Problem 21
Question
If \(\mathrm{CO}_{2}\) is passed in excess through lime water, the milkiness first formed disappears due to (a) formation of water soluble calcium bicarbonate (b) the solution getting heated by exothermic reaction (c) reversal of the original reaction (d) formation of volatile calcium derivative
Step-by-Step Solution
Verified Answer
The milkiness disappears because calcium bicarbonate, which is water-soluble, is formed.
1Step 1: Formation of Calcium Carbonate
When \( ext{CO}_2\) is initially passed through lime water (which is a solution of calcium hydroxide, \( ext{Ca(OH)}_2\)), calcium carbonate \( ext{CaCO}_3\) is formed. This reaction causes the lime water to turn milky: \[ \ ext{Ca(OH)}_2 + ext{CO}_2
ightarrow ext{CaCO}_3 ext{ (milky) } + ext{H}_2 ext{O} \]
2Step 2: Reaction with Excess CO2
When excess \( ext{CO}_2\) is continuously bubbled through the solution, the initially formed calcium carbonate reacts further with \( ext{CO}_2\) and water to form calcium bicarbonate, \( ext{Ca(HCO}_3)_2\), which is soluble in water: \[ \ ext{CaCO}_3 + ext{CO}_2 + ext{H}_2 ext{O}
ightarrow ext{Ca(HCO}_3)_2 ext{ (aqueous) } \]
Key Concepts
Lime water reactionCalcium carbonateCalcium bicarbonateSolubilityExothermic reaction
Lime water reaction
When carbon dioxide \(\text{CO}_2\) is bubbled through lime water, a fascinating chemical reaction takes place. Lime water is essentially a solution of calcium hydroxide \(\text{Ca(OH)}_2\). Initially, as the carbon dioxide passes through, it reacts to form calcium carbonate \(\text{CaCO}_3\). This newly formed compound gives the lime water a milky appearance, symbolizing the completion of what's known as the lime water reaction.
Here's how the process works step-by-step:
Here's how the process works step-by-step:
- Carbon dioxide interacts with calcium hydroxide in the solution.
- This interaction leads to the formation of a white precipitate, calcium carbonate.
Calcium carbonate
Calcium carbonate \(\text{CaCO}_3\) is a key player in the lime water reaction. It's what causes the initial milkiness when carbon dioxide is added. This compound is a white, insoluble substance that forms as a result of the chemical reaction between lime water and carbon dioxide.
In terms of its role:
In terms of its role:
- Calcium carbonate precipitates out of the solution, creating a cloudy appearance.
- It marks the first significant step in the chain of reactions that can occur when \(\text{CO}_2\) is in the mix.
Calcium bicarbonate
Calcium bicarbonate \(\text{Ca(HCO}_3)_2\) represents the next phase of the reaction. Once you pass excess carbon dioxide through the lime water, the initially formed calcium carbonate reacts further. This reaction results in calcium bicarbonate, which is different from calcium carbonate in that it is soluble in water.
The steps include:
The steps include:
- Calcium carbonate continues to react with the additional \(\text{CO}_2\) and water.
- This process forms calcium bicarbonate, which remains dissolved in the solution.
Solubility
Solubility is a crucial concept for understanding these chemical reactions. It defines the ability of a substance to dissolve in a solvent, like how calcium bicarbonate dissolves in water, unlike its predecessor, calcium carbonate.
In the context of the lime water reaction:
In the context of the lime water reaction:
- Calcium carbonate is insoluble, which explains its milky precipitate.
- Calcium bicarbonate is soluble, allowing it to dissolve and clear the solution.
Exothermic reaction
An exothermic reaction is one that releases heat as it occurs. In the context of these reactions, while heat might be produced, it is not what primarily drives the changes observed in the lime water.
Instead:
Instead:
- The visual change from milky to clear is a result of further chemical reactions, rather than heat effects.
- Understanding exothermic reactions, in general, can help appreciate how energy changes might occur in other reactions.
Other exercises in this chapter
Problem 19
What is the \(\mathrm{C}-\mathrm{C}\) bond length (in angstroms) in diamond? (a) \(5.2\) (b) \(2.0\) (c) \(1.54\) (d) \(3.35\)
View solution Problem 20
Moderate electrical conductivity is shown by (a) diamond (b) carborundum (c) graphite (d) silica
View solution Problem 22
When \(\mathrm{SiO}_{2}\) is reacted with sodium carbonate, which gas is liberated? (a) \(\mathrm{O}_{2}\) (b) \(\mathrm{O}_{3}\) (c) \(\mathrm{CO}\) (d) \(\mat
View solution Problem 23
Silica is soluble in (a) HF (b) \(\mathrm{HNO}_{3}\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (d) \(\mathrm{HCl}\)
View solution