Lecture Slides on Properties of Solutions - General Chemistry | CHEM 162, Study notes of Chemistry

Download Lecture Slides on Properties of Solutions - General Chemistry | CHEM 162 and more Study notes Chemistry in PDF only on Docsity! Chapter 17: Properties of Solutions 17.1 Solution Composition 17.2 The Thermodynamics of Solution Formation 17.3 Factors Affecting Solubility 17.4 The Vapor Pressure of Solutions 17.5 Boiling-Point Elevation and Freezing-Point Depression 17.6 Osmotic Pressure 17.7 Colligative Properties of Electrolytic Solutions 17.8 Colloids Figure 17.11: Vapor pressure for a solution of two volatile liquids. H O H H-C-C-C-H H H - - - --- H-O- H Acetone + Water H H H H H H H-C-C-C-C-C-C-H H H H H H H Hexane H H + H-C-C-O-H Ethanol H H - ----- - - - - - - -- - Figure 17.12: Phase diagrams for pure water (red lines) and for an aqueous solution containing a nonvolatile solution (blue lines). Like Example 17.3 (P847) What mass of ethanol (C2H6O) must be added to 20.0 liters of water to keep it from freezing at a temperature of -15.0oF? Solution: oC = (oF – 32)5/9 = (-15 – 32)5/9 = -26.1oC T = kf msolute msolute = = = 14.0 mol/kg 14.0 mol/kg(20 kg H2O) = 280 mol ethanol Ethanol = 2x12.01 + 6x 1.008 + 1x16.0 = 46.07 280 mol ethanol (46.07 g ethanol/mol) = 12.9 kg ethanol T kf -26.1oC 1.86 oC kg mol Determining the Boiling Point Elevation and Freezing Point Depression of an Aqueous Solution Problem: We add 475g of sucrose (sugar) to 600g of water. What will be the Freezing Point and Boiling Points of the resultant solution? Plan: We find the molality of the sucrose solution by calculating the moles of sucrose and dividing by the mass of water in kg. We then apply the equations for FP depression and BP elevation using the constants from table 12.4. Solution: Sucrose C12H22O11 has a molar mass = 342.30 g/mol 475g sucrose 342.30gsucrose/mol = 1.388 mole sucrose molality = = 2.313 m1.388 mole sucrose 0.600 kg H2O Tb = Kb x m = (2.313 m)= 1.180C BP = 100.000C + 1.180C BP = 101.180C 0.5120C m Tf = Kf x m = (2.313 m) = 4.300C FP = 0.000C - 4.300C = -4.300C 1.860C m Determining the Boiling Point Elevation and Freezing Point Depression of a Non-Aqueous Solution Problem: Calculate the effect on the Boiling Point and Freezing Point of a chloroform solution if to 500.00g of chloroform (CHCl3) 257g of napthalene (C10H8, mothballs) is dissolved. Plan: We must first calculate the molality of the cholorform solution by calculating moles of each material, then we can apply the FP and BP change equations and the contants for chloroform. Solution: napthalene = 128.16g/mol chloroform = 119.37g/mol molesnap = =2.0053 mol nap 257g nap 128.16g/mol molarity = = = 4.01 mmoles nap kg(CHCl3) 2.0053 mol 0.500 kg Tb = Kb m = (4.01m) = 14.560C normal BP = 61.70C new BP = 76.30C 3.630C m Tf = Kf m = (4.01m) =18.850C normal FP = - 63.50C new FP = - 82.40C 4.700C m Figure 17.15: The normal flow of solvent into the solution (osmosis) can be prevented by applying an external pressure to the solution. Figure 17.16: A pure solvent and its solution (containing a nonvolatile solute) are separated by a semipermeable membrane through which solvent molecules (blue) can pass but solute molecules (green) cannot. Osmotic pressure calculation Calculate the osmotic pressure generated by a sugar solution made up of 5.00 lbs of sucrose per 5.00 pints of water. Solution: 5.00 lbs ( ) = 2.27 kg Molar mass of sucrose = 342.3 g/mol 2,270g 5.00 pints H2O ( )( ) = 2.36 liters Π = MRT = ( )(0.08206 )(298 K) = 68.7 atm 1 kg 2.205 lbs 342.3g/mol = 6.63 mol sucrose 1.00 gallon 8 pints 3.7854 L 1.00 gallon 6.63 mol 2.36 L L atm mol K Figure 17.18: Reverse osmosis Reverse Osmosis for Removal of Ions Colligative Properties of Volatile Nonelectrolyte Solutions From Raoult’s law, we know that: Psolvent = Xsolvent x P0solvent and Psolute = Xsolute x P0solute Let us look at a solution made up of equal molar quantities of acetone and chloroform. Xacetone = XCHCl3 = 0.500, at 350C the vapor pressure of pure acetone = 345 torr, and pure chloroform = 293 torr. What is vapor pressure of the solution, and the vapor pressure of each component. What are the mole fractions of each component? Pacetone = Xacetone x P0acetone = 0.500 x 345 torr = 172.5 torr PCHCl3 = XCHCl3 x P0CHCl3 = 0.500 x 293 torr = 146.5 torr From Dalton’s law of partial pressures we know that XA = PA PTotal Xacetone = = = 0.541 Pacetone PTotal 172.5 torr 172.5 + 146.5 torr XCHCl3 = = = 0.459 PCHCl3 PTotal 146.5 torr 172.5 + 146.5 torr Total Pressure = 319.0 torr Colligative Properties I ) Vapor Pressure Lowering - Raoult’s Law II ) Boiling Point Elvation III ) Freezing Point Depression IV ) Osmotic Pressure Colligative Properties of Ionic Solutions For ionic solutions we must take into account the number of ions present! i = van’t Hoff factor = “ionic strength”, or the number of ions present For vapor pressure lowering: P = i XsoluteP 0solvent For boiling point elevation: Tb = i Kb m For freezing point depression: Tf = i Kf m For osmotic pressure: π = i MRT Figure 17.21: In a aqueous solution a few ions aggregate, forming ion pairs that behave as a unit.