Question

$K_{2}CO_{3}$cannot be prepared by Solvay’s process because .

• A. $KHCO_{3}$ is less soluble than $NaHCO_{3}$
• B. $KHCO_{3}$ is too soluble to precipitated by KCl and $NH_{4}HCO_{3}$
• C. $K_{2}CO_{3}$is too soluble to be precipitated by KCl
• D. $K_{2}CO_{3}$ is less soluble than $Na_{2}CO_{3}.$

Answer 1

Answer: (B)

$KHCO_{3}$ is too soluble to precipitated by KCl and $NH_{4}HCO_{3}$

Answer 2

: $KHCO_{3}$ is too soluble to be precipitated by the addition of$NH_{4}HCO_{3}$ to the saturated solution of KCl.