CHE - Lecture #1: Ch. 14 Solutions
Solubility: Key Concepts
Definition of solution
A solution is a homogeneous mixture of at least two substances. From the slides: “Solution — Homogeneous mixture of at least two substances.”
In coffee, the solution typically contains solutes like caffeine, citric acid, chlorogenic acid, and phenols in water (the solvent).
Coffee example: identifying solvent and solute
Solvent in coffee is water.
solvent: larger
solute: smaller
The coffee cup example emphasizes that the majority component is the solvent; in many aqueous solutions the solvent is water.
Ocean example: the ocean is primarily water with dissolved salts; the salt (Na⁺, Cl⁻) constitutes the solute in the aqueous solvent.
Solubility rules (memory aids)
Rule 1: All group 1 cations (Li, Na, K, Rb, Cs) and ammonium salts are soluble.
Rule 2: All nitrate (NO₃⁻) and acetate (CH₃COO⁻) salts are soluble.
Quick discussion question (from transcript): Which of the following compounds will be soluble in water? A. BaSO₄ B. CaCO₃ C. Mg(OH)₂ D. (NH₄)₂S E. All of these compounds will be soluble in water
Correct interpretation from context: Only ammonium sulfide ((NH₄)₂S) is clearly soluble; BaSO₄, CaCO₃, and Mg(OH)₂ are sparingly soluble or insoluble in water. (This aligns with typical solubility rules; ammonium salts are soluble, others are not in general.)
What ultimately determines solubility?
Non-covalent intermolecular interactions between molecules govern solubility.
Intermolecular forces determine how well solutes interact with solvent molecules.
Intermolecular forces (Quick Review)
Dispersion (London) forces
Present in all molecules and atoms.
Strength: relatively weak; increases with molecular size and surface area.
Dipole–dipole interactions
Present in polar molecules.
Strength: stronger than dispersion for many polar molecules.
Hydrogen bonding
Special case of dipole–dipole where H is bonded to F, O, or N.
Strength: strong; contributes significantly to solubility in many systems.
Ion–dipole interactions
Interactions between ions (from ionic compounds) and polar molecules.
Strength: strong, especially in aqueous solutions of salts.
Hierarchy (conceptual, from strongest to comparatively weaker interactions shown in the slides):
Ion–dipole > Hydrogen bonding > Dipole–dipole > Dispersion
How intermolecular forces relate to solutions
Solubility arises from favorable solute–solvent interactions, often driven by the same or complementary intermolecular forces.
The maxim “like dissolves like” reflects the idea that solutes and solvents with similar intermolecular forces tend to dissolve each other.
Terms introduced: miscible (complete solubility in all proportions) vs insoluble (little to no solubility).
Example statement from slide: If you’re not part of the solution, you’re part of the precipitate! (general reminder about solubility outcomes)
Specific examples of solubility concepts
CO₂ in H₂O is a common demonstration of solubility; CO₂ dissolves in water forming carbonic acid under certain conditions.
Coffee as a solution demonstration: water as solvent; solutes include caffeine, citric acid, chlorogenic acid, phenols.
Relationship to solutions: defining terminology
Solvent: The majority component of a solution; in aqueous solutions, the solvent is water.
Solute: The component dissolved in the solvent.
Aqueous solution: A solution where the solvent is water.
Example: Ocean composition; solvent = water; solutes = dissolved salts like Na⁺ and Cl⁻.
Conceptual equation: solute dissolves in solvent → formation of solution.
Concentration units you need to memorize (Table 13.5 context)
Molarity (M)
Definition: the amount of solute in moles per liter of solution.
M = rac{n{ ext{solute}}}{V{ ext{solution}}}
Molality (m)
Definition: moles of solute per kilogram of solvent.
m = rac{n{ ext{solute}}}{m{ ext{solvent}}} ext{ with } m_{ ext{solvent}} ext{ in kg}
Mole fraction (X)
Definition: ratio of moles of a component to the total moles in the mixture.
XA = rac{nA}{nA + nB}
Mole percent (mol %)
ext{mol ext{-} ext{%}A} = XA imes 100ackslash %
Percent by mass (% by mass)
Definition: mass of solute divided by mass of solution, times 100.
ext{
mass
%} = rac{m{ ext{solute}}}{m{ ext{solution}}} imes 100 ext{%}Parts per million (ppm) and parts per billion (ppb)
ext{ppm} = rac{m{ ext{solute}}}{m{ ext{solution}}} imes 10^6
ext{ppb} = rac{m{ ext{solute}}}{m{ ext{solution}}} imes 10^9
Parts by volume (% v/v) and related measures follow the same idea but use volumes when appropriate.
Memorization reminder (Unit conversions and formulas)
Between today and Friday, work on Examples 14.4 and 14.5.
You will NOT be given concentration formulas or conversion factors on the exam; you NEED to memorize them.
Focus on understanding how to convert between concentration units and perform unit conversions confidently.
Practice problem: Hydrogen peroxide molality (from 30.0% by mass, $M_{ ext{H2O2}} = 34.0 ext{ g/mol}$)
Problem statement (from transcript): An aqueous solution of H₂O₂ is 30.0% by mass. Calculate its molality.
Assumptions for a simple calculation: use a 100.0 g solution sample.
Step 1: Determine masses
Mass of solute (H₂O₂): m_{ ext{solute}} = 0.30 imes 100.0 ext{ g} = 30.0 ext{ g}
Mass of solvent (water): m_{ ext{solvent}} = 100.0 ext{ g} - 30.0 ext{ g} = 70.0 ext{ g} = 0.0700 ext{ kg}
Step 2: Convert mass of solute to moles
$$n{ ext{solute}} = rac{m{ ext{solute}}}{M_{ ext{H2O2}}} = rac{30.0 ext{ g}}{34.0 ext{ g/mol}} \