Chapter 14: Solutions – Structure, Energetics, and Concentration

Solutions

Introduction to Solutions

Solutions are homogeneous mixtures composed of two or more substances. The process of solution formation involves interactions between solute and solvent particles, which determine whether a solution will form and its properties.

Solution Interactions

Types of Intermolecular Interactions

The formation of a solution depends on the relative strengths of three types of interactions:

• Solvent–solvent interactions: Attractions between solvent molecules.

• Solute–solute interactions: Attractions between solute molecules.

• Solvent–solute interactions: Attractions between solvent and solute molecules.

The solution forms if the solvent–solute interactions are comparable to or stronger than the sum of solvent–solvent and solute–solute interactions.

Relative Interactions and Solution Formation

The likelihood of solution formation can be summarized as follows:

Solvent–solute interactions	Solvent–solvent & solute–solute interactions	Result
>		Solution forms
=		Solution forms
<		Solution may or may not form, depending on entropy

When solute-to-solvent attractions are weaker, the solution forms only if the energy difference is small enough to be overcome by the increase in entropy from mixing.

Effect of Intermolecular Forces

To mix solvent and solute, you must overcome:

• All solute–solute attractive forces (endothermic)

• Some solvent–solvent attractive forces (endothermic)

Energy is released when new solute–solvent attractions are formed (exothermic).

Will It Dissolve? – "Like Dissolves Like"

A chemical will dissolve in a solvent if it has a similar structure to the solvent:

• Polar molecules and ionic compounds are more soluble in polar solvents.

• Nonpolar molecules are more soluble in nonpolar solvents.

Energetics of Solution Formation

Steps in Solution Formation

The process of making a solution involves three steps:

1. Separating the solute into its constituent particles (endothermic, $\Delta H_{\text{solute}} > 0$)

2. Separating solvent particles from each other (endothermic, $\Delta H_{\text{solvent}} > 0$)

3. Mixing solute and solvent particles (exothermic, $\Delta H_{\text{mix}} < 0$)

The overall enthalpy change for solution formation is:

$\Delta H_{\text{soln}} = \Delta H_{\text{solute}} + \Delta H_{\text{solvent}} + \Delta H_{\text{mix}}$

Exothermic vs. Endothermic Solution Formation

If the energy cost for breaking attractions in pure solute and solvent is less than the energy released in making new solute–solvent attractions, the process is exothermic. If the energy cost is greater, the process is endothermic.

Heat of Solution and Heat of Hydration

Heat of Solution for Ionic Compounds

For ionic compounds in water, the enthalpy of solution is the difference between the heat of hydration and the lattice energy:

$\Delta H_{\text{solution}} = \Delta H_{\text{hydration}} - \Delta H_{\text{lattice}}$

Ion–Dipole Interactions

When ions dissolve in water, they become hydrated, surrounded by water molecules. The formation of ion–dipole attractions makes the heat of hydration very exothermic.

Solubility and Saturation

Solubility Limit

A solution is saturated when the solute and solvent are in dynamic equilibrium. Adding more solute will not dissolve. The saturation concentration depends on temperature and pressure (for gases).

• Unsaturated: Less solute than saturation; more solute can dissolve.

• Supersaturated: More solute than saturation; unstable and loses excess solute when disturbed.

Supersaturated Solutions

Supersaturated solutions can be made by saturating at non-room conditions and then slowly returning to room conditions. Disturbing the solution (e.g., shaking) causes excess solute to precipitate.

Temperature Dependence of Solubility

Solubility of Solids in Water

For most solids, solubility increases as temperature increases, especially when solution formation is endothermic. Solubility curves help predict whether a solution is saturated, unsaturated, or supersaturated.

Solubility of Gases in Water

Gases generally have lower solubility in water than ionic or polar covalent solids. For all gases, solubility decreases as temperature increases because solution formation is exothermic.

Henry’s Law

The solubility of a gas ($S_{\text{gas}}$) is directly proportional to its partial pressure ($P_{\text{gas}}$):

$S_{\text{gas}} = k_H P_{\text{gas}}$

Where $k_H$ is the Henry’s law constant, which varies for different gases.

Concentration Units

Describing Solution Concentration

Concentration is the amount of solute in a given amount of solution or solvent. Common units include:

Unit	Definition	Units
Molarity (M)	Amount of solute (mol) / Volume of solution (L)	mol/L
Molality (m)	Amount of solute (mol) / Mass of solvent (kg)	mol/kg
Mole fraction ($\chi$)	Amount of solute (mol) / Total moles (solute + solvent)	Unitless
Mole percent	Mole fraction × 100%	%
Percent by mass	Mass solute / Mass solution × 100%	%
Parts per million (ppm)	Mass solute / Mass solution × $10^6$	ppm
Parts per billion (ppb)	Mass solute / Mass solution × $10^9$	ppb
Parts by volume	Volume solute / Volume solution × multiplication factor	—

Preparing a Solution

To prepare a solution of known concentration:

1. Calculate the mass of solute needed using the desired concentration and volume.

2. Dissolve the solute in enough solvent to reach the total volume of solution.

Molarity (M)

Molarity is defined as moles of solute per liter of solution:

$\text{Molarity (M)} = \frac{\text{amount solute (mol)}}{\text{volume solution (L)}}$

Molality (m)

Molality is defined as moles of solute per kilogram of solvent:

$\text{Molality (m)} = \frac{\text{amount solute (mol)}}{\text{mass solvent (kg)}}$

Molality does not vary with temperature because it is based on mass, not volume.

Mole Fraction and Mole Percent

The mole fraction ($\chi$) is the fraction of moles of one component in the total moles of all components:

$\chi = \frac{\text{amount solute (mol)}}{\text{total amount of solute and solvent (mol)}}$

Mole percent is mole fraction × 100%.

Example Calculations

Example 14.4: Calculating concentrations for a solution prepared by dissolving 17.2 g ethylene glycol ($\text{C}_2\text{H}_6\text{O}_2$) in 0.500 kg water, final volume 515 mL:

• (a) Molarity

• (b) Molality

• (c) Percent by mass

• (d) Mole fraction

• (e) Mole percent

Example 14.5: Converting between concentration units for a 6.56% by mass glucose ($\text{C}_6\text{H}_{12}\text{O}_6$) solution with density 1.03 g/mL.

Additional info: The notes cover all major aspects of solution chemistry, including energetics, solubility, concentration units, and practical preparation, suitable for general chemistry students.