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# Ap chem unit 11 presentation

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AP Chem Unit 11: Solutions

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### Ap chem unit 11 presentation

1. 1. Properties of Solutions AP Chem Unit 11
2. 2. Solutions SectionsSolution CompositionThe Energies of Solution FormationFactors Affecting SolubilityThe Vapor Pressures of SolutionsBoiling-Point Elevation and Freezing-PointDepressionOsmotic PressureColligative Properties of Electrolyte SolutionsColloids
3. 3. SOLUTION COMPOSITION
4. 4. Solution CompositionSolutions are homogeneous mixtures that can begases, liquids or solids. Solutions can be described as dilute or concentrated. molarity mass percent mole fraction molality
5. 5. Concentrations molesMolarity = M = liters æ mass of solute öMass % =ç ÷ x 100% è mass of solution ø nAMole fraction of A= X A = nA + nB moles of soluteMolality = kg of solvent
6. 6. Practice Problem 1A Solution is prepared by mixing 1.00g of ethanol(C2H3OH) with 100.0g of water to give a final volume of101 ml. Calculate the molarity, mass percent, molefraction and molality of ethanol in this solution. M=.215, %=.990%, X=.00389, m=.217
7. 7. NormalityNormality is defined as the number of equivalents perliter of solution. equivalents depends on the reaction taking place in the solution example: for an acid base reaction, the equivalent is the mass of acid or base that can furnish or accept exactly 1 mole of protons (H+ ions). H2SO4 or Ca(OH)2 equivalents = molar mass/2 one equivalent of acid reacts with one equivalent of base.
8. 8. Normality• For oxidation - reduction reactions, the equivalent is defined as the quantity of oxidizing or reducing agent that can accept or furnish 1 mole of electrons. one equivalent of reducing agent will react with exactly one equivalent of oxidizing agent.
9. 9. ExampleThe equivalent mass of an oxidizing or reducingagent can be calculated from the number ofelectrons in a half reaction. MnO4- + 5e- + 8H+  Mn2+ + 4H2O Since the MnO4- ion present in 1 mole of KMnO4 in acidic solution consumes 5 moles of electrons, the equivalent mass is the molar mass divided by 5: molar mass 158g = = 31.6g Eq mass of KMnO4 = 5 5
10. 10. Practice Problem 2The electrolyte in automobile lead storage batteries is a3.75 M sulfuric acid solution that has a density of 1.230g/ml. Calculate the mass percent, molality, andnormality of the sulfuric acid. 7.50 N
11. 11. ENERGIES OF SOLUTION FORMATION
12. 12. SolubilityWhat factors affect solubility? The cardinal rule of solubility is that like dissolves like.  Polar solvents are used to dissolve a polar or ionic solute and nonpolar solvents are used to dissolve a nonpolar solute.
13. 13. SolubilitySolubility and formation of a solution takes place inthree distinct steps:1. Separating the solute into its individual components (expanding the solute).2. Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent).3. Allowing the solute and solvent to interact to form the solution.
14. 14. Solubility1. Expanding the solute: endothermic2. Expanding the solvent: endothermic3. Interaction: exothermicEnthalpy of solution = ΔHsoln =ΔH1 + ΔH2 + ΔH3
15. 15. SolubilityΔHsoln could be overall negative sign (exothermic).ΔHsoln could be overall positive sign (endothermic).
16. 16. ExampleOil slicks to not dissolve in water: 1. ΔH1 is endothermic but small. C chains held together with LDF need separated. 2. ΔH2 is endothermic and large. Hydrogen bonds in the water are difficult to separate. 3. ΔH3 will be small since polar and nonpolar interactions are minimal. ΔHsoln is an overall endothermic process and unlikely to produce a solution.
17. 17. Example 2Most ionic substances dissolve in water: 1. ΔH1 is endothermic and large. Ionic forces must be overcome. 2. ΔH2 is endothermic and large. Hydrogen bonds in the water are difficult to separate. 3. ΔH3 will be very exothermic because most ionic substances interact very well with water. ΔHsoln is usually small but can be overall exothermic or endothermic.
18. 18. Enthalpy of HydrationEnthalpy of hydration (ΔHhyd ) combines the terms ΔH2(expanding the solvent) and ΔH3 (solvent-soluteinteractions). The heat of hydration represents the enthalpy change assoicated with the dispersal of a gaseous solute in water. H2O(l) + Na+(g) + Cl-(g)  Na+(aq) + Cl-(aq)  ΔHsoln = ΔH1 + ΔHhyd
19. 19. Example 3Heat of solution of NaCl and water:1. NaCl(s)  Na+(g) + Cl-(g) ΔH1=786 kJ/mol2. & 3. H2O(l) + Na+(g) + Cl-(g)  Na+(aq) + Cl-(aq) ΔHhyd=783kJ/molΔHsoln=3kJ, endothermic but small.
20. 20. SolubilityThe dissolving process often requires a small amount ofenergy (stirring, heat), but ionic substances, eventhough their expansion is very endothermic, dissolvedue to their tendency toward increased probability. Processes that require large amounts of energy tend not to occur.
21. 21. Practice Problem 3Decide whether liquid hexane (C6H14) or liquidmethanol (CH3OH) is the more appropriate solvent forthe substances grease (C2oH42) and potassium iodide(KI). Hexane works best for grease and methanol serves as a better solvent for potassium iodide
22. 22. FACTORS AFFECTING SOLUBILITY
23. 23. Structure and SolubilitySince polarity is a big factor in solubility and molecularstructure determines polarity, structure has a lot to dowith solubility. Vitamins are divided into fat soluble and water soluble due to their structures:
24. 24. Pressure EffectsPressure has little effect on the solubilities of solids orliquids, but it does significantly increase the solubility ofa gas.
25. 25. Henry’s LawHenry’s law states that the amount of a gas dissolved ina solution is directly proportional to the pressure of thegas above the solution. The relationship between gaspressure and the concentration of dissolved gas isgiven by: C=kPC = the concentration of the dissolved gas, k is aconstant characteristic of a particular solution and Prepresents the partial pressure of the gaseous soluteabove the solution. Henry’s law is obeyed most accurately for dilute solutions that do not dissociate or react with the solvent.
26. 26. Practice Problem 4A certain soft drink is bottled so that a bottle at 25°Ccontains CO2 gas at a pressure of 5.0 atm over theliquid. Assuming that the partial pressure of CO2 in theatmosphere is 4.0 x 10-4 atm, calculate the equilibriumconcentrations of CO2 in the soda both before and afterthe bottle is opened. The Henry’s law constant for CO2in aqueous solution is 3.0 x 10-2 mol/Latm at 25°C. .16 mol/L and 1.2 x 10-5 mol/L
27. 27. Temperature EffectsSolubility doesn’t always increase with temperature.The dissolving of a solid occurs more rapidly at highertemperatures, but the amount of solid that can bedissolved may increase or decrease with increasingtemperature.Predicting the temperature dependence of solubility isvery difficult. The only sure way to determine thetemperature dependence of a solid’s solubility is byexperiment.
28. 28. Temperature and Solubility
29. 29. Temperature EffectsThe behavior of gases dissolving in water typicallydecrease with increasing temperature. Thermal pollution in lakes and rivers Boiler scale –  2Ca2+ + HCO3(aq) H2O + CO2(aq) + CaCO3(aq)
30. 30. Temperature and Solubility
31. 31. VAPOR PRESSURE OF SOLUTIONS
32. 32. Vapor Pressure and SolutionsLiquid solutions have physical properties significantlydifferent from those of the pure liquid solvent. Antifreeze with water delays freezing and boiling. Salt and water lowers freezing point.
33. 33. Vapor Pressure and SolutionsA nonvolatile solute lowers thevapor pressure of a solvent.1. Vapor pressure of the pure solvent is greater than that of the solution.2. The equilibrium vapor pressure of the pure solvent (water) is greater than that of the solution equilibrium vapor pressure.3. Water vaporizes and adds to solution.
34. 34. Vapor Pressure andRaoult’s Law: P Solubility =X P soln solvent o solvent• Psoln is the observed vapor pressure of the solution, Xsolvent is the mole fraction of solvent, and Posolvent is the vapor pressure of the pure solvent.• In a solution consisting of half nonvolatile solute molecules and half solvent molecules (typically water), the observed vapor pressure is half of that of the pure solvent, since only half as many molecules can escape.• If vapor pressures are known, moles can be determined.
35. 35. Practice Problem 5Calculate the expected vapor pressure at 25°C for asolution prepared by dissolving 158.0 g ofsucrose, molar mass= 342.3 g/mol, in 643.5cm3 ofwater. At 25°C, the density of water is 0.9971 g/cm3and the vapor pressure is 23.6 torr. 23.46 torr
36. 36. Vapor Pressure and SolubilityThe lowering of vapor pressure depends on the numberof ionic solute particles present in the solution. Example: 1 mole of sodium chloride dissolved in water lowers the vapor pressure approximately twice as much as expected because the solid has two ions per formula unit, which separates when it dissolves.
37. 37. Practice Problem 6Predict the vapor pressure of a solution prepared bymixing 35.0 g solid Na2SO4 (molar mass= 142.05g/mol) with 175 g water at 25°C. The vapor pressure ofpure water at 25°C is 23.76 torr. (how many moles ofsolute particles are present?) 22.1 torr
38. 38. Nonideal SolutionsWhen both solvent and solute are volatile, bothcontribute to the vapor pressure over the solution. Amodified form of Raoult’s law is used:Ptotal = PA + PB = XAPoA + XBPoB• Ptotal is the total vapor pressure of solution AB. XA and XB are the mole fractions of A and B. PA and PB are the partial pressures of A and B.
39. 39. Ideal vs. NonidealA liquid-liquid solution that obeys Raoult’s law is calledan ideal soution. Raoult’s law is to solutions what theideal gas law is to gases. Nearly ideal behavior is oftenobserved when solutes and solvents are similar.• When strong interactions occur (ΔHsoln is very exothermic), a negative deviation of Raoult’s law results• When weak interactions occur (ΔHsoln is endothermic), a positive deviation of Raoult’s law results.
40. 40. Behavior Summary
41. 41. Practice Problem 7A solution is prepared by mixing 5.81 g acetone(C3H6O, molar mass = 58.1 g/mol) and 11.9 gchloroform (HCCl3, molar mass = 119.4 g/mol). At35°C, this solution has a total vapor pressure of 260torr. Is this an ideal solution? The vapor pressures ofpure acetone and pure chloroform at 35°C are 345 and293 torr, respectively.Acetone (CH3)2CO and chloroform CHCl3 319 is the expected total vapor pressure; this is not an ideal solution. This is a negative deviation, therefore interactions must be strong.
42. 42. BOILING-POINT ELEVATION ANDFREEZING-POINT DEPRESSION
43. 43. Vapor Pressure and Freezing/Boiling PointsSince changes of state depend on vapor pressure, thepresence of a solute also affects the freezing point andboiling point of a solvent.• Freezing point depression, boiling point elevation and osmotic pressure are called colligative properties.• These properties are dependent on the number of solute particles and not the identity of the particles.
44. 44. Boiling Point ElevationThe normal boiling point of a liquid occurs at thetemperature at which the vapor pressure is equal to 1atmosphere. A nonvolatile solute lowers the vaporpressure of the solvent; therefore such a solution mustbe heated to a higher temperature than the ‘pure’boiling point for the vapor pressure to reach 1atmosphere.• A nonvolatile solute elevates the boiling point of the solvent.
45. 45. Boiling Point ElevationWater and a nonvolatile watersolution. The boiling pointincreases and the freezingdecreases with the solution.The effect of a nonvolatilesolute is to extend the liquidrange of a solvent.
46. 46. Boiling Point ElevationThe magnitude of the boiling point elevation dependson the concentration of the solute. The change inboiling point can be calculated by:ΔT = Kbmsolute• ΔT is the boiling point elevation, Kb is a constant that is characteristic of the solvent and is called the molal boiling-point elevation constant. msolute is the molality of the solute in the solution.• An observed boiling point elevation can determine molar mass.
47. 47. Boiling Point Elevation
48. 48. Practice Problem 8A solution was prepared by sissolving 18.00 g glucosein 150.0g water. The resulting solution was found tohave a boiling point of 100.34°C. Calculate the molarmass of glucose. Glucose is a molecular solid that ispresent as individual molecules in solution. 180 g/mol
49. 49. Freezing Point DepressionVapor pressures of ice and liquid water are the same at0°C (freezing point). When a nonvolatile solute isdissolved in water, the vapor pressure of the solutionlowers and therefore the freezing point of the solution islower than that of the pure water.ΔT = Kfmsolute• ΔT is the freezing point depression, Kf is a constant that is characteristic of the solvent and is called the molal freezing-point depressionconstant. msolute is the molality of the solute in the solution.
50. 50. Practice Problem 9What mass of ethylene glycol (C2H6O2, molar mass =62.1 g/mol), the main componenet of antifreeze, mustbe added to 10.0 L water to produce a solution for usein a car’s radiator that freezes at -10.0°F (-23.3°C)?Assume the denisty of water is exactly 1 g/mL. 7.76kg
51. 51. Practice Problem 10Mr. Wieland is trying to identify a human hormone thatcontrols metabolism by determining its molar mass. Asample weighing 0.546 g was dissolved in 15.0 gbenzene, and the freezing-point depression wasdetermine to be 0.240°C. Calculate the molar mass ofthe hormone. 776 g/mol
52. 52. OSMOTIC PRESSURE
53. 53. Osmotic PressureA solution and pure solvent are separated by asemipermeable membrane, which allows solvent butnot solute molecules to pass through. As timepasses, the volume of the solution increases andvolume of the solvent decreases.The flow of solvent into the solution through themembrane is called osmosis.Eventually the liquid levels stop changing and reachequilibrium, but there is a greater hydrostatic pressureon the solution than on the pure solvent. The excesspressure is called osmotic pressure.
54. 54. Osmotic PressureOsmosis can be prevented by applying pressure to thesolution. The minimum pressure that stops the osmosisis equal to the osmotic pressure of the solution.
55. 55. Osmotic PressureOsmotic pressure is primarily dependent on solutionconcentrations. Π = MRT• Π is the osmotic pressure in atmospheres, M is the molarity of the solution, R is the gas law constant, and T is the Kelvin temperature.
56. 56. Practice Problem 11To determine the molar mass of a certain protein, 1.00x 10-3 g of it was dissolved in enough water to make1.00 mL of solution. The osmotic pressure of thissolution was found to be 1.12 torr at 25.0°C. Calculatethe molar mass of the protein. 1.66 x 104 g/mol (proteins often have large molar masses)
57. 57. OsmosisOsmosis prevents transfer of all solute particles. Dialysis(a similar phenomenon) occurs at the walls of most plantand animal cells. In this case, the membrane allowstransfer of both solvent and small solute molecules andions.Solutions that have identical osmotic pressures are said tobe isotonic solutions.• Solutions having a higher osmotic pressure are hypertonic and can cause crenation.• Solutions having a lower osmotic pressure are hypotonic and can cause hemolysis.
58. 58. Crenation vs. Hemolysis
59. 59. Practice Problem 12What concentration of sodium chloride and water isneeded to produce an aqueous solution isotonic withblood (Π = 7.70 atm at 25°C)? .158 M
60. 60. Reverse OsmosisWhen a solution in contact with pure solvent across asemipermeable membrane is subjected to an externalpressure larger than its osmotic pressure, reverseosmosis occurs. The pressure will cause a net flow ofsolvent from the solution. This process can act as amolecular filter for many solutions.• This process is being used to produce fresh water from sea water by desalination.
61. 61. COLLIGATIVE PROPERTIES OFELECTROLYTE SOLUTIONS
62. 62. Colligative PropertiesColligative properties of solutions depend on the totalconcentration of solute particles. The relationshipbetween the moles of solute dissolved and the moles ofparticles in solution is usually expressed using the van’tHoff factor: moles of particles in solution i= moles of solute dissolved• The expected value for i can be calculated for a salt by noting the number of ions per formula unit (NaCl =2)• These calculated values assume that when a salt dissolves, it completely dissociates into its component ions.
63. 63. Colligative PropertiesNot all ions dissociate completely and act independentlyin solution; ion pairing occurs in solution and some pairswill act as a single particle.• Ion pairing is most important in concentrated solutions.• Less ion pairing occurs in more dilute solutions.• The deviation of i is greatest where the ions have multiple charges.
64. 64. Colligative Properties
65. 65. Colligative PropertiesTo adjust freezing point depression, boiling pointelevation and osmotic pressure calculations, i can beused to better account for ion pairing. ΔT = imK• K is the freezing or boiling point constant Π = iMRT
66. 66. Practice Problem 13The observed osmotic pressure for a 0.10 M solution ofFe(NH4)2(SO4)2 at 25°C is 10.8 atm. Compare theexpected and experimental values for i. i=4.4; The experimental value for i is less than the expected value.
67. 67. COLLIODS
68. 68. Colloids
69. 69. ColliodsThe Tyndall effect is when suspended particles scatterlight. This is used to distinguish between a suspensionand true solution.A suspension of tiny particles in some medium is calleda colloidial dispersion or colloid.• Suspended particles are large molecules or ions from 1-1000nm.• Suspended particles remain suspended due to electrostatic repulsion
70. 70. ColloidsColloids are classified according to the states of thedispersed phase and dipersing medium
71. 71. ColloidsColloids can be destroyed by coagulation; usually byheating or by adding an electrolyte.• Heating increases velocities of particles and allows them to collide with enough energy that particles aggregate.• Adding an electrolyte neutralizes the ion layers and particles precipitate out.
72. 72. The End