Question

$K_2CO_3$ cannot be prepared by Solvay's process because

• A. $KHCO_3$ is less soluble than $NaHCO_3$
• B. $KHCO_3$ is more soluble to be precipitated by $KCl$ and $NH_4HCO_3$
• C. $K_2CO_3$ is more soluble to be precipitated by $KCl$
• D. $K_2CO_3$ is less soluble than $Na_2CO_3$

Answer 1

Answer: (B)

$KHCO_3$ is more soluble to be precipitated by $KCl$ and $NH_4HCO_3$

Answer 2

$KHCO_3$ is more soluble to be precipitated by the addition of $NH_4HCO_3$ to the saturated solution of $KCl$.