One mole of oxalic acid reacts with two moles of NaOH. What is the product?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Sodium oxalate
Carboxylic acids form salts with alkalies. Oxalic acid (ethanedioic acid), \( \ce{HOOC-COOH} \), has two –COOH groups, and each one neutralises one NaOH. With two moles of NaOH both groups react:
\( \displaystyle \ce{HOOC-COOH + 2NaOH -> NaOOC-COONa + 2H2O} \)
The product is sodium oxalate.
Why the other options are incorrect
Option (2) Dimethyl oxalate — Incorrect. This is an ester; it forms when the acid is heated with methanol in the presence of an acid catalyst, not with NaOH.
Option (3) Oxamide — Incorrect. This is an amide; carboxylic acids give amides with ammonia (through the ammonium salt, on strong heating), not with NaOH.
Option (4) Sodium hydrogen oxalate — Incorrect. Here only one –COOH group is neutralised; that salt forms with one mole of NaOH, not two.
NCERT (2026-27 edition):
Reactions with metals and alkalies – “The carboxylic acids like alcohols evolve hydrogen with electropositive metals and form salts with alkalies similar to phenols.”