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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

If \(1000 \mathrm{NaOH}\) units were dissolved in a sample of water, the NaOH would produce ________ \(\mathrm{Na}^{+}\) ions and ________ \(\mathrm{OH}^{-}\) ions.

On the basis of the general solubility rules given in Table \(7.1,\) write a balanced molecular equation for the precipitation reactions that take place when the following aqueous solutions are mixed. Underline the formula of the precipitate (solid) that forms. If no precipitation reaction is likely for the reactants given, so indicate. a. silver nitrate and hydrochloric acid b. copper(II) sulfate and ammonium carbonate c. iron(II) sulfate and potassium carbonate d. silver nitrate and potassium nitrate e. lead(II) nitrate and lithium carbonate f. tin(IV) chloride and sodium hydroxide

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. \(\mathrm{Ni}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{NiCl}_{2}(s)\)

Without first writing a full molecular or ionic equation, write the net ionic equations for any precipitation reactions that occur when aqueous solutions of the following compounds are mixed. If no reaction occurs, so indicate. a. iron(III) nitrate and sodium carbonate b. mercurous nitrate and sodium chloride c. sodium nitrate and ruthenium nitrate d. copper(II) sulfate and sodium sulfide e. lithium chloride and lead(II) nitrate f. calcium nitrate and lithium carbonate g. gold(III) chloride and sodium hydroxide

Balance each of the following oxidation-reduction reactions. In each, indicate which substance is being oxidized and which is being reduced. a. \(\mathrm{Na}(s)+\mathrm{S}(s) \rightarrow \mathrm{Na}_{2} \mathrm{S}(s)\) b. \(\operatorname{Mg}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{MgO}(s)\) c. \(\mathrm{Ca}(s)+\mathrm{F}_{2}(g) \rightarrow \mathrm{CaF}_{2}(s)\) d. \(\operatorname{Fe}(s)+\mathrm{Cl}_{2}(g) \rightarrow \operatorname{Fe} \mathrm{Cl}_{3}(s)\)
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