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Chapter 7: Problem 8

How do chemists know that the ions behave independently of one another when an ionic solid is dissolved in water?

Short Answer

Expert verified
Chemists know that ions behave independently of one another when an ionic solid is dissolved in water due to the dissociation of the ionic compounds into individual ions, hydration of the ions by water molecules, and experimental evidence such as colligative properties and conductivity measurements. These findings show that ions in solution mainly interact with water molecules and not directly with each other.

Step by step solution

01

Understand Ionic Compounds when Dissolving in Water

Ionic compounds are compounds made up of positively charged metal atoms (cations) and negatively charged nonmetal atoms (anions). These charged atoms (ions) are attracted to each other and held together by their electrostatic forces called ionic bonds. When an ionic solid dissolves in water, it dissociates into its component ions. The ions disperse uniformly in the solution, and they are surrounded by water molecules, which facilitate the separation.
02

Role of Water Molecules in Dissolution Process

Water is a polar molecule, meaning it has a positive and negative end. Water molecules surround the ions by their opposite charge. The negatively charged end (oxygen atom) of water surrounds the positively charged cations and the positively charged end (hydrogen atoms) surrounds the negatively charged anions. This interaction is called hydration. The hydration of ions in solution weakens the ionic attractions between them, making them behave independently of one another.
03

Observations Supporting the Independent Behavior of Ions

One of the experimental evidence supporting the independent behavior of ions is known as the concept of "colligative properties". These are properties of solutions that depend only on the number of dissolved particles in the solution, not their identity. Some examples of colligative properties are freezing point depression, boiling point elevation, and osmotic pressure. The observed changes in these properties when an ionic compound dissolves in water provide indirect evidence that the ions behave independently of one another.
04

Conductivity Experiment

Another experimental evidence is the measurement of the electrical conductivity of solutions containing ionic compounds. When an ionic solid dissolves in water, the solution becomes electrolytic and conducts electricity. This is because the ions are free to move independently and carry an electrical charge from one electrode to another. The fact that the solution conducts electricity even when the ions are far apart suggests that they behave independently of one another.
05

Conclusion

In conclusion, chemists know that ions behave independently of one another when an ionic solid is dissolved in water because of the dissociation of the ionic compounds into individual ions, hydration of the ions by water molecules, and experimental evidence such as colligative properties and conductivity measurements. These findings support the idea that ions in solution interact mainly with water molecules and not directly with each other.

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Most popular questions from this chapter

Balance each of the following equations that describe combustion reactions. a. \(\mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) b. \(\mathrm{C}_{4} \mathrm{H}_{10}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) c. \(C_{6} \mathrm{H}_{14}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

Two "driving forces" for reactions discussed in this section are the formation of water in an acid-base reaction and the formation of a gaseous product. Write balanced chemical equations showing two examples of each type.

Identify each of the following unbalanced reaction equations as belonging to one or more of the following categories: precipitation, acid-base, or oxidation-reduction. a. \(\operatorname{Fe}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Fe}_{3}\left(\mathrm{SO}_{4}\right)_{2}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{HClO}_{4}(a q)+\mathrm{RbOH}(a q) \rightarrow \mathrm{RbClO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) \(\overline{\mathrm{c}} . \mathrm{Ca}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CaO}(s)\) d. \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) e. \(\operatorname{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \rightarrow\) \(\mathrm{PbCO}_{3}(s)+\mathrm{NaNO}_{3}(a q)\) f. \(\mathrm{K}_{2} \mathrm{SO}_{4}(a q)+\mathrm{CaCl}_{2}(a q) \rightarrow \mathrm{KCl}(a q)+\mathrm{CaSO}_{4}(s)\) g. \(\mathrm{HNO}_{3}(a q)+\mathrm{KOH}(a q) \rightarrow \mathrm{KNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) h. \(\mathrm{Ni}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{S}(a q) \rightarrow\) \(\mathrm{NiS}(s)+\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)\) i. \(\mathrm{Ni}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{NiCl}_{2}(s)\)

For the reaction \(2 \mathrm{Al}(s)+3 \mathrm{Br}_{2}(l) \rightarrow 2 \mathrm{AlBr}_{3}(s),\) show how electrons are gained and lost by the atoms.

On the basis of the general solubility rules given in Table \(7.1,\) predict the identity of the precipitate that forms when aqueous solutions of the following substances are mixed. If no precipitate is likely, indicate which rules apply. a. iron(III) chloride, \(\mathrm{FeCl}_{3}\), and phosphoric acid, \(\mathrm{H}_{3} \mathrm{PO}_{4}\) b. barium nitrate, \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2},\) and sodium sulfate, \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) c. potassium chloride, \(\mathrm{KCl}\), and iron(II) sulfate, \(\mathrm{FeSO}_{4}\) d. lead(II) nitrate, \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2},\) and hydrochloric acid, \(\mathrm{HCl}\) e. calcium nitrate, \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2},\) and sodium chloride, \(\mathrm{NaCl}\) f. ammonium sulfide, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{S},\) and copper(II) chloride, \(\mathrm{CuCl}_{2}\)
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