Question

Oxalic acid may be distinguished from tartaric acid by:
A. ${\text{NaHC}}{{\text{O}}_{\text{3}}}$
B. Ammonium silver nitrate
C. Litmus paper
D. phenolphthalein

Answer 1

Hint: Oxalic acid has the molecular formula ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ it is also the simplest dicarboxylic acid and the molecular formula of tartaric acid is ${{\text{C}}_{\text{4}}}{{\text{H}}_{\text{6}}}{{\text{O}}_{\text{6}}}$ . oxalic acid has the functional group of carboxylic acid (-COOH), and tartaric also has the same functional group (-COOH).

Complete Step by step answer: Now, as we have seen both tartaric acid and oxalic acid have the same functional group, tartaric acid also has a hydroxyl (-OH) group present in the adjacent carboxylic group, this is also called a $\alpha $hydroxyl group.
These hydroxyl groups will get oxidized and form corresponding ketones, the reagent used in Tollen’s test is a solution of silver, and it is called ammonium silver nitrate.
But since oxalic acid has only the carboxylic acid group, it does not oxidize.
Tollen’s test is used to distinguish between aldehyde and ketone, aldehyde and Tollens test, undergoes oxidation and forms corresponding carboxylic acid. And since it is a weak oxidizing agent it cannot oxidize ketones.
Therefore, we can use ammonium silver nitrate to distinguish oxalic acid and tartaric acid.

Hence, the correct answer is an option (B) i.e., Ammonium silver nitrate.

Note: Even though Tollen’s test does not oxidize ketone, it can however oxidize the $\alpha $hydroxyl part of a ketone. There is another exception, that formic acid (one carbon carboxylic acid), is also oxidized by the Tollen’s test, into carbon dioxide (${\text{C}}{{\text{O}}_2}$) and ${{\text{H}}_2}$