Look at Figure and answer the following questions (a) What change would you observe in the calcium hydroxide solution taken in tube B? (b) Write the reaction involved in test tubes A and B respectively. (c) If ethanol is given instead of ethanoic acid, would you expect the same change? (d) How can a solution of lime water be prepared in the laboratory?

Carbon dioxide gas Ethanoic acid Calcium hydroxide solution 

A.) Carbonation reaction:-

1. Calcium hydroxide is actually sparingly soluble in water and is producing an alkaline solution which is known as limewater.
2. So, when Carbon dioxide gas will get passed through limewater there will be a change to milky form because of the formation of Calcium carbonate $\left({\mathrm{CaCO}}_3\right)$
3. Reaction :- $\underset{\mathrm{Calcium}\mathrm{hydroxide}}{\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\left(\mathrm{aq}\right)}+\underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Calcium}\mathrm{carbonate}}{{\mathrm{CaCO}}_3\left(\mathrm{s}\right)\downarrow }+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

B.) Acid-Base Reaction:-

1. It is actually a chemical reaction which will lead to the formation of salt and water with the evolution of Carbon dioxide gas (${\mathrm{CO}}_2$)
2. When Carbon dioxide gas reacts with a base it shows acidic character because of the formation of the respective salt and water.
3. Therefore, Reaction is Test Tube A: $\underset{\mathrm{Acetic}\mathrm{acid}}{{\mathrm{CH}}_3\mathrm{COOH}\left(\mathrm{l}\right)}+\underset{\mathrm{Sodium}\mathrm{bicarbonate}}{{\mathrm{NaHCO}}_3\left(\mathrm{s}\right)}\to \underset{\mathrm{Sodium}\mathrm{acetate}}{{\mathrm{CH}}_3\mathrm{COONa}\left(\mathrm{aq}\right)+}\underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)\uparrow }+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$
4. Reaction in Test Tube B: $\underset{\mathrm{Calcium}\mathrm{hydroxide}}{\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\left(\mathrm{aq}\right)}+\underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Calcium}\mathrm{carbonate}}{{\mathrm{CaCO}}_3\left(\mathrm{s}\right)\downarrow }+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

C.) Reaction with Ethanol:-

1. Sodium bicarbonate $\left({\mathrm{NaHCO}}_3\right)$ is a weak base as it is formed by the reaction of a weak base and a weak acid.
2. Reaction :- $\underset{\mathrm{Carbonic}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{CO}}_3\left(\mathrm{aq}\right)}+\underset{\mathrm{Sodium}\mathrm{hydroxide}}{2\mathrm{NaOH}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Sodium}\mathrm{Carbonate}}{{\mathrm{Na}}_2{\mathrm{CO}}_3\left(\mathrm{aq}\right)}+\underset{\mathrm{Water}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$
3. Ethanol also can act as a weak acid/base like that of water.
4. As, weak acids and bases don't react with each other.
5. Therefore, Ethanol (${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$) will not react with Sodium bicarbonate (${\mathrm{Na}}_2{\mathrm{CO}}_3$), so same change can not be observed.

D.) Preparation of Lime Water solution :-

1. In a beaker , just take distilled water and mixing of Calcium carbonate powder $\left({\mathrm{CaCO}}_3\right)$ will have to take place.
2. So, allow the solution to settle down and decant the solution for obtaining Lime water $\mathrm{Ca}{\left(\mathrm{OH}\right)}_2$.