LO 1.16 The student can design and/or interpret the results of an experiment regarding the absorption of light to determine the concentration of an absorbing species in a solution. (Sec 11.1) LO 2.8 The student can draw and/or interpret representations of solutions that show the interactions between the solute and solvent. (Sec 11.1) LO 2.9 The student is able to create or interpret representations that link the concept of molarity with particle views of solutions. (Sec 11.2) LO 2.14 The student is able to apply Coulomb’s Law qualitatively (including using representations) to describe the interactions of ions, and the attractions between ions and solvents to explain the factors that contribute to the solubility of ionic compounds. (Sec 11.2) LO 2.15 The student is able to explain observations regarding the solubility of ionic solids and molecules in water and other solvents on the basis of particle views that include intermolecular interactions and entropic effects. (Sec 11.2-11.3)

LO 5.10 The student can support the claim about whether a process is a chemical or physical change (or may be classified as both) based on whether the process involves changes in intramolecular versus intermolecular interactions. (Sec 11.2) LO 6.24 The student can analyze the enthalpic and entropic changes associated with the dissolution of a salt, using particulate level interactions and representations. (Sec 11.2)

AP Learning Objectives, Margin Notes and References LO 1.16 The student can design and/or interpret the results of an experiment regarding the absorption of light to determine the concentration of an absorbing species in a solution. LO 2.8 The student can draw and/or interpret representations of solutions that show the interactions between the solute and solvent. AP Margin Notes Spectral analysis is a common method for analyzing the composition of a solution. See Appendix 3 “Spectral Analysis” for a discussion of the Beer-Lambert law. Additional AP References LO 1.16 (see APEC #2, “The Percentage of Copper in Brass”)

Various Types of Solutions Copyright © Cengage Learning. All rights reserved

Solution Composition Copyright © Cengage Learning. All rights reserved

Molarity Copyright © Cengage Learning. All rights reserved

EXERCISE! You have 1.00 mol of sugar in 125.0 mL of solution. Calculate the concentration in units of molarity. 8.00 M 1.00 mol / (125.0 / 1000) = 8.00 M Copyright © Cengage Learning. All rights reserved

EXERCISE! You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar? 0.200 L 2.00 mol / 10.0 M = 0.200 L Copyright © Cengage Learning. All rights reserved

EXERCISE! Consider separate solutions of NaOH and KCl made by dissolving 100.0 g of each solute in 250.0 mL of solution. Calculate the concentration of each solution in units of molarity. 10.0 M NaOH 5.37 M KCl [100.0 g NaOH / 39.998 g/mol] / [250.0 / 1000] = 10.0 M NaOH [100.0 g KCl / 74.55 g/mol] / [250.0 / 1000] = 5.37 M KCl Copyright © Cengage Learning. All rights reserved

Mass Percent Copyright © Cengage Learning. All rights reserved

EXERCISE! What is the percent-by-mass concentration of glucose in a solution made my dissolving 5.5 g of glucose in 78.2 g of water? 6.6% [5.5 g / (5.5 g + 78.2 g)] × 100 = 6.6% Copyright © Cengage Learning. All rights reserved

Mole Fraction Copyright © Cengage Learning. All rights reserved

EXERCISE! A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the mole fraction of H3PO4. (Assume water has a density of 1.00 g/mL.) 0.0145 8.00 g H3PO4 × (1 mol / 97.994 g H3PO4) = 0.0816 mol H3PO4 100.0 mL H2O × (1.00 g H2O / mL) × (1 mol / 18.016 g H2O) = 5.55 mol H2O Mole Fraction (H3PO4) = 0.0816 mol H3PO4 / [0.0816 mol H3PO4 + 5.55 mol H2O] = 0.0145 Copyright © Cengage Learning. All rights reserved

Molality Copyright © Cengage Learning. All rights reserved

EXERCISE! A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/mL.) 0.816 m 8.00 g H3PO4 × (1 mol / 97.994 g H3PO4) = 0.0816 mol H3PO4 100.0 mL H2O × (1.00 g H2O / mL) × (1 kg / 1000 g) = 0.1000 kg H2O Molality = 0.0816 mol H3PO4 / 0.1000 kg H2O] = 0.816 m Copyright © Cengage Learning. All rights reserved

AP Learning Objectives, Margin Notes and References LO 2.9 The student is able to create or interpret representations that link the concept of molarity with particle views of solutions. LO 2.14 The student is able to apply Coulomb’s Law qualitatively (including using representations) to describe the interactions of ions, and the attractions between ions and solvents to explain the factors that contribute to the solubility of ionic compounds. LO 2.15 The student is able to explain observations regarding the solubility of ionic solids and molecules in water and other solvents on the basis of particle views that include intermolecular interactions and entropic effects. LO 5.10 The student can support the claim about whether a process is a chemical or physical change (or may be classified as both) based on whether the process involves changes in intramolecular versus intermolecular interactions. LO 6.24 The student can analyze the enthalpic and entropic changes associated with the dissolution of a salt, using particulate level interactions and representations. Additional AP References LO 5.10 (see Appendix 7.6, “Distinguishing between Chemical and Physical Changes at the Molecular Level”) LO 6.24 (see Appendix 7.7, “Intermolecular Forces and Thermodynamics: Why Aren’t All Ionic Solids Soluble in Water?”)

Formation of a Liquid Solution Separating the solute into its individual components (expanding the solute). Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent). Allowing the solute and solvent to interact to form the solution. Copyright © Cengage Learning. All rights reserved

Steps in the Dissolving Process Copyright © Cengage Learning. All rights reserved

Steps in the Dissolving Process Steps 1 and 2 require energy, since forces must be overcome to expand the solute and solvent. Step 3 usually releases energy. Steps 1 and 2 are endothermic, and step 3 is often exothermic. Copyright © Cengage Learning. All rights reserved

Enthalpy (Heat) of Solution Enthalpy change associated with the formation of the solution is the sum of the ΔH values for the steps: ΔHsoln = ΔH1 + ΔH2 + ΔH3 ΔHsoln may have a positive sign (energy absorbed) or a negative sign (energy released). Copyright © Cengage Learning. All rights reserved

Enthalpy (Heat) of Solution Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Explain why water and oil (a long chain hydrocarbon) do not mix. In your explanation, be sure to address how ΔH plays a role. Oil is a mixture of nonpolar molecules that interact through London dispersion forces, which depend on molecule size. ΔH1 will be relatively large for the large oil molecules. The term ΔH3 will be small, since interactions between the nonpolar solute molecules and the polar water molecules will be negligible. However, ΔH2 will be large and positive because it takes considerable energy to overcome the hydrogen bonding forces among the water molecules to expand the solvent. Thus ΔHsoln will be large and positive because of the ΔH1 and ΔH2 terms. Since a large amount of energy would have to be expended to form an oil-water solution, this process does not occur to any appreciable extent. Copyright © Cengage Learning. All rights reserved

The Energy Terms for Various Types of Solutes and Solvents ΔHsoln Outcome Polar solute, polar solvent Large Large, negative Small Solution forms Nonpolar solute, polar solvent Large, positive No solution forms Nonpolar solute, nonpolar solvent Polar solute, nonpolar solvent Copyright © Cengage Learning. All rights reserved

In General One factor that favors a process is an increase in probability of the state when the solute and solvent are mixed. Processes that require large amounts of energy tend not to occur. Overall, remember that “like dissolves like”. Copyright © Cengage Learning. All rights reserved

AP Learning Objectives, Margin Notes and References LO 2.15 The student is able to explain observations regarding the solubility of ionic solids and molecules in water and other solvents on the basis of particle views that include intermolecular interactions and entropic effects.

Affecting aqueous solutions Structure Effects: Polarity Pressure Effects: Henry’s law Temperature Effects: Affecting aqueous solutions Copyright © Cengage Learning. All rights reserved

Structure Effects Hydrophobic (water fearing) Non-polar substances Hydrophilic (water loving) Polar substances Copyright © Cengage Learning. All rights reserved

Pressure Effects C = concentration of dissolved gas k = constant Little effect on solubility of solids or liquids Henry’s law: C = kP C = concentration of dissolved gas k = constant P = partial pressure of gas solute above the solution Amount of gas dissolved in a solution is directly proportional to the pressure of the gas above the solution. Copyright © Cengage Learning. All rights reserved

A Gaseous Solute Copyright © Cengage Learning. All rights reserved

Temperature Effects (for Aqueous Solutions) Although the solubility of most solids in water increases with temperature, the solubilities of some substances decrease with increasing temperature. Predicting temperature dependence of solubility is very difficult. Solubility of a gas in solvent typically decreases with increasing temperature. Copyright © Cengage Learning. All rights reserved

The Solubilities of Several Solids as a Function of Temperature Copyright © Cengage Learning. All rights reserved

The Solubilities of Several Gases in Water Copyright © Cengage Learning. All rights reserved