$\mathrm{Na}_2 \mathrm{CO}_3$ is prepared by Solvay process but $\mathrm{K}_2 \mathrm{CO}_3$ cannot be prepared by the same because

Question

$\mathrm{Na}_2 \mathrm{CO}_3$ is prepared by Solvay process but $\mathrm{K}_2 \mathrm{CO}_3$ cannot be prepared by the same because, $\mathrm{K}_2 \mathrm{CO}_3$ is highly soluble in $\mathrm{H}_2 \mathrm{O}$, $\mathrm{KHCO}_3$ is sparingly soluble, $\mathrm{KHCO}_3$ is appreciably soluble, $\mathrm{KHCO}_3$ decomposes

MARKS App · Accepted Answer

Hint: $\left(\mathrm{NH}_4\right) \mathrm{HCO}_3+\mathrm{KCl} \longrightarrow \mathrm{KHCO}_3(\mathrm{aq})+\mathrm{NH}_4 \mathrm{Cl}(\mathrm{aq})$ $\mathrm{KHCO}_3$ being appreciably soluble cant be isolated from reaction medium easily.

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