}^-\), which are, respectively, the conjugate base of the acid \(\textbf{H2PO4}^-\) and the conjugate acid of the base \(\textbf{CO3}^{2-}\).

Brønsted-Lowry Theory

The Brønsted-Lowry theory offers an extensive explanation for acid-base reactions. According to the theory, acids and bases are defined by their ability to donate and accept protons, respectively. It's a more comprehensive concept than the earlier Arrhenius definition because it does not require the formation of ions in water and it applies to solvents other than water as well.

Applying the Brønsted-Lowry theory to reaction (d) from the exercise, \(\textbf{HIO3}\) is the Brønsted-Lowry acid as it donates a proton, while \(\textbf{HC2O4}^-\) is the Brønsted-Lowry base, as it accepts a proton. The resulting \(\textbf{IO3}^-\) and \(\textbf{H2C2O4}\) serve as the conjugate base and conjugate acid, respectively, completing the conjugate acid-base pair.

The theory not only explains the process of proton exchange but is also crucial in predicting the direction of reaction and the strength of acids and bases by assessing their tendency to donate or accept protons.