Question

If (X) turns lime water milky, then X may be:
This question has multiple correct options
(a)- $C{{O}_{2}}$
(b)- $S{{O}_{2}}$
(c)- $N{{O}_{2}}$
(d)- ${{O}_{2}}$

Answer 1

Lime water is a colorless liquid, which is a common name for diluted solution of Calcium hydroxide $Ca{{(OH)}_{2}}$ . Calcium hydroxide is moderately soluble in water, resulting in formation of an alkaline solution known as limewater.

Complete step by step answer:
Pure lime, or quicklime, refers to calcium oxide (CaO). It is a white solid with strongly basic properties. It reacts readily with water and produces slaked lime, calcium hydroxide. A significant amount of heat is dissipated in this reaction which on further reacting with water gives lime water.
The reaction between lime water, $\left( Ca{{(OH)}_{2}} \right)$ , and carbon dioxide $\left( C{{O}_{2}} \right)$ results in the formation of an insoluble solid called calcium carbonate $\left( CaC{{O}_{3}} \right)$ . The solution will turn milky as the resulting compound calcium carbonate is a white precipitate.
$Ca{{(OH)}_{2}}+\,C{{O}_{2}}\,\to CaC{{O}_{3}}+{{H}_{2}}O$

$S{{O}_{2}}$ can also turn lime water milky due to formation of calcium sulphite $\left( CaS{{O}_{3}} \right)$.
$Ca{{(OH)}_{2}}+\,S{{O}_{2}}\,\to CaS{{O}_{3}}+{{H}_{2}}O$

Both of these reactions can be used to identify the presence of carbon dioxide and sulphur dioxide respectively. If the solution turns cloudy, it is an indicator that then the unknown gas one of them.
So, the correct options are both (a) and (b).

Now, let us see what other two reactions will give when passed through lime water.
(c)- When lime water is treated with nitrogen dioxide $\left( N{{O}_{2}} \right)$ , nitrogen dioxide gets absorbed in the neutralization reaction with lime water and forms two salts of calcium, calcium nitrate $\left( Ca{{(N{{O}_{3}})}_{2}} \right)$ and calcium nitrite $Ca{{(N{{O}_{2}})}_{2}}$ . The reaction is given as: $2Ca{{(OH)}_{2}}+4N{{O}_{2}}\to Ca{{\left( N{{O}_{3}} \right)}_{2}}+Ca{{\left( N{{O}_{2}} \right)}_{2}}+2{{H}_{2}}O$
(d)- Oxygen does not react with lime water.

So, the correct answer is “Option A and B”.

Note: If you continue passing the carbon dioxide through the limewater another acid - base reaction occurs, resulting in the precipitate getting dissolved to give soluble calcium hydrogen carbonate. This removes the milky appearance of the solution.
$CaC{{O}_{3}}+C{{O}_{2}}+{{H}_{2}}O\to Ca{{(HC{{O}_{3}})}_{2}}$

Similarly, in the case of $S{{O}_{2}}\,$, the milky appearance goes away if we continue due to formation of calcium hydrogen sulphite which is soluble in water.
$CaS{{O}_{3}}+S{{O}_{2}}+{{H}_{2}}O\to Ca{{(HS{{O}_{3}})}_{2}}$