${\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}$ can be manufactured by Solvay process but ${{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}$ cannot be prepared because:

${\rm{KHC}}{{\rm{O}}_3}$ is more soluble than ${\rm{NaHC}}{{\rm{O}}_3}$

${{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}$ is more soluble

${{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}$ is less soluble

${\rm{KHC}}{{\rm{O}}_3}$ is less soluble than ${\rm{NaHC}}{{\rm{O}}_3}$

NEET UG CHEMISTRY MCQ: S-BLOCK ELEMENTS

1 Hard

1 of 1

Understanding +4 | -1

\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}

can be manufactured by Solvay process but

{\mathrm{K}_2}\mathrm{C}{\mathrm{O}_3}

cannot be prepared because:

View answer and explanation

Correct option: C

\mathrm{KHC}{\mathrm{O}_3}

is more soluble than

\mathrm{NaHC}{\mathrm{O}_3}

Option C is correct because potassium bicarbonate (

\mathrm{KHCO_3}

) is more soluble than sodium bicarbonate (

\mathrm{NaHCO_3}

), making it difficult to precipitate

\mathrm{KHCO_3}

in the Solvay process.