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A solution of HCl in water conducts electricity, but a solution of \(\mathrm{HCl}\) in chloroform, \(\mathrm{CHCl}_{3}\), does not. What does this observation tell you about how \(\mathrm{HCl}\) exists in water and how it exists in chloroform?

Short Answer

Expert verified
HCl ionizes in water and not in chloroform, allowing water solutions to conduct electricity.

Step by step solution

01

Understanding Conductivity

Electricity conduction in solutions requires the presence of charged particles, such as ions. In aqueous (water-based) solutions, if a compound ionizes, it will potentially conduct electricity. In nonpolar substances like chloroform, this ionization usually does not occur, and hence, no conductivity is observed.
02

Ionization in Water

In water, HCl dissociates into hydrogen ions (H\(^+\)) and chloride ions (Cl\(^-\)), which are free to move. This presence of free ions causes the solution to conduct electricity effectively.
03

Molecular Dissolution in Chloroform

In contrast, when HCl is dissolved in chloroform, an organic solvent, it does not ionize but stays as neutral HCl molecules. Chloroform, being nonpolar, does not provide the environment needed for the HCl to disassociate into ions.
04

Implication of the Observation

The difference in electrical conductivity when HCl is dissolved in water versus chloroform indicates HCl's behavior: it ionizes in water, forming ions, and does not ionize in chloroform, remaining as molecules.

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Key Concepts

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

Ionization in Water
Water is a unique solvent due to its polar nature. When compounds like hydrochloric acid (HCl) are dissolved in water, they undergo a process called ionization. This means the HCl molecule splits into its constituent ions: hydrogen ions (H\(^+\)) and chloride ions (Cl\(^-\)).

Let's break it down further:
  • Splitting Molecules: The HCl molecule separates into two charged particles.
  • Positive and Negative Ions: Hydrogen (H\(^+\)) is a positive ion, while chloride (Cl\(^-\)) is negative.
  • Free Movement: these ions can move freely in the water.
This breakdown into ions is crucial because these charged particles enable the solution to conduct electricity. It's the free-moving nature of ions that helps transfer electric current through the solution. Without ionization, such conductivity wouldn't be possible.

This ability of certain substances to ionize in water is what makes water such an essential medium for many chemical reactions.
Molecular Dissolution
Chloroform, a nonpolar solvent, provides us with a contrasting behavior of HCl when dissolved. Unlike water, HCl in chloroform remains in its molecular form rather than splitting into ions. This has significant implications for how substances interact with nonpolar solvents.

Here's what happens:
  • Neutral Molecules: HCl stays as whole molecules, not forming any ions.
  • No Ionization: Without ionization, there are no charged particles in the solution.
  • Lack of Interaction: Chloroform doesn't support the separation of HCl into ions due to its nonpolar nature.
This behavior highlights how the nature of the solvent affects the solute's form—whether it remains as molecules or dissociates into ions.

Therefore, when you want to know if a solution will conduct electricity, it's essential to consider both the solvent's properties and how they interact with the solute.
Electricity Conduction
Conductivity in solutions is largely determined by the presence of ions. To conduct electricity, a solution must facilitate the movement of charged particles.

Let's dive into the main points:
  • Presence of Charged Particles: For a solution to conduct electricity, it must contain ions.
  • Movement of Ions: These ions move freely and carry electric charge through the solution.
  • Conductive Solutions: Solutions that ionize, like HCl in water, show high conductivity.
  • Non-Conductive Solutions: In nonpolar solvents like chloroform, where HCl doesn’t ionize, conductivity is absent.
Understanding this relationship between ion presence and electrical conduction helps in predicting the conductive behavior of various solutions.

Thus, knowing whether a solute can create ions in a given solvent is key to anticipating if the solution can conduct electricity or not. It's fundamental in various applications, from designing electrolytic solutions to understanding chemical reactions.

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Most popular questions from this chapter

Which element is oxidized and which is reduced in each of the following reactions? (a) \(\mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(l)\) (b) \(\mathrm{Cl}_{2}(g)+2 \mathrm{NaBr}(a q) \longrightarrow \mathrm{Br}_{2}(a q)+2 \mathrm{NaCl}(a q)\)

Write balanced net ionic equations for the following reactions: (a) \(\mathrm{LiOH}(a q)+\mathrm{HI}(a q) \rightarrow ?\) (b) \(\mathrm{HBr}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \rightarrow ?\)

(a) Use the following reactions to arrange the elements \(\mathbf{A}, \mathbf{B}, \mathbf{C}\), and \(\mathbf{D}\) in order of their decreasing ability as reducing agents: \(2 \mathrm{~A}+\mathrm{B}^{2+} \longrightarrow 2 \mathrm{~A}^{+}+\mathrm{B} \quad \mathrm{B}+\mathrm{D}^{2+} \longrightarrow \mathrm{B}^{2+}+\mathrm{D}\) \(\mathrm{A}^{+}+\mathrm{C} \longrightarrow\) no reaction \(2 \mathrm{C}+\mathrm{B}^{2+} \longrightarrow 2 \mathrm{C}^{+}+\mathrm{B}\) (b) Which of the following reactions would you expect to occur according to the activity series you established in part (a)? (1) \(2 \mathrm{~A}^{+}+\mathrm{D} \longrightarrow 2 \mathrm{~A}+\mathrm{D}^{2+}\) (2) \(\mathrm{D}^{2+}+2 \mathrm{C} \longrightarrow \mathrm{D}+2 \mathrm{C}^{+}\)

Classify each of the following reactions as a precipitation, acidbase neutralization, or oxidation-reduction: (a) \(\mathrm{S}_{8}(s)+8 \mathrm{O}_{2}(g) \rightarrow 8 \mathrm{SO}_{2}(g)\) (b) \(\mathrm{NiCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{~S}(a q) \rightarrow \mathrm{NiS}(s)+2 \mathrm{NaCl}(a q)\) (c) \(2 \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow\) \(\left(\mathrm{CH}_{3} \mathrm{CO}_{2}\right)_{2} \mathrm{Ba}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Assign oxidation numbers to each element in the following compounds: (a) \(\mathrm{VOCl}_{3}\) (b) \(\mathrm{CuSO}_{4}\) (c) \(\mathrm{CH}_{2} \mathrm{O}\) (d) \(\mathrm{Mn}_{2} \mathrm{O}_{7}\) (e) \(\mathrm{OsO}_{4}\) (f) \(\mathrm{H}_{2} \mathrm{PtCl}_{6}\)

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