Question

Place the species in each of the following groups in order of increasing acid strength. a. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2} \mathrm{Se}\) (bond energies: \(\mathrm{H}-\mathrm{O}, 467 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{S}\), \(363 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{Se}, 276 \mathrm{~kJ} / \mathrm{mol})\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{FCH}_{2} \mathrm{CO}_{2} \mathrm{H}, \mathrm{F}_{2} \mathrm{CHCO}_{2} \mathrm{H}, \mathrm{F}_{3} \mathrm{CCO}_{2} \mathrm{H}\) c. \(\mathrm{NH}_{4}^{+}, \mathrm{HONH}_{3}{ }^{+}\) d. \(\mathrm{NH}_{4}{ }^{+}, \mathrm{PH}_{4}{ }^{+}\) (bond energies: \(\mathrm{N}-\mathrm{H}, 391 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{P}-\mathrm{H}, 322\) \(\mathrm{kJ} / \mathrm{mol}\) ) Give reasons for the orders you chose.

Short answer

The species can be ordered in terms of increasing acid strength as follows: a. \(\mathrm{H}_{2}\mathrm{O} < \mathrm{H}_{2}\mathrm{S} < \mathrm{H}_{2}\mathrm{Se}\) based on bond energies. b. \(\mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{H} < \mathrm{FCH}_{2}\mathrm{CO}_{2}\mathrm{H} < \mathrm{F}_{2}\mathrm{CHCO}_{2}\mathrm{H} < \mathrm{F}_{3}\mathrm{CCO}_{2}\mathrm{H}\) based on the electron-withdrawing effect of fluorine atoms. c. \(\mathrm{NH}_{4}^{+} < \mathrm{HONH}_{3}^{+}\) due to the electron-donating effect of oxygen in hydroxylammonium. d. \(\mathrm{NH}_{4}^{+} < \mathrm{PH}_{4}^{+}\) based on the different \(\mathrm{N-H}\) and \(\mathrm{P-H}\) bond energies.

Step-by-step solution

Group a: Water, Hydrogen Sulfide, and Hydrogen Selenide

To compare the acid strength of these species, we need to consider how easily they lose a proton (\(\mathrm{H}^{+}\)). A weaker \(\mathrm{H-X}\) bond indicates that \(\mathrm{H}^{+}\) can be lost more easily and the conjugate acid is stronger. Given bond energies: \(\mathrm{H-O}\) (467 kJ/mol), \(\mathrm{H-S}\) (363 kJ/mol), and \(\mathrm{H-Se}\) (276 kJ/mol). Following the bond energies, we can order these species in terms of increasing acid strength: \(\mathrm{H}_{2}\mathrm{O} < \mathrm{H}_{2}\mathrm{S} < \mathrm{H}_{2}\mathrm{Se}\).

Group b: Carboxylic Acids

In this case, we have to consider the electron-withdrawing effect of fluorine atoms. A greater number of fluorine atoms will pull electron density away from the acid's \(\mathrm{O-H}\) bond, making it weaker, and easier to lose the \(\mathrm{H}^{+}\) ion. We can order them in terms of increasing acid strength as: \(\mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{H} < \mathrm{FCH}_{2}\mathrm{CO}_{2}\mathrm{H} < \mathrm{F}_{2}\mathrm{CHCO}_{2}\mathrm{H} < \mathrm{F}_{3}\mathrm{CCO}_{2}\mathrm{H}\).

Group c: Ammonium Ion and Hydroxylammonium Ion

To compare the acidity of these two ions, we need to consider the electron-donating effect of the oxygen in hydroxylammonium. The presence of oxygen donates electron density to the N-H bond, making it weaker and easier to lose the \(\mathrm{H}^{+}\) ion. We can order these species in terms of increasing acid strength as: \(\mathrm{NH}_{4}^{+} < \mathrm{HONH}_{3}^{+}\).

Group d: Ammonium Ion and Phosphonium Ion

In this case, we need to compare the bond energies between nitrogen and hydrogen, and phosphorus and hydrogen. This will give us an idea of the relative ease of losing the \(\mathrm{H}^{+}\) ion. Given bond energies: \(\mathrm{N-H}\) (391 kJ/mol) and \(\mathrm{P-H}\) (322 kJ/mol). Following the bond energies, we can order these species in terms of increasing acid strength: \(\mathrm{NH}_{4}^{+} < \mathrm{PH}_{4}^{+}\).

Key concepts

Bond Energy

Understanding bond energy is crucial when evaluating the strength of an acid. The concept is simple: the bond energy between hydrogen and another atom (let's use the placeholder X for any element) indicates how strongly they are bonded. The lower the bond energy, the weaker the bond between H and X, hence easier for the compound to lose a hydrogen ion (H⁺). For example, if we look at the bond energies provided for Group a of the problem, we see: - H-O: 467 kJ/mol - H-S: 363 kJ/mol - H-Se: 276 kJ/mol
From this, we can conclude that - H₂O shows stronger bonding and thus is less willing to donate H⁺ compared to H₂S and H₂Se. Thus, the acid strength increases from water to hydrogen sulfide to hydrogen selenide. Whenever you want to predict which acid is stronger, consider the bond energy first. The smaller the energy, the stronger the acid.

Electron-withdrawing Effect

The electron-withdrawing effect is a powerful concept in determining acid strength, especially in organic molecules like carboxylic acids. An atomic group known as an electron-withdrawing group (EWG) can pull electron density away from nearby atoms. The classic example is fluorine in carboxylic acids. In Group b, we see a series of carboxylic acids with increasing numbers of fluorine atoms: - CH₃CO₂H - FCH₂CO₂H - F₂CHCO₂H - F₃CCO₂H
As the number of fluorine atoms increases, their electron-withdrawing capacity increases too. This has two effects: - Weakens the O-H bond in the carboxyl group, making it easier to release H⁺. - Stabilizes the resulting negative charge on the conjugate base. Thus, more fluorine atoms mean stronger electron withdrawal, leading to stronger acid strength in the series.

Conjugate Acid

A conjugate acid is what you get when a base gains a hydrogen ion (H⁺). Understanding the concept of conjugate acids is key when comparing the acidity of different compounds. For instance, when looking at Group c's ammonium ion - NH₄⁺ - HONH₃⁺
The presence of the O atom in HONH₃⁺, which donates electron density through the N-H bond, actually weakens it. This makes HONH₃⁺ a stronger acid compared to NH₄⁺ since it can lose H⁺ more readily, thus forming its conjugate base easier. The stability and reactivity of conjugate acids can greatly affect their strength, so always consider how easily the ion is formed and stabilized.

Carboxylic Acids

Carboxylic acids are organic acids characterized by the -COOH group. Their acidity is profoundly influenced by substituents on their carbon chain. In Group b, fluorine atoms serve as electron-withdrawing groups, drastically affecting acid strength. Carboxylic acids release H⁺ ions from the -OH group, becoming negatively charged carboxylates. The substituent's ability to stabilize this carboxylate ion influences how keenly the acid releases H⁺. More electronegative atoms or more electron-withdrawing groups make the -OH bond weaker, enhancing acidity. In essence, the acidity of a carboxylic acid depends on the substituent's ability to stabilize the carboxylate ion. - The more electron-withdrawing substituents, the stronger the acid. - Substituents reduce electron cloud density on the acidic H in -OH, favoring H⁺ release. Thus, as we add electron-withdrawing fluorine in compounds like FCH₂CO₂H and F₃CCO₂H, we render the molecules more acidic compared to a simple acetic acid like CH₃CO₂H.