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On passing excess of CO(2) in lime water...

On passing excess of `CO_(2)` in lime water, its milky appearance disappears because-

A

Soluble `Ca(OH)_(2)` is formed

B

Soluble `Ca(HCO_(3))_(2)` is formed

C

Reaction becomes reversible

D

Calcium compound evaporated

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the disappearance of the milky appearance in lime water upon passing excess CO₂, we can break down the process into clear steps: ### Step-by-Step Solution: 1. **Understanding Lime Water**: - Lime water is a solution of calcium hydroxide (Ca(OH)₂) in water. 2. **Initial Reaction with CO₂**: - When carbon dioxide (CO₂) is passed through lime water, it reacts with calcium hydroxide to form calcium carbonate (CaCO₃): \[ \text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3 \downarrow + \text{H}_2\text{O} \] - Calcium carbonate is a white insoluble solid that precipitates out of the solution, causing the milky appearance. 3. **Passing Excess CO₂**: - When excess CO₂ is passed into the lime water, the previously formed calcium carbonate reacts with the additional CO₂: \[ \text{CaCO}_3 + \text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{Ca(HCO}_3)_2 \] - This reaction produces calcium bicarbonate (Ca(HCO₃)₂), which is soluble in water. 4. **Disappearance of the Milky Appearance**: - Since calcium bicarbonate is soluble, its formation leads to the dissolution of the precipitate (CaCO₃), resulting in the disappearance of the milky appearance in the solution. 5. **Conclusion**: - Therefore, the milky appearance disappears because soluble calcium bicarbonate (Ca(HCO₃)₂) is formed. ### Final Answer: The correct option is: **soluble Ca(HCO₃)₂ is formed**. ---
To solve the question regarding the disappearance of the milky appearance in lime water upon passing excess CO₂, we can break down the process into clear steps: ### Step-by-Step Solution: 1. **Understanding Lime Water**: - Lime water is a solution of calcium hydroxide (Ca(OH)₂) in water. 2. **Initial Reaction with CO₂**: ...
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