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.

Step by step solution

01

(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: \[ \mathrm{H}_{2} \mathrm{O} > \mathrm{H}_{2} \mathrm{S} > \mathrm{H}_{2}\mathrm{Se} \]

02

(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: \[ \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} < \mathrm{FCH}_{2} \mathrm{CO}_{2} \mathrm{H} < \mathrm{F}_{2} \mathrm{CHCO}_{2} \mathrm{H} < \mathrm{F}_{3} \mathrm{CCO}_{2} \mathrm{H} \]

03

(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: \[ \mathrm{NH}_{4}^{+} < \mathrm{HONH}_{3}^{+} \]

04

(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: \[ \mathrm{NH}_{4}^{+} > \mathrm{PH}_{4}^{+} \]

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