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\mathrm{Na}_{2} \mathrm{CO}_{3} can be manufactured by Solvay's process but \mathrm{K}_{2} \mathrm{CO}_{3} cannot be prepared because

Option: 1

\mathrm{K}_{2} \mathrm{CO}_{3}  is more soluble


Option: 2

\mathrm{K}_{2} \mathrm{CO}_{3} is less soluble


Option: 3

\mathrm{KHCO}_{3} is more soluble than \mathrm{NaHCO}_{3}


Option: 4

\mathrm{KHCO}_{3} is less soluble than \mathrm{NaHCO}_{3}


Answers (1)

best_answer

\mathrm{KHCO}_{3} is more soluble than \mathrm{NaHCO}_{3} because the electropositive character increases as we move down the group, the stability of bicarbonates increases in the same order.

\mathrm{NaHCO}_{3} is made by saturating a solution of sodium carbonate with carbon dioxide. The white crystalline powder of \mathrm{NaHCO}_{3},being less soluble, gets separated out. 

\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} \rightarrow 2\, \mathrm{NaHCO}_{3}

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