Question

Arrange the following in decreasing order of acidity: ${\mathrm{ClCH}}_2\mathrm{COOH},{\mathrm{Cl}}_3\mathrm{CCOOH},{\mathrm{CH}}_3\mathrm{COOH},{\mathrm{Cl}}_2\mathrm{CHCOOH}$

Answer 1

Order of acidity:

• The decreasing order of acidity will be ${\mathrm{Cl}}_3\mathrm{CCOOH}>{\mathrm{Cl}}_2\mathrm{CHCOOH}>{\mathrm{ClCH}}_2\mathrm{COOH}>{\mathrm{CH}}_3\mathrm{COOH}$.

Reason:

• Due to the fact that they only partially ionise in solution, carboxylic acids are weak acids. At the same concentration, their solutions don't have any more hydrogen ions than a strong acid solution.
• The distribution of the negative charge increases with the amount of electron-withdrawing substituents, making the acid stronger as a result.
• The ${\mathrm{Cl}}_3\mathrm{CCOOH}$ has three chlorine atoms so it is having strong electron-withdrawing power so it will have more acidity and then ${\mathrm{Cl}}_2\mathrm{CHCOOH}$ has two chlorine atoms so it is as next and then ${\mathrm{ClCH}}_2\mathrm{COOH}$ has one chlorine atom as next and then the acid without any electron withdrawing group ${\mathrm{CH}}_3\mathrm{COOH}$
• The order of acidity will be as: ${\mathrm{Cl}}_3\mathrm{CCOOH}>{\mathrm{Cl}}_2\mathrm{CHCOOH}>{\mathrm{ClCH}}_2\mathrm{COOH}>{\mathrm{CH}}_3\mathrm{COOH}$.