1. 1 SOLUTIONS AND THEIR BEHAVIOR CHAPTER 14 CHAPTER 14

2. 2 CHAPTER OVERVIEW CHAPTER OVERVIEW This chapter examines homogeneous mixtures called solutions, which are made up of a solute and a solvent.This chapter examines homogeneous mixtures called solutions, which are made up of a solute and a solvent. Concentrations of solutions can be expressed in a variety of units.Concentrations of solutions can be expressed in a variety of units. Properties of solutions that depend only on the number of solute particles and not their type are called colligative properties.Properties of solutions that depend only on the number of solute particles and not their type are called colligative properties.

3. 3 14.1 UNITS OF CONCENTRATION Molarity, M, moles solute per liter solutionMolarity, M, moles solute per liter solution Molality, m, moles of solute per kilogram solventMolality, m, moles of solute per kilogram solvent Mole fraction, X A, moles A divided by moles totalMole fraction, X A, moles A divided by moles total Weight percent (mass percent), wt.% A,Weight percent (mass percent), wt.% A, (mass A divided by mass total) x 100%

4. 4 Molarity -vs- Molality Each flask contains 19.4 g of K 2 CrO 4 Water to the 1.00 L mark Exactly 1.00 kg of water added

5. 5 UNITS OF CONCENTRATION parts per million, ppm, is calculated like percent, but multiply by 10 6parts per million, ppm, is calculated like percent, but multiply by 10 6 Remember that the mass of the solution equals the mass of the solute plus solvent.Remember that the mass of the solution equals the mass of the solute plus solvent. Conversions between molarity and the other concentration units requires the density of the solution.Conversions between molarity and the other concentration units requires the density of the solution.

6. SolutionsSolutions Why does a raw egg swell or shrink when placed in different solutions? 6

7. Some Definitions A solution is a HOMOGENEOUS mixture of 2 or more substances in a single phase. One constituent is usually regarded as the SOLVENT and the others as SOLUTES. 7

8. 8 14.2 THE SOLUTION PROCESS The key to understanding the solution process is intermolecular forces: solvent - solvent; solute - solute; solute - solvent.The key to understanding the solution process is intermolecular forces: solvent - solvent; solute - solute; solute - solvent.

9. 9 THE SOLUTION PROCESS What prevents solubility is an energy barrier when the latter interaction is significantly weaker than the former interactions.What prevents solubility is an energy barrier when the latter interaction is significantly weaker than the former interactions. Like dissolves like is a general rule, but not and explanation of the solubility phenomenon.Like dissolves like is a general rule, but not and explanation of the solubility phenomenon.

10. 10 Liquids Dissolving in LiquidsLiquids Dissolving in Liquids –Miscible liquids are soluble in all proportions. –Immiscible liquids do not mix, but form separate layers. Isopropanol is miscible with water but gasoline is not. Explain why.Isopropanol is miscible with water but gasoline is not. Explain why.

11. 11 SOLUTIONS A saturated solution is one which has reached its equilibrium solubility at that temperature.A saturated solution is one which has reached its equilibrium solubility at that temperature. An unsaturated solution is one that has not reached its equilibrium solubility.An unsaturated solution is one that has not reached its equilibrium solubility. A supersaturated solution is one in which the equilibrium solubility has been temporarily exceeded.A supersaturated solution is one in which the equilibrium solubility has been temporarily exceeded.

12. SUPERSATURATED SOLUTIONS contain more than is possible and are unstable. Definitions A saturated solution contains the maximum quantity of solute that dissolves at that temperature. Solutions can be classified as unsaturated or saturated. 12

13. Energetics of the Solution Process If the enthalpy of formation of the solution is more negative than that of the solvent and solute, the enthalpy of solution is negative. The solution process is exothermic! 13

14. 14 Solids Dissolving in Liquids The same rules apply. Compare the intermolecular forces.Compare the intermolecular forces. I 2 is quite soluble in CCl 4, but not very soluble in water. Explain why?I 2 is quite soluble in CCl 4, but not very soluble in water. Explain why?

15. Supersaturated Sodium Acetate Supersaturated Sodium Acetate One application of a supersaturated solution is the sodium acetate “heat pack.”One application of a supersaturated solution is the sodium acetate “heat pack.” Sodium acetate has an ENDOthermic heat of solution.Sodium acetate has an ENDOthermic heat of solution. 15

16. 16 Ionic Solutions The heat of solution for ionic compounds is the sum of the lattice energy (+), bonds breaking, and the hydration energy (-), bonds forming.The heat of solution for ionic compounds is the sum of the lattice energy (+), bonds breaking, and the hydration energy (-), bonds forming. It may be positive (endo) or negative (exo) depending on the relative magnitudes of these energies.It may be positive (endo) or negative (exo) depending on the relative magnitudes of these energies.

17. 17  Hsol’n can be calc’d using Hess' Law. 821 kJ/mol – 819 kJ/mol = +2 kJ/mol (endo)

18. 18  Hsol’n can be calc’d using Hess' Law.

19. 19 Ion Size also determines Solubility Remember Coulombs Law +- 2 (charge n)(charge n) Force of Attraction = d k

20. 20 Ionic Solutions Temperature has a significant effect on solubility for salts and is consistent with Le Chatelier's principle.Temperature has a significant effect on solubility for salts and is consistent with Le Chatelier's principle.

21. 21 The heat of solution for many salts is positive, endothermic, as seen by the positive slope of the graph.

22. Supersaturated Sodium Acetate Sodium acetate has an ENDOthermic heat of solution. NaCH 3 CO 2 (s) + heat ----> Na + (aq) + CH 3 CO 2 - (aq) Therefore, formation of solid sodium acetate from its ions is EXOTHERMIC. Na + (aq) + CH 3 CO 2 - (aq) ---> NaCH 3 CO 2 (s) + heat 22

23. 23 Dissolving Gases Gas solubility decreases with increasing temperature which means ΔH solution < 0, exothermic.Gas solubility decreases with increasing temperature which means ΔH solution < 0, exothermic. Gas solubilities increase with increasing pressure.Gas solubilities increase with increasing pressure. Write the general equation for the solubility of a gas showing that the process is exothermic and show how increasing the temperature decreases the solubility.Write the general equation for the solubility of a gas showing that the process is exothermic and show how increasing the temperature decreases the solubility. Gas + Solvent ⇄ Solution + Heat

24. 24

25. Colligative Properties  On adding a solute to a solvent, the properties of the solvent are modified. Vapor pressure decreasesVapor pressure decreases Melting point decreasesMelting point decreases Boiling point increasesBoiling point increases Osmosis is possible (osmotic pressure)Osmosis is possible (osmotic pressure) 25

26. Colligative Properties  These changes are called COLLIGATIVE PROPERTIES.  They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles. 26

27. An IDEAL SOLUTION is one where the properties depend only on the concentration of solute. Need concentration units to tell us the number of solute particles per solvent particle. The unit “molarity” does not do this! Concentration Units 27

28. Concentration Units WEIGHT % = grams solute per 100 g solution X A  mol fraction A= mol A + mol B+ mol C m of solute= mol solute kilograms solvent MOLALITY, m MOLE FRACTION, X For a mixture of A, B, and C 28

29. Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H 2 O. Calculate mole fraction, molality, and weight % of glycol.

30. Calculating Concentrations 250. g H 2 O = 13.9 mole Dissolve 62.1 g (1.00 mole) of ethylene glycol in 250. g of H 2 O. Calculate X, m, and % of glycol. X glycol = 1.00 mol glycol + 13.9 mol H 2 O X glycol = 0.0672 30

31. Calculating Concentrations Calculate molality Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H 2 O. Calculate X, m, and % of glycol. conc (molality)= 1.00 mol glycol 0.250 kg H 2 O  4.00 molal Calculate weight % %glycol = 62.1 g + 250. g x 100%= 19.9% 31

32. Dissolving Gases & Henry’s Law Gas solubility (M) = k H P gas k H for O 2 = 1.66 x 10 -6 M/mmHg When P gas drops, solubility drops. 32

33. Lake Nyos, Cameroon 33

34. 34 COLLIGATIVE PROPERTIES Changes in Vapor Pressure: Raoult's LawChanges in Vapor Pressure: Raoult's Law The presence of a solute in the solvent lowers the vapor pressure of the solvent.The presence of a solute in the solvent lowers the vapor pressure of the solvent. P solvent = X solvent P o solvent If the solute is also volatile, a similar equation applies to the solute. P solute = X solute P o soluteIf the solute is also volatile, a similar equation applies to the solute. P solute = X solute P o solute The total pressure for the solution is given by:P total = P solvent + P soluteThe total pressure for the solution is given by:P total = P solvent + P solute

35. 35 COLLIGATIVE PROPERTIES If the solute is nonvolatile, the total pressure is just the pressure of the solvent and is lower than that of the pure solvent.If the solute is nonvolatile, the total pressure is just the pressure of the solvent and is lower than that of the pure solvent. Study examples and exercises.Study examples and exercises.

36. Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution. H—O H H H H H surface 36

37. Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution. 37

38. Understanding Colligative Properties: Raoults’s Law VP of H 2 O over a solution depends on the number of H 2 O molecules per solute molecule. P solvent proportional to X solvent P solvent proportional to X solvent OR OR P solvent = X solvent P o solvent P solvent = X solvent P o solvent VP of solvent over solution = (Mol frac solvent)(VP pure solvent) RAOULT’S LAW 38

39. Raoult’s Law An ideal solution is one that obeys Raoult’s law. P A = X A P o A P A = X A P o A Because mole fraction of solvent, X A, is always less than 1, then P A is always less than P o A. The vapor pressure of solvent over a solution is always LOWERED! 39

40. Raoult’s Law Assume the solution containing 62.1 g of glycol in 250. g of water is ideal. What is the vapor pressure of water over the solution at 30 o C? (The VP of pure H 2 O is 31.8 mm Hg; see App.) Solution X glycol = 0.0672 and so X water = ? Because X glycol + X water = 1 X water = 1.000 - 0.0672 = 0.9328 P water = X water P o water = (0.9382)(31.8 mm Hg) P water = 29.7 mm Hg 40

41. Raoult’s Law Or (see next slide): Δ P A = VP lowering = X B P o A Δ P A = VP lowering = X B P o A VP lowering is proportional to mole fraction of the solute! For very dilute solutions, Δ P A = Kmolality B where K is a proportionality constant. where K is a proportionality constant. This helps explain changes in melting and boiling points.

42. 42 See Exercise 14.6, p. 575

43. Changes in Freezing and Boiling Points of Solvent See Figure 14.13 43

44. 44 Boiling Point Elevation If a solute is added to the pure solvent at its normal boiling point, the equilibrium vapor pressure will decrease and the liquid will no longer boil.If a solute is added to the pure solvent at its normal boiling point, the equilibrium vapor pressure will decrease and the liquid will no longer boil. To reach the new boiling point the temperature must be increased, thus boiling point elevation.To reach the new boiling point the temperature must be increased, thus boiling point elevation. Δt bp = K bp m solute Δt bp = K bp m solute

45. 45 Figure 14.13

46. The boiling point of a solution is higher than that of the pure solvent. 46

47. 47 Elevation of Boiling Point Elevation in BP =  t BP = K BP m (where K BP is characteristic of solvent) 47

48. Change in Boiling Point Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the BP of the solution? K BP = +0.512 o C/molal for water (Table 14.4). Solution 1.Calculate solution molality = 4.00 m 2.Dt BP = K BP m Dt BP = +0.512 o C/molal (4.00 molal) Dt BP = +0.512 o C/molal (4.00 molal) Dt BP = +2.05 o C Dt BP = +2.05 o C BP = 102.05 o C BP = 102.05 o C 48

49. 49 Freezing Point Depression The freezing point is lowered by the presence of a solute since these particles cannot form the solid and some of them are occupying the low energy slots needed to form the solid solvent.The freezing point is lowered by the presence of a solute since these particles cannot form the solid and some of them are occupying the low energy slots needed to form the solid solvent. Δt fp = K fp m solute Δt fp = K fp m solute Note the K is a negative value.Note the K is a negative value.

50. Change in Freezing Point The freezing point of a solution is LOWER than that of the pure solvent. FP depression = Δ t FP = K FP m Pure water Ethylene glycol/water solution 50

51. Freezing Point Depression Consider equilibrium at melting point Liquid solvent Solid solvent Rate at which molecules go from S to L depends only on the nature of the solid.Rate at which molecules go from S to L depends only on the nature of the solid. BUT — rate for L ---> S depends on how much is dissolved. This rate is SLOWED for the same reason VP is lowered.BUT — rate for L ---> S depends on how much is dissolved. This rate is SLOWED for the same reason VP is lowered. Therefore, to bring S ---> L and L ---> S rates into equilibrium for a solution, T must be lowered.Therefore, to bring S ---> L and L ---> S rates into equilibrium for a solution, T must be lowered. Thus, FP for solution < FP for solvent FP depression = Δt FP = K FP m FP depression = Δt FP = K FP m 51

52. Calculate the FP of a 4.00 molal glycol/water solution. K FP = -1.86 o C/molal (Table 14.4, p. 577) Solution Δ t FP = K FP m = (-1.86 o C/molal)(4.00 m) = (-1.86 o C/molal)(4.00 m) Δ t FP = -7.44 o C Freezing Point Depression 52

53. 53 Colligative Properties Of Ionic Solutions Ionic compounds dissociate completely into ions in water.Ionic compounds dissociate completely into ions in water. All calculations involving water and an ionic solute must account for the total number of particles present.All calculations involving water and an ionic solute must account for the total number of particles present. This factor is called the van't Hoff factor, i.This factor is called the van't Hoff factor, i.

54. How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 o C?. How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 o C?.Solution Calculate the required molality. Δ t FP = K FP m Δ t FP = K FP m -10.00 o C = (-1.86 o C/molal) Molality -10.00 o C = (-1.86 o C/molal) Molality Concentration = 5.38 molal Concentration = 5.38 molal Freezing Point Depression 54

55. How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 o C?. Solution Concentration required = 5.38 molal Freezing Point Depression This means we need 5.38 mol of dissolved particles per kg of solvent. Recognize that m represents the total conc. of all dissolved particles. Recall that 1 mol NaCl(aq) 1 mol Na + (aq) + 1 mol Cl - (aq) 55

56. How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 o C?. Solution Concentration required = 5.38 molal We need 5.38 mol of dissolved particles per kg of solvent. NaCl(aq) --> Na + (aq) + Cl - (aq) NaCl(aq) --> Na + (aq) + Cl - (aq) Freezing Point Depression To get 5.38 mol/kg of particles we need 5.38 mol / 2 = 2.69 mol NaCl / kg 5.38 mol / 2 = 2.69 mol NaCl / kg 2.69 mol NaCl / kg ---> 157 g NaCl / kg (157 g NaCl / kg)(4.00 kg) = 629 g NaCl 56

57. Boiling Point Elevation and Freezing Point Depression Δ t = K m i Δ t = K m i A generally useful equation i = van’t Hoff factor = number of particles produced per formula unit. CompoundTheoretical Value of i glycol1 glycol1 NaCl2 NaCl2 CaCl 2 3 CaCl 2 3 57

58. Osmosis 58 Salt waterPure water

59. 59 Osmosis occurs when a molecule moves from a region of high concentration to lower concentration through a semipermeable membrane.Osmosis occurs when a molecule moves from a region of high concentration to lower concentration through a semipermeable membrane.

60. 60 Osmotic pressure is defined by:  = cRT where c is the molarity of the solute and R is the gas constant.

61. 61

62. OsmosisOsmosis The semipermeable membrane should allow only the movement of solvent molecules. Therefore, solvent molecules move from pure solvent to solution. 62 Solvent Solution Semipermeable membrane

63. OsmosisOsmosis The semipermeable membrane should allow only the movement of solvent molecules. Therefore, solvent molecules move from pure solvent to solution. 63 Solvent Solution Osmotic Pressure

64. 64 OsmosisOsmosis Equilibrium is reached when pressure produced by extra solution — the OSMOTIC PRESSURE, p p = cRT (where c is conc. in mol/L) p = cRT (where c is conc. in mol/L) counterbalances pressure of solvent molecules moving thru the membrane. Solvent Solution Osmotic Pressure

65. Osmosis 65

66. Osmosis Osmosis of solvent from one solution to another can continue until the solutions are ISOTONIC — they have the same concentration.Osmosis of solvent from one solution to another can continue until the solutions are ISOTONIC — they have the same concentration. 66

67. Osmosis Calculating a Molar Mass Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. p measured to be 10.0 mm Hg at 25 ° C. Calculate molar mass of hemoglobin. Solution (a) Calculate p in atmospheres p = 10.0 mmHg (1 atm / 760 mmHg) p = 10.0 mmHg (1 atm / 760 mmHg) = 0.0132 atm (b)Calculate the concentration 67

68. Calculating a Molar Mass Concentration = 5.39 x 10 -4 mol/L Concentration = 5.39 x 10 -4 mol/L (c)Calculate the molar mass Molar mass = 35.0 g / 5.39 x 10 -4 mol/L Molar mass = 35.0 g / 5.39 x 10 -4 mol/L Molar mass = 65,100 g/mol Molar mass = 65,100 g/mol Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution.  measured to be 10.0 mm Hg at 25  C. Calculate molar mass of hemoglobin.Solution (b)Calculate concentration from  = cRT Conc = 0.0132 atm (0.0821 Latm/Kmol)(298K) 68

69. 69 Reverse Osmosis

70. 70 COLLOIDS Colloids are a suspension of very small particles that do not settle out.Colloids are a suspension of very small particles that do not settle out. –(milk, jello,..)

71. 71

72. 72 Hydrophilic and hydrophobic colloids exist and emulsions make use of molecules that contain both.

73. 73 Soap and Surfactants H2OH2O H2OH2O H2OH2O H2OH2O H2OH2O H2OH2O H2OH2O H2OH2O H2OH2O H2OH2O Dirt C O O - H O H Nonpolar tail Polar head

74. 74 Soap and Surfactants

75. 75 DetergentFabric Softener