Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 f. \({FeSO}_{4}\) b. \({MgCl}_{2}\) g. \({KMnO}_{4}\) c. \({Al}({NO}_{3})_{3}\) h. \({HClO}_{4}\) d. \(({NH}_{4})_{2} {SO}_{4}\) i. \({NH}_{4} {C}_{2} {H}_{3} {O}_{2}\) (ammonium acetate) e. \({NaOH}\)

Short Answer

Expert verified
a. NaBr \(\longrightarrow\) Na⁺ + Br⁻ b. \({MgCl}_{2}\) \(\longrightarrow\) Mg²⁺ + 2Cl⁻ c. \({Al}({NO}_{3})_{3}\) \(\longrightarrow\) Al³⁺ + 3\({NO}_{3}⁻\) d. \(({NH}_{4})_{2} {SO}_{4}\) \(\longrightarrow\) 2\({NH}_{4}⁺\) + \({SO}_{4}²⁻\) e. \({NaOH}\) \(\longrightarrow\) Na⁺ + OH⁻ f. \({FeSO}_{4}\) \(\longrightarrow\) Fe²⁺ + \({SO}_{4}²⁻\) g. \({KMnO}_{4}\) \(\longrightarrow\) K⁺ + \({MnO}_{4}⁻\) h. \({HClO}_{4}\) \(\longrightarrow\) H⁺ + \({ClO}_{4}⁻\) i. \({NH}_{4} {C}_{2} {H}_{3} {O}_{2}\) \(\longrightarrow\) \({NH}_{4}⁺\) + \({C}_{2} {H}_{3} {O}_{2}⁻\)

Step by step solution

01

a. NaBr dissociation

Sodium bromide (NaBr) dissociates into the sodium ion (Na⁺) and the bromide ion (Br⁻). Upon dissolving in water, the formula can be written as: NaBr \(\longrightarrow\) Na⁺ + Br⁻
02

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

Magnesium chloride (\({MgCl}_{2}\)) dissociates into the magnesium ion (Mg²⁺) and two chloride ions (2Cl⁻). Upon dissolving in water, the formula can be written as: \({MgCl}_{2}\) \(\longrightarrow\) Mg²⁺ + 2Cl⁻
03

c. \({Al}({NO}_{3})_{3}\) dissociation

Aluminum nitrate (\({Al}({NO}_{3})_{3}\)) dissociates into the aluminum ion (Al³⁺) and three nitrate ions (3\({NO}_{3}⁻\)). Upon dissolving in water, the formula can be written as: \({Al}({NO}_{3})_{3}\) \(\longrightarrow\) Al³⁺ + 3\({NO}_{3}⁻\)
04

d. \(({NH}_{4})_{2} {SO}_{4}\) dissociation

Ammonium sulfate (\(({NH}_{4})_{2} {SO}_{4}\)) dissociates into two ammonium ions (2\({NH}_{4}⁺\)) and one sulfate ion (\({SO}_{4}²⁻\)). Upon dissolving in water, the formula can be written as: \(({NH}_{4})_{2} {SO}_{4}\) \(\longrightarrow\) 2\({NH}_{4}⁺\) + \({SO}_{4}²⁻\)
05

e. \({NaOH}\) dissociation

Sodium hydroxide (\({NaOH}\)) dissociates into the sodium ion (Na⁺) and the hydroxide ion (OH⁻). Upon dissolving in water, the formula can be written as: \({NaOH}\) \(\longrightarrow\) Na⁺ + OH⁻
06

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

Iron(II) sulfate (\({FeSO}_{4}\)) dissociates into the iron(II) ion (Fe²⁺) and the sulfate ion (\({SO}_{4}²⁻\)). Upon dissolving in water, the formula can be written as: \({FeSO}_{4}\) \(\longrightarrow\) Fe²⁺ + \({SO}_{4}²⁻\)
07

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

Potassium permanganate (\({KMnO}_{4}\)) dissociates into the potassium ion (K⁺) and the permanganate ion (\({MnO}_{4}⁻\)). Upon dissolving in water, the formula can be written as: \({KMnO}_{4}\) \(\longrightarrow\) K⁺ + \({MnO}_{4}⁻\)
08

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

Perchloric acid (\({HClO}_{4}\)) dissociates into the hydrogen ion (H⁺) and the perchlorate ion (\({ClO}_{4}⁻\)). Upon dissolving in water, the formula can be written as: \({HClO}_{4}\) \(\longrightarrow\) H⁺ + \({ClO}_{4}⁻\)
09

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

Ammonium acetate (\({NH}_{4} {C}_{2} {H}_{3} {O}_{2}\)) dissociates into the ammonium ion (\({NH}_{4}⁺\)) and the acetate ion (\({C}_{2} {H}_{3} {O}_{2}⁻\)). Upon dissolving in water, the formula can be written as: \({NH}_{4} {C}_{2} {H}_{3} {O}_{2}\) \(\longrightarrow\) \({NH}_{4}⁺\) + \({C}_{2} {H}_{3} {O}_{2}⁻}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Ion Dissociation
When strong electrolytes dissolve in water, they undergo a process called ion dissociation. This means that the compound breaks down into its component ions. This is because water molecules are highly polar, meaning they have a positive side and a negative side. This polarity allows water to surround the ions, separating them and keeping them evenly dispersed within the solution.

Let's consider sodium bromide (NaBr). It dissociates into sodium ions (Na⁺) and bromide ions (Br⁻) when it is added to water. The interaction with water molecules results in the formula: \[\text{NaBr} \longrightarrow \text{Na}^+ + \text{Br}^-\]This principle applies to all strong electrolytes, like magnesium chloride (MgCl₂), which dissociates into one magnesium ion (Mg²⁺) and two chloride ions (Cl⁻):\[\text{MgCl}_2 \longrightarrow \text{Mg}^{2+} + 2\text{Cl}^-\]This process is crucial for understanding how solutions conduct electricity. The free movement of ions in water carries electric current, which makes solutions made from strong electrolytes excellent conductors.
Aqueous Solution
An aqueous solution is a solution in which water is the solvent. When ionic compounds dissolve in water, they form aqueous solutions by dissociating into individual ions. This is significant because it means that the original compound's structure is broken down and the ions interact independently with the water molecules.

For example, when aluminum nitrate (Al(NO₃)₃) dissolves, it yields aluminum ions (Al³⁺) and nitrate ions (NO₃⁻) in the solution.
  • Aluminum nitrate in water will dissociate like this:\[\text{Al}(\text{NO}_3)_3 \longrightarrow \text{Al}^{3+} + 3\text{NO}_3^-\]
  • This means that each aluminum ion is surrounded by water molecules, stabilizing it in the solution.
Aqueous solutions are not limited to salts. They can also include acids, such as perchloric acid (HClO₄), which dissociates into hydrogen ions (H⁺) and perchlorate ions (ClO₄⁻).\[\text{HClO}_4 \longrightarrow \text{H}^+ + \text{ClO}_4^-\]In both cases, water's role is vital for separating and keeping ions throughout the solution, enabling various chemical processes.
Chemical Equations
Chemical equations are a concise way of showing what happens in a chemical reaction, including ion dissociation. These equations use formulas and symbols to depict the substances involved and what they change into.

In the context of ion dissociation, equations show how a solid compound separates into individual ions when it is dissolved in water. For instance, look at ammonium sulfate ((NH₄)₂SO₄) when it dissociates:\[(\text{NH}_4)_2\text{SO}_4 \longrightarrow 2\text{NH}_4^+ + \text{SO}_4^{2-}\]The chemical equation indicates that two ammonium ions are created for each sulfate ion when this compound dissolves.
  • These equations guarantee that the balance of charge is maintained, meaning the number of positive and negative charges is equal on both sides.
The beauty of chemical equations is in their ability to communicate complex reactions in a straightforward manner, helping us to better understand processes such as the dissociation of electrolytes like ammonium acetate (NH₄C₂H₃O₂):\[\text{NH}_4\text{C}_2\text{H}_3\text{O}_2 \longrightarrow \text{NH}_4^+ + \text{C}_2\text{H}_3\text{O}_2^-\]This clear depiction is indispensable for studying and predicting the behavior of chemical substances in various conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

It took \(25.06 \pm 0.05 \mathrm{mL}\) of a sodium hydroxide solution to titrate a 0.4016-g sample of KHP (see Exercise 83). Calculate the concentration and uncertainty in the concentration of the sodium hydroxide solution. (See Appendix 1.5.) Neglect any uncertainty in the mass

Hydrochloric acid \((75.0 \mathrm{mL} \text { of } 0.250 \mathrm{M})\) is added to 225.0 \(\mathrm{mL}\) of 0.0550 \(\mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution. What is the concentration of the excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in this solution?

Balance the following oxidation–reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathbf{I}^{-}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{NO}(g)\) c. \(\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{Mn}^{2+}(a q)\) d. \(\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)\)

A solution is prepared by dissolving 0.6706 g oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) in enough water to make 100.0 \(\mathrm{mL}\) of solution. A 10.00-mL aliquot (portion) of this solution is then diluted to a final volume of 250.0 mL. What is the final molarity of the oxalic acid solution?

Saccharin \(\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{S}\right)\) is sometimes dispensed in tablet form. Ten tablets with a total mass of 0.5894 g were dissolved in water. The saccharin was oxidized to convert all the sulfur to sulfate ion, which was precipitated by adding an excess of barium chloride solution. The mass of BaSO\(_{4}\) obtained was 0.5032 g. What is the average mass of saccharin per tablet? What is the average mass percent of saccharin in the tablets?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free