Question

In Solvay ammonia process, sodium bicarbonate is precipitated due to:
A. presence of $N{H_3}$
B. reaction with $C{O_2}$
C. reaction with brine solution
D. reaction with $NaOH$

Answer 1

Hint: We can prepare washing soda by Solvay’s process.
The Solvay process is extensively used for industrial preparation of sodium carbonate (soda ash). The process is named after Ernest Solvay who developed the process during the$1860$.


Complete step by step solution:
Carbon dioxide reacts with the dissolved ammonia to form ammonia bicarbonate or ammonia hydrogen carbonate.
\[
  CaC{O_3}\left( {limestone} \right)\xrightarrow{{}}CaO + C{O_2} \\
  \mathop {N{H_3}}\limits_{Ammonia} + \mathop {C{O_2} + {H_2}O}\limits_{Carbon{\text{ dioxide}}} \xrightarrow{{}}\mathop {N{H_4}HC{O_3}}\limits_{Ammonia{\text{ bicarbonate}}} \\
 \]
Ammonia bicarbonate reacts with brine solution of $NaCl$ and forms sodium bicarbonate or sodium hydrogen carbonate.
$N{H_4}HC{O_3} + \mathop {NaCl}\limits_{Brine{\text{ solution}}} \xrightarrow{{}}\mathop {NaHC{O_3} + N{H_4}Cl}\limits_{Sodium{\text{ bicarbonate}}} $
Solubility of sodium bicarbonate is very less.
Sodium hydrogen carbonate crystal separates. These are heated to give sodium carbonate.
$2NaHC{O_3}\xrightarrow{{}}\underbrace {N{a_2}C{O_3} + {H_2}O + C{O_2} \uparrow }_{Wa\sin g{\text{ soda}}}$
But this process cannot be used to prepare potassium due to high solubility of potassium bicarbonate and hence ammonia bicarbonate to a saturated solution of potassium.
Sodium bicarbonate can be made in many ways but for reasons of economy most of it is now manufactured by the Solvay process. It has as its starting materials common salt $\left( {NaCl} \right)$ and limestone $\left( {CaC{O_3}} \right)$ both of which are cheap and easily available.
Natural brines occur underground, in salt lakes, or as sea water and are commercially important source of common salt $\left( {NaCl} \right)$ and other


Note: Solvay process uses salt $\left( {NaCl} \right)$, limestone $\left( {CaC{O_3}} \right)$, ammonia $\left( {N{H_3}} \right)$ and coke $\left( C \right)$ as the raw materials. Whenever we are solving reactions like these,the raw materials play a crucial role.