Question

Sodium carbonate can be manufactured by Solvay's process but potassium carbonate cannot be prepared because:

• A. ${\mathrm{K}}_2{\mathrm{CO}}_3$ is more soluble
• B. ${\mathrm{K}}_2{\mathrm{CO}}_3$is less soluble
• C. ${\mathrm{KHCO}}_3$ is more soluble than ${\mathrm{NaHCO}}_3$
• D. ${\mathrm{KHCO}}_3$ is less soluble than ${\mathrm{NaHCO}}_3$

Answer 1

The correct option is C

${\mathrm{KHCO}}_3$ is more soluble than ${\mathrm{NaHCO}}_3$

The correct option is (C):

Explanation for the correct answer:

Solvay process:

1. Sodium carbonate is basically manufactured by the Solvay process.
2. It is actually a process in which the Carbon dioxide and ammonia are passed into a cold saturated solution of the Sodium chloride.
3. In the occurring reaction the Sodium Hydrogen carbonate is formed and is only very slightly soluble and in the presence of the Sodium ions, and further is completely precipitated.
4. This is further removed by the filtration and ignited to produce the Sodium carbonate.
5. So basically the ingredients for this reaction would be salt brine $\left(\mathrm{NaCl}\right)$, Ammonia $\left({\mathrm{NH}}_3\right)$ and limestone $\left({\mathrm{CaCO}}_3\right)$ and so further in this process ${\mathrm{CaCl}}_2$ is an important by-product obtained.
6. Reactions: $$\begin{gathered} \underset{\mathrm{Ammonia}}{{\mathrm{NH}}_3\left(\mathrm{s}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Ammonium}\mathrm{carbonate}}{{\left({\mathrm{NH}}_4\right)}_2{\mathrm{CO}}_3\left(\mathrm{s}\right)} \\ \underset{\mathrm{Ammonium}\mathrm{carbonate}}{{\left({\mathrm{NH}}_4\right)}_2{\mathrm{CO}}_3\left(\mathrm{s}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}+\underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Ammonium}\mathrm{hydrogen}\mathrm{carbonate}}{2{\mathrm{NH}}_4{\mathrm{HCO}}_3} \end{gathered}$$
7. So, further the addition of the the common salt to the solution containing the ${{\mathrm{NH}}_4}^{+}\mathrm{and}{{\mathrm{HCO}}_3}^{-}$ would result in the formation of ${\mathrm{NaHCO}}_3$ which is actually least stable and further is filtered off.
8. Reactions: $$\begin{gathered} \underset{\mathrm{Ammonium}\mathrm{hydrogen}\mathrm{carbonate}}{{\mathrm{NH}}_4{\mathrm{HCO}}_3\left(\mathrm{s}\right)}+\underset{\mathrm{Sodium}\mathrm{chloride}}{\mathrm{NaCl}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Ammonium}\mathrm{chloride}}{{\mathrm{NH}}_4\mathrm{Cl}\left(\mathrm{aq}\right)}+\underset{\mathrm{Sodium}\mathrm{hydrogen}\mathrm{carbonate}}{{\mathrm{NaHCO}}_3\left(\mathrm{s}\right)\downarrow } \\ \underset{\mathrm{Sodium}\mathrm{hydrogen}\mathrm{carbonate}}{2{\mathrm{NaHCO}}_3\left(\mathrm{s}\right)}\stackrel{∆}{\to }\underset{\mathrm{Sodium}\mathrm{carbonate}}{{\mathrm{Na}}_2{\mathrm{CO}}_3\left(\mathrm{s}\right)}+\underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)} \end{gathered}$$
9. So, Solvay method can not be used for the preparation of Potassium carbonate as it is relatively soluble in water, unlike the Sodium carbonate the Potassium carbonate does not precipitate out.

Therefore, the correct option is (C): ${\mathrm{KHCO}}_3$ is more soluble than ${\mathrm{NaHCO}}_3$.