Premed · Premed · General Chemistry 1

Lecture 19: Properties of Solutions

General Chemistry I


Learning Objectives

By the end of this lecture, students will be able to:

  1. Describe the solution process at the molecular level and explain "like dissolves like"
  2. Identify the factors that affect solubility (temperature, pressure, molecular structure)
  3. Distinguish between saturated, unsaturated, and supersaturated solutions
  4. Express solution concentration using various units (molarity, molality, mole fraction, mass percent, ppm)
  5. Apply Henry's law to describe gas solubility
  6. Describe the energetics of solution formation and predict whether dissolution is exothermic or endothermic

Lecture Content

I. The Solution Process

A solution is a homogeneous mixture of two or more substances. The solvent is the component present in the largest amount (often a liquid), while the solute is the substance dissolved in it. Dissolution can be understood as a three-step process. First, solute-solute interactions must be broken, separating the solute particles from each other (endothermic, delta_H_1 > 0). Second, some solvent-solvent interactions must be disrupted to make room for the solute (endothermic, delta_H_2 > 0). Third, new solute-solvent interactions form, stabilizing the dissolved state (exothermic, delta_H_3 < 0). The overall enthalpy of solution is delta_H_soln = delta_H_1 + delta_H_2 + delta_H_3. If the energy released in forming solute-solvent interactions exceeds the energy required to break solute-solute and solvent-solvent interactions, dissolution is exothermic and the solution warms. If the reverse is true, dissolution is endothermic and the solution cools. Even endothermic dissolutions can occur spontaneously when the increase in disorder (entropy) is large enough to drive the process.

II. "Like Dissolves Like"

The single most important principle for predicting solubility is "like dissolves like." Polar solutes dissolve in polar solvents because the strong solute-solvent interactions (dipole-dipole, hydrogen bonding, ion-dipole) compensate for the energy needed to break solute-solute and solvent-solvent interactions. Sodium chloride and sugar both dissolve readily in water for this reason. Nonpolar solutes dissolve in nonpolar solvents because the London dispersion forces between solute and solvent are comparable in strength to those being disrupted. Iodine dissolves in carbon tetrachloride, and oil dissolves in hexane. However, nonpolar solutes do not dissolve well in polar solvents, and vice versa, because the strong solvent-solvent interactions (such as hydrogen bonds in water) cannot be adequately compensated by the weak solute-solvent interactions. This is why oil and water do not mix.

III. Hydration and Solvation of Ions

When an ionic compound dissolves in water, the ions become surrounded by water molecules in a process called hydration. Water molecules orient themselves with their oxygen atoms (delta-) pointing toward cations and their hydrogen atoms (delta+) pointing toward anions. The energy released during hydration, called the hydration energy or enthalpy of hydration, is always exothermic. For an ionic compound to dissolve, the hydration energy must be comparable to or greater than the lattice energy. The enthalpy of solution can be thought of as approximately the difference: delta_H_soln is roughly equal to lattice energy minus hydration energy.

<image>A molecular-level diagram of the dissolution of NaCl in water. On the left, a portion of the NaCl crystal lattice is shown at the edge of a water body. Water molecules are shown approaching the crystal surface. In the center, water molecules are pulling Na+ and Cl- ions away from the lattice: water molecules surrounding Na+ have their O atoms pointing inward (ion-dipole attraction), water molecules surrounding Cl- have their H atoms pointing inward. On the right, fully hydrated Na+(aq) and Cl-(aq) ions are shown dispersed in solution, each surrounded by a shell of oriented water molecules (hydration shell). Labels indicate: "Lattice energy must be overcome" on the left, "Ion-dipole forces stabilize ions in solution" on the right, and "delta_H_soln = small (NaCl: +3.9 kJ/mol, slightly endothermic)" at the bottom.</image>

IV. Factors Affecting Solubility

A. Temperature

For solids dissolved in liquids, solubility generally increases with increasing temperature. Since dissolving most solids is endothermic, Le Chatelier's principle predicts that higher temperatures favor greater dissolution. Exceptions are rare (Ce2(SO4)3 is one such anomaly). For gases dissolved in liquids, the trend reverses: solubility generally decreases with increasing temperature. Dissolving a gas is exothermic, so higher temperatures shift the equilibrium toward less dissolved gas. This explains why warm soda goes flat faster and why thermal pollution in lakes reduces dissolved oxygen levels, threatening aquatic life.

B. Pressure (Henry's Law -- for Gases Only)

Pressure has a negligible effect on the solubility of solids and liquids. For gases, however, solubility is directly proportional to the partial pressure of the gas above the solution, as described by Henry's law: S_gas = k_H * P_gas, where S_gas is the solubility, k_H is the Henry's law constant (which depends on the gas, solvent, and temperature), and P_gas is the partial pressure. This principle explains why carbonated beverages are bottled under high CO2 pressure and fizz when opened (the reduced pressure allows CO2 to come out of solution). It also accounts for decompression sickness in scuba diving: at depth, high pressure increases the solubility of nitrogen in the blood, and a rapid ascent allows nitrogen to form bubbles in the bloodstream.

C. Molecular Structure

Solubility increases when the solute and solvent have similar types of intermolecular forces. Functional groups such as -OH, -NH2, and -COOH are polar and capable of hydrogen bonding, increasing a molecule's solubility in water. Long hydrocarbon chains, by contrast, are nonpolar and decrease water solubility. Methanol (CH3OH) is completely miscible with water, but octanol (C8H17OH) is not because its long hydrocarbon portion overwhelms the polar hydroxyl group.

V. Saturated, Unsaturated, and Supersaturated Solutions

An unsaturated solution contains less dissolved solute than the maximum possible, and more solute can still dissolve. A saturated solution contains the maximum amount of solute that can dissolve at a given temperature; a dynamic equilibrium exists between dissolved and undissolved solute. A supersaturated solution contains more dissolved solute than the equilibrium amount, making it inherently unstable. Supersaturation is typically achieved by dissolving solute at an elevated temperature and then cooling slowly. Adding a seed crystal or even a slight disturbance can trigger rapid crystallization, as spectacularly demonstrated in the sodium acetate "hot ice" experiment.

VI. Concentration Units

A. Molarity (M)

Molarity is defined as moles of solute per liter of solution (M = mol/L). It is the most commonly used concentration unit in chemistry but is temperature-dependent because the volume of a solution changes with temperature.

B. Molality (m)

Molality is defined as moles of solute per kilogram of solvent (m = mol/kg). Because it is based on mass rather than volume, molality is temperature-independent and is the preferred unit for colligative property calculations.

C. Mole Fraction (chi)

Mole fraction is the ratio of moles of one component to the total moles of all components: chi_A = n_A / n_total. It is dimensionless, and the sum of all mole fractions in a mixture equals 1. Mole fraction is used in Raoult's law and in gas calculations.

D. Mass Percent (Weight Percent)

Mass percent is calculated as (mass of solute / mass of solution) x 100%. Like molality, it is temperature-independent.

E. Parts per Million (ppm) and Parts per Billion (ppb)

For very dilute solutions, concentration is expressed as ppm = (mass of solute / mass of solution) x 10^6 or ppb = (mass of solute / mass of solution) x 10^9. These units are standard for measuring water contaminants and atmospheric pollutants. For dilute aqueous solutions, 1 ppm is approximately equal to 1 mg/L.

VII. Interconverting Concentration Units

Converting between concentration units requires additional information. The density of the solution relates its mass to its volume. Molar mass converts between mass and moles. And the total mass of the solution equals the mass of solute plus the mass of solvent. The most common conversion -- molarity to molality or vice versa -- requires knowing the solution density.

<image>A conversion diagram showing how to interconvert between concentration units. A central box labeled "Solution Information (mass of solute, mass of solvent, volume of solution, density)" is connected by arrows to five surrounding boxes: Molarity (M = mol solute / L solution), Molality (m = mol solute / kg solvent), Mole Fraction (chi = mol solute / total mol), Mass Percent (mass solute / mass solution x 100), and ppm (mass solute / mass solution x 10^6). Each arrow is labeled with the conversion factor needed (e.g., "need molar mass" or "need density"). A worked example converts 10.0% NaCl solution with density 1.071 g/mL to molarity, molality, and mole fraction, showing each calculation step.</image>

VIII. Solubility of Ionic Compounds -- Energy Perspective

Whether an ionic compound dissolves depends fundamentally on the balance between lattice energy and hydration energy. Lattice energy must always be overcome (an endothermic input), while hydration energy is always exothermic. If the hydration energy roughly matches or exceeds the lattice energy, the compound dissolves, and the dissolution may be slightly exothermic or endothermic. If the lattice energy greatly exceeds the hydration energy, the compound is insoluble, as with BaSO4 and CaCO3. Entropy also plays a role: dissolution generally increases disorder, which thermodynamically favors solubility even when the process is slightly endothermic.

Lecture 19: Properties of Solutions — figure 1
Lecture 19: Properties of Solutions — figure 2

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