When excess of carbon dioxide is passed through lime water, the milkiness first formed disappears due to (a) the reversible reaction taking place (b) formation of water soluble calcium bicarbonate (c) huge amount of heat evolved during the reaction (d) formation of water soluble complex of calcium.

Short Answer

Expert verified
The milkiness disappears due to the formation of water soluble calcium bicarbonate.

Step by step solution

01

Understanding the Reaction

Carbon dioxide reacts with lime water, which is a solution of calcium hydroxide (Ca(OH)2) in water, to initially form calcium carbonate (CaCO3), which is what causes the milkiness as it is insoluble in water.
02

Identifying the Excess Carbon Dioxide Effect

When more carbon dioxide is passed through the mixture, the calcium carbonate reacts further with CO2 and H2O to form calcium bicarbonate [Ca(HCO3)2], which is soluble in water.
03

Result of the Continued Reaction

Because calcium bicarbonate is soluble, it dissolves in water and the solution becomes clear again as the milkiness (due to the formation of insoluble CaCO3) disappears.
04

Choosing the Correct Option

Based on the chemical process described, option (b) 'formation of water soluble calcium bicarbonate' is the correct reason for the disappearance of milkiness.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calcium Hydroxide
Calcium hydroxide, known chemically as Ca(OH)2, is a base that is highly relevant in many industrial and environmental processes. In the context of our exercise, calcium hydroxide is integral to the initial stage of the reaction we're examining, where it is dissolved in water to create what is commonly known as lime water.

When carbon dioxide (CO2) is introduced to lime water, the calcium hydroxide reacts with it to form calcium carbonate (CaCO3). This is a classic chemical reaction demonstrating the properties of calcium hydroxide as a base: it will react with a gas like CO2 which, in turn, is a weak acid once dissolved in water to form a carbonate compound.

It's important to understand that the solubility of calcium hydroxide in water is relatively low, which explains why an excess of CO2 can shift the reaction towards the formation of different products. The low solubility limit also means that even if the solution initially turns 'milky' due to the creation of calcium carbonate, further reactions can take place as more CO2 is added.
Calcium Carbonate
Calcium carbonate is a widespread substance found in rock formations all around the world and is the principal component of shells of marine organisms, snails, coal balls, pearls, and eggshells. Chemically represented by CaCO3, it initially forms when CO2 reacts with lime water, creating a milky appearance in the solution, indicative of its insolubility.

The production of calcium carbonate is evidence of a chemical change; it's the precipitate that leaves from the solution as a solid. However, the reaction does not stop there, if an excess amount of CO2 is added. In our exercise, this milky suspension that first indicates the formation of calcium carbonate will then disappear when the compound transforms into another substance which is more soluble in water.

The educational takeaway from observing calcium carbonate in this reaction is understanding its role as an intermediate: it is a pivotal point that can lead to further chemical changes under the appropriate conditions, such as the increase in CO2 concentration.
Calcium Bicarbonate Solubility
In comparison to calcium carbonate, calcium bicarbonate (also known as calcium hydrogen carbonate) does not readily precipitate from water solutions because it is highly soluble. Represented by the formula Ca(HCO3)2, this compound arises when an excess amount of carbon dioxide is passed through lime water containing calcium carbonate.

As the CO2 levels increase, the equilibrium of the earlier reaction shifts to favor the production of calcium bicarbonate. The higher solubility of this new product in water leads to the milkiness of the solution clearing up. This clear solution is the result of calcium bicarbonate completely dissolving, contrasting the previous precipitated calcium carbonate.

To fully comprehend this concept, it helps to recognize that solubility is not a static property but can be influenced by various factors including the presence of additional solutes, temperature, and the chemical composition of the solvent. The calcium bicarbonate solubility aspect in our exercise is fundamental to make sense of why the reaction behaves differently with varying amounts of carbon dioxide.

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Most popular questions from this chapter

\(\mathrm{CO}_{2}\) is not a poisonous gas but there is increase in concentration of \(\mathrm{CO}_{2}\) in the atmosphere due to burning of fossil fuels and decomposition of limestone. The increase in concentration of \(\mathrm{CO}_{2}\) may lead to (a) increase in photosynthesis in plants (b) higher concentration of \(\mathrm{CO}_{2}\) in water (c) increase in greenhouse effect, thus raising the temperature (d) increase in formation of metal carbonates.

Match the column I with column II and mark the appropriate choice. \(\begin{array}{|l|l|l|l|} \hline \text { (A) } & \text { Borax } & \text { (i) } & \mathrm{Na}_{3} \mathrm{AlF}_{6} \\ \hline \text { (B) } & \text { Inorganic benzene } & \text { (ii) } & \mathrm{N}_{2} \mathrm{~B}_{\mathbf{1}} \mathrm{O}_{7} \mathbf{~} \mathbf{1 0} \mathrm{H}_{2} \mathrm{O} \\ \hline \text { (C) } & \text { Cryolite } & \text { (iii) } & \mathbf{A}_{2} \mathrm{O}_{3} \cdot \mathbf{2 H}_{2} \mathrm{O} \\ \hline \text { (D) } & \text { Bauxite } & \text { (iv) } & \mathrm{B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6} \\ \hline \end{array}\) (a) \((\mathrm{A}) \rightarrow(\mathrm{ii}),(\mathrm{B}) \rightarrow(\mathrm{iv}),(\mathrm{C}) \rightarrow(\mathrm{i}),(\mathrm{D}) \rightarrow\) (iii) (b) \((\mathbf{A}) \rightarrow(\mathrm{i}),(\mathrm{B}) \rightarrow(\mathrm{ii}),(\mathrm{C}) \rightarrow(\mathrm{iii})\), (D) \(\rightarrow\) (iv) (c) \((A) \rightarrow(i i),(B) \rightarrow(\) iii), \((C) \rightarrow(i),(D) \rightarrow\) (iv) (d) \((A) \rightarrow(\) iii), \((B) \rightarrow(i),(C) \rightarrow(i i)\), (D) \(\rightarrow\) (iv)

Boron is unable to form \(\mathrm{BF}_{6}^{3-}\) ions due to (a) non-availability of \(d\)-orbitals (b) small size of boron atom (c) non-metallic nature (d) less reactivity towards halogens.

In diborane, (a) four bridged hydrogen atoms and two terminal hydrogen atoms are present (b) two bridged hydrogen atoms and four terminal hydrogen atoms are present (c) three bridged hydrogen atoms and three terminal hydrogen atoms are present (d) there are no bridged hydrogen atoms in diborane, only hydrogen bonds are present.

Which of the following is not a use of graphite? (a) For electrodes in batteries. (b) Crucibles made from graphite are used for its inertness to dilute acids and alkalies. (c) For adsorbing poisonous gases. (d) Lubricant at high temperature.

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