1. Chem 106, Prof. T.L. Heise 1 CHE 106: General Chemistry u CHAPTER THIRTEEN Copyright © Tyna L. Heise 2001 - 2002 All Rights Reserved

2. Chem 106, Prof. T.L. Heise 2 The Solution Process Solution Solution - homogeneous mixture of two or more substances Solvent Solvent - substance of greater amount in the homogeneous mixture (solution) Solute Solute - compounds “dissolved” in the solvent Concentration Concentration - the amount of solute dissolved in a solvent. Expressed in molarity (M) Chap 13.1

3. Chem 106, Prof. T.L. Heise 3 The Solution Process Solution Solution -may be gases, liquids or solids –Examples: air ocean water 10K gold (alloy) Chap 13.1

4. Chem 106, Prof. T.L. Heise 4 The Solution Process A solution is formed when one substance disperses uniformly through another. All solutions, except gas mixtures, involve substances in a condensed state Intermolecular forces are also going to operate between solute and solvent - interactions are known as solvation - when solvent is water, the interactions are known as hydration Chap 13.1

5. Chem 106, Prof. T.L. Heise 5 The Solution Process Chap 13.1

6. Chem 106, Prof. T.L. Heise 6 The Solution Process Energy changes during solution formation is the sum of three energy changes:  H 1  separation of solute molecules  H 2 = separation of solvent molecules  H 3 = formation of solute-solvent interactions  H 1 and  H 2 are endothermic because you are breaking or overcoming interactive forces  H 3 is exothermic due to the creation of new forces Chap 13.1

7. Chem 106, Prof. T.L. Heise 7 The Solution Process Chap 13.1

8. Chem 106, Prof. T.L. Heise 8 The Solution Process Solution Formation depends on two factors 1) energy or enthalpy changes  H 2) chaos or entropy changes  S Natural phenomenon's occur to satisfy two basic laws- energy content decreases - disorder content increases ** Formation of solutions is favored by the increase in disorder that accompanies mixing Chap 13.1

9. Chem 106, Prof. T.L. Heise 9 The Solution Process Solution Formation can occur during two basic processes1) physical changes 2) chemical changes Our focus is on physical changes, key to recognizing difference is examining whether you can get the salt BACK unchanged when reaction is done. Chap 13.1

10. Chem 106, Prof. T.L. Heise 10 Saturated Solutions and Solubility As a solid solute is dissolved, the concentration of dissolved particles increases, as does the chance of a collision between to dissolved particles. If two dissolved particles collide, the attractive forces could cause recrystallization In any solution Solute + solvent dissolve solution crystallize Chap 13.2

11. Chem 106, Prof. T.L. Heise 11 Saturated Solutions and Solubility Saturated: in equilibrium, rate of dissolving equals rate of crystallization, concentration remains constant Solubility: the amount of solute needed to form a saturated solution in a given temperature Unsaturated: dissolved less the amount needed to form a saturated solution Supersaturated: dissolving more than the amount needed Chap 13.2

12. Chem 106, Prof. T.L. Heise 12 Saturated Solutions and Solubility Chap 13.2

13. Chem 106, Prof. T.L. Heise 13 Factors Affecting Solubility Chap 13.3 Solute-Solvent Interactions » the stronger the the attractions between solute and solvent, the greater the solubility » miscible and immiscible » like dissolves like

14. Chem 106, Prof. T.L. Heise 14 Factors Affecting Solubility Chap 13.3 Pressure Effects » the solubility of a gas in any solvent is increased as the pressure of the gas over the solvent is increased

15. Chem 106, Prof. T.L. Heise 15 Factors Affecting Solubility Chap 13.3 Pressure Effects » relationship between pressure and the solubility of a gas is expressed in terms of a simple equations known as Henry’s Law

16. Chem 106, Prof. T.L. Heise 16 Factors Affecting Solubility Chap 13.3 Sample Exercise: Calculate the concentration of CO 2 in a soft drink after the bottle is opened and equilibrates at 25°C under a CO 2 partial pressure of 3.0 x 10 -4 atm.

17. Chem 106, Prof. T.L. Heise 17 Factors Affecting Solubility Chap 13.3 Sample Exercise: Calculate the concentration of CO 2 in a soft drink after the bottle is opened and equilibrates at 25°C under a CO 2 partial pressure of 3.0 x 10 -4 atm. 1) C g = kP g

18. Chem 106, Prof. T.L. Heise 18 Factors Affecting Solubility Chap 13.3 Sample Exercise: Calculate the concentration of CO 2 in a soft drink after the bottle is opened and equilibrates at 25°C under a CO 2 partial pressure of 3.0 x 10 -4 atm. 1) C g = kP g C g = x k = 3.1 x 10 -2 mol/L-atm(given page 480) P g = 3.0 x 10 -4 atm

19. Chem 106, Prof. T.L. Heise 19 Factors Affecting Solubility Chap 13.3 Sample Exercise: Calculate the concentration of CO 2 in a soft drink after the bottle is opened and equilibrates at 25°C under a CO 2 partial pressure of 3.0 x 10 -4 atm. 1) C g = kP g x = 3.1 x 10 -2 mol/L-atm(3.0 x 10 -4 atm)

20. Chem 106, Prof. T.L. Heise 20 Factors Affecting Solubility Chap 13.3 Sample Exercise: Calculate the concentration of CO 2 in a soft drink after the bottle is opened and equilibrates at 25°C under a CO 2 partial pressure of 3.0 x 10 -4 atm. 1) C g = kP g x = 3.1 x 10 -2 mol/L-atm(3.0 x 10 -4 atm) x = 9.3 x 10 -6 mol/L

21. Chem 106, Prof. T.L. Heise 21 Factors Affecting Solubility Chap 13.3 Temperature Effects » solubility of most solid solutes increases as the temperature of the solution does » solubility of most gases decreases as temperature of the solution does » look up solubility on a solubility table to find exact trend

22. Chem 106, Prof. T.L. Heise 22 Factors Affecting Solubility Chap 13.3 Temperature Effects

23. Chem 106, Prof. T.L. Heise 23 Expressing Concentration Chap 13.4 Dilute - relatively small concentration of solute in solution Concentrated - relatively large concentration of solute in solution Ways to express numerically- mass percentage mole fraction molarity molality

24. Chem 106, Prof. T.L. Heise 24 Expressing Concentration Chap 13.4 mass percentage mass of component *100 total mass of soln ppm mass of component *100,000 total mass of soln ppb mass of component *100,000,000 total mass of soln

25. Chem 106, Prof. T.L. Heise 25 Expressing Concentration Chap 13.4 Sample Exercise: Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water

26. Chem 106, Prof. T.L. Heise 26 Expressing Concentration Chap 13.4 Sample Exercise: Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water mass percent = mass of component * 100 total mass of soln

27. Chem 106, Prof. T.L. Heise 27 Expressing Concentration Chap 13.4 Sample Exercise: Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water mass percent = mass of component * 100 total mass of soln = 1.50 g NaCl *100 51.50 g soln

28. Chem 106, Prof. T.L. Heise 28 Expressing Concentration Chap 13.4 Sample Exercise: Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water mass percent = mass of component * 100 total mass of soln = 1.50 g NaCl *100 51.50 g soln =2.91 %

29. Chem 106, Prof. T.L. Heise 29 Expressing Concentration Chap 13.4 Sample Exercise: a commercial bleaching solution contains 3.62 mass percent sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution?

30. Chem 106, Prof. T.L. Heise 30 Expressing Concentration Chap 13.4 Sample Exercise: a commercial bleaching solution contains 3.62 mass percent sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution? mass percent = mass of component * 100 total mass of soln

31. Chem 106, Prof. T.L. Heise 31 Expressing Concentration Chap 13.4 Sample Exercise: a commercial bleaching solution contains 3.62 mass percent sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution? mass percent = mass of component * 100 total mass of soln 3.62 % = x * 100 2500 g soln 3.62 % * 2500g = x 100%

32. Chem 106, Prof. T.L. Heise 32 Expressing Concentration Chap 13.4 Sample Exercise: a commercial bleaching solution contains 3.62 mass percent sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution? mass percent = mass of component * 100 total mass of soln 3.62 % = x * 100 2500 g soln 3.62 % * 2500g = x 100% 90.5 g NaOCl

33. Chem 106, Prof. T.L. Heise 33 Expressing Concentration Chap 13.4 mole fraction mole fraction = moles of component total moles of all components symbol X is often used to represent mole fraction molarity = moles of solute/L of solution molality = moles of solute/kg of solvent

34. Chem 106, Prof. T.L. Heise 34 Expressing Concentration Chap 13.4 Sample Exercise: What is the molality of a solution made by dissolving 36.5 g naphthalene, C 10 H 8, in 428 g of toluene, C 7 H 8 ?

35. Chem 106, Prof. T.L. Heise 35 Expressing Concentration Chap 13.4 Sample Exercise: What is the molality of a solution made by dissolving 36.5 g naphthalene, C 10 H 8, in 428 g of toluene, C 7 H 8 ? 1) molality = moles of solute kg of solvent

36. Chem 106, Prof. T.L. Heise 36 Expressing Concentration Chap 13.4 Sample Exercise: What is the molality of a solution made by dissolving 36.5 g naphthalene, C 10 H 8, in 428 g of toluene, C 7 H 8 ? 1) molality = moles of solute kg of solvent moles = 36.5 g C 10 H 8 1 mol = 0.285 mol 128 g C 10 H 8

37. Chem 106, Prof. T.L. Heise 37 Expressing Concentration Chap 13.4 Sample Exercise: What is the molality of a solution made by dissolving 36.5 g naphthalene, C 10 H 8, in 428 g of toluene, C 7 H 8 ? 1) molality = moles of solute kg of solvent moles = 0.285 mol kg = 428g 1 kg= 0.428 kg 1000 g

38. Chem 106, Prof. T.L. Heise 38 Expressing Concentration Chap 13.4 Sample Exercise: What is the molality of a solution made by dissolving 36.5 g naphthalene, C 10 H 8, in 428 g of toluene, C 7 H 8 ? 1) molality = moles of solute kg of solvent moles = 0.285 mol =0.285 mol kg = 0.428 kg0.428 kg = 0.670 mol/kg

39. Chem 106, Prof. T.L. Heise 39 Expressing Concentration Chap 13.4 Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (a) the molality (b) the mole fraction of NaOCl

40. Chem 106, Prof. T.L. Heise 40 Expressing Concentration Chap 13.4 Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (a) the molality mass % = mass of solute *100 mass of solution

41. Chem 106, Prof. T.L. Heise 41 Expressing Concentration Chap 13.4 Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (a) the molality mass % = mass of solute *100 mass of solution 3.62% = mass of solute * 100 100 g solution* when not given an amount f solution, assume 100g.

42. Chem 106, Prof. T.L. Heise 42 Expressing Concentration Chap 13.4 Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (a) the molality mass of solute = 3.62 g molality = moles kg 3.62 g NaOCl 1 molNaOCl= 0.0486 mol 74.44 g NaOCl

43. Chem 106, Prof. T.L. Heise 43 Expressing Concentration Chap 13.4 Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (a) the molality molality = moles = 0.0486 mol = 0.505 mol kg 0.09638 kg kg *don’t forget we assumed 100 g of solution, so 3.62 g of NaOCl and 96.28 g water

44. Chem 106, Prof. T.L. Heise 44 Expressing Concentration Chap 13.4 Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (b) mole fraction of NaOCl Mole fraction = moles of solute total moles of components

45. Chem 106, Prof. T.L. Heise 45 Expressing Concentration Chap 13.4 Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (b) mole fraction of NaOCl Mole fraction = moles of solute total moles of components 3.62 g NaOCl 1 mol NaOCl = 0.0486 mol 74.44 g NaOClNaOCl 96.38 g H 2 O 1 mol H 2 O = 5.35 mol H 2 O 18 g H 2 O

46. Chem 106, Prof. T.L. Heise 46 Expressing Concentration Chap 13.4 Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (b) mole fraction of NaOCl Mole fraction = moles of solute total moles of components 0.0486 molNaOCl = 0.00900 0.0486 mol NaOCl + 5.35 mol H 2 O

47. Chem 106, Prof. T.L. Heise 47 Colligative Properties Chap 13.5 Certain physical properties of solutions differ from the pure solvent - lowering freezing point - raising boiling point - reduction of vapor pressure - alteration of osmotic pressure

48. Chem 106, Prof. T.L. Heise 48 Colligative Properties Chap 13.5 Vapor Pressure Vapor pressure is the pressure exerted by the vapor on the surface when closed flask achieves equilibrium

49. Chem 106, Prof. T.L. Heise 49 Colligative Properties Chap 13.5 Boiling Point K b is the molal boiling point constant and is equal to 0.52°C/ m - this constant is related to the number of dissolved particles

50. Chem 106, Prof. T.L. Heise 50 Colligative Properties Chap 13.5 Freezing Point K f is the molal boiling point constant and is equal to 1.87°C/ m - this constant is related to the number of dissolved particles

51. Chem 106, Prof. T.L. Heise 51 Expressing Concentration Chap 13.5 Sample Exercise: Calculate the freezing point of a solution containing 0.600 kg of CHCl 3 and 42.0 g of eucalyptol, C 10 H 18 O, a fragrant substance found in the leaves of the eucalyptus tree.

52. Chem 106, Prof. T.L. Heise 52 Expressing Concentration Chap 13.5 Sample Exercise: Calculate the freezing point of a solution containing 0.600 kg of CHCl 3 and 42.0 g of eucalyptol, C 10 H 18 O, a fragrant substance found in the leaves of the eucalyptus tree. molality = moles of solute kg of solvent

53. Chem 106, Prof. T.L. Heise 53 Expressing Concentration Chap 13.5 Sample Exercise: Calculate the freezing point of a solution containing 0.600 kg of CHCl 3 and 42.0 g of eucalyptol, C 10 H 18 O, a fragrant substance found in the leaves of the eucalyptus tree. molality = moles of solute kg of solvent 42.0 g C 10 H 18 O 1 mol C 10 H 18 O =0.273 mol 154 g C 10 H 18 O

54. Chem 106, Prof. T.L. Heise 54 Expressing Concentration Chap 13.5 Sample Exercise: Calculate the freezing point of a solution containing 0.600 kg of CHCl 3 and 42.0 g of eucalyptol, C 10 H 18 O, a fragrant substance found in the leaves of the eucalyptus tree. molality = moles of solute kg of solvent = 0.273 mol 0.600 kg = 0.455 m

55. Chem 106, Prof. T.L. Heise 55 Expressing Concentration Chap 13.5 Sample Exercise: Calculate the freezing point of a solution containing 0.600 kg of CHCl 3 and 42.0 g of eucalyptol, C 10 H 18 O, a fragrant substance found in the leaves of the eucalyptus tree.  T f = K f m = 4.68(0.455) = 2.13°C Normal T f = -63.5°C -2.13°C New T f = -65.6°C

56. Chem 106, Prof. T.L. Heise 56 Expressing Concentration Chap 13.4 Sample Exercise: Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water: 1 mol Co(NO 3 ) 2 2 mol KCl 3 mol C 2 H 6 O 2

57. Chem 106, Prof. T.L. Heise 57 Expressing Concentration Chap 13.4 Sample Exercise: Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water: 1 mol Co(NO 3 ) 2  T f = K f m = 1.87(1) = 1.87°C

58. Chem 106, Prof. T.L. Heise 58 Expressing Concentration Chap 13.4 Sample Exercise: Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water: 2 mol KCl  T f = K f m = 1.87(4) = 7.48°C

59. Chem 106, Prof. T.L. Heise 59 Expressing Concentration Chap 13.4 Sample Exercise: Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water: 3 mol C 2 H 6 O 2  T f = K f m = 1.87(3) = 5.61°C

60. Chem 106, Prof. T.L. Heise 60 Expressing Concentration Chap 13.4 Sample Exercise: Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water: 1 mol Co(NO 3 ) 2 = -1.87 °C 2 mol KCl= -7.48 °C 3 mol C 2 H 6 O 2 = -5.61 °C

61. Chem 106, Prof. T.L. Heise 61 Colligative Properties Chap 13.5 Osmotic Pressure The net movement of solvent is always toward the solution with the higher solute concentration. The pressure needed to stop such movement is called osmotic pressure

62. Chem 106, Prof. T.L. Heise 62 Colligative Properties Chap 13.5 Osmotic Pressure

63. Chem 106, Prof. T.L. Heise 63 Colloids Chap 13.6