Potassium carbonate $(K_2CO_3)$ cannot be prepared by the Solvay process. Why?
(N/A) The Solvay process involves the reaction of $NH_3$,$CO_2$,and $NaCl$ to form $NaHCO_3$,which precipitates out because it is sparingly soluble in water.
In the case of potassium,$KHCO_3$ (potassium bicarbonate) is fairly soluble in water.
Therefore,it does not precipitate out of the solution,making it impossible to isolate the intermediate product required to produce $K_2CO_3$ via this method.
In the case of potassium,$KHCO_3$ (potassium bicarbonate) is fairly soluble in water.
Therefore,it does not precipitate out of the solution,making it impossible to isolate the intermediate product required to produce $K_2CO_3$ via this method.
Explore More
Similar Questions
The correct order of increasing solubility in water is
Medium
View SolutionAn ionic hydride of an alkali metal has significant covalent character and is almost unreactive towards oxygen and chlorine. This is used in the synthesis of other useful hydrides. Write the formula of this hydride. Write its reaction with $Al_2Cl_6$.
Medium
View SolutionWhy is $Li_{2}CO_{3}$ decomposed at a lower temperature whereas $Na_{2}CO_{3}$ at higher temperature?
Medium
View SolutionMagnesium wire is heated in the atmosphere of $CO_2$ because
Medium
View SolutionThe order of negative standard electrode potential values $(E^\circ_{M^+/M})$ for $Li$,$Na$,and $K$ is:
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo