Na_2CO_3 is prepared by the Solvay process, b | vedclass

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(C) In the Solvay process, $NaHCO_3$ is precipitated because it is sparingly soluble in water and can be easily separated by filtration. However, in t

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$KHCO_3$ is appreciably soluble

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(C) In the Solvay process, $NaHCO_3$ is precipitated because it is sparingly soluble in water and can be easily separated by filtration. However, in the case of potassium, the reaction is: $(NH_4)HCO_3 + KCl \longrightarrow KHCO_3(aq) + NH_4Cl(aq)$. Since $KHCO_3$ is appreciably soluble in water, it does not precipitate out of the solution. Therefore, it cannot be isolated from the reaction medium to be converted into $K_2CO_3$ by heating.

$Na_2CO_3$ is prepared by the Solvay process, but $K_2CO_3$ cannot be prepared by the same process because:

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