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Select the (a) best and (b) poorest electrical conductors from the following solutions, and explain the reason for your choices: \(0.10 \mathrm{M} \mathrm{NH}_{3} ; 0.10 \mathrm{M} \mathrm{NaCl} ; 0.10 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH}\) (acetic acid); \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (ethanol).

Short Answer

Expert verified
The best electrical conductor among the given solutions is 0.10 M \(NaCl\) and the poorest electrical conductor is 0.10 M \(CH_{3}CH_{2}OH\).

Step by step solution

01

Identify whether the chemicals in solution ionize

The first step is to figure out which of these chemicals ionize when they are put in solution. \(NH_{3}\) produces few ions, \(NaCl\) readily ionizes into \(Na^{+}\) and \(Cl^{-}\), \(CH_{3}COOH\) (acetic acid) partially ionizes to form \(CH_{3}COO^{-}\) and \(H^{+}\), and \(CH_{3}CH_{2}OH\) (Ethanol) does not ionize in solution.
02

Rank the solutions based on ion production

Once we've identified which chemicals ionize, and how readily they ionize, we can rank the solutions from most to least conductive. In order of how many ions they would produce, our ranking looks like this: \(NaCl\) > \(CH_{3}COOH\) > \(NH_{3}\) > \(CH_{3}CH_{2}OH\).
03

Identify the best and the poorest conductors

Based on our ranking in the previous step, we can now select the best and poorest conductors from the solutions: the best electrical conductor is 0.10 M \(NaCl\) and the poorest conductor is 0.10 M \(CH_{3}CH_{2}OH\). This is because \(NaCl\) completely ionizes in solution, producing the maximum number of ions possible for maximum conductivity, while ethanol does not ionize at all, producing a solution with virtually no ions to conduct electricity.

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Key Concepts

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

Ionization in Solutions
When a solute is added to a solvent, such as water, the solute may undergo ionization. This process is where molecules split into ions, which are charged particles. The more ions generated, the stronger the solution's electrical conductivity. For example:
  • Ammonia (\(NH_{3}\) ) produces very few ions when in solution, resulting in low conductivity.
  • Sodium chloride (\(NaCl\) ) completely dissociates into \(Na^+\) and \(Cl^-\) ions, leading to high conductivity.
  • Acetic acid (\(CH_{3}COOH\)) partially ionizes, offering a moderate conductive capability.
  • Ethanol (\(CH_{3}CH_{2}OH\)) does not ionize, yielding practically no electrical conductivity.
Understanding ionization helps predict which solutions will conduct electricity effectively.
Aqueous Solutions
Solutions where water acts as the solvent are known as aqueous solutions. Water is known for its remarkable ability to dissolve many substances effectively, which is why it's often called the "universal solvent."In aqueous solutions:
  • Ionic compounds like \(NaCl\) dissolve in water and break into ions, enhancing the solution's conductivity.
  • Weak acids like acetic acid may only partially ionize, offering lesser but still significant conductivity in their solutions.
  • Nonelectrolytes, such as ethanol, do not form ions and thus do not conduct electricity in aqueous form.
Water's polar structure facilitates the dissolution and ionization of many compounds, making it essential in studying solution conductivity.
Chemical Conductivity
The ability of a solution to conduct electricity is termed as chemical conductivity. Chemical conductivity in solutions depends primarily on the presence and movement of ions.Factors influencing conductivity:
  • The number of available charged particles, or ions, in the solution.
  • The speed at which these ions can move through the solution.
Solutions with strong electrolytes, like \(NaCl\), fully dissociate into ions and thus exhibit high conductivity. Contrarily, since ethanol doesn't dissociate into ions, it displays virtually no conductivity.Key point: The greater the degree of ionization, the higher the chemical conductivity.
Electrolytes and Nonelectrolytes
Electrolytes are substances that, when dissolved in water, produce a solution that can conduct electricity. They split into ions, which are crucial for conductivity.Classification:
  • **Strong Electrolytes**: Substances like sodium chloride (\(NaCl\)) that fully ionize, forming many ions. These offer high conductivity.
  • **Weak Electrolytes**: Compounds such as acetic acid (\(CH_{3}COOH\)) that only partially ionize. These create fewer ions and show moderate conductivity.
Nonelectrolytes do not ionize and thus do not conduct. Ethanol is a typical nonelectrolyte, as it does not form ions in solution. This division between electrolytes and nonelectrolytes is fundamental to understanding electrical conductivity in solutions.

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Most popular questions from this chapter

Assuming the volumes are additive, what is the \(\left[\mathrm{Cl}^{-}\right]\) in a solution obtained by mixing \(225 \mathrm{mL}\) of \(0.625 \mathrm{M}\) \(\mathrm{KCl}\) and \(615 \mathrm{mL}\) of \(0.385 \mathrm{M} \mathrm{MgCl}_{2} ?\)

Phosphorus is essential for plant growth, but an excess of phosphorus can be catastrophic in aqueous ecosystems. Too much phosphorus can cause algae to grow at an explosive rate and this robs the rest of the ecosystem of oxygen. Effluent from sewage treatment plants must be treated before it can be released into lakes or streams because the effluent contains significant amounts of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\). (Detergents are a major contributor to phosphorus levels in domestic sewage because many detergents contain \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) ) A simple way to remove \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) from the effluent is to treat it with lime, \(\mathrm{CaO}\) which produces \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^{-}\) ions in water. The \(\mathrm{OH}^{-}\) ions convert \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) ions into \(\mathrm{PO}_{4}^{3-}\) ions and, finally, \(\mathrm{Ca}^{2+}, \mathrm{OH}^{-}\), and \(\mathrm{PO}_{4}^{3-}\) ions combine to form a precipitate of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}(\mathrm{s})\) (a) Write balanced chemical equations for the four reactions described above. [Hint: The reactants are \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O} ; \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\left.\mathrm{OH}^{-} ; \mathrm{HPO}_{4}^{2-} \text { and } \mathrm{OH}^{-} ; \mathrm{Ca}^{2+}, \mathrm{PO}_{4}^{3-}, \text { and } \mathrm{OH}^{-} .\right]\) (b) How many kilograms of lime are required to remove the phosphorus from a \(1.00 \times 10^{4}\) L holding tank filled with contaminated water, if the water contains \(10.0 \mathrm{mg}\) of phosphorus per liter?

The active ingredients in a particular antacid tablet are aluminum hydroxide, \(\mathrm{Al}(\mathrm{OH})_{3},\) and magnesium hydroxide, \(\mathrm{Mg}(\mathrm{OH})_{2} . \quad \mathrm{A} 5.00 \times 10^{2} \mathrm{mg}\) sample of the active ingredients was dissolved in \(50.0 \mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{HCl} .\) The resulting solution, which was still acidic, required \(16.5 \mathrm{mL}\) of \(0.377 \mathrm{M} \mathrm{NaOH}\) for neutralization. What are the mass percentages of \(\mathrm{Al}(\mathrm{OH})_{3}\) and \(\mathrm{Mg}(\mathrm{OH})_{2}\) in the sample?

An \(\mathrm{NaOH}(\mathrm{aq})\) solution cannot be made up to an exact concentration simply by weighing out the required mass of NaOH, because the NaOH is not pure. Also, water vapor condenses on the solid as it is being weighed. The solution must be standardized by titration. For this purpose, a 25.00 \(\mathrm{mL}\) sample of an NaOH(aq) solution requires 28.34 \(\mathrm{mL}\) of 0.1085 \(\mathrm{M}\) HCl. What is the molarity of the NaOH(aq)? \(\mathrm{HCl}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}) \longrightarrow \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)\)

In the equation \(\begin{aligned} ? \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+4 \mathrm{H}^{+}(\mathrm{aq}) & \longrightarrow ? \mathrm{Fe}^{3+}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1) \end{aligned}\) the missing coefficients (a) are each \(2 ;\) (b) are each 4; (c) can have any values as long as they are the same; (d) must be determined by experiment.

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