Give one chemical test to distinguish between the following pairs of compound:
Formic acid and acetic acid

Hint: Find a reagent which gives precipitate with either formic acid or acetic acid but not with both. Recall that the formic acid carbonyl bond is also connected to a hydrogen atom.

Complete step by step answer:
One way to distinguish between formic acid and acetic acid is Tollen’s test. This test is also known as the silver mirror test. Formic acid gives Tollens test whereas acetic acid does not give this test. Tollen’s reagent is an ammoniacal silver nitrate solution. When formic acid is heated with Tollen’s reagent, a silver mirror is formed on the inner sides of the test tube. Acetic acid does not give this test.
Write the chemical reaction for the silver mirror test.

\[\begin{array}{l}
{\rm{HCOOH + 2 }}{\left[ {{\rm{Ag}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_2}} \right]^ + }{\rm{ + 2O}}{{\rm{H}}^ - } \to {\rm{ }}2{\rm{Ag}} \downarrow {\rm{ }} + {\rm{ C}}{{\rm{O}}_2} + {\rm{ 2}}{{\rm{H}}_2}{\rm{O + 4N}}{{\rm{H}}_3}\\
{\rm{Formic acid Tollen's reagent Silver mirror}}\\
{\rm{C}}{{\rm{H}}_3}{\rm{COOH + 2 }}{\left[ {{\rm{Ag}}{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}_2}} \right]^ + }{\rm{ + 2O}}{{\rm{H}}^ - } \to {\rm{ Silver mirror is NOT formed}}
\end{array}\]

 Tollen’s test is also given by aldehydes.

Formic acid and acetic acid can also be distinguished by \[{\rm{HgC}}{{\rm{l}}_2}\] test. Formic acid forms a white precipitate with \[{\rm{HgC}}{{\rm{l}}_2}\] whereas acetic acid does not form any precipitate with \[{\rm{HgC}}{{\rm{l}}_2}\].

\[\begin{array}{l}
{\rm{HCOOH + 2 HgC}}{{\rm{l}}_2} \to {\rm{ H}}{{\rm{g}}_2}{\rm{C}}{{\rm{l}}_2} \downarrow {\rm{ }} + {\rm{ C}}{{\rm{O}}_2} + {\rm{ 2HCl}}\\
{\rm{Formic acid Mercuric Mercurous}}\\
{\rm{ chloride chloride}}\\
{\rm{ }}\left( {{\rm{white ppt}}} \right){\rm{ }}\\
{\rm{C}}{{\rm{H}}_3}{\rm{COOH + 2 HgC}}{{\rm{l}}_2} \to {\rm{ No precipitate obtained}}
\end{array}\]

Note:
Please do not use the iodoform test which is characteristic of methyl ketones. In the Iodoform Test, we obtain a yellow precipitate in case of methyl ketones. This will not work in this situation.