Question

For each of the following reactions, indicate the Brønsted-Lowry acids and bases. What are the conjugate acid/base pairs? $$ \begin{array}{l} \text { (a) } \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{H}_{2} \mathrm{O} \\ \text { (b) } \mathrm{HNO}_{2}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O} \\ \text { (c) } \mathrm{HCHO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CHO}_{2}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \end{array} $$

Short answer

Question: Identify the Brønsted-Lowry acids and bases in the following reaction and determine the conjugate acid/base pairs. Reaction: $$\mathrm{HNO}_{2}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}$$ Answer: In this reaction, the Brønsted-Lowry acid is \(\mathrm{HNO}_{2}\) and the Brønsted-Lowry base is \(\mathrm{OH}^{-}\). The conjugate acid/base pairs are \(\mathrm{HNO}_{2}/\mathrm{NO}_{2}^{-}\) and \(\mathrm{OH}^{-}/\mathrm{H}_{2}\mathrm{O}\).

Step-by-step solution

Reaction (a)

For the reaction: $$\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{H}_{2} \mathrm{O}$$ In this reaction, \(\mathrm{H}_{3}\mathrm{O}^{+}\) donates a proton to \(\mathrm{CN}^{-}\). So, \(\mathrm{H}_{3}\mathrm{O}^{+}\) is the Brønsted-Lowry acid and \(\mathrm{CN}^{-}\) is the Brønsted-Lowry base. After the proton is transferred, we have the conjugate pairs \(\mathrm{H}_{3}\mathrm{O}^{+}/\mathrm{H}_{2}\mathrm{O}\) and \(\mathrm{CN}^{-}/\mathrm{HCN}\).

Reaction (b)

For the reaction: $$\mathrm{HNO}_{2}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}$$ In this reaction, \(\mathrm{HNO}_{2}\) donates a proton to \(\mathrm{OH}^{-}\). So, \(\mathrm{HNO}_{2}\) is the Brønsted-Lowry acid and \(\mathrm{OH}^{-}\) is the Brønsted-Lowry base. After the proton is transferred, we have the conjugate pairs \(\mathrm{HNO}_{2}/\mathrm{NO}_{2}^{-}\) and \(\mathrm{OH}^{-}/\mathrm{H}_{2}\mathrm{O}\).

Reaction (c)

For the reaction: $$\mathrm{HCHO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CHO}_{2}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)$$ In this reaction, \(\mathrm{HCHO}_{2}\) donates a proton to \(\mathrm{H}_{2}\mathrm{O}\). So, \(\mathrm{HCHO}_{2}\) is the Brønsted-Lowry acid and \(\mathrm{H}_{2}\mathrm{O}\) is the Brønsted-Lowry base. After the proton is transferred, we have the conjugate pairs \(\mathrm{HCHO}_{2}/\mathrm{CHO}_{2}^{-}\) and \(\mathrm{H}_{2}\mathrm{O}/\mathrm{H}_{3}\mathrm{O}^{+}\).

Key concepts

Conjugate Acid/Base Pairs

Understanding conjugate acid/base pairs is essential to grasp the behavior of substances in acid-base reactions. In the Brønsted-Lowry concept, an acid is a proton donor, and a base is a proton acceptor. When an acid donates a proton, it transforms into its conjugate base, and when a base accepts a proton, it becomes its conjugate acid.

For instance, taking the example from the textbook exercise, when \(\mathrm{H}_{3}\mathrm{O}^{+}\) (hydronium ion) donates a proton, it becomes water (\(\mathrm{H}_{2}\mathrm{O}\)), its conjugate base. In contrast, \(\mathrm{CN}^{-}\) (cyanide ion) accepts a proton to form hydrogen cyanide (\(\mathrm{HCN}\)), its conjugate acid. Thus, we can say \(\mathrm{H}_{3}\mathrm{O}^{+}/\mathrm{H}_{2}\mathrm{O}\) and \(\mathrm{CN}^{-}/\mathrm{HCN}\) are conjugate acid/base pairs.

It's essential to identify these pairs to predict the direction of the reaction and understand the strength of the acids and bases involved. The strength of an acid is inversely related to the strength of its conjugate base. A strong acid has a weak conjugate base, and a weak acid has a relatively stronger conjugate base. This relationship is key to understanding acid-base behavior in various chemical environments.

Chemical Equilibrium

The concept of chemical equilibrium plays a central role in acid-base chemistry. Chemical equilibrium occurs when the rates of the forward and reverse reactions in a chemical process are equal, resulting in no net change in the concentration of reactants and products over time. It's represented by the double arrow (\(\rightleftharpoons\)) in the chemical equations.

For example, in the exercise provided, the reactions given are in equilibrium, which means that the formation of the products from the reactants happens at the same rate as the formation of reactants from the products. This state of balance does not imply that the reactants and products are present in equal amounts, but rather that their ratios remain constant.

Chemical equilibrium is influenced by a variety of factors, including concentration, temperature, and pressure. According to Le Chatelier's principle, if an external condition is changed, the equilibrium will shift to minimize that change. Understanding equilibrium is crucial for predicting how the concentration of acids and bases will alter under different conditions and how this will affect the acidity or basicity of a solution.

Acid-Base Reactions

In acid-base reactions, the transfer of protons is the signature event. Each reaction typically involves an acid and a base, each of which becomes its conjugate after the reaction, as highlighted in the step by step solutions for the exercise. The direction in which an acid-base reaction proceeds can depend on various factors including the relative strengths of the acids and bases.

Using the reactions listed in the exercise, we can understand this process. In reaction (b), \(\mathrm{HNO}_{2}\) donates a proton to \(\mathrm{OH}^{-}\) resulting in the formation of \(\mathrm{NO}_{2}^{-}\) and water. These acid-base reactions are not only critical in laboratory chemistry but also play a significant role in biological systems and the environment.

Understanding how acids and bases react with each other is fundamental for predicting the outcomes of many chemical processes, such as titration, buffer system formation, and even the functioning of metabolism in living organisms. The interplay between acids and bases can define the pH of a system, which has profound implications for chemical reactivity and biological compatibility.