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What must be present in an aqueous solution for it to conduct an electric current? What general classes of compounds form solutions that conduct?

Short Answer

Expert verified
An aqueous solution must contain ions to conduct electricity. Ionic compounds, acids, and bases form ions in solution and conduct electricity.

Step by step solution

01

Identify key components for conductivity

For a solution to conduct electricity, it must contain charged particles that can move. In an aqueous solution, these charged particles are typically ions. Therefore, the presence of ions is necessary for the conduction of electricity in an aqueous solution.
02

Understand formation of ions

When certain substances dissolve in water, they dissociate into their constituent ions. This process is called ionization. Substances that can dissociate into ions are capable of conducting electricity in an aqueous solution.
03

Classify compounds that form ions in solution

The general classes of compounds that form solutions that conduct electricity include ionic compounds and specific molecular compounds: 1. Ionic compounds: These compounds, such as salts (e.g., sodium chloride), dissociate completely into cations and anions when dissolved in water. 2. Acids and Bases: Strong acids (e.g., hydrochloric acid) and strong bases (e.g., sodium hydroxide) ionize completely in water, providing ions that can conduct electricity.

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Key Concepts

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

ions in solution
When we talk about ions in a solution, we refer to charged particles that result from the dissolving of certain substances in water. These substances break down into their constituent ions, a process known as ionization. The movement of these ions allows the solution to conduct electricity. For instance, when table salt (sodium chloride) dissolves in water, it separates into sodium ions (Na+) and chloride ions (Cl-).
Conductivity in an aqueous solution directly depends on the presence of these free-moving ions. The more ions present, the better the solution can conduct electricity. This is why pure water is a poor conductor; it has very few ions.
On the contrary, when substances like salts or acids dissolve in water, they release a considerable number of ions, enhancing the solution's ability to conduct electricity. Understanding the role of ions in solution is fundamental to grasping why certain solutions conduct electricity while others do not.
ionic compounds
Ionic compounds are a key category of substances that contribute to solution conductivity. These compounds are formed from metals and nonmetals. In these compounds, atoms transfer electrons to form positive and negative ions, creating a lattice structure in the solid state.
When ionic compounds dissolve in water, they dissociate completely into their respective ions. For example, sodium chloride (NaCl) breaks down into sodium ions (Na+) and chloride ions (Cl-). This release of ions into the solution is what makes it conductive.
Another example is potassium nitrate (KNO3). When it dissolves, it separates into potassium ions (K+) and nitrate ions (NO3-). These free ions can move freely in the solution, allowing it to conduct an electric current.
It's important to understand that the ability of ionic compounds to form conductive solutions is due to their complete dissociation in water.
acids and bases
Acids and bases also play a significant role in the conductivity of aqueous solutions. Strong acids and bases ionize completely in water, releasing ions that enhance conductivity.
A strong acid, like hydrochloric acid (HCl), dissociates fully into hydrogen ions (H+) and chloride ions (Cl-). These ions are free to move in the solution, making it highly conductive.
Similarly, a strong base, such as sodium hydroxide (NaOH), dissociates into sodium ions (Na+) and hydroxide ions (OH-), contributing to the solution's conductivity.
It's worth mentioning that not all acids and bases are strong. Weak acids and bases do not dissociate completely in water, resulting in fewer ions and thus lower conductivity.
An example of a weak acid is acetic acid (CH3COOH), which only partially dissociates in solution. Therefore, understanding the strength of acids and bases is crucial to predicting their effect on solution conductivity.

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Most popular questions from this chapter

Give the oxidation number of carbon in each of the following: (a) \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) (b) \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (c) HCO \(_{3}^{-}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{6}\)

The mass percent of \(\mathrm{Cl}^{-}\) in a seawater sample is determined by titrating \(25.00 \mathrm{~mL}\) of seawater with \(\mathrm{AgNO}_{3}\) solution, causing a precipitation reaction. An indicator is used to detect the end point, which occurs when free \(\mathrm{Ag}^{+}\) ion is present in solution after all the \(\mathrm{Cl}^{-}\) has reacted. If \(53.63 \mathrm{~mL}\) of \(0.2970 \mathrm{M} \mathrm{AgNO}_{3}\) is required to reach the end point, what is the mass percent of \(\mathrm{Cl}^{-}\) in the seawater \((d\) of seawater \(=1.024 \mathrm{~g} / \mathrm{mL}) ?\)

The amount of ascorbic acid (vitamin \(\left.\mathrm{C}, \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\right)\) in tablets is determined by reaction with bromine and then titration of the hydrobromic acid with standard base: $$ \begin{array}{l} \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}(a q)+\mathrm{Br}_{2}(a q) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}(a q)+2 \mathrm{HBr}(a q) \\ \mathrm{HBr}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{NaBr}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ A certain tablet is advertised as containing \(500 \mathrm{mg}\) of vitamin \(\mathrm{C}\). One tablet was dissolved in water and reacted with \(\mathrm{Br}_{2}\). The solution was then titrated with \(43.20 \mathrm{~mL}\) of \(0.1350 \mathrm{M} \mathrm{NaOH}\). Did the tablet contain the advertised quantity of vitamin C?

Limestone \(\left(\mathrm{CaCO}_{3}\right)\) is used to remove acidic pollutants from smokestack flue gases. It is heated to form lime \((\mathrm{CaO})\), which reacts with sulfur dioxide to form calcium sulfite. Assuming a \(70 . \%\) yield in the overall reaction, what mass of limestone is required to remove all the sulfur dioxide formed by the combustion of \(8.5 \times 10^{4} \mathrm{~kg}\) of coal that is 0.33 mass \% sulfur?

A mixture of \(\mathrm{KClO}_{3}\) and \(\mathrm{KCl}\) with a mass of \(0.950 \mathrm{~g}\) was heated to produce \(\mathrm{O}_{2}\). After heating, the mass of residue was 0.700 g. Assuming all the \(\mathrm{KClO}_{3}\) decomposed to \(\mathrm{KCl}\) and \(\mathrm{O}_{2}\), calculate the mass percent of \(\mathrm{KClO}_{3}\) in the original mixture.

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