Match the ${\mathbf{pK}}_{\mathbf{a}}$values to the appropriate compound. ${\mathbf{pK}}_{\mathbf{a}}$values: 0.28, 1.24, 2.66, 2.86, and 3.12. Compounds: (a) ${\mathbf{FCH}}_{\mathbf{2}}\mathbf{COOH}$; (b) ${\mathbf{CF}}_{\mathbf{3}}\mathbf{COOH}$; (c) ${\mathbf{F}}_{\mathbf{2}}\mathbf{CHCOOH}$; (d) ${\mathbf{ICH}}_{\mathbf{2}}\mathbf{COOH}$; (e) ${\mathbf{BrCH}}_{\mathbf{2}}\mathbf{COOH}$

Electron-withdrawing groups increase the acidity of carboxylic acids.

More the number of electron-withdrawing groups on an acid molecule, higher is its acidity.

${\mathbf{pK}}_{\mathbf{a}}$ is inversely related to the acidity of a compound, i.e., a strong acid has lower ${\mathbf{pK}}_{\mathbf{a}}$, while a weak acid has higher ${\mathbf{pK}}_{\mathbf{a}}$.

Among the halogen atoms, fluorine is the most electronegative (highly electron-withdrawing) and hence the carboxylic acids containing fluorine atoms are more acidic, followed by the one containing bromine and then the one containing iodine.

Higher the number of electron-withdrawing groups, higher is the acidity.

Therefore, the order of acidity of the given acids are:

${\mathrm{CF}}_3\mathrm{COOH}>{\mathrm{F}}_2\mathrm{CHCOOH}>{\mathrm{FCH}}_2\mathrm{COOH}>{\mathrm{BrCH}}_2\mathrm{COOH}>{\mathrm{ICH}}_2\mathrm{COOH}$

Therefore, the of the given acids are:

${\mathrm{CF}}_3\mathrm{COOH}\left({\mathrm{pK}}_{\mathrm{a}}=0.28\right),{\mathrm{F}}_2\mathrm{CHCOOH}\left({\mathrm{pK}}_{\mathrm{a}}=1.24\right),{\mathrm{FCH}}_2\mathrm{COOH}\left({\mathrm{pK}}_{\mathrm{a}}=2.66\right),{\mathrm{BrCH}}_2\mathrm{COOH}\left({\mathrm{pK}}_{\mathrm{a}}=2.86\right),{\mathrm{ICH}}_2\mathrm{COOH}\left({\mathrm{pK}}_{\mathrm{a}}=3.12\right)$