Question

$\text{ KOH }$ is preferably used to absorb $\text{ C}{{\text{O}}_{\text{2}}}\text{ }$ gas because:
A) $\text{ KHC}{{\text{O}}_{3}}\text{ }$ is soluble in water and $\text{ NaHC}{{\text{O}}_{3}}\text{ }$ is sparingly soluble in water.
B) $\text{ KOH }$ is cheaper than $\text{ NaOH }$
C) $\text{ KOH }$ is more soluble than $\text{ NaOH }$ in water
D) $\text{ KOH }$ is a stronger base than $\text{ NaOH }$

Answer 1

Hint: The acid and base react with each other and generate salt and water. This chemical reaction is known as the neutralization reaction. The general reaction is depicted as follows,
$\text{ Acid + Base }\to \text{ Salt + }{{\text{H}}_{\text{2}}}\text{O }$

Complete Solution :
- In a neutralization reaction, acid and base react with each other to form salt and water molecules. The nature of salt formed by the neutralization reaction depends on the strength of an acid.
- Carbon dioxide is an acidic oxide. When dissolved in water it forms carbonic acid. The potassium or sodium hydroxide is a strong base.
- When carbon dioxide is exposed to the potassium hydroxide they react with each other and undergo the neutralization reaction. the chemical reaction between the carbon dioxide and the potassium hydroxide is depicted as follows,
$\text{ 2KOH + C}{{\text{O}}_{2}}\to {{\text{K}}_{2}}\text{C}{{\text{O}}_{3}}\text{ + }{{\text{H}}_{2}}\text{O }$

- So, here the potassium hydroxide absorbs the carbon dioxide liberating in the reaction and forms the potassium carbonate $\text{ }{{\text{K}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$.
So, with the help of this reaction, we can estimate the percentage of carbon present in the organic compound.
- The potassium carbonate on further exposure to the carbon dioxide forms the potassium bicarbonate. The reaction of the formation of potassium bicarbonate is as shown below: $\text{ }{{\text{K}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ + C}{{\text{O}}_{2}}\text{ + }{{\text{H}}_{\text{2}}}\text{O }\to 2\text{KHC}{{\text{O}}_{3}}\text{ }$

- The potassium bicarbonate formed is soluble in water. The weight of the carbon dioxide can be calculated by the potassium hydroxide.
- The sodium hydroxide generates the sodium carbonate and on further exposure to the carbon dioxide forms the sodium bicarbonate. The reaction of the formation of sodium bicarbonate is as shown below,
$\text{ N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ + C}{{\text{O}}_{2}}\text{ + }{{\text{H}}_{\text{2}}}\text{O }\to 2\text{NaHC}{{\text{O}}_{3}}\text{ }$

- But, the $\text{ NaHC}{{\text{O}}_{3}}\text{ }$ cannot be used to absorb the carbon dioxide. Because, the sodium ion is smaller than the potassium ion. Larger the ion increases the solvation of ions. Thus, potassium carbonate is easily soluble in water.
- The $\text{ NaHC}{{\text{O}}_{3}}\text{ }$ is a sparingly soluble salt.
Thus, we can conclude that $\text{ KOH }$ is preferable to absorb $\text{ C}{{\text{O}}_{2}}\text{ }$ because $\text{ KHC}{{\text{O}}_{3}}\text{ }$ is soluble in water and $\text{ NaHC}{{\text{O}}_{3}}\text{ }$ is sparingly soluble in water.
So, the correct answer is “Option A”.

Note: Note that the potassium hydroxide has a wide application in The Liebig’s Method. The method is used for the estimation of elements in an organic compound. The organic compound is heated in presence of dry cupric oxide. The heating results in the oxidation of elements into $\text{ C}{{\text{O}}_{2}}\text{ }$ and $\text{ }{{\text{H}}_{\text{2}}}\text{O }$ .
- These obtained oxidized products are then absorbed by the caustic soda $\text{ KOH }$ and anhydrous $\text{ CaC}{{\text{l}}_{\text{2}}}\text{ }$ respectively. The mass adsorbed can be used to calculate the percentage of an element in the given known mass.