Problem 37 - Chemistry An Atoms First Approach

Question

For each of the following aqueous reactions, identify the acid, the base, the conjugate base, and the conjugate acid. a. \(\mathrm{H}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{CO}_{3} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HCO}_{3}^{-}\) b. \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}+\mathrm{H}_{3} \mathrm{O}^{+}\) c. \(\mathrm{HCO}_{3}^{-}+\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}+\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)

Step-by-Step Solution

Answer

a. Acid: \(\mathrm{H}_{2}\mathrm{CO}_{3}\), Base: \(\mathrm{H}_{2}\mathrm{O}\), Conjugate Base: \(\mathrm{HCO}_{3}^{-}\), Conjugate Acid: \(\mathrm{H}_{3}\mathrm{O}^{+}\) b. Acid: \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{NH}^{+}\), Base: \(\mathrm{H}_{2}\mathrm{O}\), Conjugate Base: \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}\), Conjugate Acid: \(\mathrm{H}_{3}\mathrm{O}^{+}\) c. Acid: \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{NH}^{+}\), Base: \(\mathrm{HCO}_{3}^{-}\), Conjugate Base: \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}\), Conjugate Acid: \(\mathrm{H}_{2}\mathrm{CO}_{3}\)

Step 1: a. \(\mathrm{H}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{CO}_{3} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HCO}_{3}^{-}\)#

Here, \(\mathrm{H}_{2}\mathrm{O}\) accepts a proton from \(\mathrm{H}_{2}\mathrm{CO}_{3}\) and forms \(\mathrm{H}_{3}\mathrm{O}^{+}\). Therefore, - Acid: \(\mathrm{H}_{2}\mathrm{CO}_{3}\) - Base: \(\mathrm{H}_{2}\mathrm{O}\) - Conjugate Base: \(\mathrm{HCO}_{3}^{-}\) - Conjugate Acid: \(\mathrm{H}_{3}\mathrm{O}^{+}\)

Step 2: b. \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}+\mathrm{H}_{3} \mathrm{O}^{+}\)#

In this reaction, \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{NH}^{+}\) donates a proton to \(\mathrm{H}_{2}\mathrm{O}\) and forms \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}\). Therefore, - Acid: \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{NH}^{+}\) - Base: \(\mathrm{H}_{2}\mathrm{O}\) - Conjugate Base: \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}\) - Conjugate Acid: \(\mathrm{H}_{3}\mathrm{O}^{+}\)

Step 3: c. \(\mathrm{HCO}_{3}^{-}+\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}+\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)#

In this case, \(\mathrm{HCO}_{3}^{-}\) accepts a proton from \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{NH}^{+}\) and forms \(\mathrm{H}_{2}\mathrm{CO}_{3}\). Therefore, - Acid: \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{NH}^{+}\) - Base: \(\mathrm{HCO}_{3}^{-}\) - Conjugate Base: \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}\) - Conjugate Acid: \(\mathrm{H}_{2}\mathrm{CO}_{3}\)

Key Concepts

Conjugate Acid-Base Pairs

In acid-base reactions, conjugate acid-base pairs play a crucial role. These pairs are formed when acids donate and bases accept protons. A conjugate acid forms when a base gains a proton, while a conjugate base is what remains after an acid has donated a proton.
For example, in the reaction \(\mathrm{H}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{CO}_{3} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HCO}_{3}^{-}\), \(\mathrm{H}_{2} \mathrm{CO}_{3}\) is the acid and \(\mathrm{H}_{2} \mathrm{O}\) is the base. When \(\mathrm{H}_{2} \mathrm{CO}_{3}\) donates a proton, it converts into its conjugate base \(\mathrm{HCO}_{3}^{-}\), and \(\mathrm{H}_{2} \mathrm{O}\) becomes \(\mathrm{H}_{3} \mathrm{O}^{+}\), which is the conjugate acid.
Recognizing these pairs helps predict the direction of proton transfer and understand chemical equilibria in different reactions.

Aqueous Reactions

Aqueous reactions are chemical processes that occur in water as the solvent. Water is an excellent medium for chemical reactions due to its ability to dissolve ionic compounds and facilitate proton transfer.
In aqueous solutions, water can act both as an acid and a base, referred to as being amphoteric. For instance, during the reaction of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}+\mathrm{H}_{3} \mathrm{O}^{+}\), water acts as a base by accepting a proton.
Aqueous reactions are significant in various chemical processes and biological systems, impacting phenomena like pH balance and buffer systems. Knowing how these reactions work can help in predicting outcomes and controlling reaction rates.

Proton Transfer

Proton transfer is the fundamental process of acid-base reactions, where an acid donates a proton (\(\text{H}^+\)) to a base. This process is a reversible exchange and is central to the behavior and interaction of acids and bases.
In the reaction \(\mathrm{HCO}_{3}^{-}+\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}+\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\), \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) donates a proton to \(\mathrm{HCO}_{3}^{-}\), demonstrating how protons are transferred between reactants.
The position of equilibrium in a proton transfer reaction depends on the relative strengths of the acids and bases involved. Understanding proton transfer is essential for predicting which species will be the predominant form in a solution.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products. In an equilibrium state, both reactions continue to occur, but at the same rate.
For the given reactions, equilibrium is signified by the double arrows (\(\rightleftharpoons\)). This indicates that both the forward and reverse reactions are happening simultaneously. For example, in the reaction \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}+\mathrm{H}_{3} \mathrm{O}^{+}\), the conversion between reactants and products is in a state of balance.
Understanding chemical equilibrium is crucial for manipulating and predicting the behavior of chemical reactions, especially in industrial and laboratory settings where control over reaction rates and extents is necessary.