1. Solutions

2. Solution Concentration Although molarity (M) is used for stoichiometry calculations, there are many other ways to express the concentration of a solution. Molarity will vary slightly with changes in temperature as the volume expands or contracts. Units such as mass percent, mole fraction, or molality remain constant as temperature changes.

3. Solution Concentration Mass percent = (mass of solute) (100%) (mass of solution) (mass of solution) Mole fraction (X A ) = (moles of A) total # of moles Molality (m) = moles of solute kg of solvent kg of solvent

4. The Solution Process We will focus on solid or liquid solutes dissolved in a liquid solvent. Since all particles are in contact with each other, the solute-solute and solvent-solvent forces of attraction are disrupted, and new, solute-solvent forces of attraction are created.

5. The Solution Process The disruption of solute-solute and solvent- solvent forces of attraction requires energy, and is endothermic. The interaction of solvent and solute usually releases energy. The sum of the energy of all three steps is called the enthalpy of solution, ΔH o soln. Note that solutions may form whether the net process is endothermic or exothermic.

6. The Solution Process

7. The general rule on solution formation is: Like dissolves like. Polar and ionic compounds dissolve in polar solvents. Non-polar compounds dissolve in non-polar solvents.

8. Like Dissolves Like Vitamin A consists almost entirely of carbon and hydrogen, and is non-polar. As a result, vitamin A is fat-soluble, and can be stored in the body.

9. Like Dissolves Like Vitamin C contains polar C-O and O-H bonds. It is water soluble, and must be consumed often, as it is excreted easily. O-H bonds C-O bond

10. Like Dissolves Like Ionic Compounds

11. Like Dissolves Like

12. The Solution Process In addition to the enthalpy of solution, we must also consider the entropy of mixing. Entropy is a measure of randomness or disorder. An increase in entropy makes a process more likely to occur. Since mixing pure substances increases entropy, this factor makes processes that are slightly endothermic favorable.

13. Entropy of Mixing

14. The Solution Process Disrupt- ion of solute Disrupt- ion of solvent Solute/ Solvent interact -ion

15. Factors Affecting Solubility Molecular Structure Molecular Structure Pressure (for gaseous solutes) Pressure (for gaseous solutes) Temperature Temperature

16. Pressure Effects Gases dissolved in a liquid solute obey Henry’s Law: C = kP where C is the concentration, k is a constant specific to solute and solvent, and P is the pressure of the gas above the solution

17. Pressure Effects Gases dissolved in a liquid solute obey Henry’s Law: C = kP The amount of a gas dissolved in a solution is directly proportional to the pressure of the gas above the solution.

18. Henry’s Law

19. Pressure Effects

20. Temperature Effects For gases dissolved in liquids, the solubility decreases as temperature increases. That is, gases dissolve better in cold liquids than in warmer liquids.

21. Temperature Effects For solid solutes dissolved in water, the effect of temperature on solubility is difficult to predict.

22. The Colligative Properties The colligative properties are properties that depend upon the concentration of particles (molecules or ions) dissolved in a volatile solvent, and not on the nature of the particles. They include: vapor pressure vapor pressure freezing point freezing point boiling point boiling point osmotic pressure osmotic pressure

23. The Colligative Properties Relatively simple mathematical relationships can be used to predict the changes in vapor pressure, freezing and boiling point, etc. The properties can be predicted for dilute solutions (<0.1M) of non-volatile solute (usually solids) dissolved in a volatile solvent (usually a liquid).

24. Vapor Pressure The addition of a non-volatile solute to a volatile solvent lowers the vapor pressure of the solvent.

26. Vapor Pressure The decrease in vapor pressure can be understood by looking at the evaporation process. We need to compare the enthalpy change (ΔH vap ) and entropy change of evaporation.

27. Vapor Pressure The vapor pressure of the pure solvent or the solution is the result of solvent molecules escaping the liquid surface and becoming gaseous. Since the solute is non-volatile, it does not evaporate. Since only solvent molecules evaporate, the enthalpy change for pure solvent or the solution is the same.

28. Vapor Pressure The decrease in vapor pressure of the solution is the result of changes in entropy. The vapor in either container is disordered, due to the random motion of gaseous solvent.

29. Vapor Pressure The liquid phases differ in entropy. The pure solvent is relatively ordered since all of the molecules are the same (solvent).

30. Vapor Pressure The liquid phase of the solution is much more random, since it is a mixture.

31. Vapor Pressure Upon evaporation, the pure solvent undergoes a greater increase in entropy than the solution.

32. Vapor Pressure Systems tend to maximize entropy. The pure solvent evaporates more readily, because it undergoes a greater increase in entropy.

33. Boiling Point Elevation

34. Vapor Pressure Lowering The change in vapor pressure can be calculated as follows: ∆vp = -X solute P solvent where X is the mole fraction of solute particles P o solvent is the vapor pressure of the pure solvent o

35. Vapor Pressure Lowering ∆vp = -X solute P o solvent The sign is negative because the vapor pressure decreases.

36. Vapor Pressure Lowering P soln = X solvent P o solvent The mole fraction of solvent, X solvent, = moles of solvent/total moles of particles and solvent.

37. Problem – Vapor Pressure Water has a vapor pressure of 92.6 mmHg at 50 o C. a) Compare the vapor pressure of two aqueous solutions at 50 o C. One contains.100 mole of sucrose dissolved in 1.00 mol of water. The other contains.100 moles of CaCl 2 dissolved in 1.00 mol of water. b) Calculate the vapor pressure of the CaCl 2 solution.

38. Solution Phase Diagrams The lowering of the vapor pressure due to the presence of a non-volatile solute affects several properties. The phase diagram for the solution will be shifted, due to the lower vapor pressure of the solution.

39. Solution Phase Diagrams

40. As a result of the lower vapor pressure, the boiling point of the solution is greater than that of pure solvent.

41. Solution Phase Diagrams Since the liquid- solid line is shifted to a lower temperature, the freezing point of the solution is lowered.

42. Properties of Solutions Solutions of non-volatile solutes in a volatile solvent have - higher boiling points and - lower freezing points than the pure solvent.

43. Boiling Point Elevation The size of the increase in boiling point depends upon the concentration of solute particles. ∆T b = K b m(i) where K b is the solvent dependent boiling point elevation constant, m = molality of the solute i = van’t Hoff factor

44. The van’t Hoff Factor, i The van’t Hoff factor is the number of particles in solution compared to the number dissolved. If an ionic compound forms two ions per formula unit, its i value = 2.

45. The van’t Hoff Factor, i If a molecule “pairs up” in solution, with two molecules uniting to form one molecule, then the i factor will be 0.5. For non-electrolytes, the i factor is usually 1, and is often ignored.

46. Freezing Point Depression The size of the decrease in freezing point depends upon the concentration of solute particles. ∆T f = -K f m(i) where K f is the solvent dependent freezing point depression constant, m = molality of the solute i = van’t Hoff factor

47. Constants for Common Solvents

48. Applications Solutions of sugar in water or maple syrup (sap) have boiling points that are higher than 100 o C.

49. Applications Salt is spread on roads to lower the freezing point of ice and keep the roads from icing up at temperatures below 0 o C.

50. Applications Antifreeze keeps the radiators in cars from freezing during the winter and overheating in the summer.

51. Problem Which of the following aqueous solutions will have the lowest freezing point? Which of the following aqueous solutions will have the lowest freezing point? 0.015m calcium nitrate 0.040m sodium chloride 0.040m sucrose 0.020m hydrochloric acid

52. Problem The solubility of NaNO 3 in water at 0 o C is 75 grams per 100g of water. Calculate the freezing point of the solution. K f for water = 1.86 o C/m (or o C-kg/mol). The solubility of NaNO 3 in water at 0 o C is 75 grams per 100g of water. Calculate the freezing point of the solution. K f for water = 1.86 o C/m (or o C-kg/mol).

53. Applications – Molar Mass Since boiling point or freezing point changes are proportional to concentration (molality), it is possible to calculate molar masses of unknown solutes using a measured temperature change. Solvents with greater values of K f or K b will provide the largest change in temperature for a given concentration.

54. Constants for Common Solvents

55. Applications – Molar Mass ∆T b = K b m(i) where m = molality = moles of solute/kg of solvent ∆T b = K b m(i) = K b (moles solute/kg solvent)(i) or ∆T f = -K f m(i) = K f (moles solute/kg solvent)(i)

56. Applications – Molar Mass ∆T b = K b m(i) = K b (moles solute/kg solvent)(i) or ∆T f = -K f m(i) = K f (moles solute/kg solvent)(i) Using either relationship, moles of solute can be calculated. If the mass of the solute is also known, the molar mass is easily calculated.

57. Problem – Molar Mass A solution of 2.50g of a compound with an empirical formula of C 6 H 5 P in 25.0 g of benzene has a freezing point of 4.3 o C. Calculate the molar mass of the solute and its molecular formula. [The normal freezing point of benzene is 5.5 o C, and K f for benzene = 5.12 o C/m. Assume i =1] A solution of 2.50g of a compound with an empirical formula of C 6 H 5 P in 25.0 g of benzene has a freezing point of 4.3 o C. Calculate the molar mass of the solute and its molecular formula. [The normal freezing point of benzene is 5.5 o C, and K f for benzene = 5.12 o C/m. Assume i =1]

58. Osmotic Pressure Osmosis is the flow of solvent across a semipermeable membrane. The membrane allows solvent molecules to pass through, but not solute particles.

59. Osmotic Pressure

60. The minimum pressure needed to just stop osmosis is called the osmotic pressure.

61. Osmotic Pressure Π = MRT(i) where Π is the osmotic pressure M is molarity (mol solute/L of soln) R = 0.08206 L-atm/mol-K T is temperature in Kelvins

62. Osmotic Pressure - Applications A relatively dilute solution provides a fairly large osmotic pressure. As a result, osmotic pressure is an excellent way to obtain molar masses of very dilute solutes such a proteins.

63. Problem 0.8750 g of a protein is dissolved in enough water to make 100. ml of solution. The solution has an osmotic pressure of 3.8 mm Hg at 25 o C. Calculate the molar mass of the protein. 0.8750 g of a protein is dissolved in enough water to make 100. ml of solution. The solution has an osmotic pressure of 3.8 mm Hg at 25 o C. Calculate the molar mass of the protein.

64. Osmotic Pressure - Applications Renal dialysis uses osmosis to rid the blood of waste products in people with kidney failure.

65. Osmotic Pressure - Applications Isotonic saline is a salt solution with the same osmotic pressure as blood cells. This maintains a fluid balance within the cell.

66. Osmotic Pressure - Applications If a solution of saline is too concentrated, the cell will become dehydrated and shrink (crenate).

67. Osmotic Pressure - Applications If a solution of saline is too dilute, the red blood cells become swollen with excess water and eventually burst (hemolysis).

68. Osmotic Pressure - Applications In reverse osmosis, a pressure greater than the osmotic pressure is applied to a solution. Pure solvent can be obtained on the other side of the membrane.

69. Osmotic Pressure - Applications

70. Behavior of Electrolytes The van’t Hoff factor, i, represents the number of particles formed in solution for each solute particle dissolved. i = moles of particles in solution moles of solute dissolved

71. Behavior of Electrolytes For ionic solutes, we expect the value of i to be 2 for NaCl, 3 for MgCl 2, and 4 for FeCl 3, etc. In extremely dilute solutions, the observed value of i is very close to these expected values.

72. Behavior of Electrolytes However, as solutions become more concentrated, ion pairing occurs, and some of the ions formed in solution pair up and behave like a single particle.

73. Behavior of Electrolytes

74. Problem When a 0.00500 moles of acetic acid is dissolved in 100 grams of benzene, the change in the freezing point of benzene is half of the expected value. Explain why. When a 0.00500 moles of acetic acid is dissolved in 100 grams of benzene, the change in the freezing point of benzene is half of the expected value. Explain why.

75. Liquid-Liquid Solutions When two volatile liquids mix, they form a solution. An ideal solution, similar to an ideal gas, will exert a vapor pressure which is related to the vapor pressures of the pure liquids and their relative abundance in the mixture.

76. Liquid-Liquid Solutions The solution obeys Raoult’s law: P A = χ A P o A P B = χ B P o B where χ A is the mole fraction of component A and P o A is the vapor pressure of pure A

77. Liquid-Liquid Solutions Raoult’s law is best seen graphically. The vapor pressure of the mixture is the sum of the vapor pressures of each component.

78. Liquid-Liquid Solutions Ideal solutions typically involve non- polar molecules with similar structures. Mixtures of liquid hydrocarbons often form ideal solutions.

79. Liquid-Liquid Solutions If the two components of the mixture are strongly attracted to each other, such as two polar molecules, the vapor pressure of the mixture is often less than that predicted by Raoult’s law.

80. Liquid-Liquid Solutions This is known as a negative deviation from Raoult’s law. It occurs with mixtures of liquid acids and water. As the acid ionizes, the forces of attraction in the mixture increase.

81. Liquid-Liquid Solutions If a mixture contains liquids that have stronger attractive forces when pure than when mixed, the mixture will exert a vapor pressure that is greater than that predicted by Raoult’s law. This is called a positive deviation from Raoult’s law.

82. Liquid-Liquid Solutions A mixture of ethanol and water exhibits a positive deviation from Raoult’s law. The hydrogen bonding of each pure liquid is disrupted when the two liquids are mixed.

83. Application – Fractional Distillation Mixtures of volatile liquids can sometimes be separated by a technique called fractional distillation. If the mixture is boiled, the vapor is often enriched in the more volatile component. Collection of the vapor provides the more volatile component, and the liquid remaining in the flask will be enriched in the less volatile component. If the mixture is boiled, the vapor is often enriched in the more volatile component. Collection of the vapor provides the more volatile component, and the liquid remaining in the flask will be enriched in the less volatile component.

84. Fractional Distillation