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Identify each of the following substances as a strong electrolyte, weak electrolyte, or nonelectrolyte: (a) \(\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{KCl}\) (c) \(\mathrm{HNO}_{3}\), (d) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (e) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\)

Short Answer

Expert verified
(a) nonelectrolyte, (b) strong electrolyte, (c) strong electrolyte, (d) weak electrolyte, (e) nonelectrolyte.

Step by step solution

01

Analyzing Water

Water ( H_{2}O ) is a molecular compound that, under most conditions, does not conduct electricity as it does not dissociate into ions. Therefore, it is classified as a nonelectrolyte.
02

Analyzing Potassium Chloride

Potassium chloride ( KCl ) in solution dissociates completely into its ions ( K^+ and Cl^- ). Therefore, KCl is classified as a strong electrolyte because it conducts electricity well.
03

Analyzing Nitric Acid

Nitric acid ( HNO_3 ) is known to dissociate completely in water to produce H^+ and NO_3^- ions. This complete dissociation makes HNO_3 a strong electrolyte.
04

Analyzing Acetic Acid

Acetic acid ( HC_2H_3O_2 ), unlike strong acids, partially ionizes in solution. This partial dissociation results in acetic acid being classified as a weak electrolyte.
05

Analyzing Sucrose

Sucrose ( C_{12}H_{22}O_{11} ) is a sugar that dissolves in water without dissociating into ions. Therefore, it does not conduct electricity and is classified as a nonelectrolyte.

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Key Concepts

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

strong electrolyte
Strong electrolytes are substances that completely dissociate into ions when dissolved in water. This means that the compound breaks apart into its charged components. Because of this complete ionization, strong electrolytes conduct electricity very efficiently in solution.

Common examples of strong electrolytes include strong acids like nitric acid (\( \mathrm{HNO}_{3} \)) and salts like potassium chloride (\( \mathrm{KCl} \)). Both of these substances dissociate completely when they dissolve in water:
  • \( \mathrm{HNO}_{3} \) breaks down into \( \mathrm{H}^{+} \) and \( \mathrm{NO}_{3}^{-} \).
  • \( \mathrm{KCl} \) dissociates into \( \mathrm{K}^{+} \) and \( \mathrm{Cl}^{-} \).
Strong electrolytes are distinguished by their high conductivity in solution due to the free movement of ions.
weak electrolyte
Weak electrolytes are substances that only partially dissociate into ions when dissolved in water. This partial dissociation means that only a small fraction of the molecules break into ions, while the rest remain intact.

Acetic acid (\( \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \)), a common weak electrolyte, dissociates only slightly in water. As a result:
  • Only a few \( \mathrm{H}^{+} \) and \( \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-} \) ions form.
  • Most acetic acid remains in its molecular form.
The limited number of ions limits the conductance of electricity, making weak electrolytes less conductive compared to strong electrolytes.
nonelectrolyte
Nonelectrolytes are substances that do not dissociate into ions at all when they dissolve in water. Because they do not form ions, nonelectrolytes do not contribute to electrical conductivity in solution.

Examples include water itself (\( \mathrm{H}_{2} \mathrm{O} \)) and sucrose (\( \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \)), which dissolve by dispersing individual molecules throughout the water:
  • \( \mathrm{H}_{2} \mathrm{O} \) remains in its molecular form in solution.
  • \( \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \) has no charge and does not ionize when dissolved.
Thus, nonelectrolytes are characterized by their inability to conduct electricity in solution.
ion dissociation
Ion dissociation refers to the process by which ionic compounds separate into individual ions when dissolved in a solvent like water. This is a crucial concept for understanding how substances conduct electricity in solution.

In a strong electrolyte, every molecule of the substance dissociates completely into its constituent ions:
  • Each \( \mathrm{KCl} \) unit breaks apart into \( \mathrm{K}^{+} \) and \( \mathrm{Cl}^{-} \) ions.
In contrast, a weak electrolyte like \( \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \) only partially dissociates:
  • Only some molecules break into \( \mathrm{H}^{+} \) and \( \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-} \).
This dissociation degree directly affects the conductivity of the solution.
conductivity in solution
The conductivity in solution is the ability to allow electricity to pass through it and is directly related to the presence of ions. It depends on how well a substance can dissociate into ions when dissolved in a solvent.

Strong electrolytes contribute to high conductivity because they produce a large number of ions. For example:
  • \( \mathrm{KCl} \) and \( \mathrm{HNO}_{3} \) dissociate fully, allowing for efficient current flow.
On the other hand, weak electrolytes like \( \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \) produce fewer ions, resulting in lower conductivity.

Nonelectrolytes such as \( \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \) do not produce ions at all and, therefore, do not conduct electricity. This clear distinction helps us classify substances based on their ability to conduct electrical current when in solution.

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Most popular questions from this chapter

Calculate the molarity of each of the following solutions: (a) \(29.0 \mathrm{~g}\) of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) in \(545 \mathrm{~mL}\) of solution, (b) \(15.4 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) in \(74.0 \mathrm{~mL}\) of solution, (c) \(9.00 \mathrm{~g}\) of sodium chloride \((\mathrm{NaCl})\) in \(86.4 \mathrm{~mL}\) of solution.

Calculate the volume in milliliters of a \(1.420 \mathrm{M} \mathrm{NaOH}\) solution required to titrate the following solutions: a) \(25.00 \mathrm{~mL}\) of a \(2.430 \mathrm{M} \mathrm{HCl}\) solution b) \(25.00 \mathrm{~mL}\) of a \(4.500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution c) \(25.00 \mathrm{~mL}\) of a \(1.500 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) solution

Calculate the molarity of each of the following solutions: (a) \(6.57 \mathrm{~g}\) of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) in \(1.50 \times 10^{2} \mathrm{~mL}\) of solution, (b) \(10.4 \mathrm{~g}\) of calcium chloride \(\left(\mathrm{CaCl}_{2}\right)\) in \(2.20 \times 10^{2} \mathrm{~mL}\) of solution, \((\mathrm{c}) 7.82 \mathrm{~g}\) of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) in \(85.2 \mathrm{~mL}\) of benzene solution.

Water is added to \(25.0 \mathrm{~mL}\) of a \(0.866 \mathrm{M} \mathrm{KNO}_{3}\) solution until the volume of the solution is exactly \(500 \mathrm{~mL}\). What is the concentration of the final solution?

The following "cycle of copper" experiment is performed in some general chemistry laboratories. The series of reactions starts with copper and ends with metallic copper. The steps are as follows: (1) A piece of copper wire of known mass is allowed to react with concentrated nitric acid [the products are copper(II) nitrate, nitrogen dioxide, and water]. (2) The copper(II) nitrate is treated with a sodium hydroxide solution to form copper(II) hydroxide precipitate. (3) On heating, copper(II) hydroxide decomposes to yield copper(II) oxide. (4) The copper(II) oxide is combined with concentrated sulfuric acid to yield copper(II) sulfate. (5) Copper(II) sulfate is treated with an excess of zinc metal to form metallic copper. (6) The remaining zinc metal is removed by treatment with hydrochloric acid, and metallic copper is filtered, dried, and weighed. (a) Write a balanced equation for each step and classify the reactions. (b) Assuming that a student started with \(65.6 \mathrm{~g}\) of copper, calculate the theoretical yield at each step. (c) Considering the nature of the steps, comment on why it is possible to recover most of the copper used at the start.

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