General Chemistry II: Solutions, Intermolecular Forces, and Gases – Final Exam Review Notes

Solutions and Solution Stoichiometry

Concentration Units: Molarity

Concentration describes the amount of solute dissolved in a given quantity of solvent or solution. Molarity (M) is the most common unit, defined as moles of solute per liter of solution.

• Molarity (M): $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

• Simple Dilutions: $M_1V_1 = M_2V_2$ (where 1 = initial, 2 = final)

Example: To prepare 250 mL of 0.10 M NaCl from a 1.0 M stock solution, use $V_1 = \frac{M_2V_2}{M_1} = \frac{0.10 \times 0.250}{1.0} = 0.025$ L = 25 mL of stock solution.

Titration and Neutralization Reactions

Titration is a technique to determine the concentration of an unknown solution by reacting it with a solution of known concentration.

• Equivalence Point: The point at which stoichiometrically equivalent quantities of acid and base have reacted.

• Stoichiometry: Use the balanced chemical equation to relate moles of reactants and products.

Example: 15.00 mL of HClO4 titrated with 17.03 mL of 0.1000 M NaOH. The reaction is:

• $\mathrm{HClO_4(aq) + NaOH(aq) \rightarrow H_2O(l) + NaClO_4(aq)}$

• Since the ratio is 1:1, $M_1V_1 = M_2V_2$

• $M_{\text{HClO}_4} = \frac{17.03 \times 0.1000}{15.00} = 0.1135$ M

Limiting Reactant, Theoretical Yield, and Percent Yield

In precipitation and other reactions, the limiting reactant is the one that is completely consumed first, determining the maximum amount of product formed.

• Theoretical Yield: The maximum amount of product possible, calculated from the limiting reactant.

• Percent Yield: $\%\text{Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$

Example: Mixing 25.0 mL of 1.20 M KCl with 15.0 mL of 0.900 M Ba(NO3)2 forms BaCl2:

• Limiting reactant: Ba(NO3)2 (13.5 mmol)

• Theoretical yield of BaCl2: 2.81 g

• Percent yield if 2.45 g collected: $\frac{2.45}{2.81} \times 100 = 87.2\%$

Solubility and Precipitation Reactions

Solubility Rules for Ionic Compounds in Water

Solubility rules help predict whether an ionic compound will dissolve in water or form a precipitate.

Compound of	Rule
Li+, Na+, K+, NH4+	Always soluble
NO3-, C2H3O2-	Always soluble
Cl-, Br-, I-	Insoluble with Ag+, Hg22+, or Pb2+; soluble with other ions
SO42-	Insoluble with Ba2+, Ca2+, Sr2+, Pb2+, Ag+, Hg22+; soluble with other ions
CO32-, PO43-	Soluble with Li+, Na+, K+, NH4+; insoluble with other ions
OH-, S2-	Soluble with Li+, Na+, K+, NH4+; OH- also soluble with Ca2+, Sr2+, Ba2+

Precipitation Reactions

When two aqueous solutions are mixed, an insoluble product (precipitate) may form if the combination of ions produces an insoluble compound.

• Example: $2\mathrm{Na_3PO_4(aq)} + 3\mathrm{CuCl_2(aq)} \rightarrow Cu_3(PO_4)_2(s) + 6\mathrm{NaCl(aq)}$

• To find the required volume of one reactant, use stoichiometry and molarity relationships.

Electrolytes and Conductivity

Compounds that dissociate into ions in water conduct electricity (electrolytes). Molecular compounds that do not ionize are non-electrolytes.

• Strong electrolytes: Ionic compounds (e.g., NaOH, MgSO4)

• Non-electrolytes: Most molecular compounds (e.g., C12H22O11)

Intermolecular Forces and Properties of Liquids/Solids

Types of Intermolecular Forces (IMFs)

IMFs are forces of attraction between molecules, affecting physical properties like boiling and melting points.

• Dispersion (London) Forces: Present in all molecules, especially significant in nonpolar molecules.

• Dipole-Dipole Forces: Occur between polar molecules.

• Hydrogen Bonding: Strong dipole-dipole interaction when H is bonded to N, O, or F.

Example Table:

Compound	IMFs Present
Xe	Dispersion
HF	Dispersion, Dipole-Dipole, Hydrogen Bonding
NF3	Dispersion, Dipole-Dipole
SiBr4	Dispersion
CH3OH	Dispersion, Dipole-Dipole, Hydrogen Bonding

Boiling Point Comparisons

Boiling point increases with stronger IMFs.

• Hydrogen bonding > Dipole-dipole > Dispersion

• Example: NH3 (hydrogen bonding) has a higher boiling point than CH4 (dispersion only).

Thermochemistry: Heating, Cooling, and Phase Changes

Heat Calculations

Heat (q) is the energy transferred due to temperature difference. For temperature changes and phase transitions:

• Heating/Cooling: $q = m \cdot C_s \cdot \Delta T$

• Phase Transitions: $q = n \cdot \Delta H$ (per mole), or $q = m \cdot \Delta H$ (per gram)

Example: To calculate the heat required to melt 10 g of ice, use the enthalpy of fusion for water.

Heating Curve for Water

The heating curve shows temperature changes and phase transitions as heat is added to water. Plateaus represent phase changes (melting, boiling) where temperature remains constant.

Exothermic and Endothermic Processes

• Condensation: Releases heat (exothermic), $\Delta H$ is negative.

• Melting, Boiling: Absorb heat (endothermic), $\Delta H$ is positive.

Example: Condensation of 10 g of water vapor releases more heat than freezing or melting the same mass of water.

Gases and Gas Laws

Ideal Gas Law

The behavior of gases is described by the ideal gas law:

• $PV = nRT$

• P = pressure (atm), V = volume (L), n = moles, R = 0.08206 L·atm/(mol·K), T = temperature (K)

Example: To find the mass of N2 in a 7.5 L cylinder at 1.00 atm and 298 K:

• $n = \frac{PV}{RT} = \frac{1.00 \times 7.5}{0.08206 \times 298} = 0.307$ mol

• Mass = $0.307 \text{ mol} \times 28.02 \text{ g/mol} = 8.60$ g

Stoichiometry of Gaseous Reactions

Use the ideal gas law and stoichiometry to relate volumes and moles in reactions involving gases.

• Example: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ at STP. 2.00 mol N2 and 6.00 mol H2 produce 4.00 mol NH3, which at STP occupies $4.00 \times 22.4 = 89.6$ L.

Kinetic Molecular Theory and Molecular Velocities

The distribution of molecular velocities depends on molar mass and temperature. Lighter molecules move faster on average.

• Graham's Law of Effusion: $\frac{\text{Rate}_A}{\text{Rate}_B} = \sqrt{\frac{M_B}{M_A}}$

• At the same temperature, molecules with lower molar mass have higher average velocities and effusion rates.

Example: In a velocity distribution graph, the curve with the lower peak velocity corresponds to the heavier molecule.

Summary Table: Key Equations and Concepts

Concept	Equation
Molarity	$M = \frac{\text{moles}}{\text{liters}}$
Dilution	$M_1V_1 = M_2V_2$
Ideal Gas Law	$PV = nRT$
Heat (q) for temp. change	$q = mC_s\Delta T$
Heat (q) for phase change	$q = n\Delta H$ or $q = m\Delta H$
Percent Yield	$\%\text{Yield} = \frac{\text{Actual}}{\text{Theoretical}} \times 100$
Graham's Law	$\frac{\text{Rate}_A}{\text{Rate}_B} = \sqrt{\frac{M_B}{M_A}}$

Additional info:

• These notes cover topics from General Chemistry chapters on solutions, intermolecular forces, thermochemistry, and gases, as well as laboratory and mathematical operations relevant to these topics.

• Some context and explanations have been expanded for clarity and completeness.