Question

For each conjugate acid-base pair, identify the first species as an acid or a base and the second species as its conjugate acid or conjugate base. In addition, draw Lewis structures for each species, showing all valence electrons and any formal charges. (a) \(\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{HSO}_{4}^{-}\) (b) \(\mathrm{NH}_{3^{\prime}} \mathrm{NH}_{2}^{-}\) (c) \(\mathrm{CH}_{3} \mathrm{OH}, \mathrm{CH}_{3} \mathrm{O}^{-}\)

Short answer

Question: Identify the conjugate acid-base pairs and draw the Lewis structures for the following species: a) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) / \(\mathrm{HSO}_{4}^{-}\) b) \(\mathrm{NH}_{3}\) / \(\mathrm{NH}_{2}^{-}\) c) \(\mathrm{CH}_{3} \mathrm{OH}\) / \(\mathrm{CH}_{3} \mathrm{O}^{-}\) Answer: a) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is the acid, \(\mathrm{HSO}_{4}^{-}\) is its conjugate base. Lewis structures: \(\mathrm{H}_{2} \mathrm{SO}_{4}\): Sulfur atom with four outer orbitals: two double-bonded oxygens and two singly-bonded \(\mathrm{OH}\) groups. \(\mathrm{HSO}_{4}^{-}\): Sulfur atom with four outer orbitals: two double-bonded oxygens, one singly-bonded \(\mathrm{OH}\) group, and one singly-bonded oxygen with a negative charge. b) \(\mathrm{NH}_{3}\) is the base, \(\mathrm{NH}_{2}^{-}\) is its conjugate acid. Lewis structures: \(\mathrm{NH}_{3}\): Nitrogen atom bonded to three hydrogen atoms, with a lone pair of electrons on the nitrogen. \(\mathrm{NH}_{2}^{-}\): Nitrogen atom bonded to two hydrogen atoms, with a lone pair of electrons on the nitrogen, and a negative charge on the nitrogen. c) \(\mathrm{CH}_{3} \mathrm{OH}\) is the acid, \(\mathrm{CH}_{3} \mathrm{O}^{-}\) is its conjugate base. Lewis structures: \(\mathrm{CH}_{3} \mathrm{OH}\): Carbon atom bonded to three hydrogen atoms, single bond to an oxygen atom (with two lone pairs) which is bonded to a hydrogen atom. \(\mathrm{CH}_{3} \mathrm{O}^{-}\): Carbon atom bonded to three hydrogen atoms, single bond to an oxygen atom (with two lone pairs), and a negative charge on the oxygen.

Step-by-step solution

(a) Identify conjugate acid-base pair and their species

Firstly, let's look at \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{HSO}_{4}^{-}\). Notice that \(\mathrm{HSO}_{4}^{-}\) is formed by the loss of a proton (H\(^+\)) from \(\mathrm{H}_{2} \mathrm{SO}_{4}\). This makes \(\mathrm{H}_{2} \mathrm{SO}_{4}\) an acid, and \(\mathrm{HSO}_{4}^{-}\) its conjugate base.

(a) Draw Lewis structures for each species

Next, we will draw the Lewis structures for each species. For \(\mathrm{H}_{2} \mathrm{SO}_{4}\), draw a sulfur atom with four outer orbitals containing two double-bonded oxygens and two singly bonded \(\mathrm{OH}\) groups. For the \(\mathrm{HSO}_{4}^{-}\) ion, draw a similar structure, but remove one hydrogen from an \(\mathrm{OH}\) group and add a negative charge to the remaining oxygen atom.

(b) Identify conjugate acid-base pair and their species

Now let's look at the second pair: \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{2}^{-}\). Notice that \(\mathrm{NH}_{2}^-\) is formed by the addition of a proton (H\(^+\)) to \(\mathrm{NH}_{3}\). This makes \(\mathrm{NH}_{3}\) a base, and \(\mathrm{NH}_{2}^{-}\) its conjugate acid.

(b) Draw Lewis structures for each species

For \(\mathrm{NH}_{3}\), draw a nitrogen atom bonded to three hydrogen atoms, with a lone pair of electrons on the nitrogen. For the \(\mathrm{NH}_{2}^{-}\) ion, draw the same structure but with only two hydrogen atoms bonded to the nitrogen, and add a negative charge to the nitrogen atom.

(c) Identify conjugate acid-base pair and their species

Finally, let's look at the third pair: \(\mathrm{CH}_{3} \mathrm{OH}\) and \(\mathrm{CH}_{3} \mathrm{O}^{-}\). Notice that \(\mathrm{CH}_{3} \mathrm{O}^{-}\) is formed by the loss of a proton (H\(^+\)) from \(\mathrm{CH}_{3} \mathrm{OH}\). This makes \(\mathrm{CH}_{3} \mathrm{OH}\) an acid, and \(\mathrm{CH}_{3} \mathrm{O}^{-}\) its conjugate base.

(c) Draw Lewis structures for each species

For \(\mathrm{CH}_{3} \mathrm{OH}\), draw a carbon atom bonded to three hydrogen atoms, and a single bond to an oxygen atom, which also has two lone pairs and is bonded to a hydrogen atom. For the \(\mathrm{CH}_{3} \mathrm{O}^{-}\) ion, draw the same structure but remove the hydrogen from the oxygen atom and add a negative charge to the oxygen atom.

Key concepts

Lewis Structures

Understanding Lewis structures is fundamental to the study of chemistry as they provide a visual representation of the bonds between atoms in a molecule as well as any lone pairs of electrons. A Lewis structure is essentially a diagram that shows the individual valence electrons as dots surrounding the chemical symbols of the elements, indicating how atoms are bonded in a molecule.

When creating a Lewis structure, one must know the total number of valence electrons available for bonding, which is crucial to ensure all atoms achieve a stable electron configuration. Valence electrons are placed around atoms such that each atom, if possible, follows the octet rule, having eight electrons in its valence shell. However, there are exceptions, such as hydrogen (which follows the duet rule with two electrons) and some transitional compounds. In the context of acids and bases, as seen in the provided exercise, drawing Lewis structures helps to identify where a proton is lost or gained, indicating the formation of a conjugate acid or base.

To improve comprehension, it's important to emphasize the step-by-step process of constructing the Lewis structure, starting from counting valence electrons to arranging them to satisfy the octet rule. Additionally, using simplified diagrams with clear indications of electron pairs can make the concept more accessible.

Valence Electrons

The concept of valence electrons is integral to understanding chemical bonding and reactions. Valence electrons are the electrons in an atom's outermost shell that are involved in forming bonds with other atoms. For main group elements, the number of valence electrons is directly related to the group number in the periodic table. For example, oxygen in group 16 has six valence electrons, while hydrogen in group 1 has one valence electron.

In the context of a Lewis structure, having a clear count of valence electrons is crucial because it determines how atoms bond within a compound and whether any formal charges are present. In the textbook exercise examples, knowing that sulfur has six valence electrons or that nitrogen has five is necessary when constructing Lewis structures and identifying acid-base pairs. Enhancements to student understanding can be made by mapping this concept to real-world examples and applications, such as how the valence electron count influences the acidity or basicity of a substance.

Formal Charges

The notion of formal charges helps chemists understand the distribution of electrons in a molecule and which structures are more stable. A formal charge is the hypothetical charge on an atom in a molecule if all bonding electrons are equally shared between the bonded atoms. It's calculated by subtracting the number of nonbonding electrons and half the number of bonding electrons from the number of valence electrons an atom has in its isolated state.

When determining the most accurate Lewis structure, the goal is typically to have formal charges that are as close to zero as possible. In the exercise, drawing Lewis structures while paying attention to formal charges helps clarify why certain species can act as acids or bases. For instance, a negative formal charge on oxygen in \(\mathrm{HSO}_4^{-}\) indicates a higher tendency to gain a proton, thus behaving as a base. To enhance student understanding, it is beneficial to relate the concept of formal charges to the stability of structures, emphasizing that the sum of the formal charges in a molecule should equal the molecule's overall charge.