Read the passage carefully and answer the Questions.
Carboxylic acids are the earliest organic compounds isolated from nature. These are also known as organic acids but are weaker acids as compared to mineral acids. However, they are more acidic than alcohols and phenols because of greater resonance stabilization of the parent acid and its conjugate base. The acidic strength is determined by the Ka, which is called the acidity constant. The greater the value of Ka, the greater the tendency of the acid to ionize and hence the stronger the acid. The acidic strength of the saturated aliphatic carboxylic acid depends upon the inductive effect and its position with respect to - COOH groups. Electron-donating substituents tend to decrease whereas electron -withdrawing substituents tend to increase the acidic strength. The acidic strength of the aromatic carboxylic acid, on the other hand depends upon both the inductive effect and the resonance effect of the substituents.
How can we express the (acidity constant) '\( K_a \)' of RCOOH?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(K_a = [RCOO^-][H_3O^+]/ [RCOOH]\)
The acidity constant is the equilibrium constant for the ionisation of the acid in water. A carboxylic acid gives a proton to water, forming the carboxylate ion and the hydronium ion:
\( \displaystyle \ce{RCOOH + H2O <=> RCOO- + H3O+} \)
As in any equilibrium constant, the products are in the numerator and the un-ionised acid in the denominator. Water is the solvent, present in large excess, so its practically constant concentration is taken into the constant.
A larger Ka means a greater tendency to ionise and so a stronger acid. Acid strength is usually quoted as pKa = −log Ka: the smaller the pKa, the stronger the acid.
Why the other options are incorrect
Option (1) \(K_a = [RCOO^-] [RCOOH] / [H_3O^+]\) — Incorrect. It puts the un-ionised acid (a reactant) in the numerator and one product in the denominator.
Option (3) \(K_a = [H_3O^+] [RCOOH]/[RCOO^-]\) — Incorrect. Here the acid and its conjugate base have changed places; the carboxylate ion, a product, belongs in the numerator.
Option (4) \(K_a = 1/ [RCOOH][RCOO^-]\) — Incorrect. It has no hydronium ion at all and multiplies a reactant by a product, so it does not describe the ionisation equilibrium.
NCERT (2026-27 edition):
Reactions with Metals and Alkalies – …
