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Chapter 13: Problem 3

What is solvation? What are the factors that influence the extent to which solvation occurs? Give two examples of solvation, including one that involves ion-dipole interaction and another in which dispersion forces come into play.

Short Answer

Expert verified
Solvation is the process where solvent molecules surround and interact with solute particles. Factors affecting solvation include the nature of solute and solvent, temperature, and pressure. An example of ion-dipole solvation is NaCl in water, while an example of solvation involving dispersion forces is the dissolution of helium gas in hexane.

Step by step solution

01

Define Solvation

Solvation is the process by which solvent molecules surround and interact with solute particles. This interaction leads to stabilization and dispersion of solute particles in the solvent.
02

Factors Influencing Solvation

Factors that influence the extent of solvation include: \n1. The nature of the solute and the solvent: Similar substances tend to dissolve in each other. This is due to the 'Like dissolves Like' rule, meaning polar solutes dissolve in polar solvents, while nonpolar solutes dissolve in nonpolar solvents. \n2. Temperature: Increasing temperature generally increases solubility as it provides kinetic energy which helps overcome the forces holding the solute particles together. \n3. Pressure: The effect of pressure on solubility of solids and liquids is negligible. However, for gases, an increase in pressure increases solubility according to Henry's Law.
03

Examples of Solvation

Example 1 - Ion-Dipole Interaction: Sodium chloride (NaCl) in water demonstrates this interaction. Na and Cl ions interact with the polar water molecules, with the Na+ ions being attracted to the oxygen end of water, while the the Cl- ions are attracted to the Hydrogen end. This interaction leads to dissolving of NaCl in water. \nExample 2 - Dispersion Forces: The solvation of helium gas in a nonpolar solvent like Hexane demonstrates this. Here, helium, a noble gas with a filled valence shell, can form temporary induced dipole interactions with the electrons of nonpolar Hexane molecules, leading to its solvation.

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Most popular questions from this chapter

Calculate the molarity and the molality of \(\mathrm{NH}_{3}\) for a solution of \(30.0 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) in \(70.0 \mathrm{~g}\) of water. The density of the solution is \(0.982 \mathrm{~g} / \mathrm{mL}\).

The solubility of \(\mathrm{KNO}_{3}\) is \(155 \mathrm{~g}\) per \(100 \mathrm{~g}\) of water at \(75^{\circ} \mathrm{C}\) and \(38.0 \mathrm{~g}\) at \(25^{\circ} \mathrm{C}\). What mass (in grams) of \(\mathrm{KNO}_{3}\) will crystallize out of solution if exactly \(100 \mathrm{~g}\) of its saturated solution at \(75^{\circ} \mathrm{C}\) are cooled to \(25^{\circ} \mathrm{C} ?\)

Why is naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) more soluble than \(\mathrm{CsF}\) in benzene?

Calculate the molalities of these aqueous solutions: (a) \(1.22 M\) sugar \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) solution (density of solution \(=1.12 \mathrm{~g} / \mathrm{mL}\) ), (b) \(0.87 \mathrm{M} \mathrm{NaOH}\) solution (density of solution \(=1.04 \mathrm{~g} / \mathrm{mL}),(\mathrm{c})\) \(5.24 \mathrm{M} \mathrm{NaHCO}_{3}\) solution (density of solution \(=\) \(1.19 \mathrm{~g} / \mathrm{mL})\)

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