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.

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 )

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
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EXERCISE! Consider separate solutions of NaOH and KCl made by dissolving g of each solute in 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 / g/mol] / [250.0 / 1000] = 10.0 M NaOH [100.0 g KCl / g/mol] / [250.0 / 1000] = 5.37 M KCl 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 g)] × 100 = 6.6% Copyright © Cengage Learning. All rights reserved

EXERCISE! A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 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 / g H3PO4) = mol H3PO4 100.0 mL H2O × (1.00 g H2O / mL) × (1 mol / g H2O) = 5.55 mol H2O Mole Fraction (H3PO4) = mol H3PO4 / [ mol H3PO mol H2O] = Copyright © Cengage Learning. All rights reserved

EXERCISE! A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 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 / g H3PO4) = mol H3PO4 100.0 mL H2O × (1.00 g H2O / mL) × (1 kg / 1000 g) = kg H2O Molality = mol H3PO4 / kg H2O] = m Copyright © Cengage Learning. All rights reserved

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

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

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

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

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

The Solubilities of Several Gases in Water

An Aqueous Solution and Pure Water in a Closed Environment

Liquid/Vapor Equilibrium
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Vapor Pressure Lowering: Addition of a Solute

Vapor Pressures of Solutions
Nonvolatile solute lowers the vapor pressure of a solvent. Raoult’s Law: Psoln = observed vapor pressure of solution solv = mole fraction of solvent = vapor pressure of pure solvent Copyright © Cengage Learning. All rights reserved

A Solution Obeying Raoult’s Law

Nonideal Solutions Liquid-liquid solutions where both components are volatile. Modified Raoult’s Law: Nonideal solutions behave ideally as the mole fractions approach 0 and 1. Copyright © Cengage Learning. All rights reserved

Vapor Pressure for a Solution of Two Volatile Liquids

Summary of the Behavior of Various Types of Solutions
Interactive Forces Between Solute (A) and Solvent (B) Particles ΔHsoln ΔT for Solution Formation Deviation from Raoult’s Law Example A  A, B  B  A  B Zero None (ideal solution) Benzene-toluene A  A, B  B < A  B Negative (exothermic) Positive Negative Acetone-water A  A, B  B > A  B Positive (endothermic) Ethanol-hexane Copyright © Cengage Learning. All rights reserved

Hexane (C6H14) and chloroform (CHCl3) Ethyl alcohol (C2H5OH) and water
CONCEPT CHECK! For each of the following solutions, would you expect it to be relatively ideal (with respect to Raoult’s Law), show a positive deviation, or show a negative deviation? Hexane (C6H14) and chloroform (CHCl3) Ethyl alcohol (C2H5OH) and water Hexane (C6H14) and octane (C8H18) a) Positive deviation; Hexane is non-polar, chloroform is polar. b) Negative deviation; Both are polar, and the ethyl alcohol molecules can form stronger hydrogen bonding with the water molecules than it can with other alcohol molecules. c) Ideal; Both are non-polar with similar molar masses. Copyright © Cengage Learning. All rights reserved

Colligative Properties
Depend only on the number, not on the identity, of the solute particles in an ideal solution: Boiling-point elevation Freezing-point depression Osmotic pressure Copyright © Cengage Learning. All rights reserved

Boiling-Point Elevation
Nonvolatile solute elevates the boiling point of the solvent. ΔT = Kbmsolute ΔT = boiling-point elevation Kb = molal boiling-point elevation constant msolute = molality of solute Copyright © Cengage Learning. All rights reserved

Boiling Point Elevation: Liquid/Vapor Equilibrium

Boiling Point Elevation: Addition of a Solute

Boiling Point Elevation: Solution/Vapor Equilibrium

Freezing-Point Depression
When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent. ΔT = Kfmsolute ΔT = freezing-point depression Kf = molal freezing-point depression constant msolute = molality of solute Copyright © Cengage Learning. All rights reserved

Freezing Point Depression: Solid/Liquid Equilibrium

Freezing Point Depression: Addition of a Solute

Freezing Point Depression: Solid/Solution Equilibrium

Changes in Boiling Point and Freezing Point of Water

EXERCISE! A solution was prepared by dissolving g of glucose in g water. The molar mass of glucose is g/mol. What is the boiling point of the resulting solution (in °C)? Glucose is a molecular solid that is present as individual molecules in solution °C The change in temperature is ΔT = Kbmsolute. Kb is 0.51 °C·kg/mol. To solve formsolute, use the equation m = moles of solute/kg of solvent. Moles of solute = (25.00 g glucose)(1 mol / g glucose) = mol glucose Kg of solvent = (200.0 g)(1 kg / 1000 g) = kg water msolute = ( mol glucose) / ( kg water) = mol/kg ΔT = (0.51 °C·kg/mol)( mol/kg) = 0.35 °C. The boiling point of the resulting solution is °C °C = °C. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

EXERCISE! You take 20.0 g of a sucrose (C12H22O11) and NaCl mixture and dissolve it in 1.0 L of water. The freezing point of this solution is found to be °C. Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture. 72.8% sucrose and 27.2% sodium chloride; mole fraction of the sucrose is 0.313 The solution is 72.8% sucrose and 27.2% sodium chloride. The mole fraction of the sucrose is To solve this problem, the students must assume that i = 2 for NaCl. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

EXERCISE! A plant cell has a natural concentration of m. You immerse it in an aqueous solution with a freezing point of –0.246°C. Will the cell explode, shrivel, or do nothing? The cell will explode (or at least expand). The concentration of the solution is m. Thus, the cell has a higher concentration, and water will enter the cell. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

M = molarity of the solution R = gas law constant
Osmosis – flow of solvent into the solution through a semipermeable membrane. = MRT = osmotic pressure (atm) M = molarity of the solution R = gas law constant T = temperature (Kelvin) Copyright © Cengage Learning. All rights reserved

EXERCISE! When 33.4 mg of a compound is dissolved in mL of water at 25°C, the solution has an osmotic pressure of 558 torr. Calculate the molar mass of this compound. 111 g/mol The molar mass is 111 g/mol. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

Examples The expected value for i can be determined for a salt by noting the number of ions per formula unit (assuming complete dissociation and that ion pairing does not occur). NaCl i = 2 KNO3 i = 2 Na3PO4 i = 4 Copyright © Cengage Learning. All rights reserved

Ion Pairing Ion pairing is most important in concentrated solutions.
As the solution becomes more dilute, the ions are farther apart and less ion pairing occurs. Ion pairing occurs to some extent in all electrolyte solutions. Ion pairing is most important for highly charged ions. Copyright © Cengage Learning. All rights reserved

A suspension of tiny particles in some medium.
Tyndall effect – scattering of light by particles. Suspended particles are single large molecules or aggregates of molecules or ions ranging in size from 1 to 1000 nm. Copyright © Cengage Learning. All rights reserved

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.

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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.

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