Question

In each reaction, identify the Bronsted-Lowry acid, the Bronsted-Lowry base, the conjugate acid, and the conjugate base. $$\begin{array}{l}{\text { a. } \mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q)} \\ {\text { b. } \mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q)} \\ {\text { c. } \mathrm{HNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q)} \\ {\text { d. } \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q)}\end{array}$$

Short answer

a. Acid: \(\mathrm{H}_{2}\mathrm{CO}_{3}\), Base: \(\mathrm{H}_{2}\mathrm{O}\), Conjugate acid: \(\mathrm{H}_{3}\mathrm{O}^{+}\), Conjugate base: \(\mathrm{HCO}_{3}^{-}\). b. Acid: \(\mathrm{H}_{2}\mathrm{O}\), Base: \(\mathrm{NH}_{3}\), Conjugate acid: \(\mathrm{NH}_{4}^{+}\), Conjugate base: \(\mathrm{OH}^{-}\). c. Acid: \(\mathrm{HNO}_{3}\), Base: \(\mathrm{H}_{2}\mathrm{O}\), Conjugate acid: \(\mathrm{H}_{3}\mathrm{O}^{+}\), Conjugate base: \(\mathrm{NO}_{3}^{-}\). d. Acid: \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}\), Base: \(\mathrm{H}_{2}\mathrm{O}\), Conjugate acid: \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{NH}^{+}\), Conjugate base: \(\mathrm{OH}^{-}\).

Step-by-step solution

Identify the Bronsted-Lowry Acid and Base in Reaction a

In reaction a, the Bronsted-Lowry acid is the species that donates a proton, which is \(\mathrm{H}_{2}\mathrm{CO}_{3}\). The Bronsted-Lowry base is the species that accepts a proton, which is \(\mathrm{H}_{2}\mathrm{O}\).

Identify the Conjugate Acid and Base in Reaction a

The conjugate acid is the species formed after the base accepts a proton, which is \(\mathrm{H}_{3}\mathrm{O}^{+}\). The conjugate base is the species formed after the acid donates a proton, which is \(\mathrm{HCO}_{3}^{-}\).

Identify the Bronsted-Lowry Acid and Base in Reaction b

In reaction b, the Bronsted-Lowry acid is \(\mathrm{H}_{2}\mathrm{O}\) because it donates a proton, and the Bronsted-Lowry base is \(\mathrm{NH}_{3}\) because it accepts a proton.

Identify the Conjugate Acid and Base in Reaction b

The conjugate acid for reaction b is \(\mathrm{NH}_{4}^{+}\) and the conjugate base is \(\mathrm{OH}^{-}\).

Identify the Bronsted-Lowry Acid and Base in Reaction c

In reaction c, \(\mathrm{HNO}_{3}\) is the Bronsted-Lowry acid as it donates a proton, and \(\mathrm{H}_{2}\mathrm{O}\) acts as the Bronsted-Lowry base as it accepts a proton.

Identify the Conjugate Acid and Base in Reaction c

The conjugate acid in reaction c is \(\mathrm{H}_{3}\mathrm{O}^{+}\), and the conjugate base is \(\mathrm{NO}_{3}^{-}\).

Identify the Bronsted-Lowry Acid and Base in Reaction d

For reaction d, the Bronsted-Lowry acid is \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}\) because it donates a proton, while the Bronsted-Lowry base is \(\mathrm{H}_{2}\mathrm{O}\).

Identify the Conjugate Acid and Base in Reaction d

The resulting conjugate acid in reaction d is \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{NH}^{+}\) and the conjugate base is \(\mathrm{OH}^{-}\).

Key concepts

Conjugate Acid-Base Pairs

Understanding the concept of conjugate acid-base pairs is essential in the study of chemical reactions in water solutions, particularly those involving acids and bases. The conjugate acid-base pair refers to two substances related by the loss or gain of a single hydrogen ion (proton). When an acid donates a proton, it transforms into its conjugate base. On the flip side, when a base accepts a proton, it forms its conjugate acid.

Take the reaction \(\mathrm{H}_2\mathrm{CO}_3(aq) + \mathrm{H}_2\mathrm{O}(l) \rightleftharpoons \mathrm{H}_3\mathrm{O}^+(aq) + \mathrm{HCO}_3^-(aq)\) for example. The carbonic acid, \(\mathrm{H}_2\mathrm{CO}_3\), donates a proton and becomes the bicarbonate ion, \(\mathrm{HCO}_3^-\), which is its conjugate base. Conversely, water, \(\mathrm{H}_2\mathrm{O}\), accepts a proton, becoming the hydronium ion, \(\mathrm{H}_3\mathrm{O}^+\), thus acting as the conjugate acid. These pairs are critical to understanding the reversible nature of acid-base reactions and alignment with equilibrium principles.

Proton Donor and Acceptor

The Bronsted-Lowry acid-base theory introduces the idea of a proton donor (acid) and proton acceptor (base). In simple terms, an acid is any compound that can donate a 'proton'—in chemistry, this is a hydrogen ion, \(H^+\), to another compound called the base. It's a way of describing a substance’s ability to increase the concentration of \(H^+\) ions in a solution.

When examining acid-base reactions, we can use the example of ammonia, \(\mathrm{NH}_3\), reacting with water. Here, water acts as a proton donor by giving away a \(H^+\), converting into a hydroxide ion, \(\mathrm{OH}^-\), while ammonia accepts the \(H^+\) to become ammonium, \(\mathrm{NH}_4^+\). This exchange of a proton is the heart of an acid-base reaction and helps us to predict the behavior of substances in various chemical environments.

Acid-Base Reaction

An acid-base reaction involves the transfer of a proton from the acid to the base. This transaction determines the direction of the reaction and the resulting products. In the example of \(\mathrm{NH}_3(aq) + \mathrm{H}_2\mathrm{O}(l) \rightleftharpoons \mathrm{NH}_4^+(aq) + \mathrm{OH}^-(aq)\), the reaction illustrates the proton transfer where \(\mathrm{NH}_3\), the base, gains a proton becoming \(\mathrm{NH}_4^+\), and the \(\mathrm{H}_2\mathrm{O}\), the acid, loses a proton to form \(\mathrm{OH}^-\).

This process of proton transfer is central to the understanding of countless chemical processes that happen around us, from digestion in our body to the manufacturing of many everyday products.

Equilibrium in Acid-Base Reactions

In the discussion of acid-base reactions, equilibrium plays a fundamental role. Equilibrium in this context refers to the state of balance in a reversible chemical reaction, where the rate of the forward reaction equals the rate of the reverse reaction. Acid-base reactions often reach equilibrium, meaning both the formation of products and the reverse process of products re-forming reactants occur at the same rate.

For instance, in the reaction \(\mathrm{C}_5\mathrm{H}_5\mathrm{N}(aq) + \mathrm{H}_2\mathrm{O}(l) \rightleftharpoons \mathrm{C}_5\mathrm{H}_5\mathrm{NH}^+(aq) + \mathrm{OH}^-(aq)\), the reaction reaches a state where the concentrations of reactants and products stay constant over time. This equilibrium concept explains why we often observe both reactants and products present in a solution even after the reaction seems complete. The position of the equilibrium, leaning towards either products or reactants, can be affected by various factors including concentration, temperature, and pressure, giving insight into a substance’s acid or base strength.