Chapter 11. Solutions and Their Properties

1. Chapter 11. Solutions and Their Properties

2. Solutions Solute: material present in least amount Solvent: material present in most amount Solution = solvent + solute(s)

3. What criteria must be satisfied in order to form a solution?

6. An artist’s conception of how a salt dissolves in water How could we identify the cation and anion?

8. Dissolving of sugar

10. Solution energy = M+X- (s) + H2O M+(aq) + X-(aq) Lattice energy = M+(g) + X-(g) M+X- (s)

11. Enthalpy, H Solvent aggregated Solute aggregated Hinitial A Exothermic solution process

12. Enthalpy, H Solvent aggregated Solute aggregated Hinitial Solution DHsoln < 0 Hfinal A Exothermic solution process

13. Solvent separated Enthalpy, H Solvent aggregated Solute aggregated DHsolvent Hinitial Solution DHsoln < 0 Hfinal A Exothermic solution process

14. Solute separated Solvent separated Enthalpy, H Solvent aggregated Solute aggregated DHsolvent DHsolute Hinitial Solution DHsoln < 0 Hfinal A Exothermic solution process

15. Solute separated Solvent separated DHsolute + DHsolvent Enthalpy, H Solvent aggregated Solute aggregated DHsolvent DHsolute Hinitial Solution DHsoln < 0 Hfinal A Exothermic solution process

16. Solute separated Solvent separated DHsolute + DHsolvent Enthalpy, H Solvent aggregated Solute aggregated DHsolvent DHsolute DHmix Hinitial Solution DHsoln < 0 Hfinal A Exothermic solution process

17. Enthalpy, H Solvent aggregated Solute aggregated Hinitial B Endothermic solution process

18. Solution Enthalpy, H Solvent aggregated Solute aggregated Hfinal DHsoln > 0 Hinitial B Endothermic solution process

19. Solvent separated Solution Enthalpy, H Solvent aggregated Solute aggregated Hfinal DHsolvent DHsoln > 0 Hinitial B Endothermic solution process

20. Solvent separated Solute separated Solution Enthalpy, H Solvent aggregated Solute aggregated Hfinal DHsolvent DHsolute DHsoln > 0 Hinitial B Endothermic solution process

21. Solvent separated Solute separated DHsolute + DHsolvent Solution Enthalpy, H Solvent aggregated Solute aggregated Hfinal DHsolvent DHsolute DHsoln > 0 Hinitial B Endothermic solution process

22. Solvent separated DHmix Solute separated DHsolute + DHsolvent Solution Enthalpy, H Solvent aggregated Solute aggregated Hfinal DHsolvent DHsolute DHsoln > 0 Hinitial B Endothermic solution process

23. Interactions that must be overcome and those that form Why do salts with a positive enthalpy of solution form solutions?

24. Variation of solubility of solids and liquids with temperature

25. Solubility of gases with temperature Solvent : H2O

26. Solubility of gases as a function of pressure Le Chatelier’s principle

27. B is a non-volatile component Vapor pressure of a pure component Pure solvent A Solution of A and B air + vapor

28. Mole fraction of a = na/(na +nb +nc + ...)

29. in the presence of a non-volatile solute Colligative property: a physical property that depends on how the amount present

30. Distillation of volatile materials Raoult’s Law: PA = xAPAo; PB = xBPBo PTobs = xA PAo + xBPBo for two volatile components where PAo and PBo are the vapor pressures of the pure components

31. Raoult’s Law: PA = XAPAo; PB = XBPBo PTobs = XA PAo + XBPBo for two volatile components 25 °C mol fraction

33. 760 Boiling occurs when the total pressure = 760 mm For a 1:1 mol mixture at 93 ° C 93 °C 300 450 760 1000 560 200 PA = XAPAo; PB = XBPBo PTobs = XA PAo + XBPBo

34. Two volatile liquids For an initial 1:1 mixture of benzene-toluene, the initial composition of the vapor based on its vapor pressure is approximately Ptoluene = 200 mm Pbenzene = 560 mm Treating the vapor as an ideal gas, if we were to condense the vapor: PbV/PtV = nbRT/ntRT Pb/Pt = nb/nt nb/nt = 56/20; the vapor is enriched in the more volatile component xB = 56/(56+20) = 0.73

35. Starting with a 50:50 mixture: (the composition of the first drop)

36. A Volatile Liquid and Non-volatile Solid • A recipe for making maple syrup requires adding two cups sugar (sucrose, mw 342 g/mol) for every cup of water. Assuming a cup of water contains 300 mL of water and a cup of sugar contains 300 g of sucrose (C12H22O11), at what temperature would you expect the syrup to boil given the following vapor pressures of pure water? • Temp (°C) Vapor pressure (mm) pure H2O • 100 760 • 101 787 • 102 815 • 103 845 • 104 875 • 105 906 Raoult’s Law Pobs = xH2OPoH2O Pobs = 760 mm for boiling to occur 300g/18 g/mol = 16.7 mol H2O 600g/342 g/mol = 1.75 mol sugar xH2O =16.7/(16.7 + 1.75)= 0.91 PoH20 = Pobs/xH20; 760 /0.91 = 835 mm

37. Freezing point depression and boiling point elevation (using non-volatile solutes) T = Km K is a constant characteristic of each substance m = molality (mols/1000g solvent)

38. T = Kmm = molality, mol/1000g solvent The boiling point elevation is for non-volatile solutes

39. Boiling Point Elevation for A Volatile Liquid and Non-volatile Solid A recipe for making maple syrup requires adding two cups sugar (sucrose, mw 342 g/mol) for every cup of water. Assuming a cup of water contains 300 mL of water and a cup of sugar contains 300 g of sucrose (C12H22O11), at what temperature would you expect the syrup to boil given the following vapor pressures of pure water? sucrose, 600/342 g/mol = 1.75 mol 1.75 mol/300 g H2O = x mol/1000gH2O x = 5.83 m T = Kf m; T = 0.51*5.83m; T = 3 °C Boiling point of water = 103 °C

40. It is found that 1 g of naphthalene in 100 g of camphor depresses the freezing point of camphor by 2.945 °C. What is the molecular weight of naphthalene? T = Kf m T= 37.7 °C kg/mol*m 2.945 °C = 37.7 °C /molal*m; m = 0.0781molal If 1 g of naphthalene is dissolved in 10 g of the solvent camphor, 10 g of naphthalene is dissolved in 1000 g camphor molality is equal to the number of moles/1000 g of solvent therefore, 10g/MW = 0.0781 moles; MW = 10/0.0781 = 128 g /mol

41. Osmosis and osmotic pressure A simple example of osmosis: evaporation of water from a salt solution What acts as the semipermeable membrane?

42. Osmotic pressure

43. Osmotic Pressure π = MRT where π is the osmotic pressure between a pure solvent and a solution containing that solvent; M is the molarity of the solution (mols /liter), R is the gas constant, and T = temperature (K)

44. The total concentration of dissolved particle in red blood cells in 0.3 M. What would be the osmotic pressure between red cells and pure water if the blood cells were to be placed in pure water at 37 °C? π = 0.3moles/liter*0.0821liter atm/(mol K)*(273.15+37 )K π = 0.3 mol/l*0.0821l atm/(K mol)*298 K π = 7.6 atm In what direction would the pressure be directed? What would happen to the cells?

45. What is the osmotic pressure developed from a solution of 0.15 M NaCl in contact with a semipermeable membrane of pure water at T = 37 °C? NaCl + H2O = Na+ + Cl- π = nRT π = 0.3 mol/l*0.0821l atm/(K mol)*310.2 K π = 7.6 atm

46. Osmotic pressure Applied pressure needed to prevent volume increase Pure solvent Solution Net movement of solvent Semipermeable membrane Solute molecules Solvent molecules Reverse Osmosis P P A B C

47. De-salination of sea water

49. What happens when polar molecules meet non-polar ones?