1. CHE1102, Chapter 12 Learn, 1 Chapter 12 Mixtures at the Molecular Level: Properties of Solutions Practice Exercises 12.1, 10-11, 14-15, 17, 22-23, 26, 28-29 Examples: 12.1-6, 8-12 In Text Problems

2. CHE1102, Chapter 12 Learn, 2 Solutions Solutions are homogeneous mixtures of two or more pure substances. In a solution, the solute is dispersed uniformly throughout the solvent. The ability of substances to form solutions depends on  natural tendency toward mixing.  intermolecular forces.

3. CHE1102, Chapter 12 Learn, 3 solvent – greatest component of the solution solute – lesser components of the solution Solutions

4. CHE1102, Chapter 12 Learn, 4 Common Types of Solutions A solution may be composed of a solid and a liquid, a gas an a liquid, or other combinations.

5. CHE1102, Chapter 12 Learn, 5 1. Entropy – dissolution always results in an increase in disorder Ability of Substance to Form Solutions Depends On:

6. CHE1102, Chapter 12 Learn, 6 2. Solvation – interaction between solute and solvent Ability of Substance to Form Solutions Depends On:

7. CHE1102, Chapter 12 Learn, 7 Intermolecular Forces of Attraction Any intermolecular force of attraction can be the attraction between solute and solvent molecules.

8. CHE1102, Chapter 12 Learn, 8 Attractions Involved When Forming a Solution Solute–solute interactions must be overcome to disperse these particles when making a solution. Solvent–solvent interactions must be overcome to make room for the solute. Solvent–solute interactions occur as the particles mix. The dissolution process may be exothermic or endothermic.

9. CHE1102, Chapter 12 Learn, 9 Opposing Processes The solution-making process and crystallization are opposing processes. When the rate of the opposing processes is equal, additional solute will not dissolve unless some crystallizes from solution. This is a saturated solution. If we have not yet reached the amount that will result in crystallization, we have an unsaturated solution.

10. CHE1102, Chapter 12 Learn, 10 solubility – the extent to which a solute dissolves saturated solution – no more solute will dissolve temperature dramatically affects solubility Solubility

11. CHE1102, Chapter 12 Learn, 11 For solids and liquids, an increase in temperature generally results in an increase in solubility

12. CHE1102, Chapter 12 Learn, 12 For gases, an increase in temperature generally results in a decrease in solubility

13. CHE1102, Chapter 12 Learn, 13 Temperature Dependence of Solubility of Gases in Water

14. CHE1102, Chapter 12 Learn, 14 Case Study: Dead Zones During the industrial revolution, factories were built on rivers so that the river water could be used as a coolant for the machinery. The hot water was dumped back into the river and cool water recirculated. After some time, the rivers began to darken and many fish died. The water was not found to be contaminated by the machinery. What was the cause of the mysterious fish kills?

15. CHE1102, Chapter 12 Learn, 15 Solubility miscible – liquids that are soluble in each other in all proportions immiscible – liquids that are NOT soluble in each other soluble – solids that are soluble in a liquid insoluble – solids that are NOT soluble in a liquid

16. CHE1102, Chapter 12 Learn, 16 In General ….“Like dissolves Like” Polar species tend to be miscible/soluble with other polar species Nonpolar species tend to be miscible/soluble with other nonpolar species Polar and Nonpolar species tend to be immiscible/insoluble with each other

17. CHE1102, Chapter 12 Learn, 17

18. CHE1102, Chapter 12 Learn, 18 Solubility and Biological Importance Fat-soluble vitamins (like vitamin A) are nonpolar; they are readily stored in fatty tissue in the body. Water-soluble vitamins (like vitamin C) need to be included in the daily diet.

19. CHE1102, Chapter 12 Learn, 19 Pressure Effects The solubility of solids and liquids are not appreciably affected by pressure. Gas solubility is affected by pressure.

20. CHE1102, Chapter 12 Learn, 20 William Henry 1774 – 1836 – the amount of gas that is dissolved in a liquid is directly proportional to the pressure applied to the solution Henry’s Law

21. CHE1102, Chapter 12 Learn, 21 Henry’s Law The solubility of a gas is proportional to the partial pressure of the gas above the solution. C g = kP g

22. CHE1102, Chapter 12 Learn, 22 Using Henry’s Law Calculate the concentration of CO 2 in a soft drink that is bottled with a partial pressure of CO 2 of 5 atm over the liquid at 25 °C. The Henry’s Law constant for CO 2 in water at this temperature is 3.12  10 –2 mol L –1 atm –1. = (3.12 × 10 –2 mol/L atm) × (5.0 atm) = 0.156 mol/L  0.16 mol/L

23. CHE1102, Chapter 12 Learn, 23 Using Henry’s Law (cont.) Calculate the concentration of CO 2 in a soft drink after the bottle is opened and it equilibrates at 25 °C under a partial pressure of CO 2 of 4.0  10 –4 atm. C 2 = 1.2  10 –4 mol CO 2 /L Compared to the concentration of CO 2 at 5 atm: 0.16 mol CO 2 /L

24. CHE1102, Chapter 12 Learn, 24 Expressing Solution Concentration mass of solute total mass of solution  100 Mass % = 2. 1. moles of solute L solution Molarity, M = 1.92 g NaCl (s) is dissolved into 50.0 g H 2 O. What is the mass % NaCl (aq) ?

25. CHE1102, Chapter 12 Learn, 25 Preparing a Solution

26. CHE1102, Chapter 12 Learn, 26 What mass of NH 4 Cl must be dissolved in water to make 100 mL of a 0.250 M solution? A.1340 g B.21.40 g C.13.4 g D. 1.34 g

27. CHE1102, Chapter 12 Learn, 27 mass of solute total mass of solution  1,000,000 ppm = 3. parts per million, ppm OSHA mandates a limit of 35 ppm for CO gas exposure in the workplace. The tunnel through the Cumberland Gap (on 25E) was found to contain 23 g CO in a 325 kg sample of air. Is the air in the tunnel legal? Expressing Solution Concentration

28. CHE1102, Chapter 12 Learn, 28

29. CHE1102, Chapter 12 Learn, 29 moles of solute kg of solvent molality, m = 4. 18.5 g Na 2 SO 4 (s) is dissolved into 221.1 g H 2 O. What is the molality of the resulting solution ? Expressing Solution Concentration

30. CHE1102, Chapter 12 Learn, 30 moles of substance total moles of mixture mole fraction, X = 5. 10.27 g glucose, C 6 H 12 O 6 (s) is dissolved into 22.62 g ethanol, CH 3 CH 2 OH. What is the mole fraction of glucose ? Expressing Solution Concentration

31. CHE1102, Chapter 12 Learn, 31 Converting Between Concentrations Calculate the molality and the molarity of a 40.0% HBr solution. The density of this solution is 1.38 g/mL. Assume 100.0 g of solution; then 40.0 g = HBr, 60.0 g = H 2 O.

32. CHE1102, Chapter 12 Learn, 32 Now calculate molarity of 40% HBr mol HBr = 0.494 mol = 72.46 mL = 6.82 M Converting Between Concentrations

33. CHE1102, Chapter 12 Learn, 33 – Properties which depend only on the quantity, not on the identity of the solute particles. VP solution < VP pure solvent Colligative Properties

34. CHE1102, Chapter 12 Learn, 34 Among colligative properties are: –Vapor-pressure lowering –Boiling-point elevation –Freezing-point depression –Osmotic pressure Colligative Properties

35. CHE1102, Chapter 12 Learn, 35 Francois-Marie Raoult 1830 – 1901 Raoult’s Law – VP of a solution is lower than VP of pure solvent Colligative Properties

36. CHE1102, Chapter 12 Learn, 36 Vapor Pressure Because of solute–solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase. Therefore, the vapor pressure of a solution is lower than that of the pure solvent.

37. CHE1102, Chapter 12 Learn, 37 When solute is nonvolatile and nonionic, the extent the VP is lowered depends on mole fraction of the solvent, X solvent P A = X A P  A your text book VP soln = (X solvent ) (VP  solvent ) me VP soln = vapor pressure of resulting solution VP  solvent = vapor pressure of pure solvent X solvent = mole fraction of solvent in solution

38. CHE1102, Chapter 12 Learn, 38 38.4 g urea, CH 4 N 2 O (s) is dissolved into 102.3 g water at 25 °C. What is the vapor pressure of the resulting solution ? (VP pure water = 23.8 torr @ 25 o C)

39. CHE1102, Chapter 12 Learn, 39 Addition of Solute Lowers the VP of Resulting Solution

40. CHE1102, Chapter 12 Learn, 40 BP of a soln is higher than BP of pure solvent BP of solution > BP of pure solvent If the solute is nonvolatile, nonionic then:  T b = (K b )(m)  T b = (BP solution – BP pure solvent ) K b = BP elevation constant m = molality of solute in solution always +# Boiling Point Elevation

41. CHE1102, Chapter 12 Learn, 41 K b and K f

42. CHE1102, Chapter 12 Learn, 42 76.01 g sugar, C 12 H 22 O 11 (s) is dissolved into 95.510 g water. What temperature will this solution boil at ? (P = 1 atm)

43. CHE1102, Chapter 12 Learn, 43 Addition of Solute Raises the BP of Resulting Solution

44. CHE1102, Chapter 12 Learn, 44 Freezing-Point Depression The construction of the phase diagram for a solution demonstrates that the freezing point is lowered while the boiling point is raised.

45. CHE1102, Chapter 12 Learn, 45 FP of a soln is lower than FP of pure solvent FP of solution < FP of pure solvent If the solute is nonvolatile, nonionic then:  T f = (K f ) ( m)  T f = (FP pure solvent – FP solution ) K f = FP depression constant m = molality of solute in solution always +# Freezing Point Depression

46. CHE1102, Chapter 12 Learn, 46 3.461 g of an unknown compound was dissolved into 85.022 g benzene, C 6 H 6. The resulting solution froze at 4.08 °C. What is the MW of the unknown compound ?

47. CHE1102, Chapter 12 Learn, 47 Colligative Properties of Electrolytes Since the colligative properties of electrolytes depend on the number of particles dissolved, solutions of electrolytes (which dissociate in solution) should show greater changes than those of nonelectrolytes.

48. CHE1102, Chapter 12 Learn, 48 However, a 1M solution of NaCl does not show twice the change in freezing point that a 1M solution of methanol does. Colligative Properties of Electrolytes

49. CHE1102, Chapter 12 Learn, 49 van’t Hoff Factor One mole of NaCl in water does not really give rise to two moles of ions.

50. CHE1102, Chapter 12 Learn, 50 Some Na + and Cl  reassociate for a short time, so the true concentration of particles is somewhat less than two times the concentration of NaCl. van’t Hoff Factor

51. CHE1102, Chapter 12 Learn, 51 Reassociation is more likely at higher concentration. Therefore, the number of particles present is concentration- dependent. van’t Hoff Factor

52. CHE1102, Chapter 12 Learn, 52 We modify the previous equations by multiplying by the van’t Hoff factor, i:  T f = K f  m  i van’t Hoff Factor

53. CHE1102, Chapter 12 Learn, 53 If all of the following generic compounds have the same K b, which will cause a greater change in the boiling point of a 0.25 m nonelectrolyte solution? (Assume all dissociate completely in solution) A. MX B. MX 2 C. M 2 X 3 D. MX 3 E. Not enough information given

54. CHE1102, Chapter 12 Learn, 54 – the movement of solvent molecules but NOT solute particles through a semipermeable membrane semipermeable membrane – a barrier with very tiny holes that selectively allow only small molecules such as solvent molecules to pass through Osmosis

55. CHE1102, Chapter 12 Learn, 55 Only Solvent Molecules Pass Through

56. CHE1102, Chapter 12 Learn, 56 Solvent molecules always move through membrane toward side with the largest solute concentration

57. CHE1102, Chapter 12 Learn, 57 Osmosis and Blood Cells Red blood cells have semipermeable membranes. If stored in a hypertonic solution, they will shrivel as water leaves the cell; this is called crenation. If stored in a hypotonic solution, they will grow until they burst; this is called hemolysis.

58. CHE1102, Chapter 12 Learn, 58 Osmotic Pressure The pressure required to stop osmosis, known as osmotic pressure, , is  = ( ) RT = MRT nVnV where M is the molarity of the solution. If the osmotic pressure is the same on both sides of a membrane (i.e., the concentrations are the same), the solutions are isotonic.

59. CHE1102, Chapter 12 Learn, 59 Using  to determine MM The osmotic pressure of an aqueous solution of a certain protein was measured to determine its molar mass. The solution contained 3.50 mg of protein in sufficient H 2 O to form 5.00 mL of solution. The measured osmotic pressure of this solution was 1.54 torr at 25 °C. Calculate the molar mass of the protein.

60. CHE1102, Chapter 12 Learn, 60 – a homogeneous mixture where the solute is dispersed in the solvent Solute particles of a colloid are larger than individual ions or molecules, but too small to be settled out by gravity Colloids

61. CHE1102, Chapter 12 Learn, 61 Types of Colloidal Suspensions

62. CHE1102, Chapter 12 Learn, 62 Colloids can aid in the emulsification of fats and oils in aqueous solutions. An emulsifier causes something that normally does not dissolve in a solvent to do so. Colloids in Biological Systems

63. CHE1102, Chapter 12 Learn, 63 Brownian Motion Brownian motion keeps particles from settling

64. CHE1102, Chapter 12 Learn, 64 Tyndall Effect Colloidal suspensions can scatter rays of light. (Solutions do not.) This phenomenon is known as the Tyndall effect.