• # question_answer 74)   In the Solvay process, can we obtain sodium carbonate directly by treating the solution containing ${{(N{{H}_{4}})}_{2}}C{{O}_{3}}$ with sodium chloride? Explain.

In solvay process, when $C{{O}_{2}}$ is passed through ammonical sodium chloride solution, the sparingly soluble $NaHC{{O}_{3}}$ is formed which on heating gives$N{{a}_{2}}C{{O}_{3}}$. $NaCl+N{{H}_{3}}+{{H}_{2}}O+C{{O}_{2}}\to NaHC{{O}_{3}}+N{{H}_{4}}Cl$$2NaHC{{O}_{3}}\xrightarrow{Heat}N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O+C{{O}_{2}}$ The solution which reacts with $NaCl$ consists $N{{H}_{4}}HC{{O}_{3}}$ and not $(N{{H}_{4}})C{{O}_{3}}.$ $N{{H}_{3}}+{{H}_{2}}O+C{{O}_{2}}\to N{{H}_{4}}HC{{O}_{3}}$ Thus, in solvay process, Na2C03 cannot be obtained directly.