The weakest acid among those given is (a) \(\mathrm{Cl}_{3} \mathrm{C}-\mathrm{COOH}\) (b) \(\mathrm{Br} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{COOH}\) (d) \(\mathrm{FCH}_{2} \mathrm{COOH}\)

We have four given acids: (a) \(\mathrm{Cl}_{3} \mathrm{C}-\mathrm{COOH}\) (b) \(\mathrm{Br} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{COOH}\) (d) \(\mathrm{FCH}_{2} \mathrm{COOH}\)

We need to analyze the inductive effect, which is the effect of electron-withdrawing or electron-donating groups, in each of the given acids. This helps us understand the stabilization of the conjugate base formed by the loss of a hydrogen ion from the acid. (a) In \(\mathrm{Cl}_{3}\mathrm{C}-\mathrm{COOH}\), the \(\mathrm{Cl}\) atoms are electron-withdrawing, which means they can stabilize the negative charge on the conjugate base after the acid donates its hydrogen ion. (b) In \(\mathrm{Br}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{COOH}\), the \(\mathrm{Br}\) atom is also electron-withdrawing but less effective than \(\mathrm{Cl}\). Also, it is further away from the carboxyl group. (c) In \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{C}-\mathrm{COOH}\), the \(\mathrm{CH}_{3}\) groups are electron-donating, which destabilizes the negative charge on the conjugate base. (d) In \(\mathrm{FCH}_{2}\mathrm{COOH}\), the \(\mathrm{F}\) atom is electron-withdrawing and the most electronegative element, which helps stabilize the negative charge on the conjugate base.

Based on the analysis, we can determine that the weakest acid is the one with the least ability to stabilize the negative charge on its conjugate base after losing a hydrogen ion. In this case, it is the acid with electron-donating groups: (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{COOH}\)