Solveeit Logo
Login

Question

Chemistry Question on Acids and Bases

A sodium salt on treatment with MgCl2 MgCl_{2} gives white precipitate on heating. The anion of sodium salt is

A

HCO3−HCO_{3}^{-}

B

CO32−CO_{3}^{2-}

C

NO3−NO_{3}^{-}

D

SO42−SO_{4}^{2-}

Answer

HCO3−HCO_{3}^{-}

Explanation

Solution

Since the sodium salt on treatment with MgCl2MgCl _{2} gives white ppt on heating, the salt must be sodium bicarbonate and the reactions are as follows 2NaHCO3 sodium bicarbonate +MgCl2→Mg(HCO3)2 magnesium bicarbonate +2NaCl\underset{\text { sodium bicarbonate }}{2 NaHCO _{3}}+ MgCl _{2} \rightarrow \underset{\text { magnesium bicarbonate }}{ Mg \left( HCO _{3}\right)_{2}}+2 NaCl Mg(HCO3)2→ΔMgCO3 white ppt ↓+H2O+CO2Mg \left( HCO _{3}\right)_{2} \xrightarrow{\Delta} \underset{\text { white ppt }}{ MgCO _{3}} \downarrow+ H _{2} O + CO _{2}