1. Chapter 11 Properties of Solutions

2. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 2 Various Types of Solutions

3. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 3 Solution Composition

4. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 4 Molarity

5. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 5 You have 1.00 mol of sugar in 125.0 mL of solution. Calculate the concentration in units of molarity. 8.00 M EXERCISE!

6. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 6 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 EXERCISE!

7. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 7 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 EXERCISE!

8. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 8 Mass Percent

9. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 9 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% EXERCISE!

10. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 10 Mole Fraction

11. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 11 A solution of phosphoric acid was made by dissolving 8.00 g of H 3 PO 4 in 100.0 mL of water. Calculate the mole fraction of H 3 PO 4. (Assume water has a density of 1.00 g/mL.) 0.0145 EXERCISE!

12. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 12 Molality

13. Section 11.1 Solution Composition Copyright © Cengage Learning. All rights reserved 13 A solution of phosphoric acid was made by dissolving 8.00 g of H 3 PO 4 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 EXERCISE!

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 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 CONCEPT CHECK!

20. Section 11.2 The Energies of Solution Formation The Energy Terms for Various Types of Solutes and Solvents Copyright © Cengage Learning. All rights reserved 20 ΔH1ΔH1 ΔH2ΔH2 ΔH3ΔH3 ΔH soln Outcome Polar solute, polar solventLarge Large, negativeSmallSolution forms Nonpolar solute, polar solventSmallLargeSmallLarge, positiveNo solution forms Nonpolar solute, nonpolar solvent Small Solution forms Polar solute, nonpolar solventLargeSmall Large, positiveNo solution forms

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 Copyright © Cengage Learning. All rights reserved 30 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 HERECLICK HERE

31. Section 11.4 The Vapor Pressures of Solutions Vapor Pressure Lowering: Addition of a Solute Copyright © Cengage Learning. All rights reserved 31 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 HERECLICK HERE

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 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:  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 Copyright © Cengage Learning. All rights reserved 36 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 None (ideal solution) Benzene- toluene A  A, B  B < A  B Negative (exothermic) PositiveNegative Acetone- water A  A, B  B > A  B Positive (endothermic) NegativePositive Ethanol- hexane

37. Section 11.4 The Vapor Pressures of Solutions 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 CONCEPT CHECK!

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 Copyright © Cengage Learning. All rights reserved 40 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 HERECLICK HERE

41. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Boiling Point Elevation: Addition of a Solute Copyright © Cengage Learning. All rights reserved 41 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 HERECLICK HERE

42. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Boiling Point Elevation: Solution/Vapor Equilibrium Copyright © Cengage Learning. All rights reserved 42 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 HERECLICK HERE

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 = freezing-point depression K f = 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 Copyright © Cengage Learning. All rights reserved 44 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 HERECLICK HERE

45. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Freezing Point Depression: Addition of a Solute Copyright © Cengage Learning. All rights reserved 45 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 HERECLICK HERE

46. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression Freezing Point Depression: Solid/Solution Equilibrium Copyright © Cengage Learning. All rights reserved 46 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 HERECLICK HERE

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 A solution was prepared by dissolving 25.00 g of glucose in 200.0 g water. The molar mass of glucose is 180.16 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. 100.35 °C Copyright © Cengage Learning. All rights reserved 48 EXERCISE!

49. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression 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 -0.426°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 Copyright © Cengage Learning. All rights reserved 49 EXERCISE!

50. Section 11.5 Boiling-Point Elevation and Freezing-Point Depression A plant cell has a natural concentration of 0.25 m. You immerse it in an aqueous solution with a freezing point of –0.246°C. Will the cell explode, shrivel, or do nothing? Copyright © Cengage Learning. All rights reserved 50 EXERCISE!

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 Copyright © Cengage Learning. All rights reserved 53 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 HERECLICK HERE

54. Section 11.6 Osmotic Pressure Copyright © Cengage Learning. All rights reserved 54

55. Section 11.6 Osmotic Pressure 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 EXERCISE!

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: 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).  NaCli = 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 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