Problem 17 - Chemistry: An Atoms-Focused Approach

Question

Identify the conjugate base of each of the following compounds: \(\mathrm{HNO}_{2}, \mathrm{HClO}, \mathrm{H}_{3} \mathrm{PO}_{4},\) and \(\mathrm{NH}_{3}\)

Step-by-Step Solution

Answer

Answer: The conjugate bases of the given compounds are: - \(\mathrm{HNO}_{2}\): \(\mathrm{NO}_{2}^{-}\) - \(\mathrm{HClO}\): \(\mathrm{ClO}^{-}\) - \(\mathrm{H}_{3} \mathrm{PO}_{4}\): \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) - \(\mathrm{NH}_{3}\): \(\mathrm{NH}_{2}^{-}\)

Step 1: Identify the conjugate base of \(\mathrm{HNO}_{2}\)

To identify the conjugate base of \(\mathrm{HNO}_{2}\), we remove one \(\mathrm{H}^{+}\) ion. \(\mathrm{HNO}_{2} \rightarrow \mathrm{NO}_{2}^{-}\) The conjugate base of \(\mathrm{HNO}_{2}\) is \(\mathrm{NO}_{2}^{-}\).

Step 2: Identify the conjugate base of \(\mathrm{HClO}\)

To identify the conjugate base of \(\mathrm{HClO}\), we remove one \(\mathrm{H}^{+}\) ion. \(\mathrm{HClO} \rightarrow \mathrm{ClO}^{-}\) The conjugate base of \(\mathrm{HClO}\) is \(\mathrm{ClO}^{-}\).

Step 3: Identify the conjugate base of \(\mathrm{H}_{3} \mathrm{PO}_{4}\)

To identify the conjugate base of \(\mathrm{H}_{3} \mathrm{PO}_{4}\), we remove one \(\mathrm{H}^{+}\) ion. \(\mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) The conjugate base of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\).

Step 4: Identify the conjugate base of \(\mathrm{NH}_{3}\)

To identify the conjugate base of \(\mathrm{NH}_{3}\), we remove one \(\mathrm{H}^{+}\) ion. \(\mathrm{NH}_{3} \rightarrow \mathrm{NH}_{2}^{-}\) The conjugate base of \(\mathrm{NH}_{3}\) is \(\mathrm{NH}_{2}^{-}\). To conclude, the conjugate bases of the given compounds are: - \(\mathrm{HNO}_{2}\): \(\mathrm{NO}_{2}^{-}\) - \(\mathrm{HClO}\): \(\mathrm{ClO}^{-}\) - \(\mathrm{H}_{3} \mathrm{PO}_{4}\): \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) - \(\mathrm{NH}_{3}\): \(\mathrm{NH}_{2}^{-}\)

Key Concepts

Understanding Acid-Base Chemistry

Acid-base chemistry is a fundamental topic in chemistry that deals with the properties and reactions of acids and bases. An acid is a substance that donates hydrogen ions (\(\mathrm{H}^{+}\)), while a base is a substance that accepts them.
This concept is part of the Bronsted-Lowry acid-base theory, which is very useful in understanding chemical reactions.

• **Acids**: They release \(\mathrm{H}^{+}\) ions when dissolved in water.
• **Bases**: They accept \(\mathrm{H}^{+}\) ions, often producing hydroxide ions (\(\mathrm{OH}^{-}\)) in solutions.

A special concept within this is the conjugate acid-base pair. When an acid loses an \(\mathrm{H}^{+}\), it becomes its conjugate base, and when a base gains one, it becomes its conjugate acid.
This relationship helps chemists predict the behavior of substances in different reactions.

The Process of Proton Removal

Proton removal is key to identifying a conjugate base. It involves removing a hydrogen ion (\(\mathrm{H}^{+}\)) from an acid to form its conjugate base.
To find the conjugate base of a compound, simply subtract an \(\mathrm{H}^{+}\) ion from the molecular formula. This transition shows the essence of the acid's role in reactions.

Examples of Proton Removal

• For \(\mathrm{HNO}_{2}\), removing an \(\mathrm{H}^{+}\) results in \(\mathrm{NO}_{2}^{-}\).
• With \(\mathrm{HClO}\), removing an \(\mathrm{H}^{+}\) leads to \(\mathrm{ClO}^{-}\).
• In \(\mathrm{H}_{3}\mathrm{PO}_{4}\), removing an \(\mathrm{H}^{+}\) gives \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\).
• For \(\mathrm{NH}_{3}\), removing an \(\mathrm{H}^{+}\) results in \(\mathrm{NH}_{2}^{-}\).

This step is crucial in changing the acid to its conjugate base form, illustrating its potential to behave differently in reactions.

Chemical Equilibrium in Acid-Base Reactions

Chemical equilibrium occurs when the forward and reverse reactions occur at the same rate, leading to stable concentrations of reactants and products. In acid-base chemistry, this concept is important to understand how acids and bases interact.

Equilibrium in Acid-Base Reactions

During an acid-base reaction, an equilibrium can be reached:

• The acid donates a proton to the base, forming its conjugate base and conjugate acid.
• The reverse process can occur, where the conjugate base can accept a proton, reverting to the original acid.

For example, in the reaction between \(\mathrm{HCl}\) and water, \(\mathrm{HCl}\) donates a proton to form \(\mathrm{Cl}^{-}\) and \(\mathrm{H}_{3}\mathrm{O}^{+}\).
The chemical equilibrium law helps calculate equilibrium constants (K_a and K_b) to determine the strength of acids and bases.