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  • 1. SOLUTIONS II • COLLIGATIVE PROPERTIES • Chapter 13.6 Silberberg 13-1
  • 2. Chapter 13 The Properties of Mixtures: Solutions and Colloids 13-2
  • 3. The Properties of Mixtures: Solutions and Colloids 13.1 Types of Solutions: Intermolecular Forces and Solubility 13.2 Intermolecular Forces and Biological Macromolecules 13.3 Why Substances Dissolve: Understanding the Solution Process 13.4 Solubility as an Equilibrium Process 13.5 Concentration Terms 13.6 Colligative Properties of Solutions 13.7 Structure and Properties of Colloids 13-3
  • 4. Goals & Objectives • See the following Learning Objectives on pages 543-544. • Understand these Concepts: • 13.15-20. • Master these Skills: • 13.5-10. 13-4
  • 5. Colligative Properties • Colligative properties depend on the number, rather than the kind, of solute particles. • Colligative properties are physical properties of solutions. 13-5
  • 6. Colligative Properties • Examples of colligative properties include: – change in vapor pressure if solute is volatile – lowering of vapor pressure if solute is not volatile – freezing point depression – boiling point elevation if solute is not volatile 13-6 – osmotic pressure
  • 7. Colligative Properties of Solutions Colligative properties are properties that depend on the number of solute particles, not their chemical identity. The number of particles in solution can be predicted from the formula and type of the solute. An electrolyte separates into ions when it dissolves in water. Strong electrolytes dissociate completely while weak electrolytes dissociate very little. A nonelectrolyte does not dissociate to form ions. Each mole of dissolved compound yields 1 mole of particles in solution. 13-7
  • 8. Figure 13.22 Conductivity of three types of electrolyte solutions. strong electrolyte 13-8 weak electrolyte nonelectrolyte
  • 9. Volatile Nonelectrolyte Solutions For a volatile nonelectrolyte, the vapor of the solution contains both solute and solvent. The presence of each volatile component lowers the vapor pressure of the other, since each one lowers the mole fraction of the other. For such a solution, the vapor will have a higher mole fraction of the more volatile component. The vapor has a different composition than the solution. 13-9
  • 10. Colligative Properties of Electrolyte Solutions For vapor pressure lowering: For boiling point elevation: For freezing point depression: For osmotic pressure: 13-10 P = i( solute x Tb = i( Posolvent) bm) Tf = i( fm) Π = i(MRT)
  • 11. Figure 13.26 Nonideal behavior of strong electrolyte solutions. Ions in solution may remain clustered near ions of opposite charge, creating an ionic atmosphere. The ions do not act independently, and the effective concentration of dissolved particles is less than expected. 13-11
  • 12. Vapor Pressure Lowering The vapor pressure of a solution containing a nonvolatile nonelectrolyte is always lower than the vapor pressure of the pure solvent. Raoult’s law states that the vapor pressure of the solvent above the solution is proportional to the mole fraction of the solvent present: Psolvent = Xsolvent x P solvent The vapor pressure lowering is proportional to the mole fraction of the solute present. Psolvent = Xsolute x P 13-12 solvent
  • 13. Figure 13.23 Effect of solute on the vapor pressure of solution. Equilibrium is reached with a given number of particles in the vapor. 13-13 Equilibrium is reached with fewer particles in the vapor.
  • 14. Sample Problem 13.6 Using Raoult’s Law to Find ΔP PROBLEM: Find the vapor pressure lowering, P, when 10.0 mL of glycerol (C3H8O3) is added to 500. mL of water at 50. C. At this temperature, the vapor pressure of pure water is 92.5 torr and its density is 0.988 g/mL. The density of glycerol is 1.26 g/mL. PLAN: We are given the vapor pressure of pure H2O, so to calculate P we just need the mole fraction of glycerol, Xglycerol. volume (mL) of glycerol multiply by density mass (g) of glycerol multiply by M amount (mol) of glycerol multiply by P mole fraction (X) of glycerol 13-14 glycerol vapor pressure lowering
  • 15. Sample Problem 13.6 SOLUTION: 10.0 mL C3H8O3 x 1.26 g C3H8O3 1 mL C3H8O3 x 1 mol C3H8O3 92.09 g C3H8O3 = 0.137 mol C3H8O3 500.0 mL H2O x Xglycerol = 0.988 g H2O 1 mL H2O x 1 mol H2O 18.02 g H2O 0.137 mol C3H8O3 = 0.00498 0.137 mol C3H8O3 + 27.4 mol H2O P = 0.00498 x 92.5 torr = 0.461 torr 13-15 = 27.4 mol H2O
  • 16. • Vapor Pressure & Raoult's Law the vapor Raoult’s Law states that pressure of a volatile component in an ideal solution decreases as its mole fraction decreases. – Psolution = (XA)(P A)+(XB)(P B)+(XC)(P C)… • When a nonvolatile solute is dissolved in a liquid, the vapor pressure of the liquid is always lowered. – Psolution = (Xsolvent)(P solvent) + (Xsolute)(P solute) 13-16
  • 17. Vapor Pressure & Raoult's Law • When an ionic solute is dissolved in a liquid, the vapor pressure of the liquid is lowered in proportion to the number of ions present. NaCl (s) + excess H2O Na1+(aq) + Cl1 (aq) • The vapor pressure is lower for a 1.0M solution of NaCl than for a 1.0 M solution of glucose due to the number of ions. 13-17
  • 18. • Vapor Pressure & Raoult's Determine the vapor Law of a solution pressure prepared by dissolving 18.3 g of sucrose in 500g of water at 70 C. The vapor pressure of water at 70 C is 233.7 torr. The molar mass of sucrose is 342.3g/mole. • What is the vapor pressure of a mixture that is 25% by mass acetone in water at 25 C? The vapor pressure of water at 25 C is 23.8 torr, and the vapor pressure of acetone at that temperature is 200 torr. The molar mass of acetone is 58.0 g/mole. 13-18
  • 19. 13-19
  • 20. 13-20
  • 21. Boiling Point Elevation A solution always boils at a higher temperature than the pure solvent. This colligative property is a result of vapor pressure lowering. The boiling point elevation is proportional to the molality of the solution. Tb = Kbm Kb is the molal boiling point elevation constant for the solvent. 13-21
  • 22. Boiling Point Elevation • kb = the molal boiling point elevation constant kb for water, H2O kb for benzene, C6H6 kb for camphor 13-22 0.512 C/m 2.53 C/m 5.95 C/m
  • 23. Boiling Point Elevation • Determine the normal boiling point of a 2.50 m glucose, C6H12O6, solution. 13-23
  • 24. 13-24
  • 25. Figure 13.24 13-25 Boiling and freezing points of solvent and solution.
  • 26. Freezing Point Depression A solution always freezes at a lower temperature than the pure solvent. The freezing point depression is proportional to the molality of the solution. Tf = Kfm Kf is the molal freezing point depression constant for the solvent. 13-26
  • 27. Table 13.6 Molal Boiling Point Elevation and Freezing Point Depression Constants of Several Solvents Kb ( C/m) Melting Point ( C) Kf ( C/m) 117.9 3.07 16.6 3.90 Benzene 80.1 2.53 5.5 4.90 Carbon disulfide 46.2 2.34 -111.5 3.83 Carbon tetrachloride 76.5 5.03 -23 Chloroform 61.7 3.63 -63.5 4.70 Diethyl ether 34.5 2.02 -116.2 1.79 Ethanol 78.5 1.22 -117.3 1.99 100.0 0.512 0.0 1.86 Solvent Boiling Point (oC)* Acetic acid Water *At 13-27 1 atm. 30.
  • 28. Sample Problem 13.7 Determining Boiling and Freezing Points of a Solution PROBLEM: You add 1.00 kg of ethylene glycol (C2H6O2) antifreeze to 4450 g of water in your car’s radiator. What are the boiling and freezing points of the solution? PLAN: We need to find the molality of the solution and then calculate the boiling point elevation and freezing point depression. mass (g) of solute divide by M amount (mol) of solute divide by kg of solvent molality (m) Tb(solution) 13-28 Tf = Kfm Tb = Kbm Tb + Tb Tb Tf T f - Tf Tf(solution)
  • 29. Sample Problem 13.7 SOLUTION: 1.00x103 g C2H6O2 x molality = 1 mol C2H6O2 62.07 g C2H6O2 16.1 mol C2H6O2 4.450 kg H2O = 16.1 mol C2H6O2 = 3.62 m C2H6O2 Tb = 3.62 m x 0.512 C/m = 1.85 C Tb(solution) = 100.00 + 1.85 = 101.85 C Tf = 3.62 m x 1.86 C/m = 6.73 C Tb(solution) = 0.00 - 6.73 = -6.73 C 13-29
  • 30. Freezing Point Depression Tf = kfm • kf = the molal freezing point depression constant kf kf kf kf 13-30 for water, H2O for benzene, C6H6 for t-butanol, (CH3)3COH for camphor 1.86 5.12 8.09 37.7 C/m C/m C/m C/m
  • 31. Freezing Point Depression • Calculate the freezing point of 2.50m glucose solution. • Calculate the freezing point of a solution that contains 8.50g of benzoic acid, C6H5COOH, in 75.0g of benzene, C6H6. kf for benzene, C6H6 = 5.12 C/m MW for benzoic acid = 122 g/mole Freezing point of benzene = 5.5 C 13-31
  • 32. 13-32
  • 33. Determination of Molar Masses by Freezing Point Depression • A 37.0g sample of a new covalent compound, a nonelectrolyte, was dissolved in 200g of water. The resulting solution froze at 5.58 C. • Determine the molecular weight of the compound. (V-3) 13-33
  • 34. 13-34
  • 35. 13-35
  • 36. Osmotic Pressure Osmosis is the movement of solvent particles from a region of higher to a region of lower concentration through a semipermeable membrane. Solvent will always flow from a more dilute solution to a more concentrated one. Osmotic pressure is the pressure that must be applied to prevent the net flow of solvent. Π = MRT 13-36 M = molarity R = 0.0821 atm·L/mol·K T = Kelvin temperature
  • 37. Figure 13.25 13-37 The development of osmotic pressure.
  • 38. Osmotic Pressure • Determine the osmotic pressure of a 1.00 M solution of glucose at 25OC. 13-38
  • 39. 13-39
  • 40. Sample Problem 13.8 Determining Molar Mass from Osmotic Pressure PROBLEM: Biochemists have discovered more than 400 mutant varieties of hemoglobin, the blood protein that carries O2. A physician dissolves 21.5 mg of one variety in water to make 1.50 mL of solution at 5.0 C. She measures an osmotic pressure of 3.61 torr. What is the molar mass of the protein? PLAN: We convert Π to atm and T to degrees K and calculate molarity from osmotic pressure. We can then determine the molar mass using the number of moles and the known mass. Π (atm) Π = MRT M (mol/L) convert to mol amount (mol) of solute 13-40 divide mass (g) by moles M (g/mol)
  • 41. Sample Problem 13.8 SOLUTION: M= Π = RT 3.61 torr x 1 atm 760 torr = 2.08 x10-4 M (0.0821 L·atm/mol·K)(278.15 K) 2.08 x10-4 mol x 1L = 3.12x10-7 mol 1L 103 mL 21.5 mg x M= 13-41 1g = 0.0215 g 103 mg 0.0215 g 3.12x10-7 mol = 6.89x104 g/mol
  • 42. Osmotic Pressure • A 1.00 g sample of a protein was dissolved in enough water to give 100 mL of solution. The osmotic pressure of the solution was 2.80 torr at 25 C. Calculate the molarity and the approximate molecular weight of the material. • 2.80 torr/ 760 torr = 0.00368 atm 13-42
  • 43. 13-43
  • 44. Figure 13.29a Osmotic pressure and cell shape. A red blood cell in an isotonic solution has its normal shape. 13-44
  • 45. Figure 13.29b Osmotic pressure and cell shape. A hypotonic solution has a lower concentration of particles than the cell. A cell in a hypotonic solution absorbs water and swells until it bursts. 13-45
  • 46. Figure 13.29c Osmotic pressure and cell shape. A hypertonic solution has a higher concentration of dissolved particles than the cell. If a cell is placed in a hypertonic solution, water moves out of the cell, causing it to shrink. 13-46
  • 47. Table 13.7 Types of Colloids Colloid Type Dispersed Substance Dispersing Medium Example(s) Aerosol Liquid Gas Fog Aerosol Solid Gas Smoke Foam Gas Liquid Whipped cream Solid foam Gas Solid Marshmallow Emulsion Liquid Liquid Milk Solid emulsion Liquid Solid Butter Sol Solid Liquid Paint; cell fluid Solid sol Solid Solid Opal 13-47
  • 48. Figure 13.30 Light scattering and the Tyndall effect. The narrow, barely visible light beam that passes through a solution (left), is scattered and broadened by passing through a colloid (right). 13-48 Sunlight is scattered by dust in air.
  • 49. Figure 13.31 The Nile delta (reddish-brown area). At the mouths of rivers, colloidal clay particles coalesce into muddy deltas. 13-49
  • 50. Chemical Connections Figure B13.1 Figure B13.1 13-50 The typical steps in municipal water treatment.
  • 51. Chemical Connections Figure B13.2 Ion exchange to remove hard-water cations. 13-51
  • 52. Chemical Connections Figure B13.3 Reverse osmosis to remove ions. 13-52