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Classify each of the following aqueous solutions as a nonelectrolyte, weak electrolyte, or strong electrolyte: (a) \(\mathrm{HClO}_{4}\), (b) \(\mathrm{HNO}_{3}\), (c) \(\mathrm{NH}_{4} \mathrm{Cl}\), (d) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) (acetone), (e) \(\operatorname{CoSO}_{4}\), (f) \(C_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) (sucrose).

Short Answer

Expert verified
(a) \(\mathrm{HClO}_{4}\) is a strong electrolyte. (b) \(\mathrm{HNO}_{\mathrm{3}}\) is a strong electrolyte. (c) \(\mathrm{NH}_{\mathrm{4}} \mathrm{Cl}\) is a strong electrolyte. (d) \(\mathrm{CH}_{\mathrm{3}} \mathrm{COCH}_{\mathrm{3}}\) (acetone) is a nonelectrolyte. (e) \(\operatorname{CoSO}_{\mathrm{4}}\) is a strong electrolyte. (f) \(C_{\mathrm{12}} \mathrm{H}_{\mathrm{22}} \mathrm{O}_{\mathrm{11}}\) (sucrose) is a nonelectrolyte.

Step by step solution

01

(a) Classifying \(\mathrm{HClO}_{4}\)

HClO4 (perchloric acid) is a strong acid, which means it will completely dissociate into ions in an aqueous solution: \[HClO_{\mathrm{4}}(aq) \rightarrow H^{+}(aq) + ClO_{4}^{-}(aq) \] Since it completely dissociates into ions, \(\mathrm{HClO}_{4}\) is classified as a strong electrolyte.
02

(b) Classifying \(\mathrm{HNO}_{\mathrm{3}}\)

HNO3 (nitric acid) is also a strong acid, which means it will completely dissociate into ions in an aqueous solution: \[HNO_{\mathrm{3}}(aq) \rightarrow H^{+}(aq) + NO_{\mathrm{3}}^{-}(aq) \] Since it completely dissociates into ions, \(\mathrm{HNO}_{\mathrm{3}}\) is classified as a strong electrolyte.
03

(c) Classifying \(\mathrm{NH}_{\mathrm{4}} \mathrm{Cl}\)

NH4Cl (ammonium chloride) is an ionic compound that will dissociate into ions in an aqueous solution: \[NH_{\mathrm{4} \mathrm{Cl}}(aq) \rightarrow NH_{\mathrm{4}}^{+}(aq) + Cl^{-}(aq) \] Since it completely dissociates into ions, \(\mathrm{NH}_{\mathrm{4}} \mathrm{Cl}\) is classified as a strong electrolyte.
04

(d) Classifying \(\mathrm{CH}_{\mathrm{3}} \mathrm{COCH}_{\mathrm{3}}\) (acetone)

Acetone is a non-polar covalent compound, and when dissolved in water, it does not dissociate into ions: \[CH_{3}COCH_3 \nrightarrow H^{+} + CH_{3}COCH_{2}^{-} \] Since it does not dissociate into ions, acetone is classified as a nonelectrolyte.
05

(e) Classifying \(\operatorname{CoSO}_{\mathrm{4}}\)

CoSO4 (cobalt(II) sulfate) is an ionic compound that will dissociate into ions in an aqueous solution: \[CoSO_{\mathrm{4}}(aq) \rightarrow Co^{2+}(aq) + SO_{\mathrm{4}}^{2-}(aq) \] Since it completely dissociates into ions, \(\operatorname{CoSO}_{\mathrm{4}}\) is classified as a strong electrolyte.
06

(f) Classifying \(C_{\mathrm{12}} \mathrm{H}_{\mathrm{22}} \mathrm{O}_{\mathrm{11}}\) (sucrose)

Sucrose is a covalent compound and, when dissolved in water, it does not dissociate into ions: \[C_{\mathrm{12}}H_{\mathrm{22}}O_{\mathrm{11}} \nrightarrow H^{+} + C_{\mathrm{12}}H_{\mathrm{21}}O_{\mathrm{11}}^{-} \] Since it does not dissociate into ions, sucrose 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
A strong electrolyte is a substance that dissolves in water to yield free and mobile ions, resulting in a solution that conducts electricity very well. Examples of strong electrolytes include many acids, bases, and salts that ionize completely when dissolved in water.

For instance, the acids \(\textrm{HClO}_{4}\) (perchloric acid) and \(\textrm{HNO}_{3}\) (nitric acid) mentioned in the exercise are strong electrolytes. They dissociate completely in aqueous solutions to form \(\textrm{H}^{+}\) ions and their respective anions. This is one aspect of the step-by-step solution that could be further clarified by explaining the significance of complete dissociation and its role in electrical conductivity.

How Strong Electrolytes Behave in Water

When strong electrolytes dissolve, they break apart into individual ions that are surrounded by water molecules, a process called solvation. The presence of these ions allows the solution to carry an electrical current efficiently. Therefore, when testing the electrical conductivity of a solution, a strong electrolyte will show a bright bulb or a high reading on a conductivity meter.
Weak Electrolyte
In contrast to strong electrolytes, a weak electrolyte partially dissociates into ions in aqueous solution. This partial dissociation produces a smaller number of ions, resulting in moderate electrical conductivity. Weak electrolytes include weak acids and weak bases; they establish an equilibrium between the un-dissociated molecules and the ions produced from dissociation.

The exercise provided doesn't list a weak electrolyte, but it is important to recognize them as they are common in many chemical contexts. An example not mentioned in the problem is acetic acid \(\textrm{CH}_{3}COOH\), which only partially ionizes in water.

Equilibrium in Weak Electrolyte Solutions

For weak electrolytes, an important concept is the dynamic equilibrium between ionized and unionized forms in solution. This balance is often represented by a reversible arrow in the chemical equation, indicative of the constant interconversion between forms. Weak electrolytes produce fewer ions, resulting in weaker electrical conductivity when compared to strong electrolytes.
Nonelectrolyte
A nonelectrolyte is a compound that does not ionize at all when dissolved in water, resulting in a non-conductive solution. This occurs primarily with covalent compounds where the molecules are composed of atoms sharing electrons rather than exchanging them to form ions.

From the exercise, acetone \(\textrm{CH}_{3}COCH_{3}\) and sucrose \(\textrm{C}_{12}\textrm{H}_{22}\textrm{O}_{11}\) are prime examples of nonelectrolytes. Since they do not produce ions in solution, they do not conduct electricity.

Physical Dissolution vs. Ionic Dissociation

While nonelectrolytes like sugar do dissolve in water, they do so by dispersing as whole molecules throughout the solvent. This contrasts with ionic compounds, which separate into individual anions and cations. It is essential to distinguish between physical dissolution, which occurs even in nonelectrolytes, and ionic dissociation, which is specific to electrolytes.
Ionic Dissociation in Water
Ionic dissociation in water refers to the process by which ionic compounds separate into their respective positive and negative ions when they dissolve in water. This process is crucial for the conductivity of the solution, as free-moving ions are required to transport electric charge.

For example, \(\textrm{NH}_{4}\textrm{Cl}\) (ammonium chloride) and \(\textrm{CoSO}_{4}\) (cobalt(II) sulfate), as referred to in the exercise, both undergo ionic dissociation in water. The dissolution and dissociation of \(\textrm{CoSO}_{4}\) into \(\textrm{Co}^{2+}\) and \(\textrm{SO}_{4}^{2-}\) ions indicate its behavior as a strong electrolyte.

Role of Water in Ionic Dissociation

The polarity of water molecules plays a fundamental role in the dissociation process. The partial positive charge on the hydrogen atoms in water molecules attracts anions, while the partial negative charge on the oxygen atom attracts cations, helping them to separate and become solvated. This interaction is pivotal in the dissolution process and is particularly effective with ionic compounds, showcasing water's solvent capabilities.

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Most popular questions from this chapter

The average concentration of bromide ion in seawater is \(65 \mathrm{mg}\) of bromide ion per \(\mathrm{kg}\) of seawater. What is the molarity of the bromide ion if the density of the seawater is \(1.025 \mathrm{~g} / \mathrm{mL}\) ?

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Calculate (a) the number of grams of solute in \(0.250 \mathrm{~L}\) of \(0.175 \mathrm{M} \mathrm{KBr},(\mathrm{b})\) the molar concentration of a solution containing \(14.75 \mathrm{~g}\) of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) in \(1.375 \mathrm{~L},(\mathrm{c})\) the volume of \(1.50 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) in milliliters that contains \(2.50 \mathrm{~g}\) of solute.

Can oxidation occur without accompanying reduction? Explain.

Federal regulations set an upper limit of 50 parts per million (ppm) of \(\mathrm{NH}_{3}\) in the air in a work environment [that is, 50 molecules of \(\mathrm{NH}_{3}(g)\) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing \(1.00 \times 10^{2} \mathrm{~mL}\) of \(0.0105 \mathrm{M}\) HCl. The \(\mathrm{NH}_{3}\) reacts with HCl as follows: $$ \mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$ After drawing air through the acid solution for \(10.0 \mathrm{~min}\) at a rate of \(10.0 \mathrm{~L} / \mathrm{min}\), the acid was titrated. The remaining acid needed \(13.1 \mathrm{~mL}\) of \(0.0588 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point. (a) How many grams of \(\mathrm{NH}_{3}\) were drawn into the acid solution? (b) How many ppm of \(\mathrm{NH}_{3}\) were in the air? (Air has a density of \(1.20 \mathrm{~g} / \mathrm{L}\) and an average molar mass of \(29.0 \mathrm{~g} / \mathrm{mol}\) under the conditions of the experiment.) (c) ls this manufacturer in compliance with regulations?

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