Question

$K_{2}C{O}_3$ cannot be prepared by Solvay's process because:

• A. $KHC{O}_3$ is less soluble than $NaHC{O}_3$
• B. $KHC{O}_3$ is too soluble to be precipitated by $KCl$ and $N{H}_{4}HC{O}_3$
• C.
  $K_{2}C{O}_3$ is insoluble to be precipitated by $KCl$
  
• D. $K_{2}C{O}_3$ is less soluble than $N{a}_{2}C{O}_3$

Answer 1

The correct option is B $KHC{O}_3$ is too soluble to be precipitated by $KCl$ and $N{H}_{4}HC{O}_3$

Solvay process cannot be used to prepare potassium carbonate. This is because, unlike sodium bicarbonate, potassium bicarbonate is fairly soluble in water and does not precipitate out i.e., $KHC{O}_3$ is too soluble to be precipitated by $KCl$ and $N{H}_{4}HC{O}_3$.