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

a. ${\mathrm{H}}_2\mathrm{Se}<{\mathrm{H}}_2\mathrm{S}<{\mathrm{H}}_2\mathrm{O}$ (due to bond energies) b. ${\mathrm{CH}}_3{\mathrm{CO}}_2\mathrm{H}<{\mathrm{FCH}}_2{\mathrm{CO}}_2\mathrm{H}<{\mathrm{F}}_2{\mathrm{CHCO}}_2\mathrm{H}<{\mathrm{F}}_3{\mathrm{CCO}}_2\mathrm{H}$ (due to electron-withdrawing F atoms) c. ${\mathrm{NH}}_4^{+}<{\mathrm{HONH}}_3^{+}$ (due to O atom pulling electron density) d. ${\mathrm{PH}}_4^{+}<{\mathrm{NH}}_4^{+}$ (due to bond energies)

Step-by-step solution

(a. Rank species by acid strength)

We have ${\mathrm{H}}_2\mathrm{O},{\mathrm{H}}_2\mathrm{S},{\mathrm{H}}_2\mathrm{Se}$. Analyzing their bond energies, we notice that they decrease from H-O to H-Se. Since lower bond energy means it's easier to break the bond, and thereby release a proton, the species can be ordered as follows: $${\mathrm{H}}_2\mathrm{O}>{\mathrm{H}}_2\mathrm{S}>{\mathrm{H}}_2\mathrm{Se}$$

(b. Rank species by acid strength)

We have ${\mathrm{CH}}_3{\mathrm{CO}}_2\mathrm{H},{\mathrm{FCH}}_2{\mathrm{CO}}_2\mathrm{H},{\mathrm{F}}_2{\mathrm{CHCO}}_2\mathrm{H},{\mathrm{F}}_3{\mathrm{CCO}}_2\mathrm{H}$. As $\mathrm{F}$ is an electron-withdrawing group, it increases the stability of the conjugate base, which in turn increases the acid strength. The more F atoms attached, the higher the acid strength. Therefore, the order is: $${\mathrm{CH}}_3{\mathrm{CO}}_2\mathrm{H}<{\mathrm{FCH}}_2{\mathrm{CO}}_2\mathrm{H}<{\mathrm{F}}_2{\mathrm{CHCO}}_2\mathrm{H}<{\mathrm{F}}_3{\mathrm{CCO}}_2\mathrm{H}$$

(c. Rank species by acid strength)

We have ${\mathrm{NH}}_4^{+},{\mathrm{HONH}}_3^{+}$. Comparing these two species, we notice that ${\mathrm{HONH}}_3^{+}$ has an additional O atom, which is more electronegative and can pull electron density away from the N-H bond. This effect makes it easier to lose the proton, making it a stronger acid. The order is: $${\mathrm{NH}}_4^{+}<{\mathrm{HONH}}_3^{+}$$

(d. Rank species by acid strength)

We have ${\mathrm{NH}}_4^{+},{\mathrm{PH}}_4^{+}$. Analyzing their bond energies, we notice that N-H bond energy is higher than the P-H bond energy. Since lower bond energy means it's easier to break the bond and release a proton, the species can be ordered as follows: $${\mathrm{NH}}_4^{+}>{\mathrm{PH}}_4^{+}$$

Key concepts

Bond Energy

Bond energy is crucial when determining acid strength, especially in hydrides like ${\mathrm{H}}_2\mathrm{O}$, ${\mathrm{H}}_2\mathrm{S}$, and ${\mathrm{H}}_2\mathrm{Se}$. The bond energy refers to the amount of energy required to break a bond between two atoms. For acids, breaking this bond means releasing a proton (${\mathrm{H}}^{+}$). The lower the bond energy, the easier it is to release a proton, making the substance a stronger acid. For example, $\mathrm{H}-\mathrm{O}$ has a bond energy of 467 kJ/mol, whereas $\mathrm{H}-\mathrm{Se}$ is only 276 kJ/mol. Thus, ${\mathrm{H}}_2\mathrm{Se}$ is a stronger acid because it stands to release protons more readily due to its weaker bond. By understanding the concept of bond energy, one can predict and compare the relative acid strength of different molecules.

Electron-Withdrawing Groups

Electron-withdrawing groups (EWGs) play a significant role in determining the strength of an acid. These groups attract electrons towards themselves, which can stabilize the conjugate base of an acid by delocalizing the negative charge. Fluorine ($\mathrm{F}$) is a classic example of a strong electron-withdrawing group due to its high electronegativity. In the series ${\mathrm{CH}}_3{\mathrm{CO}}_2\mathrm{H},{\mathrm{FCH}}_2{\mathrm{CO}}_2\mathrm{H},{\mathrm{F}}_2{\mathrm{CHCO}}_2\mathrm{H},{\mathrm{F}}_3{\mathrm{CCO}}_2\mathrm{H}$, increasing the number of fluorine atoms leads to stronger acids.This happens because the more $\mathrm{F}$ atoms present, the better the stabilization of the conjugate base.Thus, ${\mathrm{F}}_3{\mathrm{CCO}}_2\mathrm{H}$ is the strongest acid as it has the most electron-withdrawing fluorine atoms stabilizing its conjugate base.

Conjugate Base Stability

The stability of the conjugate base directly impacts the acid strength. When an acid donates a proton, it becomes its conjugate base.The more stable this conjugate base is, the stronger the acid is considered. Stability often comes through resonance or electron-withdrawing effects that spread out or reduce the negative charge on the conjugate base.For instance, fluorinated carboxylic acids such as ${\mathrm{F}}_3{\mathrm{CCO}}_2\mathrm{H}$ derive their high acid strength from the stable conjugate base, ${\mathrm{F}}_3{\mathrm{CCO}}_2^{-}$.The negative charge is stabilized via the electronegative fluorine atoms mitigating charge concentration. Hence, acid strength is inherently tied to how well the conjugate base can spread out or stabilize its extra electron.

Electronegativity Effects

Electronegativity is the tendency of an atom to attract a bonding pair of electrons towards itself. Higher electronegativity often increases acid strength due to electronegative atoms pulling electron density and weakening competing bonds. Consider ${\mathrm{NH}}_4^{+}$ and ${\mathrm{HONH}}_3^{+}$. The oxygen atom in ${\mathrm{HONH}}_3^{+}$ is more electronegative than nitrogen and pulls electron density. This helps in bond weakening and makes it easier for the ${\mathrm{H}}^{+}$ ion to be released, rendering ${\mathrm{HONH}}_3^{+}$ a stronger acid compared to ${\mathrm{NH}}_4^{+}$.Electronegativity also explains trends in periodic tables, with acids becoming stronger as one moves across a period due to increasing electronegativity.