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Aldehydes, Ketones and Carboxylic Acids - Acidity of carboxylic acids

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Carboxylic acids are acidic because they undergo ionization in aqueous solution to produce a hydronium ion (H3O+H_3O^+) and a carboxylate ion (RCOO−RCOO^-). The acidity is primarily due to the resonance stabilization of the carboxylate ion, where the negative charge is delocalized over two electronegative oxygen atoms.

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The strength of the acid is measured by the acid dissociation constant (KaK_a) or the pKapK_a value. A higher KaK_a value or a lower pKapK_a value indicates a stronger acid.

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Carboxylic acids are stronger acids than alcohols and phenols. Unlike the phenoxide ion, the negative charge in the carboxylate ion is shared by two equivalent oxygen atoms, making it more stable than the phenoxide ion where the charge is delocalized over less electronegative carbon atoms in the benzene ring.

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Electron Withdrawing Groups (EWG) such as −Cl-Cl, −NO2-NO_2, and −CN-CN increase acidity by stabilizing the carboxylate ion through the −I-I (inductive) effect and dispersing the negative charge.

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Electron Donating Groups (EDG) such as alkyl groups (−CH3-CH_3, −C2H5-C_2H_5) decrease acidity by destabilizing the carboxylate ion through the +I+I effect, which intensifies the negative charge on the oxygen.

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The effect of substituents decreases as the distance between the substituent and the carboxyl group increases (distance-dependent inductive effect).

📐Formulae

RCOOH+H2O⇌RCOO−+H3O+RCOOH + H_2O \rightleftharpoons RCOO^- + H_3O^+

Ka=[RCOO−][H3O+][RCOOH]K_a = \frac{[RCOO^-][H_3O^+]}{[RCOOH]}

pKa=−log⁡KapK_a = -\log K_a

Acidity∝−I effect∝1+I effect\text{Acidity} \propto -I \text{ effect} \propto \frac{1}{+I \text{ effect}}

💡Examples

Problem 1:

Arrange the following in increasing order of acidity: CH3COOHCH_3COOH, ClCH2COOHClCH_2COOH, Cl2CHCOOHCl_2CHCOOH, and Cl3CCOOHCl_3CCOOH.

Solution:

CH3COOH<ClCH2COOH<Cl2CHCOOH<Cl3CCOOHCH_3COOH < ClCH_2COOH < Cl_2CHCOOH < Cl_3CCOOH

Explanation:

Chlorine is an electron-withdrawing group (−I-I effect). As the number of chlorine atoms increases on the α\alpha-carbon, the stabilization of the carboxylate ion increases due to greater dispersal of the negative charge, thereby increasing the acidity.

Problem 2:

Between HCOOHHCOOH and CH3COOHCH_3COOH, which is a stronger acid and why?

Solution:

HCOOHHCOOH is stronger than CH3COOHCH_3COOH.

Explanation:

In CH3COOHCH_3COOH, the methyl group (−CH3-CH_3) exerts a +I+I (electron-donating) effect, which increases the electron density on the carboxylate ion, making it less stable compared to the formate ion (HCOO−HCOO^-) where no such alkyl group is present.

Problem 3:

Which is more acidic: p−nitrobenzoicp-nitrobenzoic acid or benzoic acid?

Solution:

p−nitrobenzoicp-nitrobenzoic acid is more acidic.

Explanation:

The nitro group (−NO2-NO_2) is a strong electron-withdrawing group. It exerts both −I-I and −R-R (resonance) effects at the para position, which stabilizes the p−nitrobenzoatep-nitrobenzoate ion more effectively than the benzoate ion.

Acidity of carboxylic acids - Revision Notes & Key Formulas | ICSE Class 12 Chemistry