1. Notes One Unit Seven– Chapter 13 Solutions Definitions Types of Mixtures Example Solutions Factors Affecting Solubility Like Dissolves Like Solubility of Solids Changes with Temperature Solubility of Gases Changes with Temperature Pressure Factor Molar Concentration Finding Molarity From Mass and Volume Finding Mass from Molarity and Volume Finding Volume from Molarity and Mass Pages 466-486

2. Definitions Solutions are homogeneous mixtures. Uniform throughout. Solvent. Determines the state of solution Largest component Solute. Dissolved in solvent

3. Common Mixtures liquid emulsion mayonnaise gas liquidliquid foam whipped cream liquid gasaerosol hair spray solid ruby glass SOLUTE SOLVENTType EXAMPLE solid gasaerosol dust in air liquid solidemulsion pearl gas solidsolid foam Styrofoam

4. Solution Types gas a i r gas liquid soda pop liquid antifreeze solid liquid seawater solid brass SOLUTE SOLVENTPHASE EXAMPLE liquid solid filling

5. Factors Affecting Solubility 1. Nature of Solute / Solvent1. Nature of Solute / Solvent. 2. Temperature Increase2. Temperature Increase i) Solid/Liquid ii) gas 3. Pressure Factor -3. Pressure Factor - i) Solids/Liquids - Very little ii) gas iii) squeezes gas into solution.

6. Like Dissolves Like Non-polar in Non-polar Butter in Oil Non-polar in polar Oil in H 2 O Polar in Polar C 2 H 5 OH in H 2 O Ionic compounds in polar solvents NaCl in H 2 O

7. Solubility of solids Changes with Temperature 19gx _____ 250g 100g =48g solidsgases 101g 82g How does the solubility Δ with temperature Increase? How many grams of potassium chromate will dissolve in100g water at 70 o C? 70g How many grams of lead(II) nitrate will precipitate from 250g water cooling from 70 o C to 50 o C?

8. Solubility of Gases Changes with Temperature a) Why are fish stressed, if the temperature of the water increases? How much does the solubility of oxygen change, for a 20 o C to 60 o C change? 0.90-0.60=0.30mg 0.60mg 0.90mg

9. Pressure Factor Greater pressure…more dissolved gas

10. Pressure Factor

11. Molar Concentration M=n/V n=MxV V=n/M

12. Calculate molarity for 25.5 g of NH 3 in 600. mL solution. 1) Calculate Formula Mass: 2) Calculate the moles of solute: 3) Calculate the Moles/Liters Ratio Finding Molarity From Mass and Volume 25.5g÷17.0g/m=1.50m N H 1x 3x 14.0 = 1.0 = 14.0 3.0 17.0g/m E # Mass M =1.50m / 0.600L M = 2.52 mol/L

13. Finding Volume from Molarity and Mass How many milliliters of 2.50M solution can be made using 25.5grams of NH 3 ? 1)Calculate formula mass: 2)Calculate the moles of solute: 3)Calculate Volume: V= N H 1x 3x 14.0 = 1.0 = 14.0 3.0 17.0g/m E# Mass 25.5g÷17.0g/m=1.50m V=V=(1.50m) / (2.50M) V=n/M 0.600L solution 600.mL

14. Finding Mass from Molarity and Volume How many grams of NH 3 are in 600. mL solution at 2.50M? 1) Calculate formula mass: 2) Calculate moles n=1.50m 3) Calculate mass g=25.5g NH 3 n=2.50M x 0.600L g=g=(17.0g/m) x (1.50m) N H 1x 3x 14.0 = 1.0 = 14.0 3.0 17.0g/m E# Mass g ÷fm=mol M xLn = fm x ng =

15. Notes Two Unit Seven– Chapter 13 Solutions Saturated versus Unsaturated Colligative properties of water Forming a Saturated Solution How Does a Solution Form? Colligative Properties Vapor Pressure Boiling and Freezing Point BP Elevation and Freezing FP Depression Calculating Freezing Point Depression Mass Pages 487-501

16. Characteristics of Saturated Solutions Solid dissolve water dissolve precipitate dissolve precipitate Unsaturated Saturated Dynamic Equilibrium Cooling causes precipitation. Warming causes dissolving.

17. As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them. Solvation

18. Colligative Properties Colligative properties depend on moles dissolved particles.  Vapor pressure lowering  Boiling point elevation  Melting point depression  Osmotic pressure

19. How do you get from this…

20. …to this?

21. Add an ionic compound!

22. Vapor Pressure

23. Vapor Pressure Lowering The particles of solute are surrounded by and attracted to particles of solvent. Now the solvent particles have less kinetic energy and tend less to escape into the space above the liquid. So the vapor pressure is less.

24. Ionic vs Molecular Solutes Ionic solutes produce two or more ion particles in solution. They affect the colligative properties proportionately more than molecular solutes (that do not ionize). The effect is proportional to the number of particles of the solute in the solution.

25. How many particles do each of the following give upon solvation? NaCl CaCl 2 Glucose

26. Freezing Point Depression

27. Example Salt is added to melt ice by reducing the freezing point of water.

28. Boiling Point Elevation

29. Example Addition of ethylene glycol C 2 H 6 O 2 (antifreeze) to car radiators.

30. Freezing Point Depression and Boiling Point Elevation Boiling Point Elevation ∆T b =imk b (for water k b =0.51 o C/m) Freezing Point Depression ∆T f =imk f (for water k f =1.86 o C/m) Note: m is the molality of the particles, so if the solute is ionic, multiply by the #of particles it dissociates to. (i is # of particles called van hoff factor)

31. BP Elevation Constants (K b ) FP Depression Constants( K f )

32. Which is more effective for lowering the freezing point of water? NaCl or CaCl 2

33. Example 1: Find the new freezing point of 3m NaCl in water.

34. Example 2: Find the new boiling point of 3m NaCl in water.

35. Molarity versus Molality Molality (m) = ________________ moles of solute kilograms solvent Molarity (M) = ________________ moles of solute liters of solution

36. Calculating T f andT b Calculate the freezing and boiling points of a solution made using 1000.g antifreeze (C 2 H 6 O 2 ) in 4450g water. 1) Calculate Moles 2) Calculate molality 3) Calculate Temperature Change Δt=Kxm ΔT f = T f = ΔT b = T b = 1000.g ÷62.0g/mol =16.1 moles 16.1 mole ÷4.45 Kg water = 3.62m (1.858 o C/m)(3.62 m) = 6.73 o C 0.000 o C-6.73 o C=-6.73 o C (0.512 o C/m)(3.62 m) = 1.96 o C 100.000 o C +1.96 o C =101.96 o C C H 2x 6x 12.0 = 1.0 = 24.0 6.0 E# Mass O2x16.0 = 32.0 62.0g/m

37. Calculating Boiling Point Elevation Mass A solution containing 18.00 g of glucose in 150.0 g of water boils at 100.34 o C. Calculate the molecular weight of glucose. 1.)Calculate Temperature Change ΔT b = 2.)Calculate moles per Kilograms ΔT b = K b x m  m = ΔT b /K b m =0.67m/kg 3.)Calculate grams / kilograms g = g =120 g/kg MW=120 g/0.67m 180g/m 100.34 o C-100.00 o C=0.34 o C 0.34÷0.512 o C/ m =m 18.00 g ÷0.1500kg

38. One Molal Solution of Water Pressure Temperature solid Liquid gas KfKf KbKb 1 atm 1.858 o C 0.512 o C

39. Notes Three Unit Seven Ice-cream Lab A Calculating Freezing Point Depression Mass Colligative Properties of Electrolytes Distillation Osmotic Pressure Dialysis Pages 487-501

40. Ice-cream

41. Calculating Freezing Point Depression Mass 1.)Calculate Temperature Change ΔT f = 2.)Calculate moles per Kilograms ΔT f = K f x m  m = ΔT f /K f m =1.83mol/kg 3.)Calculate grams / kilograms g = g =36.4 g/kg MW=36.4 g/1.83m 19.9g/m 1.1 o C-(-2.3 o C)=3.4 o C 3.4 o C÷1.858 o C/ m =m 1.89 g ÷ 0.05196kg

42. Colligative Properties of Electrolytes Colligative properties depend on the number of particles dissolved. NaCl  Na +1 +Cl -1 CH 3 OH Al 2 (SO 4 ) 3  2Al +3 + 3SO 4 -2 C 6 H 12 O 6

43. Distillation

45. Osmotic Pressure Hypertonic > 0.92% (9.g/L) Crenation Isotonic Saline = 0.92% (9.g/L) Hypotonic < 0.92% (9.g/L) Rupture

46. Dialysis

47. Kidney

48. Dialysis

49. Final Quiz Notes

50. Finding Molarity From Mass and Volume Calculate molarity for 14.0 g of sodium peroxide(Na 2 O 2 ) in 615 mL solution. 1) Calculate Formula Mass: 2) Calculate the moles of solute: 3) Calculate the Moles/Liters Ratio 14.0g÷78.0g/m=0.179m Na2x23.0 = 46.0 E# Mass O2x16.0 = 32.0 78.0g/m M = 0. 179moles ÷ 0.615L = 0.289M M = n/ v

51. Finding Volume From Mass and Molarity Find the volume for a solution having 14.0 g of sodium peroxide(Na 2 O 2 ) in a 0.289M solution. 1) Calculate Formula Mass: 2) Calculate the moles of solute: 3) Calculate the Liters. 14.0g÷78.0g/m=0.179m Na2x23.0 = 46.0 E# Mass O2x16.0 = 32.0 78.0g/m v = 0. 179moles ÷ 0.289M= 0.615L v = n / M

52. Finding Mass from concentration and Volume How many grams of sodium peroxide(Na 2 O 2 ) would be needed for a 0.289M solution of 615mL volume? 1) Calculate Formula Mass: 2) Calculate the moles of solute: M=n/V  MxV=n 3) Calculate the Grams. 0.289mol/LX0.615L =0.179m Na2x23.0 = 46.0 E# Mass O2x16.0 = 32.0 78.0g/m g = 0. 179moles x 78.0g/m= 14.0g g = n x fm

53. Calculating Freezing Point Depression Mass A solution containing 7.67 g of ethanol in 333.0 g of water freezes at -0.929 o C. Calculate the molecular weight of ethanol. 1.)Calculate Temperature Change ΔT f = 2.)Calculate moles per Kilograms ΔT f = K f x m  m = ΔT f /K f m =0.500m/kg 3.)Calculate grams / kilograms g/kg = g/Kg =23.0g/kg fm= 46.0g/m 0.000 o C-( - 0.929 o C)=0.929 o C 0.929÷1.858 o C/ m =m 7.67 g ÷0.3330kg 23.0 g/ 0.500m

54. Solubility of solids Changes with Temperature How many grams Cs 2 SO 4 will precipitate from 267g water as it cools from 60 o C to 25 o C? 18gx _____ 267g 100g =48g 200g 182g

55. Phase Diagram Pressure Temperature solid Liquid gas melting   freezing NFP NBP 1 atm Triple point  Critical point  0.0 o C 100.0 o C vaporizing   condensing sublimation   depostion

56. end

57. 0 1.0 2.03.04.0 5.0 6.07.09.010.0 0 10 20 30 40 50 60 70 8.0 80 Mg of gas per 100 grams of water Gas pressure in atmospheres

58. Calculating T f andT b Calculate the freezing and boiling points of a solution made using 36.0g glucose(C 6 H 12 O 6 ) in 225.0g water. 1) Calculate Moles 2) Calculate molality of H 2 O 3) Calculate Temperature Change Δt= ΔT f = T f = ΔT b = T b = 36.0g ÷180.0g/mol =0.200 moles 0.200 mol ÷0.2250 Kg= 0.889m (1.858 o C/m)(0.889m) = 1.65 o C 0.000 o C-1.65 o C= -1.65 o C (0.512 o C/m)(0.889 m) = 0.455 o C 100.000 o C +0.455 o C =100. 455 o C C H 6x 12x 12.0 = 1.0 = 72.0 12.0 E# Mass O6x16.0 = 96.0 180.0g/m K xmxm m = n/ Kg

59. Finding Molarity From Mass and Volume Calculate molarity for 24.0 g of antifreeze(C 2 H 6 O 2 ) in 445. mL solution. 1) Calculate Formula Mass: 2) Calculate the moles of solute: 3) Calculate the Moles/Liters Ratio 24.0g÷62.0g/m=0.387m C H 2x 6x 12.0 = 1.0 = 24.0 6.0 E# Mass O2x16.0 = 32.0 62.0g/m M = 0. 387moles ÷ 0.445L = 0.870M M = n/ v

60. Seven/Eight Rows

61. Calculating Freezing Point Depression Mass A solution containing 1.89 g of methanol in 51.96 g of water freezes at -3.4 o C. Calculate the molecular weight of methanol. 1.)Calculate Temperature Change ΔT b = 2.)Calculate moles per Kilograms ΔT f = K f x m  m = ΔT f /K f m =0.500m/kg 3.)Calculate grams / kilograms g = g =23.0g/kg fm= 46.0g/m 0.000 o C-0.929 o C=0.929 o C 0.929÷1.858 o C/ m =m 7.67 g ÷0.3330kg 23.0 g/ 0.500m

62. Calculating Freezing Point Depression Mass A solution containing 1.89 g of ethanol in 51.96 g of water freezes at -3.4 o C. Calculate the molecular weight of ethanol. 1.)Calculate Temperature Change ΔT b = 2.)Calculate moles per Kilograms ΔT f = K f x m  m = ΔT f /K f m =0.500m/kg 3.)Calculate grams / kilograms g = g =23.0g/kg fm= 46.0g/m 0.000 o C-0.929 o C=0.929 o C 0.929÷1.858 o C/ m =m 7.67 g ÷0.3330kg 23.0 g/ 0.500m

63. Calculating Freezing Point Depression Mass A solution containing 1.89 g of methanol in 51.96 g of water freezes at -3.4 o C. Calculate the molecular weight of methanol. 1.)Calculate Temperature Change ΔT b = 2.)Calculate moles per Kilograms ΔT f = K f x m  m = ΔT f /K f m =0.500m/kg 3.)Calculate grams / kilograms g = g =23.0g/kg fm= 46.0g/m 0.000 o C-0.929 o C=0.929 o C 0.929÷1.858 o C/ m =m 7.67 g ÷0.3330kg 23.0 g/ 0.500m