1. Chapter 11 Properties of Solutions Solutions – homogeneous mixtures that could be gasses, liquids, or solids. Let’s remember the terms… Dilute – relatively little solute present Concentrated – relatively large amount of solute.

2. More precisely… We talk about solution composition in terms of Molarity – moles of solute per liter of solution. And Mass Percent -

3. And even… The mole fraction – the ratio of the number of moles of a component to the total number of moles of a solution. And molality – number of moles of solute per kilogram of solvent.

4. Let’s see how they work A solution is prepared by mixing 1.00g ethanol (C 2 H 5 OH) with 100.0g water to give a final volume 101 mL. Calculate the molarity, mass percent, mole fraction and molality of ethanol.

5. Molarity =0.215M

6. Mass percent =0.990% C 2 H 5 OH

7. Mole fraction

8. Molality

9. How about Normality Symbolized by N It is defined as the number of equivalents per liter of solution. Where the definition of equivalent depends on the reaction taking place in the solution. THIS IS NOT ON THE AP TEST!!

10. You try one… The electrolyte in automobile lead storage batteries is a 3.75M sulfuric acid solution that has a density of 1.230 g/mL. Calculate the mass percent and molality of the sulfuric acid.

11. 11.2 The energies of solution formation. Solubility is important! Real life examples… – Vitamin solubility is correct in determining dosages. – Barium is not soluble, which is why it is used to improve the x-rays of the gratrointestinal tract (although Ba is toxic)

12. The cardinal rule of solubility… LIKE DISSOLVES LIKE Polar solvents dissolve polar or ionic solutes Nonpolar sovents dissolve nonpolar solutes BUT WHY?

13. Assume the formation of a liquid solution occurs in 3 distinct steps 1.Separating the solute into its individual components (expanding the solute) 2.Overcoming the 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. (mixing) What does this all mean????

14. Energy of making solutions Heat of solution (  H soln ) is the energy change for making a solution. The energy comes from the 3 steps since forces must be overcome to expand the solute and the solvent.

15. Here’s how… 1.Break apart the solvent Have to overcome attractive forces  H 1 >0(endothermic) 2.Break apart the solute Have to overcome attractive forces  H 2 >0(endothermic)

16. 3. Mixing solvent and solute  H 3 depends on what you are mixing. If molecules can attract each other  H 3 is large and negative. Molecules can’t attract-  H 3 is small and negative. This explains the rule “Like dissolves Like”

18. Heat of solution endo or exothermic?

19. Generally speaking…Processes that required LARGE amounts of energy tend NOT to occur. H1H1 H2H2 H3H3  H soln Outcome Polar solvent, polar solute Large Large, negative SmallSolution forms Polar solvent, nonpolar solute SmallLargeSmallLarge, positive No solution forms Nonpolar solvent, nonpolar solute Small Solution forms Nonpolar solvent, polar solute LargeSmall Large, positive No solution forms

20. Types of Solvent and solutes If  H soln is small and positive, a solution will still form because of entropy. There are many more ways for them to become mixed than there is for them to stay separate.

21. 11.3 Factors affecting solubility Structure – Since molecular structure is what determines polarity, there is a definite connection between structure and solubility – Hydrophobic: water-fearing…nonpolar materials – Hydrophilic: water-loving…polar substances

22. Let’s think Why does dish detergent (soap) help dissolve grease in water?

23. Soap PO-O- CH 3 CH 2 O-O- O-O-

24. Soap Hydrophobic non- polar end PO-O- CH 3 CH 2 O-O- O-O-

25. Soap Hydrophilic polar end PO-O- CH 3 CH 2 O-O- O-O-

26. PO-O- CH 3 CH 2 O-O- O-O- _

27. A drop of grease in water Grease is non-polar Water is polar Soap lets you dissolve the non-polar in the polar.

28. Hydrophobic ends dissolve in grease

29. Hydrophilic ends dissolve in water

30. Water molecules can surround and dissolve grease. Helps get grease out of your way.

31. Pressure effects Changing the pressure doesn’t affect the amount of solid or liquid that dissolves because liquids and solids are incompressible. However, pressure DOES affect gases.

32. Dissolving Gases Pressure affects the amount of gas that can dissolve in a liquid. The dissolved gas is at equilibrium with the gas above the liquid.

33. The gas is at equilibrium with the dissolved gas in this solution. The equilibrium is dynamic.

34. If you increase the pressure the gas molecules dissolve faster. The equilibrium is disturbed. More gas enters the solution at a higher rate than it leaves.

35. The system reaches a new equilibrium with more gas dissolved. Henry’s Law. C=kP Concentration of gas = constant times pressure. Amount of gas dissolved in a solution is directly proportional to the pressure of the gas above the solution.

36. Temperature Effects It is true that an Increased temperature will increase the rate at which a solid dissolves. BUT We can’t predict whether it will increase the amount of solid that dissolves.

37. Solubility vs Temperature Graph

38. Predicting solid solubility with respect to temperature This is NOT someething that is easy to too. The ONLY SURE WAY to determine the temperature dependence of a solid’s solubility is by experiment!!

39. Gasses on the other hand… Are much less complex.. As temperature increases, solubility decreases because the gas molecules can move fast enough to escape.

40. 11.4 Vapor pressures of solutions. Remember from chapter 10 that liquids with high vapor pressures are said to be volatile. A nonvolatile solvent lowers the vapor pressure of the solution

41. Aqueous Solution Pure water Water has a higher vapor pressure than a solution.. So VP to achieve equilibrium with the water > than that required to achieve equilibrium with the solution.

42. Aqueous Solution Pure water Water emits vapor to attempt to reach equilibrium and solution absorbs vapor to lower the VP towards equilibrium.

43. The water condenses faster in the solution. Thus the solution takes on water. This will happen and equilibrium VP will be reached only when all the water is transferred to the solution. Aqueous Solution

44. Raoult’s Law P soln =  solvent ● P 0 solvent Vapor pressure of the solution = mole fraction of solvent x vapor pressure of the pure solvent Applies only to an ideal solution where the solute doesn’t contribute to the vapor pressure.

45. Try this problem. A solution of cyclopentane with a nonvolatile compound has vapor pressure of 211 torr. If vapor pressure of the pure liquid is 313 torr, what is the mole fraction of the cyclopentane?

46. Another? Determine the vapor pressure of a solution at 25  C that has 45 grams of C 6 H 12 O 6, glucose, dissolved in 72 grams of H 2 O. The vapor pressure of pure water at 25  C is 23.8 torr.

47. Vapor Pressure Lowering: Addition of a Solute

48. By lowering the VP, we now have a way to count molecules and then determine molar masses. Additionally, we can use VP to characterize solutions. For example, NaCl lowers vapor pressure almost twice as much as expected due to the fact that there are 2 ions per formula unit which separate when dissolved. Thus, lowering of vapor pressure depends on the number of solute particle present in the solution.

49. What if... …the solution is a liquid liquid solution and they are both volatile? We need to modify Raoult’s law to account for the partial pressures of both liquids. P total = P A + P B =  A P A 0 +  B P B 0

50. Ideal solutions If the liquid-liquid solution obey’s Raoult’s law, it is called an ideal solution. As with gasses, ideal solutions are rarely achieved, but often closely approached. Nearly ideal solutions occur when the solute and solvent are very similar and the solute basically just dilutes the solvent.

52. Deviations If solvent has a strong affinity for solute (H bonding). Lowers solvent’s ability to escape. Lower vapor pressure than expected by Raoult’s law. Causing a negative deviation from Raoult’s law.  H soln is large and negative ~exothermic. There is a strong interaction between solute and solvent.

53. Negative deviations Strong attraction between solute and solvent Negative ΔH soln

54. Deviations If solvent has a weak affinity for solute, interactions between the solute and solvent are less than among the molecules of the pure liquid. More energy required to expand the liquids than is released when the liquids mix. Molecules of the solution have a higher tendency to escape. Higher vapor pressure than expected by Raoult’s law. Causing a positive deviation from Raoult’s law.  H soln is positive --- endothermic.

55. Positive deviations- Weak attraction between solute and solvent Positive ΔH soln

56. P total = P A + P B =  A P A 0 +  B P B 0 A solution is prepared by mixing 5.81g acetone (C 3 H 6 O molar mass=58.1 g/mol) & 11.9g chloroform (CHCl 3 molar mass 119.4 g/mol). At 35 o C, this solution has a total vapor pressure of 260. torr. Is this an ideal solution? The vapor pressures of pure acetone and pure chloroform at 35 o C are 345 and 293 torr respectively. 319 torr

57. 11.5 Boiling-Point elevation and Freezing-Point Depression Changes of state depend on vapor pressure. So adding a solute will affect vapor pressure of solvent and thus the FP and BP of the solvent (the phase changes). Colligative properties depend only on the number - not the kind of solute particles present Useful for determining molar mass

58. Boiling point elevation Rember from chapter 10 that the NORMAL BOILING POINT of a liquid occurs at the temperature where the vapor pressure is equal to 1 atm. However… Because a non-volatile solute lowers the vapor pressure it raises the boiling point.

60. Molal boiling point elevation constant The change in the boiling point can be represented by the following equation  T = K b m solute Where:  T is the change in the boiling point K b is a constant determined by the solvent. m solute is the molality of the solute in the solution. Using this information we can determine the molar mass of a solute.

61. Here’s a problem. A solution was prepared by dissolving 18.00 g glucose in 150.0 g water. The resulting solution was found to have a boiling point of 100.34 o C. Calculate the molar mass of glucose. Where K b =0.51 o Ckg/mol

62. Freezing point depression Because a non-volatile solute lowers the vapor pressure of the solution it lowers the freezing point.  T = K f m solute  T is the change in the freezing point K f is a constant determined by the solvent m solute is the molality of the solute

64. Try a freezing-point depression problem A chemist is trying to identify a human hormone that controls metabolism by determining its molar mass. A sample weighing 0.546g was dissolved in 15.0g benzene, and the freezing-point depression was determined to be 0.240 o C. Calculate the molar mass of the hormone. If K f =5.12 o Ckg/mol

65. Soooo…adding a solute… Will lower the freezing point of a liquid and raise the boiling point causing the liquid state to have an extended range of temperatures.

66. 11.6 Osmotic Pressure Osmosis - the flow of solvent into a solution through a semi-permeable membrane

68. Osmotic pressure Excess hydrostatic pressure exerted on the solution after the solvent enters the solution and changes the levels of liquid. We can prevent osmosis by applying pressure to the solution…the minimum pressure that stops osmosis is equal to osmotic pressure. Osmotic pressure can be used to characterize solutions & determine molar masses. A small concentration of solute=large osmotic pressure.

69. Osmotic pressure M is the molarity of the solution R is the ideal gas constant T is the temperature (in K)

70. Try this To determine the molar mass of a certain protein, 1.00 x 10 -3 g of it was dissolved in enough water to make 1.00mL of solution. The osmotic pressure of this solution was found to be 1.12 torr at 25 o C. Calculate the molar mass of the protein. Remember R is in

71. Using osmotic pressure generally gives much more accurate molar masses, rather than freezing point or boiling point changes.

72. Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. Vapor-Pressure Lowering P 1 = X 1 P 1 0 Boiling-Point Elevation  T b = K b m Freezing-Point Depression  T f = K f m Osmotic Pressure (  )  = MRT

73. Electrolyte Solutions van’t Hoff factor i = actual number of particles in soln after dissociation number of formula units initially dissolved in soln Expected i values assume that when a salt dissolves, it completely dissociates into its component ions. This is not always true due to ion pairing. (ions pair up and act as a single particle) Ion pairing is most important in concentrated solutions because ions are closer together.

74. Colligative Properties of Electrolyte Solutions Boiling-Point Elevation  T b = i K b m Freezing-Point Depression  T f = i K f m Osmotic Pressure (  )  = iMRT

75. One last problem The observed osmotic pressure for a 0.10M solution of Fe(NH 4 ) 2 (SO 4 ) 2 at 25 o C is 10.8 atm. Compare the expected and experimental values for i.