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Chapter 4: Problem 27

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: \({NH}_{4} {C}_{2} {H}_{3} {O}_{2}\) \(\longrightarrow\) \({NH}_{4}⁺\) + \({C}_{2} {H}_{3} {O}_{2}⁻}\)

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Most popular questions from this chapter

What mass of silver chloride can be prepared by the reaction of 100.0 mL of 0.20 M silver nitrate with 100.0 mL of 0.15 M calcium chloride? Calculate the concentrations of each ion remaining in solution after precipitation is complete

Assign oxidation numbers to all the atoms in each of the following. a. \(\mathrm{SrCr}_{2} \mathrm{O}_{7} \quad\) g. \(\mathrm{PbSO}_{3}\) b. \(\mathrm{CuCl}_{2} \quad \quad\) h. \(\mathrm{PbO}_{2}\) c. \(\mathrm{O}_{2} \quad\quad\quad\) i. $\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$ d. \(\mathrm{H}_{2} \mathrm{O}_{2} \quad\quad \mathrm{j} . \mathrm{CO}_{2}\) e. \(\mathrm{MgCO}_{3} \quad\) k. $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}$ f. \(\mathrm{Ag} \quad\quad\quad \)l. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)

Balance the following oxidation–reduction reactions that occur in basic solution. a. $\mathrm{Cr}(s)+\mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(s)$ b. $\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s)$ c. $\mathrm{CN}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)+\mathrm{MnO}_{2}(s)$

A 6.50-g sample of a diprotic acid requires 137.5 mL of a 0.750 M NaOH solution for complete neutralization. Determine the molar mass of the acid.

You made 100.0 mL of a lead(II) nitrate solution for lab but forgot to cap it. The next lab session you noticed that there was only 80.0 mL left (the rest had evaporated). In addition, you forgot the initial concentration of the solution. You decide to take 2.00 mL of the solution and add an excess of a concentrated sodium chloride solution. You obtain a solid with a mass of 3.407 g. What was the concentration of the original lead(II) nitrate solution?
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