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Show how each of the following strong electrolytes "breaks up" into its component ions upon dissolving in water by drawing molecular-level pictures. a. NaBr b. \(\mathrm{MgCl}_{2}\) c. \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) d. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) e, \(\mathrm{NaOH}\) f. \(\mathrm{FeSO}_{4}\) g. \(\mathrm{KMnO}_{4}\) h. \(\mathrm{HClO}_{4}\) i. \(\mathrm{NH}_{4} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

Short Answer

Expert verified
a. NaBr (s) ⟶ Na⁺ (aq) + Br⁻ (aq) b. \(\mathrm{MgCl}_{2}\) (s) ⟶ Mg²⁺ (aq) + 2Cl⁻ (aq) c. \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) (s) ⟶ Al³⁺ (aq) + 3NO₃⁻ (aq) d. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) (s) ⟶ 2NH₄⁺ (aq) + SO₄²⁻ (aq) e. \(\mathrm{NaOH}\) (s) ⟶ Na⁺ (aq) + OH⁻ (aq) f. \(\mathrm{FeSO}_{4}\) (s) ⟶ Fe²⁺ (aq) + SO₄²⁻ (aq) g. \(\mathrm{KMnO}_{4}\) (s) ⟶ K⁺ (aq) + MnO₄⁻ (aq) h. \(\mathrm{HClO}_{4}\) (s) ⟶ H⁺ (aq) + ClO₄⁻ (aq) i. \(\mathrm{NH}_{4} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (s) ⟶ NH₄⁺ (aq) + C₂H₃O₂⁻ (aq)

Step by step solution

01

a. NaBr

When NaBr dissolves in water, it breaks up into its individual ions: sodium ion (Na⁺) and bromide ion (Br⁻). Therefore, the molecular-level picture is: NaBr (s) ⟶ Na⁺ (aq) + Br⁻ (aq)
02

b. \(\mathrm{MgCl}_{2}\)

When \(\mathrm{MgCl}_{2}\) dissolves in water, it breaks up into its individual ions: magnesium ion (Mg²⁺) and two chloride ions (2Cl⁻). Therefore, the molecular-level picture is: \(\mathrm{MgCl}_{2}\) (s) ⟶ Mg²⁺ (aq) + 2Cl⁻ (aq)
03

c. \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\)

When \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) dissolves in water, it breaks up into its individual ions: aluminum ion (Al³⁺) and three nitrate ions (3NO₃⁻). Therefore, the molecular-level picture is: \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) (s) ⟶ Al³⁺ (aq) + 3NO₃⁻ (aq)
04

d. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)

When \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) dissolves in water, it breaks up into its individual ions: two ammonium ions (2NH₄⁺) and one sulfate ion (SO₄²⁻). Therefore, the molecular-level picture is: \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) (s) ⟶ 2NH₄⁺ (aq) + SO₄²⁻ (aq)
05

e. \(\mathrm{NaOH}\)

When \(\mathrm{NaOH}\) dissolves in water, it breaks up into its individual ions: sodium ion (Na⁺) and hydroxide ion (OH⁻). Therefore, the molecular-level picture is: \(\mathrm{NaOH}\) (s) ⟶ Na⁺ (aq) + OH⁻ (aq)
06

f. \(\mathrm{FeSO}_{4}\)

When \(\mathrm{FeSO}_{4}\) dissolves in water, it breaks up into its individual ions: iron (II) ion (Fe²⁺) and sulfate ion (SO₄²⁻). Therefore, the molecular-level picture is: \(\mathrm{FeSO}_{4}\) (s) ⟶ Fe²⁺ (aq) + SO₄²⁻ (aq)
07

g. \(\mathrm{KMnO}_{4}\)

When \(\mathrm{KMnO}_{4}\) dissolves in water, it breaks up into its individual ions: potassium ion (K⁺) and permanganate ion (MnO₄⁻). Therefore, the molecular-level picture is: \(\mathrm{KMnO}_{4}\) (s) ⟶ K⁺ (aq) + MnO₄⁻ (aq)
08

h. \(\mathrm{HClO}_{4}\)

When \(\mathrm{HClO}_{4}\) dissolves in water, it breaks up into its individual ions: hydrogen ion (H⁺) and perchlorate ion (ClO₄⁻). Therefore, the molecular-level picture is: \(\mathrm{HClO}_{4}\) (s) ⟶ H⁺ (aq) + ClO₄⁻ (aq)
09

i. \(\mathrm{NH}_{4} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

When \(\mathrm{NH}_{4} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) dissolves in water, it breaks up into its individual ions: ammonium ion (NH₄⁺) and acetate ion (C₂H₃O₂⁻). Therefore, the molecular-level picture is: \(\mathrm{NH}_{4} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (s) ⟶ NH₄⁺ (aq) + C₂H₃O₂⁻ (aq)

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Key Concepts

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

Molecular-Level Representation
Understanding dissolution at a molecular level is like peeking into a hidden world. Imagine starting with a solid piece of a compound, such as sodium bromide (NaBr), and watching as it vanishes into water. What happens in this "molecular-level" view is that water molecules begin interacting with the NaBr. They surround the individual ions in the solid.
  • The solid breaks apart into ions: Na⁺ and Br⁻.
  • These ions become dispersed within the water. This results in an aqueous solution.

Picture this process as individual dancers leaving a tightly packed group to dance freely around the room. Each dancer (ion) is still distinguishable within the broader flow (solution). So, a molecular-level representation captures these details, showing how each unit of the solid separates into distinct ions as it dissolves.
Ionic Dissociation
Ionic dissociation is the term used to describe this separation process. It involves the breaking of ionic bonds in a compound when dissolved in water, forming charged particles or ions. For example, when magnesium chloride (MgCl₂) is introduced into water:
  • MgCl₂ splits into one magnesium ion (Mg²⁺) and two chloride ions (Cl⁻).
  • Water molecules stabilize these ions, preventing them from immediately recombining.

This transformation is crucial because ions are essential to carrying electrical current through the solution. Unlike non-dissociating molecules, these charged entities make strong electrolytes conductive. Each must be accounted for in solutions, as the number and types of ions can affect properties like acidity, basicity, and conductivity.
Aqueous Solution Chemistry
Aqueous solution chemistry revolves around the interactions in water-based solutions. It focuses on how substances dissolve and behave when submerged in water. In these solutions, water acts as a solvent, dissolving solid, liquid, or gaseous solutes to form a mixture. For example, when a substance like aluminum nitrate (Al(NO₃)₃) goes into water:
  • The compound dissociates into aluminum ions (Al³⁺) and nitrate ions (NO₃⁻).
  • These ions intermingle uniformly throughout the solution, forming an ionic mixture.

Aqueous solutions are prevalent in nature and industry. They enable vital biological functions and industrial processes. Thus, understanding how solutes behave in water is important for applications ranging from medicine to environmental science. The principles governing these solutions explain vital phenomena, like why salts dissolve in water yet not in oils.

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Most popular questions from this chapter

Acetylsalicylic acid is the active ingredient in aspirin. It took \(35.17 \mathrm{~mL}\) of \(0.5065 M\) sodium hydroxide to react completely with \(3.210 \mathrm{~g}\) of acetylsalicylic acid. Acetylsalicylic acid has one acidic hydrogen. What is the molar mass of acetylsalicylic acid?

In most of its ionic compounds, cobalt is either \(\mathrm{Co}(\mathrm{II})\) or \(\mathrm{Co}(\mathrm{III})\). One such compound, containing chloride ion and waters of hydration, was analyzed, and the following results were obtained. A \(0.256-\mathrm{g}\) sample of the compound was dissolved in water, and excess silver nitrate was added. The silver chloride was filtered, dried, and weighed, and it had a mass of \(0.308 \mathrm{~g}\). A second sample of \(0.416 \mathrm{~g}\) of the compound was dissolved in water, and an excess of sodium hydroxide was added. The hydroxide salt was filtered and heated in a flame, forming cobalt(III) oxide. The mass of cobalt(III) oxide formed was \(0.145 \mathrm{~g}\). a. What is the percent composition, by mass, of the compound? b. Assuming the compound contains one cobalt ion per formula unit, what is the formula? c. Write balanced equations for the three reactions described.

A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) hydrochloric acid, \(100.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) of nitric acid, \(500.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M}\) calcium hydroxide, and \(200.0 \mathrm{~mL}\) of \(0.100 M\) rubidium hydroxide. Did the acids and bases exactly neutralize each other? If not, calculate the concentration of excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in solution.

A stream flows at a rate of \(5.00 \times 10^{4}\) liters per second \((\mathrm{L} / \mathrm{s})\) upstream of a manufacturing plant. The plant discharges \(3.50 \times 10^{3} \mathrm{~L} / \mathrm{s}\) of water that contains \(65.0 \mathrm{ppm} \mathrm{HCl}\) into the stream. (See Exercise 121 for definitions.) a. Calculate the stream's total flow rate downstream from this plant. b. Calculate the concentration of \(\mathrm{HCl}\) in ppm downstream from this plant. c. Further downstream, another manufacturing plant diverts \(1.80 \times 10^{4} \mathrm{~L} / \mathrm{s}\) of water from the stream for its own use. This plant must first neutralize the acid and does so by adding lime: $$ \mathrm{CaO}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ What mass of \(\mathrm{CaO}\) is consumed in an \(8.00-\mathrm{h}\) work day by this plant? d. The original stream water contained \(10.2 \mathrm{ppm} \mathrm{Ca}^{2+}\). Although no calcium was in the waste water from the first plant, the waste water of the second plant contains \(\mathrm{Ca}^{2+}\) from the neutralization process. If \(90.0 \%\) of the water used by the second plant is returned to the stream, calculate the concentration of \(\mathrm{Ca}^{2+}\) in ppm downstream of the second plant.

Suppose \(50.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{CoCl}_{2}\) solution is added to \(25.0 \mathrm{~mL}\) of \(0.350 \mathrm{M} \mathrm{NiCl}_{2}\) solution. Calculate the concentration, in moles per liter, of each of the ions present after mixing.

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