Question

Predict the stronger acid in each pair: (a) \(\mathrm{HNO}_{3}\) or HNO \(_{2}\) (b) \(\mathrm{H}_{2} \mathrm{~S}\) or \(\mathrm{H}_{2} \mathrm{O} ;\) ; (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) or \(\mathrm{H}_{2} \mathrm{SeO}_{4}\) (d) \(\mathrm{CH}_{3} \mathrm{COOH}\) or \(\mathrm{CCl}_{3}\) ?OOH.

Short answer

(a) \(HNO_3\) is the stronger acid. (b) \(H_{2}S\) is the stronger acid. (c) \(H_{2}SeO_{4}\) is the stronger acid. (d) \(CCl_{3}COOH\) is the stronger acid.

Step-by-step solution

(a) HNO\({}_{3}\) or HNO\({}_{2}\)

We will compare the stability of the resulting anions after donating a proton: - HNO\({}_{3}\) donates a proton, forming NO\({}_{3}^{-}\) anion. - HNO\({}_{2}\) donates a proton, forming NO\({}_{2}^{-}\) anion. The NO\({}_{3}^{-}\) anion is more stable due to resonance structures, which allow the negative charge to be delocalized over the entire anion. On the other hand, NO\({}_{2}^{-}\) has only one resonance structure and is less stable. Thus, \(HNO_3\) is the stronger acid.

(b) H\({}_{2}\)S or H\({}_{2}\)O

We will compare the strength of the O-H and S-H bonds, as well as the stability of the resulting anions: - H\({}_{2}\)O donates a proton, forming OH\(^{-}\) anion. - H\({}_{2}\)S donates a proton, forming HS\(^{-}\) anion. Sulfur is larger and less electronegative than oxygen, therefore the S-H bond is weaker than the O-H bond, making it easier for H\({}_{2}\)S to donate a proton. Additionally, HS\(^{-}\) is more stable due to the lower electronegativity and larger size of sulfur compared to oxygen. H\({}_{2}\)S is the stronger acid.

(c) H\({}_{2}\)SO\({}_{4}\) or H\({}_{2}\)SeO\(_{4}\)

We will compare the stability of the resulting anions after donating a proton: - H\({}_{2}\)SO\({}_{4}\) donates a proton, forming HSO\({}_{4}^{-}\) anion. - H\({}_{2}\)SeO\({}_{4}\) donates a proton, forming HSeO\({}_{4}^{-}\) anion. Selenium is larger and less electronegative than sulfur, resulting in weaker O-Se-H bonds compared to O-S-H bonds. Also, the HSeO\({}_{4}^{-}\) anion is more stable due to the larger size and lower electronegativity of selenium compared to sulfur. H\({}_{2}\)SeO\({}_{4}\) is the stronger acid.

(d) CH\({}_{3}\)COOH or CCl\(_{3}\)COOH

We will analyze the inductive effect of electron-withdrawing groups (EWGs) on the stability of the resulting anions: - CH\({}_{3}\)COOH donates a proton, forming CH\({}_{3}\)COO\(^{-}\) anion. - CCl\({}_{3}\)COOH donates a proton, forming CCl\({}_{3}\)COO\(^{-}\) anion. The CCl\({}_{3}\) group is a stronger electron-withdrawing group (EWG) than the CH\({}_{3}\) group due to the higher electronegativity of chlorine. The strong EWG effect of the CCl\({}_{3}\) group increases the stability of the CCl\({}_{3}\)COO\(^{-}\) anion, making it easier for CCl\({}_{3}\)COOH to donate a proton. CCl\({}_{3}\)COOH is the stronger acid.

Key concepts

Resonance Structures

Resonance structures help us understand the stability of molecules by showing different arrangements of electrons. In the case of acids, molecules like \(\text{HNO}_3\) can donate a proton to form an anion such as \(\text{NO}_3^-\). This anion can be represented by multiple resonance forms, meaning the negative charge can spread out across different atoms.

This dispersion of charge leads to increased stability. On the other hand, anions such as \(\text{NO}_2^-\) have fewer resonance structures. Less distribution of the negative charge makes them less stable.

Thus, acids that can form anions with more resonance structures are generally stronger, as seen with \(\text{HNO}_3\) being stronger than \(\text{HNO}_2\).

Inductive Effect

The inductive effect refers to the way electronegative atoms can influence electron distribution through sigma bonds. Electron-withdrawing groups (EWGs) create a strong inductive effect, pulling electrons toward themselves.

This effect stabilizes the anion formed when an acid donates a proton. It makes the anion more stable by spreading out the negative charge.

• For example, \(\text{CCl}_3\text{COOH}\) has chlorine atoms, which are highly electronegative EWGs.
• This increases the stability of the resulting \(\text{CCl}_3\text{COO}^-\) anion.

Due to this, \(\text{CCl}_3\text{COOH}\) is a stronger acid compared to \(\text{CH}_3\text{COOH}\), which has fewer electronegative elements.

Electron-Withdrawing Groups

Electron-withdrawing groups (EWGs) play a crucial role in determining acid strength. EWGs attract electrons towards themselves, enhancing the stability of anions formed after losing a proton.

When an acid has strong EWGs, the negative charge of the resulting anion is more evenly distributed. This stabilization makes the acid stronger. Consider \(\text{CCl}_3\) in \(\text{CCl}_3\text{COOH}\), which is a strong EWG due to chlorine's electronegativity.

In contrast, \(\text{CH}_3\text{COOH}\) lacks such EWGs and forms less stable anions, making it a weaker acid. The presence of strong EWGs directly correlates with increased acid strength.

Anion Stability

Anion stability is a key factor in determining the strength of an acid. The more stable the anion, the stronger the acid is likely to be. Anions achieve stability through various means such as resonance structures, inductive effect, and the presence of electron-withdrawing groups.

For instance, \(\text{HSO}_4^{-}\) and \(\text{HSeO}_4^{-}\) show differences in stability. Selenium in \(\text{HSeO}_4^{-}\), being less electronegative and larger than sulfur, offers unique stabilization. This allows the negative charge to be delocalized, enhancing stability.

Similarly, larger atoms like sulfur and selenium help disperse charge effectively, contributing to greater anion stability.