11 - Properties of Solutions

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Unformatted text preview: Properties of Solutions Chapter 11 Zumdahl Solutions Gas Gas in gas Gas in liquid Gas in solid Liquid in liquid Liquid in solid Solid in liquid Solid in solid Liquid Solid Solution Composition Mass Percent: discussed in Ch. 4 Mass % = mass of solute x 100% mass of solution Molarity: moles of solute per liter of solution Normality: equivalents per liter of solution Molality: moles of solute per kg of solvent Mole Fraction: moles of component per mole of solution Molarity Number of moles of solute per liter of solution. Units: mol/L moles of solute molarity = liters of solution mol M = V Molarity Example What is the molarity of a solution made by dissolving 2.355 g of sulfuric acid (H2SO4) in water and diluting to a final volume of 50.0 mL? Molar mass of H2SO4 is 98.1 g/mol. Molarity Example What is the molarity of a solution made by dissolving 2.355 g of sulfuric acid (H2SO4) in water and diluting to a final volume of 50.0 mL? Molar mass of H2SO4 is 98.1 g/mol. Know mass of H2SO4 = 2.355g Volume of solution = 50.0 mL Molarity Example Know mass of H2SO4 Molar mass of H2SO4 = 2.355g is 98.1 g/mol. Volume of solution = 50.0 mL Moles H 2SO 4 Molarity = Liters of solution Molarity Example Know mass of H2SO4 Molar mass of H2SO4 = 2.355g is 98.1 g/mol. = 0.0240 mol H 2SO 4 Volume of solution = 50.0 mL mol H 2SO 4 1 mol of H 2SO 4 = (2.355 g H 2SO 4 ) 98.1 g H 2SO 4 Molarity Example Know mass of H2SO4 Molar mass of H2SO4 = 2.355g is 98.1 g/mol. = 0.0240 mol H 2SO 4 Volume of solution = 50.0 mL mol H 2SO 4 1 mol of H 2SO 4 = (2.355 g H 2SO 4 ) 98.1 g H 2SO 4 1L Liter of solution = (50.0 mL) = 0.0500 L 1000 mL Molarity Example mol H 2SO 4 1 mol of H 2SO 4 = (2.355 g H 2SO 4 ) 98.1 g H 2SO 4 = 0.0240 mol H 2SO 4 1L Liter of solution = (50.0 mL) = 0.0500 L 1000 mL 0.0240 mol H 2SO 4 Molarity = = 0.480 M 0.0500 L Normality (N) # of equivalents per liter of solution Acid or base HCl H+ + Cl H2SO4 2 H+ + SO42 There are two equivalents of H per molecule of H2SO4 MnO4- + 5e- + 8H+ Mn2+ + 4H2O KMnO4 1M 5N Redox Molality # of mol of solute molality = # of kg of solvent mol m= kg solvent Molality Example mol m= kg solvent Calculate the molality of 25.0 grams of KBr dissolved in 750.0 mL pure water. Molality Example mol m= kg solvent Calculate the molality of 25.0 grams of KBr dissolved in 750.0 mL pure water. KBr mol = ( 25.0 g KBr ) 0.210 mol KBr 119.0 g Molality Example mol m= kg solvent 0.210 mol KBr Calculate the molality of 25.0 grams of KBr dissolved in 750.0 mL pure water. 1 g 1 kg = ( 750.0 mL water ) 0.750 kg water 1 mL 1000 g Molality Example mol m= kg solvent 0.210 mol KBr 0.750 kg water Calculate the molality of 25.0 grams of KBr dissolved in 750.0 mL pure water. 0.210 mol KBr m= = 0.280 m 0.750 kg water Mole Fraction XA = __nA__ nA + nB Calculate the mole fraction of 25.0 grams of KBr dissolved in 750.0 mL pure water. Mole Fraction XA = __nA__ nA + nB Calculate the mole fraction of 25.0 grams of KBr dissolved in 750.0 mL pure water. 25.0 g KBr 0.210 mol KBr 750.0 mL H2O 41.63 mol H2O Mole Fraction XA = __nA__ = nA + nB ___0.210____ 0.210 + 41.63 Calculate the mole fraction of 25.0 grams of KBr dissolved in 750.0 mL pure water. 25.0 g KBr 0.210 mol KBr 750.0 mL H2O 41.63 mol H2O Mole Fraction XA = ___0.210____ 0.210 + 41.63 = 5.02 x 10-3 Calculate the mole fraction of 25.0 grams of KBr dissolved in 750.0 mL pure water. 25.0 g KBr 0.210 mol KBr 750.0 mL H2O 41.63 mol H2O Solution Process Like dissolves like Solvent Solute Solvent Solvent Solute Solute Solutions form when these three kinds of forces are similar. Like Dissolves Like Polar solvent Polar solute Water and ethyl alcohol Ionic solute Water and sodium chloride Nonpolar solvent Nonpolar solute Hexane and oils Pressure Effect on Solubility Henry's Law C = kP Solubility vs. Temperature Solids dissolved in liquids Gases dissolved in liquids Sol. Sol. To To As To , solubility As To , solubility Solubility vs. Temperature for Solids 140 130 KI Solubility (grams of solute/100 g H2O) Solubility Table shows the dependence of solubility on temperature 120 110 100 90 80 70 60 50 40 30 20 10 NaNO3 gases solids KNO3 HCl NH3 NH4Cl KCl NaCl KClO3 SO2 0 10 20 30 40 50 60 70 80 90 100 LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517 Colligative Properties properties of a solution that depends only on the number of dissolved particles, not on their chemical identity Properties Lower Vapor Pressure Boiling Point Elevation Freezing Point Depression Osmosis Vapor Pressure Depression equilibrium between molecules entering and leaving the liquid surface Raoult's Law Psoln = solventP0solvent observed vapor pressure of solution mole fraction of solvent Psoln solvent P0solvent vapor pressure of pure solvent Raoult's Law Vapor pressure of a solution is directly proportional to the mole fraction of solvent Raoult's Law Example Determine the vapor pressure of a solution at 25C that has 45 grams of C6H12O6, glucose, dissolved in 72 grams of H2O. The vapor pressure of pure water at 25C is 23.8 torr. Raoult's Law Example Determine the vapor pressure of a solution at 25C that has 45 grams of C6H12O6, glucose, dissolved in 72 grams of H2O. The vapor pressure of pure water at 25C is 23.8 torr. 45 g glucose(1 mol /180 g glucose) = .25 mol Raoult's Law Example Determine the vapor pressure of a solution at 25C that has 45 grams of C6H12O6, glucose, dissolved in 72 grams of H2O. The vapor pressure of pure water at 25C is 23.8 torr. 72 g H2O(1 mole/18 g H2O) = 4 mol H2O Raoult's Law Example Determine the vapor pressure of a solution at 25C that has 45 grams of C6H12O6, glucose, dissolved in 72 grams of H2O. The vapor pressure of pure water at 25C is 23.8 torr. water = mol water / total moles = 4 / 4.25 = 0.9412 Raoult's Law Example Determine the vapor pressure of a solution at 25C that has 45 grams of C6H12O6, glucose, dissolved in 72 grams of H2O. The vapor pressure of pure water at 25C is 23.8 torr. Psoln = solventP0solvent = (0.9412)(23.8 torr) = 22.4 torr 22 torr Nonideal Solutions Molecules are breaking away more easily than they do in the pure liquids. Intermolecular forces between molecules of A and B are less than they are in the pure liquids. Nonideal Solutions Molecules break away from the mixture less easily than they do from the pure liquids Stronger forces must exist in the mixture than in the original Modified Raoult's Law Ptotal Ptotal = PA + PB = AP0A + BP0B = total vapor pressure of solution A and B = mole fractions of A and B P0A and P0B = vapor pressures of pure A & B Raoult's Law Example #2 In a mixture of 86.0 g C6H6 (P = 93.96 torr) and 90.0 g C2H4Cl2 (P = 224.9 torr), what is the total vapor pressure? = PA + PB = AP0A + BP0B Ptotal Raoult's Law Example #2 In a mixture of 86.0 g C6H6 (P = 93.96 torr) and 90.0 g C2H4Cl2 (P = 224.9 torr), what is the total vapor pressure? mol of C6H6: 86.0 g(1 mol/78.11 g) = 1.10 mol mol of C2H4Cl2: 90.0 g(1 mol/98.94 g) = 0.910 mol Raoult's Law Example #2 In a mixture of 86.0 g C6H6 (P = 93.96 torr) and 90.0 g C2H4Cl2 (P = 224.9 torr), what is the total vapor pressure? C6H6 = 1.10/(1.10 + 0.910) = 0.547 C2H4Cl2 = 0.910/(1.10 + 0.910) = 0.453 Raoult's Law Example #2 In a mixture of 86.0 g C6H6 (P = 93.96 torr) and 90.0 g C2H4Cl2 (P = 224.9 torr), what is the total vapor pressure? = C6H6P0C6H6 = (0.547)(93.96 torr) = 51.2 torr = C2H4Cl2P0C2H4Cl2 = (0.453)(224.9) = 102 torr PC6H6 PC2H4Cl2 Raoult's Law Example #2 In a mixture of 86.0 g C6H6 (P = 93.96 torr) and 90.0 g C2H4Cl2 (P = 224.9 torr), what is the total vapor pressure? PC6H6 + PC2H4Cl2 Ptotal = = 51.2 + 102 = 153 torr Properties of Solutions Water Boiling point Freezing point 100.0C 0.0C 1.0 M NaCl Boiling point Freezing point 101.0C -3.7C Boiling Point Elevation T = K bm Freezing Point Depression Antifreeze Road Salt Freezing Point Depression Tf = Kfm Molar Mass Determination When 2.04 g of a substance was dissolved in 7.84 g of solvent, the solution froze at 10.8C. The pure solvent was found to freeze at 14.3C. The freezing-point depression constant for the solvent is 2.8C kg/mol. Calculate the molar mass of the substance. Molar Mass Determination When 2.04 g of a substance was dissolved in 7.84 g of solvent, the solution froze at 10.8C. The pure solvent was found to freeze at 14.3C. The freezing-point depression constant for the solvent is 2.8C kg/mol. Tf = Kfm m = Tf/Kf Molar Mass Determination When 2.04 g of a substance was dissolved in 7.84 g of solvent, the solution froze at 10.8C. The pure solvent was found to freeze at 14.3C. The freezing-point depression constant for the solvent is 2.8C kg/mol. m = Tf/Kf = (14.3-10.8C) = 1.25 mol/kg 2.8C kg/mol Molar Mass Determination When 2.04 g of a substance was dissolved in 7.84 g of solvent, the solution froze at 10.8C. The pure solvent was found to freeze at 14.3C. The freezing-point depression constant for the solvent is 2.8C kg/mol. m= mol substance = 1.25 mol/kg 7.84 g(1 kg/1000g) Calculate total moles of solute Molar Mass Determination When 2.04 g of a substance was dissolved in 7.84 g of solvent, the solution froze at 10.8C. The pure solvent was found to freeze at 14.3C. The freezing-point depression constant for the solvent is 2.8C kg/mol. m = mol substance = 1.25 mol/kg 7.84 g(1 kg/1000g) mol = 9.80 x 10-3 mol Molar Mass Determination When 2.04 g of a substance was dissolved in 7.84 g of solvent, the solution froze at 10.8C. The pure solvent was found to freeze at 14.3C. The freezing-point depression constant for the solvent is 2.8C kg/mol. 2.04 g 9.80 x 10-3 mol = 208 g/mol Osmosis Semipermeable Osmosis Osmosis Osmotic Pressure = MRT = osmotic pressure in atm M = molarity of the solution R = gas constant T = temperature in Kelvin Osmotic Pressure = MRT What is the molar mass of a nonelectrolyte if 1.50 g in 250.0 mL of solution has an osmotic pressure of 0.521 atm at 15 C? Osmotic Pressure = MRT What is the molar mass of a nonelectrolyte if 1.50 g in 250.0 mL of solution has an osmotic pressure of 0.521 atm at 15 C? molar mass = g/mol = MRT = 0.521 atm R = 0.0821 Latm/molK T = 15 C + 273.15 = 288 K g = 1.50 g Osmotic Pressure = MRT What is the molar mass of a nonelectrolyte if 1.50 g in 250.0 mL of solution has an osmotic pressure of 0.521 atm at 15 C? = MRT M = /RT M = 0.521 atm/(0.0821 Latm/molK)(288 K) M = 0.0220 M Osmotic Pressure = MRT What is the molar mass of a nonelectrolyte if 1.50 g in 250.0 mL of solution has an osmotic pressure of 0.521 atm at 15 C? M = mol/L mol = (M)(L) mol = (0.0220 M)(0.250 L) mol = 0.00551 mol Osmotic Pressure = MRT What is the molar mass of a nonelectrolyte if 1.50 g in 250.0 mL of solution has an osmotic pressure of 0.521 atm at 15 C? mol = 0.00551 mol Molar mass = g/mol = 1.50 g/ 0.00551 mol MM = 272 g/mol Osmotic Pressure Dialysis Isotonic solutions: identical osmotic pressure Osmotic Pressure Reverse osmosis van't Hoff Factor i = moles of particles in solution moles of solute dissolved = iMRT T = iKm van't Hoff Factor Example A 1.00 m solution of acetic acid (CH3COOH) in benzene has a freezing point depression of 2.6 K. Calculate the value for i. T = i Kf m 2.6 K = (i)(5.10 K/m)(1.00m) i = 0.51 ...
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