
When carbon dioxide gas is passed through lime water, lime water becomes milky. This is an example of:
A. Physical change
B. Chemical change
C. Reversible change
D. No change
Answer
593.7k+ views
Hint: Here the lime water is calcium hydroxide. And the calcium hydroxide gives a chemical reaction with carbon dioxide and gives white precipitate.
Step by step solution:
Lime water is calcium hydroxide. When carbon dioxide is passed through lime water then it gives a chemical reaction which yields a whitish or “milky” solution. This milkyness is due to the insoluble suspension of calcium carbonate formed. Below given is the reaction of sodium hydroxide and carbon dioxide:
\[Ca{{(OH)}_{2}}+C{{O}_{2}}\to CaC{{O}_{3}}+{{H}_{2}}O\]
Later on, that milkiness disappears leaving a clear liquid. We can observe that this reaction is a neutralization reaction. Where base is \[Ca{{(OH)}_{2}}\] and acid is \[{{H}_{2}}C{{O}_{3}}\] and a salt \[CaC{{O}_{3}}\] and \[{{H}_{2}}O\] are formed. \[C{{O}_{2}}\] is no more \[C{{O}_{2}}\] after dissolving in water. It becomes \[{{H}_{2}}C{{O}_{3}}\]via addition reaction:
\[C{{O}_{2}}+{{H}_{2}}O\to {{H}_{2}}C{{O}_{3}}\]
\[CaC{{O}_{3}}\] is insoluble in the water. However if excess carbon dioxide is passed, the milkiness disappears. This means that\[CaC{{O}_{3}}\]is more in the solution. Because excess of \[C{{O}_{2}}\] is the cause of the reaction for making calcium bicarbonate:
\[C{{O}_{2}}+{{H}_{2}}O\to {{H}_{2}}C{{O}_{3}}\]
\[CaC{{O}_{3}}+{{H}_{2}}C{{O}_{3}}\to Ca{{(HC{{O}_{3}})}_{2}}\]
All these changes in the solution are only because of chemical reactions, and properties of initial compounds disappear. So, this is a chemical change example.
So, the correct option is “B”.
Note: This reaction is an acid base reaction. And here \[{{H}_{2}}C{{O}_{3}}\] act as an acid and \[C{{O}_{3}}^{2-}\] act as base or electron accepting.
Step by step solution:
Lime water is calcium hydroxide. When carbon dioxide is passed through lime water then it gives a chemical reaction which yields a whitish or “milky” solution. This milkyness is due to the insoluble suspension of calcium carbonate formed. Below given is the reaction of sodium hydroxide and carbon dioxide:
\[Ca{{(OH)}_{2}}+C{{O}_{2}}\to CaC{{O}_{3}}+{{H}_{2}}O\]
Later on, that milkiness disappears leaving a clear liquid. We can observe that this reaction is a neutralization reaction. Where base is \[Ca{{(OH)}_{2}}\] and acid is \[{{H}_{2}}C{{O}_{3}}\] and a salt \[CaC{{O}_{3}}\] and \[{{H}_{2}}O\] are formed. \[C{{O}_{2}}\] is no more \[C{{O}_{2}}\] after dissolving in water. It becomes \[{{H}_{2}}C{{O}_{3}}\]via addition reaction:
\[C{{O}_{2}}+{{H}_{2}}O\to {{H}_{2}}C{{O}_{3}}\]
\[CaC{{O}_{3}}\] is insoluble in the water. However if excess carbon dioxide is passed, the milkiness disappears. This means that\[CaC{{O}_{3}}\]is more in the solution. Because excess of \[C{{O}_{2}}\] is the cause of the reaction for making calcium bicarbonate:
\[C{{O}_{2}}+{{H}_{2}}O\to {{H}_{2}}C{{O}_{3}}\]
\[CaC{{O}_{3}}+{{H}_{2}}C{{O}_{3}}\to Ca{{(HC{{O}_{3}})}_{2}}\]
All these changes in the solution are only because of chemical reactions, and properties of initial compounds disappear. So, this is a chemical change example.
So, the correct option is “B”.
Note: This reaction is an acid base reaction. And here \[{{H}_{2}}C{{O}_{3}}\] act as an acid and \[C{{O}_{3}}^{2-}\] act as base or electron accepting.
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