1. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Properties of Solutions Chapter 11

2. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 2 Overview Introduce student to solution composition and energy of solution formation. Factor affecting solubilities will be discussed: structure, pressure and temperature effects. Vapor pressure of solutions, boiling point, freezing point affected by solute addition. Colligative properties of electrolyte solutions and colloids.

3. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 3 A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount

4. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 4 A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature. A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature. Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate.

5. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 5 Solution Composition 1.Molarity (M) = 2.Mass (weight) percent = 3.Mole fraction (  A ) = 4.Molality (m) =

6. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 6 Concentration Units The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Percent by Mass % by mass = x 100% mass of solute mass of solute + mass of solvent = x 100% mass of solute mass of solution Mole Fraction (X) X A = moles of A sum of moles of all components

7. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 7 Concentration Units Continued M = moles of solute liters of solution Molarity (M) Molality (m) m = moles of solute mass of solvent (kg)

8. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 8 N = Number of equivalent liters of solution Normality (N) Acid-Base reaction: is the amount needed to accept one mole of H + Number of equivalent = mass Equivalent mass x 1 V Equivalent mass of H 2 SO 4 = MM 2 Equivalent mass of Ca(OH) 2 = MM 2

9. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 9 Redox reaction : is the amount needed to accept exactly one mole e - MnO 4 - + 5e - + 8H + Mn 2+ + 4H 2 O Equivalent mass of KMnO 4 = MM 5

10. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 10

11. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 11 S.Ex 11.1 1.0 g C 2 H 5 OH dissolved in 100.0 g water and the final volume of solution is 101 mL. Calculate M, mass%, mole fraction, molality Moles of C 2 H 5 OH = 1.00/46.07 = 2.17x10 -2 M = 2.17x10 -2 mol/0.101L = 0.215 M Mass % = 1.00 x100% = 0.990 % C 2 H 5 OH 100+1 Mols of H 2 O = 100/18.0 = 5.56 χ C 2 H 5 OH = 2.17x10 -2 / ( 2.17x10 -2 + 5.56) = 0.00389 m = 2.17x10 -2 mol/0.1000 kg = 0.217 m

12. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 12 What is the molality of a 5.86 M ethanol (C 2 H 5 OH) solution whose density is 0.927 g/mL? m =m = moles of solute mass of solvent (kg) M = moles of solute liters of solution Assume 1 L of solution: 5.86 moles ethanol = 270 g ethanol 927 g of solution (1000 mL x 0.927 g/mL) mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = 0.657 kg m =m = moles of solute mass of solvent (kg) = 5.86 moles C 2 H 5 OH 0.657 kg solvent = 8.92 m

13. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 13 S.Ex 11.2 Lead storage battery has 3.75 M sulfuric acid solution with a density of 1.230 g/mL.Calculate mass %, molality, and Normality. Consider 1 L solution, …. 1230 g mass of solution mass of H 2 SO 4 = 3.75 x 98.1 = 368 g mass of water = 1230 – 368 = 862 g

14. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 14 Steps in Solution Formation Step 1 -Expanding the solute (endothermic) Step 2 -Expanding the solvent (endothermic) Step 3 -Allowing the solute and solvent to interact to form a solution (exothermic)  H soln =  H step 1 +  H step 2 +  H step 3

15. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 15 The formation of a liquid solution can be divided into three steps: (1) expanding the solute, (2) expanding the solvent, and (3) combining the expanded solute and solvent to form the solution.

16. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 16 Figure 11.2 a&b (a) Exothermic and (b) Endothermic

17. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 17 HH < 0Exothermic Solution will occur

18. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 18 The driving factor that favor a process of solution formation is an increase in disorder

19. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 19

20. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 20 “like dissolves like” Two substances with similar intermolecular forces are likely to be soluble in each other. non-polar molecules are soluble in non-polar solvents CCl 4 in C 6 H 6 polar molecules are soluble in polar solvents C 2 H 5 OH in H 2 O ionic compounds are more soluble in polar solvents NaCl in H 2 O or NH 3 (l)

21. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 21 The molecular structures of (a) vitamin A (nonpolar, fat-soluble) and (b) vitamin C (polar, water-soluble). The circles in the structural formulas indicate polar bonds. Note that vitamin C contains far more polar bonds than vitamin A. Fat Soluble Hydrophobic Water Soluble Hydrophilic

22. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 22 Ex 11.3 Liquid hexane C 6 H 14, or liquid methanol CH 3 OH is the most appropriate solvent for grease C 20 H 42 and KI

23. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 23 Hexane, Liquid Methanol, Grease

24. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 24 Factor Affecting Solubility 1.Structure 2.Pressure 3.Temperature

25. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 25 Pressure Effects Pressure has little effect on solids and liquids. It increases the solubility of gases.

26. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 26 (a) A gaseous solute in equilibrium with a solution. (b) The piston is pushed in, increasing the pressure of the gas and number of gas molecules per unit volume. This causes an increase in the rate at which the gas enters the solution, so the concentration of dissolved gas increases. (c) The greater gas concentration in the solution causes an increase in the rate of escape. A new equilibrium is reached.

27. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 27

28. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 28 Henry’s Law C = kP P = partial pressure of gaseous solute above the solution C = concentration of dissolved gas k = a constant The amount of a gas dissolved in a solution is directly proportional to the pressure of the gas above the solution.

29. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 29 Henry’s Law Applied for Dilute solutions Gases that do not dissociate or react with solvent: –O 2 /waterApplied –HCl/waterIs not applied

30. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 30 S.Ex 11.4 A certain soft drink bottle at 25 o C contains CO 2 gas at 5.0 atm over the liquid. Assuming partial pressure of CO 2 in the atmosphere is 4.0x10 -4 atm, calculate the equilibrium concentration of CO 2 before and after the bottle is opened. Kco 2 = 3.1x10 -2 mol/L.atm at 25 o C »C CO2 = K CO2. P CO2 Unopened bottle; C CO2 = 3.1x10 -2 mol/L.atm x 5atm = 0.16 mol/L Opened bottle; C CO2 = 3.1x10 -2 mol/L.atmx 4.0x10 -4 atm = 1.2x10 -5 mol/L

31. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 31 Temperature Effect on Gases The solubilities of several gases in water as a function of temperature at a constant pressure of 1 atm of gas above the solution.

32. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 32 The solubilities of several solids as a function of temperature. Temperature Effect on Solids

33. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 33 The Vapor Pressure of Solutions Solutions have different physical properties from pure solvent. Solutions of nonvolatile solutes differ from solutions of volatile solvents.

34. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 34 An aqueous solution and pure water in a closed environment. (a) Initial stage. (b) After a period of time, the water is transferred to the solution. Vapor pressure of pure water is higher VP of solution is lower

35. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 35 Raoult’s Law P soln =  solvent P  solvent P soln = vapor pressure of the solution  solvent = mole fraction of the solvent P  solvent = vapor pressure of the pure solvent The presence of a nonvolatile solute lowers the vapor pressure of a solvent.  solvent = Moles of solvents Moles of solvent + Moles of solute

36. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 36 For a solution that obeys Raoult's law, a plot of P soln versus x solvent gives a straight line.

37. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 37 Note Solutions that obey Raoult’s Law are called Ideal Solutions. Can be used to determine the molar mass of unknown. For ionic compounds you should multiply by the total number of ions per molecule e.g. Na 2 SO 4 n = 3 x Na 2 SO 4

38. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 38 11.5 158 g sugar (MM=342.3 g/mol) dissolved in 643.5 ml of water. At 25 o C, density of water= 0.9971 g/ml and v.p. = 23.76 torr. Calculate v.p. of the sugar solution.

39. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 39 11.6 35.0 g solid Na 2 SO 4 (MM= 142 g/mol) in 175 g water at 25 o C. P o (water). = 23.76 torr Calc v.p. of solution. The lowering of v.p. depends on the number of solute particles present in the solution.

40. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 40 Non-Ideal Solutions When a solution contains two volatile components, both contribute to the total vapor pressure.

41. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 41 P A = X A P A 0 P B = X B P B 0 P T = P A + P B P T = X A P A 0 + X B P B 0

42. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 42 (a) ideal liquid-liquid solution by Raoult's law. (b) This solution shows a positive deviation from Raoult's law. (c) This solution shows a negative deviation from Raoult's law.

43. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 43

44. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 44 P T is greater than predicted by Raoults’s law P T is less than predicted by Raoults’s law Force A-B Force A-A Force B-B <& Force A-B Force A-A Force B-B >&

45. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 45 11.7 5.81 g acetone + 11.9 g Chloroform MM=58.1MM = 119.4 P o = 345 torrP o = 293 torr n = 0.100 moln = 0.100 mol At 35 o C, total V.P. of solution = 260 torr Is this an ideal solution ? X ac = 0.5 X ch = 0.5 Raoult’s Law; P T = X A P A + X B P B P total = 0.5 x 345 torr + 0.5 x 293 torr = 319 torr 260 torr < 319 (ideal) ; Negative deviation

46. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 46 Colligative Properties of Non-Electrolyte Solutions Depend only on the number, not on the identity, of the solute particles in an ideal solution. 4 Boiling point elevation 4 Freezing point depression 4 Osmotic pressure

47. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 47 Boiling Point Elevation A nonvolatile solute elevates the boiling point of the solvent.  T = K b m solute K b = molal boiling point elevation constant m = molality of the solute

48. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 48 Boiling-Point Elevation  T b = T b – T b 0 T b > T b 0  T b > 0 T b is the boiling point of the pure solvent 0 T b is the boiling point of the solution  T b = K b m m is the molality of the solution K b is the molal boiling-point elevation constant ( 0 C/m)

49. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 49 11.8 18.00 g glucose in 150 g water. B.P. = 100.34 o C Calc molar mass of glucose ΔT = K b m solute K b = 0.51 for water ΔT = 100.34 – 100.00 = 0.34 o C m solute = ΔT /K b = 0.34 o C = 0.67 mol/kg 0.51 o C.kg/mol m solute = n glucose /0.1500 kg n glucose = 0.67 mol/kg x 0.1500 kg = 0.10 mol MM = mass/mol = 18.00g / 0.10 mol = 180 g/mol

50. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 50 Freezing Point Depression A nonvolatile solute depresses the freezing point of the solvent.  T = K f m solute K f = molal freezing point depression constant m = molality of the solute

51. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 51 Freezing-Point Depression  T f = T f – T f 0 T f > T f 0  T f > 0 T f is the freezing point of the pure solvent 0 T f is the freezing point of the solution  T f = K f m m is the molality of the solution K f is the molal freezing-point depression constant ( 0 C/m)

52. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 52 S.Ex. 11.9 What mass of glycol(MM=62.1 g/mol) must be added to 10.0 L of water to produce a solution that freezes at -23.3 o C? (d =1 g/ml) ∆T = 0.0 –(-23.3) = 23.3 o C= K f m solute m solute = 23.3 o C = 12.5 mol/kg 1.86 o C.kg/mol n = 12.5 mol/kg x 10.0 kg = 1.25x10 2 mol Mass of glycol = 1.25x10 2 mol x 62.1g/mol =7.76x10 3 g = 7.76 kg

53. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 53 Osmotic Pressure Osmosis: The flow of solvent into the solution through the semipermeable membrane. Osmotic Pressure: The excess hydrostatic pressure on the solution compared to the pure solvent.

54. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 54 A tube with a bulb on the end that is covered by a semipermeable membrane.

55. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 55 Osmotic Pressure (  ) Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. Osmotic pressure (  ) is the pressure required to stop osmosis.  = MRT M is the molarity of the solution R is the gas constant T is the temperature (in K)

56. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 56 Ex 11.1 1.00x10 -3 g of protein dissolved to make 1.00 mL solution. Osmotic pressure was found to be 1.12 torr at 25 o C. Calculate the MM of protein. π = MRT =1.12 torr/760 = 1.47x10 -3 atm R = 0.08206 L atm/K.mol T = 25.0 + 273 = 298 K M = 1.47x10-3 atm = 6.01x10 -5 mol/L (0.08206 L.atm/K.mol)(298 K) 1.00x10 -3 g in 1 ml = 1.00 g /L MM = 1.00 g x L = 1.66x10 4 g/mol –L 6.01x10 -5 mol

57. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 57 (a) A pure solvent and its solution (containing a nonvolatile solute) are separated by a semipermeable membrane through which solvent molecules (blue) can pass but solute molecules (green) cannot. The rate of solvent transfer is greater from solvent to solution than from solution to solvent. (b) The system at equilibrium, where the rate of solvent transfer is the same in both directions.

58. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 58 The normal flow of solvent into the solution (osmosis) can be prevented by applying an external pressure to the solution. The minimum pressure required to stop the osmosis is equal to the osmotic pressure of the solution.

59. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 59 If the external pressure is larger than the osmotic pressure, reverse osmosis occurs. One application is desalination of seawater.

60. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 60 Reverse osmosis.

61. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 61 A cell in an: isotonic Solution (identical  ) hypotonic Solution (water in) hypertonic Solution (water out)

62. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 62 S. Ex. 11.12 What concentration of NaCl in water is needed to produce an aqueous solution isotonic with blood (π = 7.70 atm at 25 o C)

63. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 63 Colligative Properties of Electrolyte Solutions  T = imK  = iMRT van’t Hoff factor, “i”, relates to the number of ions per formula unit. NaCl = 2, K 2 SO 4 = 3

64. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 64 Boiling-Point Elevation  T b = i K b m Freezing-Point Depression  T f = i K f m Osmotic Pressure (  )  = iMRT Colligative Properties of Electrolyte Solutions

65. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 65 In an aqueous solution a few ions aggregate, forming ion pairs that behave as a unit: Ion Pairing

66. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 66 S. Ex 11.13 The observed osmotic pressure for a 0.10 M solution of Fe(NH 4 ) 2 (SO 4 ) 2 at 25 o C is 10.8 atm. Compare the expected and experimental value of i. Fe(NH 4 ) 2 (SO 4 ) 2 → Fe 2+ + 2NH 4+ + 2SO 4 2-

67. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 67 The Cleansing Action of Soap

68. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 68 The Cottrell precipitator installed in a smokestack. The charged plates attract the colloidal particles because of their ion layers and thus remove them from the smoke.