Problem 21
Question
If \(\mathrm{CO}\), 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
(a) formation of water soluble calcium bicarbonate
1Step 1: Understanding Lime Water Reaction
When lime water, which is a solution of calcium hydroxide (
Ca(OH)_2
), reacts with carbon monoxide (
CO
), it typically does not react directly. However, the question likely implies
CO_2
(carbon dioxide), a common gas that reacts with lime water. Initially,
CO_2
reacts with
Ca(OH)_2
to form calcium carbonate (
CaCO_3
), which is insoluble and creates a milky suspension.
2Step 2: Formation of Calcium Carbonate
As
CO_2
is bubbled through lime water, the reaction
Ca(OH)_2 + CO_2
ightarrow CaCO_3 + H_2O
occurs, resulting in calcium carbonate. This calcium carbonate is an insoluble white solid, making the solution appear milky.
3Step 3: Effect of Excess CO_2
When an excess of
CO_2
is bubbled through the mixture, the initially formed calcium carbonate reacts further. This second reaction is
CaCO_3 + CO_2 + H_2O
ightarrow Ca(HCO_3)_2
, where calcium bicarbonate, which is soluble in water, is formed. As a result, the milkiness disappears.
Key Concepts
Lime Water ReactionCalcium Carbonate FormationCalcium Bicarbonate Solubility
Lime Water Reaction
Lime water is an aqueous solution containing calcium hydroxide or \( \text{Ca(OH)}_2 \). In the context of this exercise, the term 'lime water reaction' usually refers to the interaction between lime water and carbon dioxide \( \text{CO}_2 \), not carbon monoxide \( \text{CO} \), which was mentioned. This is a critical correction because carbon dioxide is the reactive gas that participates in this classic chemistry demonstration.
When \( \text{CO}_2 \) is introduced to lime water, a chemical reaction occurs:
When \( \text{CO}_2 \) is introduced to lime water, a chemical reaction occurs:
- Initially, the \( \text{CO}_2 \) reacts with \( \text{Ca(OH)}_2 \) in the solution.
- This reaction forms calcium carbonate \( \text{CaCO}_3 \) and water \( \text{H}_2\text{O} \):
\[ \text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3 + \text{H}_2\text{O} \]
Calcium Carbonate Formation
The formation of calcium carbonate \( \text{CaCO}_3 \) is a key part of the lime water reaction. This compound is a white, insoluble solid. So when it is produced, it causes the once clear lime water solution to become cloudy or "milky" in appearance. This phenomenon is used as an indicator in various chemistry demonstrations and experiments.
- Calcium carbonate itself is a common substance found naturally. It's present in rocks like limestone and chalk, as well as in seashells and eggshells.
- The creation of \( \text{CaCO}_3 \) can be explained by considering the balanced chemical equation:
\[ \text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3 + \text{H}_2\text{O} \] - This process illustrates how a gas can react with a liquid to produce a solid precipitate, a key concept in chemistry known as a precipitation reaction.
Calcium Bicarbonate Solubility
Excess carbon dioxide introduces an interesting twist in the lime water reaction. When \( \text{CO}_2 \) is in surplus, it causes calcium carbonate to react further, forming calcium bicarbonate \( \text{Ca(HCO}_3)_2 \), which is soluble in water. This subsequent reaction explains why the initial milkiness caused by \( \text{CaCO}_3 \) disappears.
- In the chemical context, this reaction is represented by:
\[ \text{CaCO}_3 + \text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{Ca(HCO}_3)_2 \] - The resulting compound, calcium bicarbonate, dissolves in the solution, making it clear once more.
- This process illustrates important concepts of solubility and the dynamic nature of chemical equilibria.
Other exercises in this chapter
Problem 18
The use of diamond as a gem depends on its (a) chemical inertness (b) hardness (c) purest form of carbon (d) high refractive index
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) \(\mathrm{O}_{2}\) (c) \(\mathrm{CO}^{2}\) (b) \(\mathrm{O}_{3}\) (d) \(
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