1. II III I I. Types of Mixtures Ch. 14 – Mixtures & Solutions

2. A. Definitions  Mixture = Variable combination of 2 or more pure substances  Homogeneous = uniform composition throughout  Heterogeneous = variable composition HeterogeneousHomogeneous

3. A. Definitions  Solution –  Solution – homogeneous mixture Solvent Solvent – dissolves the solute Solute Solute – substance being dissolved

4. B. Mixtures  Gases can also mix with liquids  Gases are usually dissolved in water  Examples are carbonated drinks Homogeneous mixtures (solutions) Contain sugar, flavorings and carbon dioxide dissolved in water

5. B. Mixtures  Solution homogeneous very small particles no Tyndall effect particles don’t settle Ex: rubbing alcohol Tyndall Effect

6. B. Mixtures  Colloid heterogeneous medium-sized particles  Tyndall effect :the scattering of light by colloidal particles. particles don’t settle Ex: milk

7. B. Mixtures  Suspension Heterogeneous large particles usually > 1000nm particles settle Tyndall effect Ex:fresh-squeezed lemonade

8. B. Mixtures  Examples: mayonnaise muddy water fog saltwater Italian salad dressing colloid suspension colloid true solution suspension

9. II III I II. Factors Affecting Solvation (p. 489 – 496) Ch. 14 –Solutions

10. A. Solvation  Solvation –  Solvation – the process of dissolving solute particles are separated and pulled into solution solute particles are surrounded by solvent particles

11. Solution Process For a solute to be dissolved in a solvent, the attractive forces between the solute and solvent particles must be great enough to overcome the attractive forces within the pure solvent & pure solute. The solute & the solvent molecules in a solution are expanded compared to their position within the pure substances.

12. B. Solvation  Dissociation Separation/Solvation of an ionic solid into aqueous ions  For ionic solids, the lattice energy describes the attractive forces between the solute molecules (i.e. ions)  For an ionic solid to dissolve in water, the water- solute attractive forces has to be strong enough to overcome the lattice energy NaCl(s)  Na + (aq) + Cl – (aq)

13. B. Solvation  Ionization breaking apart of some polar molecules into aqueous ions HNO 3 (aq) + H 2 O(l)  H 3 O + (aq) + NO 3 – (aq)

14. B. Solvation  Molecular Solvation molecules stay intact C 6 H 12 O 6 (s)  C 6 H 12 O 6 (aq)

15. B. Solvation  Soap/Detergent polar “head” with long nonpolar “tail” dissolves nonpolar grease in polar water

16. B. Factors Affecting Solvation  Molecules are constantly in motion according to… Kinetic Theory  When particles collide, energy is transferred

17. B. Factors Affecting Solvation  Solubility = max. amount of a solute that will dissolve in a solvent @ a specific T  Smaller pieces of a substance dissolve faster b/c of larger surface area  Stirring or shaking speeds dissolving b/c particles are moving faster and colliding more  Heating speeds dissolving of solids  Not all substances dissolve

18. C. Solubility  Water is universal solvent b/c of its polarity  If something can dissolve in something else, it is said to be soluble or miscible  If it cannot dissolve, it is said to be insoluble or immiscible  “Like dissolves like”

19. C. Solubility NONPOLAR POLAR “Like Dissolves Like”

20. Saturated Solutions A solution that can contain the maximum amount of solute at a given temperature (if the pressure is constant). Solution said to be at a dynamic equilibrium Any point on the line Ex: At 90 ° C 40 g of NaCl (s) in 100 g of H 2 O represent a saturated solution

21. Unsaturated Solution A solution that can contain less than the maximum amount of solute at a given temperature (if the pressure is constant). It is any value under the solid line on the solubility graph

22. Supersaturated Solutions A solution that can contain greater than the maximum amount of solute at a given temperature (if the pressure is constant). A supersaturated solution is very unstable & the amount of solute in excess can precipitate or crystallize. It is any value above the solid line on the solubility graph

23. Solubility SATURATED SOLUTION no more solute dissolves UNSATURATED SOLUTION more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form concentration

24. Solubility Any solution can be made Saturated, Unsaturated, or Supersaturated by changing the Temperature.

25. C. Solubility  Solubility Curves maximum grams of solute that will dissolve in 100 g of solvent at a given temperature varies with temp based on a saturated soln

26. C. Solubility  Solubility Curve shows the dependence of solubility on temperature

27. C. Solubility  Solids are more soluble at... high temperatures.  Gases are more soluble at... low temperatures high pressures (Henry’s Law). With larger mass (LDF) EX: nitrogen narcosis, the “bends,” soda

28. Strong Electrolyte Non- Electrolyte solute exists as ions only - + salt - + sugar solute exists as molecules only - + acetic acid Weak Electrolyte solute exists as ions and molecules DISSOCIATIONIONIZATION Although H 2 O is a poor conductor of electricity, dissolved ions in an aqueous solution can conduct electricity. Ionic aqueous solutions are known as electrolytes.

29. II III I II. Solution Concentration (p. 480 – 486) Ch. 16 – Solutions

30. A. Concentration  The amount of solute in a solution  Describing Concentration % by mass - medicated creams % by volume - rubbing alcohol ppm, ppb - water contaminants molarity - used by chemists molality - used by chemists

31. B. Percent Solutions  Percent By Volume (%(v/v)) Concentration of a solution when both solute and solvent are liquids often expressed as percent by volume total combined volume substance being dissolved

32. B. Percent Solutions  Find the percent by volume of ethanol (C 2 H 6 O) in a 250 mL solution containing 85 mL ethanol. 85 mL ethanol 250 mL solution = 34% ethanol (v/v)  Solute = 85 mL ethanol  Solution = 250 mL % (v/v) = x 100

33. C. Molarity  Concentration of a solution most often used by chemists total combined volume substance being dissolved

34. C. Molarity 2M HCl What does this mean?

35. Molar Mass (g/mol) 6.02  10 23 particles/mol MASS IN GRAMS MOLES NUMBER OF PARTICLES Molar Volume (22.4 L/mol) LITERS OF GAS AT STP LITERS OF SOLUTION Molarity (mol/L) D. Molarity Calculations

36.  How many moles of NaCl are required to make 0.500L of 0.25M NaCl? 0.500 L sol’n 0.25 mol NaCl 1 L sol’n = 0.013 mol NaCl

37. D. Molarity Calculations  How many grams of NaCl are required to make 0.500L of 0.25M NaCl? 0.500 L sol’n 0.25 mol NaCl 1 L sol’n = 7.3 g NaCl 58.44 g NaCl 1 mol NaCl

38. D. Molarity Calculations  Find the molarity of a 250 mL solution containing 10.0 g of NaF. 10.0 g NaF 1 mol NaF 41.99 g NaF = 0.24 mol NaF 0.24 mol NaF 0.25 L = 0.95 M NaF

39. E. Dilution  Preparation of a desired solution by adding water to a concentrate  Moles of solute remain the same

40. E. Dilution  What volume of 15.8M HNO 3 is required to make 250 mL of a 6.0M solution? GIVEN: M 1 = 15.8M V 1 = ? M 2 = 6.0M V 2 = 250 mL WORK: M 1 V 1 = M 2 V 2 (15.8M) V 1 = (6.0M)(250mL) V 1 = 95 mL of 15.8M HNO 3

41. F. Molality mass of solvent only 1 kg water = 1 L water

42. G. Molality Calculations  Find the molality of a solution containing 75 g of MgCl 2 in 250 mL of water. 75 g MgCl 2 1 mol MgCl 2 95.21 g MgCl 2 = 3.2 m MgCl 2 0.25 kg water =.79 mol MgCl 2.79 mol MgCl 2

43. G. Molality Calculations  How many grams of NaCl are req’d to make a 1.54m solution using 0.500 kg of water? 0.500 kg water1.54 mol NaCl 1 kg water = 45.0 g NaCl 58.44 g NaCl 1 mol NaCl

44. H. Preparing Solutions  500 mL of 1.54M NaCl 500 mL water 45.0 g NaCl mass 45.0 g of NaCl add water until total volume is 500 mL mass 45.0 g of NaCl add 0.500 kg of water 500 mL mark 500 mL volumetric flask  1.54m NaCl in 0.500 kg of water

45. H. Preparing Solutions  250 mL of 6.0M HNO 3 by dilution measure 95 mL of 15.8M HNO 3 95 mL of 15.8M HNO 3 water for safety 250 mL mark combine with water until total volume is 250 mL Safety: “Do as you oughtta, add the acid to the watta!” or AA – add acid!

46. II III I IV. Colligative Properties of Solutions (p. 498 – 504) Ch. 14 – Mixtures & Solutions

47. A. Definition  Colligative Property property that depends on the concentration of solute particles, not their identity Examples: vapor pressure, freezing point, boiling point

48. B. Types

49.  Freezing Point Depression  Freezing Point Depression (  T f ) f.p. of a solution is lower than f.p. of the pure solvent  Boiling Point Elevation  Boiling Point Elevation (  T b ) b.p. of a solution is higher than b.p. of the pure solvent  Vapor Pressure Lowering lower number of solvent particles at the surface of the solution; therefore, this lowers the tendency for the solvent particles to escape into the vapor phase.

50. B. Types View Flash animation.Flash animation Freezing Point Depression

51. B. Types Solute particles weaken IMF in the solvent Boiling Point Elevation

52. B. Types  Applications salting icy roads making ice cream antifreeze cars (-64°C to 136°C)

53. C. Calculations  T : change in temperature (° C ) i : Van’t Hoff Factor (VHF), the number of particles into which the solute dissociates m : molality ( m ) K : constant based on the solvent (° C·kg/mol ) or (°C/ m )  T = i · m · K

54. C. Calculations   T Change in temperature Not actual freezing point or boiling point Change from FP or BP of pure solvent Freezing Point (FP) i  T F is always subtracted from FP of pure solvent Boiling Point (BP) i  T B is always added to BP of pure solvent

55. C. Calculations  i – VHF Nonelectrolytes (covalent) remain intact when dissolved 1 particle Electrolytes (ionic) dissociate into ions when dissolved 2 or more particles

56. C. Calculations  i – VHF Examples CaCl 2 Ethanol C 2 H 5 OH Al 2 (SO 4 ) 3 Methane CH 4 i = 3 1 5 1

57. C. Calculations  K – molal constant K F K F – molal freezing point constant Changes for every solvent 1.86 °C·kg/mol (or °C/ m ) for water K B K B – molal boiling point constant Changes for every solvent 0.512 °C·kg/mol (or °C/ m ) for water

58. C. Calculations: Recap! : subtract from F.P.   T : subtract from F.P. add to B.P. add to B.P.  i – VHF : covalent = 1 ionic > 2 ionic > 2  K : K F water =  K : K F water = 1.86 °C·kg/mol K B water = K B water = 0.512 °C·kg/mol  T = i · m · K

59.  At what temperature will a solution that is composed of 0.730 moles of glucose in 225 g of water boil? C. Calculations m = 3.24 m K B = 0.512°C/ m  T B = i · m · K B WORK: m = 0.730 mol ÷ 0.225 kg GIVEN: b.p. = ?  T B = ? i = 1  T B = (1)(3.24 m )(0.512°C/ m )  T B = 1.66°C b.p. = 100.00°C + 1.66°C b.p. = 101.66°C 100 +  T b

60. C. Calculations  Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water. i = 2 m = 4.8 m K F = 1.86°C/ m  T F = i · m · K F WORK: m = 0.48mol ÷ 0.100kg GIVEN: f.p. = ?  T F = ?  T F = (2)(4.8 m )(1.86°C/ m )  T F = 18°C f.p. = 0.00°C – 18°C f.p. = -18°C 0 –  T F

61. D. Osmotic Pressure  Osmosis: The flow of solvent into a solution through a semipermeable membrane  Semipermeable Membrane: membrane that allows solvent to pass through but not solute

62. D. Osmotic Pressure  Net transfer of solvent molecules into the solution until the hydrostatic pressure equalizes the solvent flow in both directions

63.  Because the liquid level for the solution is higher, there is greater hydrostatic pressure on the solution than on the pure solvent  Osmotic Pressure: The excess hydrostatic pressure on the solution compared to the pure solvent D. Osmotic Pressure

64. Osmotic Pressure: Minimum Pressure required to stop flow of solvent into the solution D. Osmotic Pressure

65. Osmosis at Equilibrium

66.   = i M R T where: osmotic pressure (atm) π = osmotic pressure (atm) VHF i = VHF M = Molarity (moles/L) R = Gas Law Constant T = Temperature (Kelvin) E. Osmotic Pressure Calculations 0.08206 L atm/mol K

67. E. Osmotic Pressure Calculations  Calculate the osmotic pressure (in torr) at 25 o C of aqueous solution containing 1.0g/L of a protein with a molar mass of 9.0 x 10 4 g/mol. i = 1 M = 1.11 x 10 -5 M R = 0.08206 L atm/mol K T = 25 o C = 298 K WORK: M = 1.0 g prot. GIVEN:  = ? 1.11 x 10 -5 M  = (1)(1.11x10 -5 )(.08206)(298)  = 2.714 x 10 -4 atm  = 0.21 torr 1 mol prot. 1 L sol’n 9.0 x 10 4 g =

68. ▲Colligative Properties useful for: ▲characterizing the nature of a solute after it is dissolved in a solvent ▲determining molar masses of substances D. Osmotic Pressure

69.  If the external pressure is larger than the osmotic pressure, reverse osmosis occurs  One application is desalination of seawater F. Reverse Osmosis

70. Net flow of solvent from the solution to the solvent