Chapter 11. Properties of Solutions
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1 Chapter 11 Properties of Solutions
2 Section 11.1 Solution Composition Various Types of Solutions Copyright Cengage Learning. All rights reserved 2
3 Section 11.1 Solution Composition Solution Composition moles of solute Molarity ( M) = liters of solution mass of solute Mass (weight) percent = 100% mass of solution Mole fraction ( moles A A ) = total moles of solution moles of solute Molality ( m) = kilogram of s olvent Copyright Cengage Learning. All rights reserved 3
4 Section 11.1 Solution Composition Molarity moles of solute Molarity ( M) = liters of solution Copyright Cengage Learning. All rights reserved 4
5 Section 11.1 Solution Composition EXERCISE! You have 1.00 mol of sugar in ml of solution. Calculate the concentration in units of molarity M Copyright Cengage Learning. All rights reserved 5
6 Section 11.1 Solution Composition EXERCISE! You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar? L Copyright Cengage Learning. All rights reserved 6
7 Section 11.1 Solution Composition 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 M NaOH 5.37 M KCl Copyright Cengage Learning. All rights reserved 7
8 Section 11.1 Solution Composition Mass Percent mass of solute Mass (weight) percent = 100% mass of solution Copyright Cengage Learning. All rights reserved 8
9 Section 11.1 Solution Composition 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% Copyright Cengage Learning. All rights reserved 9
10 Section 11.1 Solution Composition Mole Fraction Mole fraction ( moles A A ) = total moles of solution Copyright Cengage Learning. All rights reserved 10
11 Section 11.1 Solution Composition EXERCISE! A solution of phosphoric acid was made by dissolving 8.00 g of H 3 PO 4 in ml of water. Calculate the mole fraction of H 3 PO 4. (Assume water has a density of 1.00 g/ml.) Copyright Cengage Learning. All rights reserved 11
12 Section 11.1 Solution Composition Molality moles of solute Molality ( m) = kilogram of solvent Copyright Cengage Learning. All rights reserved 12
13 Section 11.1 Solution Composition EXERCISE! A solution of phosphoric acid was made by dissolving 8.00 g of H 3 PO 4 in ml of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/ml.) m Copyright Cengage Learning. All rights reserved 13
14 Section 11.2 The Energies of Solution Formation Formation of a Liquid Solution 1. Separating the solute into its individual components (expanding the solute). 2. Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent). 3. Allowing the solute and solvent to interact to form the solution. Copyright Cengage Learning. All rights reserved 14
15 Section 11.2 The Energies of Solution Formation Steps in the Dissolving Process Copyright Cengage Learning. All rights reserved 15
16 Section 11.2 The Energies of Solution Formation 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 16
17 Section 11.2 The Energies of Solution Formation Enthalpy (Heat) of Solution Enthalpy change associated with the formation of the solution is the sum of the ΔH values for the steps: ΔH soln = ΔH 1 + ΔH 2 + ΔH 3 ΔH soln may have a positive sign (energy absorbed) or a negative sign (energy released). Copyright Cengage Learning. All rights reserved 17
18 Section 11.2 The Energies of Solution Formation Enthalpy (Heat) of Solution Copyright Cengage Learning. All rights reserved 18
19 Section 11.2 The Energies of Solution Formation 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. Copyright Cengage Learning. All rights reserved 19
20 Section 11.2 The Energies of Solution Formation The Energy Terms for Various Types of Solutes and Solvents ΔH 1 ΔH 2 ΔH 3 ΔH soln Outcome Polar solute, polar solvent Large Large Large, negative Small Solution forms Nonpolar solute, polar solvent Small Large Small Large, positive No solution forms Nonpolar solute, nonpolar solvent Small Small Small Small Solution forms Polar solute, nonpolar solvent Large Small Small Large, positive No solution forms Copyright Cengage Learning. All rights reserved 20
21 Section 11.2 The Energies of Solution Formation 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 21
22 Section 11.3 Factors Affecting Solubility Structure Effects: Polarity Pressure Effects: Henry s law Temperature Effects: Affecting aqueous solutions Copyright Cengage Learning. All rights reserved 22
23 Section 11.3 Factors Affecting Solubility Structure Effects Hydrophobic (water fearing) Non-polar substances Hydrophilic (water loving) Polar substances Copyright Cengage Learning. All rights reserved 23
24 Section 11.3 Factors Affecting Solubility Pressure Effects 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 24
25 Section 11.3 Factors Affecting Solubility A Gaseous Solute Copyright Cengage Learning. All rights reserved 25
26 Section 11.3 Factors Affecting Solubility 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 26
27 Section 11.3 Factors Affecting Solubility The Solubilities of Several Solids as a Function of Temperature Copyright Cengage Learning. All rights reserved 27
28 Section 11.3 Factors Affecting Solubility The Solubilities of Several Gases in Water Copyright Cengage Learning. All rights reserved 28
29 Section 11.4 The Vapor Pressures of Solutions An Aqueous Solution and Pure Water in a Closed Environment Copyright Cengage Learning. All rights reserved 29
30 Section 11.4 The Vapor Pressures of Solutions Liquid/Vapor Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 30
31 Section 11.4 The Vapor Pressures of Solutions Vapor Pressure Lowering: Addition of a Solute To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 31
32 Section 11.4 The Vapor Pressures of Solutions Vapor Pressures of Solutions Nonvolatile solute lowers the vapor pressure of a solvent. Raoult s Law: P soln = observed vapor pressure of solution solv = mole fraction of solvent = vapor pressure of pure solvent P solv P = P soln solv solv Copyright Cengage Learning. All rights reserved 32
33 Section 11.4 The Vapor Pressures of Solutions A Solution Obeying Raoult s Law Copyright Cengage Learning. All rights reserved 33
34 Section 11.4 The Vapor Pressures of Solutions Nonideal Solutions Liquid-liquid solutions where both components are volatile. Modified Raoult s Law: PTotal = APA + BPB Nonideal solutions behave ideally as the mole fractions approach 0 and 1. Copyright Cengage Learning. All rights reserved 34
35 Section 11.4 The Vapor Pressures of Solutions Vapor Pressure for a Solution of Two Volatile Liquids Copyright Cengage Learning. All rights reserved 35
36 Section 11.4 The Vapor Pressures of Solutions Summary of the Behavior of Various Types of Solutions Interactive Forces Between Solute (A) and Solvent (B) Particles ΔH soln ΔT for Solution Formation Deviation from Raoult s Law Example A A, B B A B Zero Zero None (ideal solution) Benzenetoluene A A, B B < A B Negative (exothermic) Positive Negative Acetonewater A A, B B > A B Positive (endothermic) Negative Positive Ethanolhexane Copyright Cengage Learning. All rights reserved 36
37 Section 11.4 The Vapor Pressures of Solutions 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? a) Hexane (C 6 H 14 ) and chloroform (CHCl 3 ) b) Ethyl alcohol (C 2 H 5 OH) and water c) Hexane (C 6 H 14 ) and octane (C 8 H 18 ) Copyright Cengage Learning. All rights reserved 37
38 Section 11.5 Boiling-Point Elevation and Freezing-Point Depression 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 38
39 Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Boiling-Point Elevation Nonvolatile solute elevates the boiling point of the solvent. ΔT = K b m solute ΔT = boiling-point elevation K b = molal boiling-point elevation constant m solute = molality of solute Copyright Cengage Learning. All rights reserved 39
40 Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Boiling Point Elevation: Liquid/Vapor Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 40
41 Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Boiling Point Elevation: Addition of a Solute To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 41
42 Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Boiling Point Elevation: Solution/Vapor Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 42
43 Section 11.5 Boiling-Point Elevation and Freezing-Point Depression 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 = K f m solute ΔT K f = freezing-point depression = molal freezing-point depression constant m solute = molality of solute Copyright Cengage Learning. All rights reserved 43
44 Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Freezing Point Depression: Solid/Liquid Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 44
45 Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Freezing Point Depression: Addition of a Solute To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 45
46 Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Freezing Point Depression: Solid/Solution Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 46
47 Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Changes in Boiling Point and Freezing Point of Water Copyright Cengage Learning. All rights reserved 47
48 Section 11.5 Boiling-Point Elevation and Freezing-Point Depression 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 Copyright Cengage Learning. All rights reserved 48
49 Section 11.5 Boiling-Point Elevation and Freezing-Point Depression EXERCISE! You take 20.0 g of a sucrose (C 12 H 22 O 11 ) 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 Copyright Cengage Learning. All rights reserved 49
50 Section 11.5 Boiling-Point Elevation and Freezing-Point Depression EXERCISE! A plant cell has a natural concentration of 0.25 m. You immerse it in an aqueous solution with a freezing point of C. Will the cell explode, shrivel, or do nothing? Copyright Cengage Learning. All rights reserved 50
51 Section 11.6 Osmotic Pressure 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 51
52 Section 11.6 Osmotic Pressure Copyright Cengage Learning. All rights reserved 52
53 Section 11.6 Osmotic Pressure Osmosis To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 53
54 Section 11.6 Osmotic Pressure Copyright Cengage Learning. All rights reserved 54
55 Section 11.6 Osmotic Pressure EXERCISE! When 33.4 mg of a compound is dissolved in 10.0 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 Copyright Cengage Learning. All rights reserved 55
56 Section 11.7 Colligative Properties of Electrolyte Solutions van t Hoff Factor, i The relationship between the moles of solute dissolved and the moles of particles in solution is usually expressed as: i = moles of particles in solution moles of solute dissolved Copyright Cengage Learning. All rights reserved 56
57 Section 11.7 Colligative Properties of Electrolyte Solutions Ion Pairing At a given instant a small percentage of the sodium and chloride ions are paired and thus count as a single particle. Copyright Cengage Learning. All rights reserved 57
58 Section 11.7 Colligative Properties of Electrolyte Solutions 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 KNO 3 i = 2 Na 3 PO 4 i = 4 Copyright Cengage Learning. All rights reserved 58
59 Section 11.7 Colligative Properties of Electrolyte Solutions 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 59
60 Section 11.7 Colligative Properties of Electrolyte Solutions Modified Equations T = imk = imrt Copyright Cengage Learning. All rights reserved 60
61 Section 11.8 Colloids 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 61
62 Section 11.8 Colloids Types of Colloids Copyright Cengage Learning. All rights reserved 62
63 Section 11.8 Colloids Coagulation Destruction of a colloid. Usually accomplished either by heating or by adding an electrolyte. Copyright Cengage Learning. All rights reserved 63
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