Question

Sodium salt of an organic acid ‘X’ produces effervescence with conc. H₂SO₄. ‘X’ reacts with the acidified aqueous CaCl₂ solution to give a white precipitate which decolourizes the acidic solution of KMnO₄. ‘X’ is :

• A. CH₃COONa
• B. Na₂C₂O₄
• C. C₆H₅COONa
• D. HCOONa

Answer 1

The correct option is B

Na₂C₂O₄

The correct answer is option B.

Explanation of correct answer :-

• The compound that produces effervescence on reaction with conc. Sulphuric acid $\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$ is Oxalic acid.
• So, the sodium salt of Oxalic acid given is ${\mathrm{N}}_2{\mathrm{C}}_2{\mathrm{O}}_4$ i.e. Sodium Oxalate.

Step 1 :- Evolution of effervescence.

• When Sodium oxalate reacts with conc. Sulphuric acid, it produces effervescence due to evolution of Carbon dioxide and Carbon monoxide gas from the solution.

$\underset{\mathrm{Sodium}\mathrm{oxalate}}{{\mathrm{Na}}_2{\mathrm{C}}_2{\mathrm{O}}_4}+\underset{\mathrm{conc}.\mathrm{Sulphuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4}\to \underset{\mathrm{Sodium}\mathrm{sulphate}}{{\mathrm{Na}}_2{\mathrm{SO}}_4}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}}+\underset{\mathrm{Carbon}\mathrm{monoxide}}{\mathrm{CO}\uparrow }+\underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\uparrow }$

Step 2 :- Formation of precipitate.

• Sodium oxalate reacts with calcified aqueous ${\mathrm{CaCl}}_2$ solution, it gives Calcium oxalate precipitates which is white in colour.

$\underset{\mathrm{Sodium}\mathrm{oxalate}}{{\mathrm{Na}}_2{\mathrm{CO}}_4}+\underset{\mathrm{Calcium}\mathrm{chloride}}{{\mathrm{CaCl}}_2}\to \underset{\mathrm{Calcium}\mathrm{oxalate}}{{\mathrm{CaC}}_2{\mathrm{O}}_4\downarrow }+\underset{\mathrm{Sodium}\mathrm{chloride}}{2\mathrm{NaCl}}$

Step 3 :- Decolorisation of solution in an acidic medium.

• Again, when this calcium oxalate is made to react with an acidic solution of ${\mathrm{MnO}}_4$_, it decolorizes the solution.

$5{\mathrm{C}}_2{\mathrm{O}}_4+16{\mathrm{H}}^{+}+2{\mathrm{MnO}}^{4-}\to 2{\mathrm{Mn}}^{2+}+10{\mathrm{CO}}_2+8{\mathrm{H}}_2\mathrm{O}$

Explanation of incorrect answers:-

Option A :

• Sodium acetate $\left({\mathrm{CH}}_3\mathrm{COONa}\right)$ on reaction with conc. Sulphuric acid $\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$, forms Sodium bisulphate $\left({\mathrm{NaHSO}}_4\right)$ displacing Acetic acid $\left({\mathrm{CH}}_3\mathrm{COOH}\right)$.

$\underset{\mathrm{cocn}.\mathrm{Sulphuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4}+\underset{\mathrm{Sodium}\mathrm{acetate}}{{\mathrm{CH}}_3\mathrm{COONa}}\to \underset{\mathrm{Sodium}\mathrm{bisulphate}}{{\mathrm{NaHSO}}_4}+\underset{\mathrm{Acetic}\mathrm{acid}}{{\mathrm{CH}}_3\mathrm{COOH}}$

Option C :

• Sodium benzoate $\left({\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}\right)$ reacts with conc. Sulphuric acid $\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$ , gives Benzoic acid$\left({\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}\right)$ displacing Sodium bisulphate $\left({\mathrm{NaHSO}}_4\right)$.

$\underset{\mathrm{Sodium}\mathrm{benzoate}}{{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}}+\underset{\mathrm{Sulphuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4}\to \underset{\mathrm{Benzoic}\mathrm{acid}}{{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}}+\underset{\mathrm{Sodium}\mathrm{bisulphate}}{{\mathrm{NaHSO}}_4}$

Option D :-

• When a conc. Sulphuric acid$\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$is made to react with Sodium formate$(\mathrm{HCOONa})$, it forms Formic acid $\left(\mathrm{HCOOH}\right)$ alongside with Sodium sulfate(Na₂SO₄).

$\underset{\mathrm{Sodium}\mathrm{formate}}{2\mathrm{HCOONa}}+\underset{\mathrm{Sulphuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4}\to \underset{\mathrm{Formic}\mathrm{acid}}{2\mathrm{HCCOH}}+\underset{\mathrm{Sodium}\mathrm{sulphate}}{{\mathrm{Na}}_2{\mathrm{SO}}_4}$