Problem 11 - Chemistry The Molecular Science

Question

Write a chemical equation to describe the proton transfer that occurs when each of these bases is added to water. (a) \(\mathrm{HSO}_{4}^{-}\) (b) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) (c) I (d) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)

Step-by-Step Solution

Answer

(a) \( \mathrm{HSO}_4^- + \mathrm{H_2O} \rightleftharpoons \mathrm{H_2SO_4} + \mathrm{OH}^- \); (b) \( \mathrm{CH}_3\mathrm{NH}_2 + \mathrm{H_2O} \rightarrow \mathrm{CH}_3\mathrm{NH}_3^+ + \mathrm{OH}^- \); (c) \( \mathrm{I}^- \) has no reaction; (d) \( \mathrm{H_2PO_4^-} + \mathrm{H_2O} \rightarrow \mathrm{H_3PO_4} + \mathrm{OH}^- \).

Step 1: Understanding Proton Transfer

When a base is added to water, it can accept a proton from a water molecule. This results in the formation of hydroxide ions, \( \mathrm{OH}^- \), and the conjugate acid of the base.

Step 2: Proton Transfer for \( \mathrm{HSO}_4^- \)

\( \mathrm{HSO}_4^- \) acts as a base by accepting a proton from water: \[ \mathrm{HSO}_4^- + \mathrm{H_2O} \rightleftharpoons \mathrm{H_2SO_4} + \mathrm{OH}^- \] Here, \( \mathrm{HSO}_4^- \) becomes \( \mathrm{H_2SO_4} \), and \( \mathrm{OH}^- \) is formed.

Step 3: Proton Transfer for \( \mathrm{CH}_3\mathrm{NH}_2 \)

\( \mathrm{CH}_3\mathrm{NH}_2 \) accepts a proton from water: \[ \mathrm{CH}_3\mathrm{NH}_2 + \mathrm{H_2O} \rightleftharpoons \mathrm{CH}_3\mathrm{NH}_3^+ + \mathrm{OH}^- \] \( \mathrm{CH}_3\mathrm{NH}_2 \) becomes \( \mathrm{CH}_3\mathrm{NH}_3^+ \), and \( \mathrm{OH}^- \) is formed.

Step 4: Proton Transfer for Iodide Ion (No Reaction)

Iodide ion, \( \mathrm{I}^- \), is the conjugate base of a strong acid (\( \mathrm{HI} \)), hence it does not accept protons in water. No reaction occurs.

Step 5: Proton Transfer for \( \mathrm{H_2PO_4^-} \)

\( \mathrm{H_2PO_4^-} \) can accept a proton from water: \[ \mathrm{H_2PO_4^-} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3PO_4} + \mathrm{OH}^- \] \( \mathrm{H_2PO_4^-} \) becomes \( \mathrm{H_3PO_4} \), and \( \mathrm{OH}^- \) is formed.

Key Concepts

Bronsted-Lowry acid-base theory

The Bronsted-Lowry acid-base theory is one of the most fundamental concepts in chemistry for understanding proton transfer reactions. According to this theory, acids are substances that donate protons (\( \mathrm{H}^+ \)) while bases are substances that accept protons. This definition is broader than that of the Arrhenius theory, allowing substances in a wider range of environments to act as acids or bases.

This theory is exemplified when a base such as hydroxide ion (\( \mathrm{OH}^- \)) is formed as a base accepts a proton from water. Comparing it to Lewis theory, which defines acids and bases in terms of electron pair donation, Bronsted-Lowry focuses specifically on proton transfer.

Understanding this theory aids in predicting the behavior of acids and bases, even in complex reactions, and is crucial for grasping the nature of equilibrium in acid-base chemistry.

Conjugate acid-base pair

A conjugate acid-base pair consists of two species that transform into each other by the gain or loss of a proton. In a Bronsted-Lowry acid-base reaction, the acid donates a proton to form its conjugate base, while the base accepts a proton to form its conjugate acid.

For instance, when \( \mathrm{HSO}_4^- \) acts as a base and accepts a proton from water, it forms \( \mathrm{H_2SO_4} \), its conjugate acid. Similarly, in the reaction involving \( \mathrm{CH}_3\mathrm{NH}_2 \), as it accepts a proton, it transitions to its conjugate acid, \( \mathrm{CH}_3\mathrm{NH}_3^+ \).

This pairing is crucial for maintaining chemical equilibrium in solutions and for understanding the behavior of substances in different environments. Conjugate pairs illustrate the reversible nature of acid-base interactions, emphasizing that every acid has a conjugate base, and every base has a conjugate acid.

Chemical equilibrium

Chemical equilibrium is the state in a chemical reaction where the concentrations of reactants and products remain constant over time. This condition occurs when the forward and reverse reactions take place at the same rate.

In the context of proton transfer reactions, such as those involving \( \mathrm{HSO}_4^- \) and \( \mathrm{CH}_3\mathrm{NH}_2 \), equilibrium is established when the rate of the base accepting a proton equals the rate of the conjugate acid donating a proton back to water. The equation reaches a point where no further net change is observed, illustrating the dynamic nature of equilibrium.

This balance is vital in many biological and chemical processes and influences reaction yield and direction. Understanding equilibrium enables chemists to manipulate conditions to favor either the forward or reverse reaction, highlighting the concept's importance in both theoretical and applied chemistry.

Bases in water

When bases are added to water, they participate in proton transfer reactions, often resulting in the production of hydroxide ions (\( \mathrm{OH}^- \)). This occurs because the base accepts a proton (\( \mathrm{H}^+ \)) from water (\( \mathrm{H_2O} \)), leaving behind \( \mathrm{OH}^- \).

For instance, when \( \mathrm{CH}_3\mathrm{NH}_2 \) is added to water, it accepts a proton forming \( \mathrm{CH}_3\mathrm{NH}_3^+ \) and leaving hydroxide ions in the solution, which can increase the pH. Conversely, iodide ion (\( \mathrm{I}^- \)), as a conjugate base of a strong acid, does not react with water to accept protons.

This behavior underscores the nature of bases in an aqueous environment and is essential for understanding pH changes, buffer systems, and chemical reactivity in aqueous solutions. The concept of bases in water also highlights the broader Bronsted-Lowry idea that bases are proton acceptors, as opposed to solely focusing on hydroxide ion production as per the Arrhenius definition.