Question

Which of the following is the strongest acid?
(a) $C{{H}_{3}}COOH$
(b) $C{{H}_{2}}ClCOOH$
(c) $CHC{{l}_{2}}COOH$
(d) $CC{{l}_{3}}COOH$

Answer 1

Hint: Electron donating group decreases the acidic strength of the acid and the electron withdrawing group increases the acidic strength of the acid. Now on this basis, you can easily identify the strongest acid from the above given acids.

Complete step by step answer:
First of let’s discuss what is the strong acid. By the strong acid , we mean that acid which has a strong tendency to lose the proton.
Whenever an electron donating group is attached to the acid, the acidic character of that acid reduces because the tendency it loses the proton decreases and hence, such acid acts as weak acid.
On the other hand, whenever an electron withdrawing group is attached to the acid, the acidic character of that acid increases because the tendency to lose the proton increases and hence, such acid acts as strong acid.
Now considering the statement;
In option (a), there is a methyl group which is an electron donating group and hence, it is a weak acid.
In option (b), there is one electron withdrawing group i.e. $-Cl$ group is present and hence is stronger acid than the option(a).
In option (c), there are two electron withdrawing groups i.e. two $-Cl$ groups are present and hence is stronger acid than the both (a) and (b)
In option (d), there are three electron withdrawing groups i.e. three $-Cl$ groups are present and hence is the strongest acid of all.
So, thus from all the above given acids, $CC{{l}_{3}}COOH$ is the strongest acid.

Hence, option (d) is correct.

Note: The more the number of electron withdrawing groups attached to the acid, more is the dispersal of the charge and more is the tendency of the acid to lose the proton and hence, stronger is the acid. On the contrary, the more the number of electron donating groups attached to the acid, more is the accumulation of the charge and less is the tendency of the acid to lose the proton and hence, weaker is the acid.