1. Solutions

2. Solution Formation The compositions of the solvent and the solute determine whether a substance will dissolve. The factors that determine how fast a substance dissolves are: stirring (agitation) temperature the surface area of the dissolving particles

3. Solution Formation A cube of sugar in cold tea dissolves slowly.

4. Solution Formation Granulated sugar dissolves in cold water more quickly than a sugar cube, especially with stirring.

5. Solution Formation Granulated sugar dissolves very quickly in hot tea.

6. Solution Formation Stirring and Solution Formation –Stirring speeds up the dissolving process because fresh solvent (the water in tea) is continually brought into contact with the surface of the solute (sugar). –Reminder: - Solute – substance being dissolved - Solvent – substance doing the dissolving

7. Solution Formation Temperature and Solution Formation –At higher temperatures, the kinetic energy of water molecules is greater than at lower temperatures, so they move faster. As a result, the solvent molecules collide with the surface of the sugar crystals more frequently and with more force.

8. Solution Formation Particle Size and Solution Formation –A spoonful of granulated sugar dissolves more quickly than a sugar cube because the smaller particles in granulated sugar expose a much greater surface area to the colliding water molecules.

9. Solubility Saturated solution- contains the maximum amount of solute for a given quantity of solvent at a given temperature and pressure Unsaturated solution- contains less solute than a saturated solution at a given temperature and pressure

10. Solubility In a saturated solution, the rate of dissolving equals the rate of crystallization, so the total amount of dissolved solute remains constant.

11. Solubility Solubility- is the amount of solute that dissolves in a given quantity of a solvent at a specified temperature and pressure to produce a saturated solution. Solubility is often expressed in grams of solute per 100 g of solvent.

12. Solubility Some liquids combine in all proportions, while others don’t mix at all. – Miscible- when two liquids dissolve in each other in all proportions – Immiscible- when two liquids are insoluble in each other

13. Solubility Oil and water are immiscible.

14. Solubility Vinegar and oil are immiscible.

15. Factors Affecting Solubility Temperature affects the solubility of solid, liquid, and gaseous solutes in a solvent –Both temperature and pressure affect the solubility of gaseous solutes.

16. Factors Affecting Solubility Temperature The solubility of most solid substances increases as the temperature of the solvent increases. The solubilities of most gases are greater in cold water than in hot.

17. Factors Affecting Solubility Example: The mineral deposits around hot springs result from the cooling of the hot, saturated solution of minerals emerging from the spring.

18. Factors Affecting Solubility

19. Supersaturated solution- contains more solute than it can theoretically hold at a given temperature The crystallization of a supersaturated solution can be initiated if a very small crystal, called a seed crystal, of the solute is added.

20. Factors Affecting Solubility A supersaturated solution is clear before a seed crystal is added.

21. Factors Affecting Solubility Crystals begin to form in the solution immediately after the addition of a seed crystal.

22. Factors Affecting Solubility Excess solute crystallizes rapidly.

23. Factors Affecting Solubility

24. Pressure –Changes in pressure have little effect on the solubility of solids and liquids, but pressure strongly influences the solubility of gases. –Gas solubility increases as the partial pressure of the gas above the solution increases.

25. Factors Affecting Solubility Henry’s law states that at a given temperature, the solubility (S) of a gas in a liquid is directly proportional to the pressure (P) of the gas above the liquid.

26. Calculating the Solubility of a Gas Example: If the solubility of a gas in water is 0.77 g/L at 3.5 atm of pressure, what is the solubility (in g/L) at 1.0 atm of pressure? (The temperature is held at 25°C.) Known: P 1 = 3.5 atm, S 1 = 0.77 g/L, P 2 = 1.0 atm Remember Henry’s Law:

27. Calculating the Solubility of a Gas Unknown : S 2 = ? g/L Solve for S 2 : S 2 = (S 1 X P 2 )/P 1 = (0.77 g/L X 1.0 atm) / 3.5 atm = 0.22 g/L

28. Molarity Concentration - a measure of the amount of solute that is dissolved in a given quantity of solvent in a solution – Dilute solution- is one that contains a small amount of solute – Concentrated solution- contains a large amount of solute

29. Molarity Molarity (M)- is the number of moles of solute dissolved in one liter of solution. To calculate the molarity of a solution, divide the moles of solute by the volume of the solution.

30. Molarity To make a 0.5-molar (0.5M) solution, first add 0.5 mol of solute to a 1-L volumetric flask half filled with distilled water.

31. Molarity Swirl the flask carefully to dissolve the solute.

32. Molarity Fill the flask with water exactly to the 1-L mark.

33. Calculating the Molarity of a Solution Intravenous (IV) saline solutions are often administered to patients in the hospital. One saline solution contains 0.90 g NaCl in exactly 100 mL of solution. What is the molarity of the solution? Knowns: solution concentration = 0.90 g NaCl/100 mL molar mass NaCl = 58.5 g/mol

34. Calculating the Molarity of a Solution Unknown: solution concentration = ? M Remember: Solution concentration = 0.90 g NaClX 1 mol NaCl X 1,000 mL 100 mL 58.5 g NaCl 1 L = 0.15 mol/L = 0.15 M

35. Finding the Moles of Solute in a Solution Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO). How many moles of solute are present in 1.5 L of 0.70 M NaClO? Knowns: volume of solution = 1.5 L solution concentration = 0.70 M NaClO Unknown: moles solute = ? mol

36. Finding the Moles of Solute in a Solution moles solute = M X L = (mol/L) X L moles solute = (0.70 mol NaClO / 1 L) X 1.5 L = 1.1 mol NaClO

37. Making Dilutions Diluting a solution reduces the number of moles of solute per unit volume, but the total number of moles of solute in solution does not change.

38. Making Dilutions The total number of moles of solute remains unchanged upon dilution, so you can write this equation. M 1 and V 1 are the molarity and volume of the initial solution, and M 2 and V 2 are the molarity and volume of the diluted solution.

39. Making Dilutions Making a Dilute Solution

40. Making Dilutions To prepare 100 ml of 0.40M MgSO 4 from a stock solution of 2.0M MgSO 4, a student first measures 20 mL of the stock solution with a 20-mL pipet.

41. Making Dilutions She then transfers the 20 mL to a 100-mL volumetric flask.

42. Making Dilutions Finally she carefully adds water to the mark to make 100 mL of solution.

43. Making Dilutions Volume-Measuring Devices:

44. Preparing a Dilute Solution How many milliliters of aqueous 2.00M MgSO 4 solution must be diluted with water to prepare 100.0 mL of aqueous 0.400M MgSO 4 ? Knowns: M 1 = 2.00M MgSO 4, M 2 = 0.400M MgSO 4, V 2 = 100.0 mL of 0.400M MgSO 4

45. Preparing a Dilute Solution Unknown: V 1 = ? mL of 2.00 MgSO 4 Remember: V 1 = (M 2 X V 2 )/M 1 = (0.400M X 100.0 mL)/2.00M = 20.0 mL

46. Percent Solutions The concentration of a solution in percent can be expressed in two ways: -as the ratio of the volume of the solute to the volume of the solution or as the ratio of the mass of the solute to the mass of the solution

47. Percent Solutions Concentration in Percent (Volume/Volume)

48. Percent Solutions Isopropyl alcohol (2-propanol) is sold as a 91% solution. This solution consist of 91 mL of isopropyl alcohol mixed with enough water to make 100 mL of solution.

49. Calculating Percent (Volume/Volume) What is the percent by volume of ethanol (C 2 H 6 O, or ethyl alcohol) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water? Knowns: volume of ethanol = 85 mL volume of solution = 250 mL Unknown: % ethanol (v/v) = ? %

50. Calculating Percent (Volume/Volume) Remember: % (v/v) = (85 mL ethanol / 250 mL) X 100% = 34% ethanol (v/v)

51. Percent Solutions Concentration in Percent (Mass/Mass)

52. Vapor-Pressure Lowering Colligative property- a property that depends only upon the number of solute particles, and not upon their identity

53. Vapor-Pressure Lowering Three important colligative properties of solutions are: –vapor-pressure lowering –boiling-point elevation –freezing-point depression

54. Vapor-Pressure Lowering In a pure solvent, equilibrium is established between the liquid and the vapor.

55. Vapor-Pressure Lowering In a solution, solute particles reduce the number of free solvent particles able to escape the liquid. Equilibrium is established at a lower vapor pressure.

56. Vapor-Pressure Lowering The decrease in a solution’s vapor pressure is proportional to the number of particles the solute makes in solution.

57. Vapor-Pressure Lowering Three moles of glucose dissolved in water produce 3 mol of particles because glucose does not dissociate.

58. Vapor-Pressure Lowering Three moles of sodium chloride dissolved in water produce 6 mol of particles because each formula unit of NaCl dissociates into two ions.

59. Vapor-Pressure Lowering Three moles of calcium chloride dissolved in water produce 9 mol of particles because each formula unit of CaCl 2 dissociates into three ions.

60. Freezing-Point Depression Freezing-point depression - the difference in temperature between the freezing point of a solution and the freezing point of the pure solvent

61. Freezing-Point Depression The magnitude of the freezing-point depression is proportional to the number of solute particles dissolved in the solvent and does not depend upon their identity.

62. Freezing-Point Depression The freezing-point depression of aqueous solutions makes walks and driveways safer when people sprinkle salt on icy surfaces to make ice melt. The melted ice forms a solution with a lower freezing point than that of pure water.

63. Boiling-Point Elevation Boiling-point elevation - the difference in temperature between the boiling point of a solution and the boiling point of the pure solvent The same antifreeze added to automobile engines to prevent freeze-ups in winter, protects the engine from boiling over in summer.

64. Boiling-Point Elevation The magnitude of the boiling-point elevation is proportional to the number of solute particles dissolved in the solvent. Example: The boiling point of water increases by 0.512°C for every mole of particles that the solute forms when dissolved in 1000 g of water.

65. Molality and Mole Fraction The unit molality and mole fractions are two additional ways in which chemists express the concentration of a solution.

66. Molality and Mole Fraction molality (m) is the number of moles of solute dissolved in 1 kilogram (1000 g) of solvent Molality is also known as molal concentration.

67. Molality and Mole Fraction To make a 0.500m solution of NaCl, use a balance to measure 1.000 kg of water and add 0.500 mol (29.3 g) of NaCl.

68. Molality and Mole Fraction Ethlylene Glycol (EG) is added to water as antifreeze.

69. Using Solution Molality How many grams of potassium iodide must be dissolved in 500.0 g of water to produce a 0.060 molal KI solution? Known: mass of water = 500.0g = 0.5000kg solution concentration = 0.060m molar mass KI = 166.0 g/mol

70. Using Solution Molality Unknown: mass of solute = ? g KI 0.5000 kg H 2 0 X 0.0600 mol KI X 166.0 g KI 1.000 kg H 2 O 1 mol KI = 5.0 g KI

71. Molality and Mole Fraction Mole fraction - is the ratio of the moles of that solute to the total number of moles of solvent and solute in a solution

72. Molality and Mole Fraction In a solution containing n A mol of solute A and n B mol of solvent B (X B ), the mole fraction of solute A (X A ) and the mole fraction of solvent B (X B ) can be expressed as follows.

73. Calculating Mole Fractions Ethylene glycol (C 2 H 6 O 2 ) is added to automobile cooling systems to protect against cold weather. What is the mole fraction of each component in a solution containing 1.25 mol of ethylene glycol (EG) and 4.00 mol of water? Known: moles of ethylene glycol (n EG ) = 1.25 mol EG moles of water (n H 2 O ) = 4.00 mol H 2 O

74. Calculating Mole Fractions Unknown: mole fraction EG (X EG ) = ? mole fraction H 2 O (X H 2 O ) = ? X EG = (n EG )/ (n EG + n H 2 O ) = (1.25 mol) / (1.25 mol + 4.00 mol) = 0.238

75. Calculating Mole Fractions X H 2 O = (n H 2 O )/(n EG + n H 2 O ) = (4.00 mol)/(1.25 mol + 4.00 mol) = 0.762

76. Freezing-Point Depression and Boiling-Point Elevation The magnitudes of the freezing-point depression and the boiling-point elevation of a solution are directly proportional to the molal concentration (m), when the solute is molecular, not ionic.

77. Freezing-Point Depression and Boiling-Point Elevation molal freezing-point depression constant- is equal to the change in freezing point for a 1-molal solution of a nonvolatile molecular solute - represented by the constant K f

78. Freezing-Point Depression and Boiling-Point Elevation

79. molal boiling-point elevation constant- is equal to the change in boiling point for a 1-molal solution of a nonvolatile molecular solute - represented by constant K b

80. Freezing-Point Depression and Boiling-Point Elevation

82. Calculating the Freezing-Point Depression of a Solution Antifreeze protects a car from freezing. It also protects it from overheating. Calculate the freezing-point depression and the freezing point of a solution containing 100 g of ethylene glycol (C 2 H 6 O 2 ) antifreeze in 0.500 kg of water. Known: mass of solute = 100 g C 2 H 6 O 2 mass of solvent = 0.500 kg H 2 O K f for H 2 O = 1.86°C/m ΔT f = K f X m

83. Calculating the Freezing- Depression of a Solution Unknown: ΔT f = ? °C freezing point = ? °C moles C 2 H 6 O 2 = 100 g C 2 H 6 O 2 X 1 mol 62.0 g C 2 H 6 O 2 = 1.61 mol

84. Calculating the Freezing-Point Depression of a Solution m = (mol solute) / (kg solvent) = (1.61 mol)/(0.500 kg) = 3.22 m ΔT f = K f X m = 1.86°C/m X 3.22 m = 5.99°C The freezing point of the solution is: 0.00 °C – 5.99°C = -5.99°C

85. Calculate the Boiling Point of an Ionic Solution What is the boiling point of a 1.50m NaCl solution? Known: K b for H 2 O = 0.512°C/m ΔT b = K b X m Unknown: boiling point = ? °C

86. Calculate the Boiling Point of an Ionic Solution ΔT b = K b X m = (0.512 °C)/ (m) X 3.00m = 1.54 °C The boiling point of the solution is: 100 °C + 1.54 °C = 101.54 °C