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Transcript of Solutions: AP Notes Use Pre-AP Notes for background solution info Colligative Properties

• Solutions: AP NotesUse Pre-AP Notes for background solution infoColligative Properties

Mullis

• Colligative PropertiesDepend on NUMBER of solute particles, not the kind of particles in a solution.FOUR typesBoiling point elevationFreezing point depressionOsmotic pressureVapor pressure lowering

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• Boiling Point Elevation T = Kb miKb is a proportionality constant that is given or retrieved from a table. It is unique for a particular solvent. (Usually C/m) T is the change in temp, final initial (Usually C)i is the vant Hoff factor, a measure of ionization or dissociation. The simplest assumption for i is to assume very dilute solutions and assume that the ionic compound completely dissociates.

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• Freezing Point Depression T = Kf miKf is a proportionality constant that is given or retrieved from a table. It is unique for a particular solvent. (Usually C/m)

T is the change in temp, final initial (Usually C)i is the vant Hoff factor, a measure of ionization or dissociation. The simplest assumption for i is to assume very dilute solutions and assume that the ionic compound completely dissociates.

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• When i is not ideal

For solutions which are not dilute, i is the ratio of the actual colligative property to the value that would be observed if no dissociation occurred.

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• Calculating i for weak electrolytesi = Tf (actual) = Kfmeffective = meffective Tf (if nonelectrolyte) Kfmstated mstated

A problem that asks for the calculation of i will provide an actual freezing or boiling point at the given concentration. First find T assuming that the solvent is pure water.

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• Example: Calculating ii = Tf (actual) = Kfmeffective = meffective Tf (if nonelectrolyte) Kfmstated mstatedThe freezing point of a 0.100m Na2SO4 (aq) solution is -0.0349C. Calculate i.Use the ratio above.meffective = .0349C = 0.188 m1.86C/mmeffective = 0.0188 m = 1.88mstated0.100 m

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• But I thought i = 3 for Na2SO4!In very dilute solutions, no significant amount of ion association occurs to lessen the effect of all three available ions being separated from one another. This is when i is assumed to be 3. Compare the following example to the first one (0.100m Na2SO4 (aq)).The freezing point of a 1.00m Na2SO4 (aq) solution is -3.29C. Calculate i.meffective = 3.29C = 1.77 m 1.86C/mmeffective = 0.0188 m = 1.77 Value of i is smaller since mstated 1.00 m concentration increased.

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• Osmotic Pressurehttp://www.wasanlab.ubc.ca/assets/flash/osmosis.swfClick the above link for osmosis simulation.Sugar solution is placed in the bulb of the thistle tube and a semipermeable membrane is placed over the end of this tube. The tube is then submerged in water.The sugar solution becomes diluted as water moves into the tube, pushing the solution level up until the pressure in the tube is just sufficient to prevent solvent flow from the pure solvent side of the membrane to the solution side.This pressure is called osmotic pressure.

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• Usefulness of Osmotic PressureThis technique is a relatively simple way to determine the molecular weight using very little solute. Even a dilute solution may be used, so the method is advantageous when:Only a small amount of solute or solution is available.The solute is expensiveThe solute does not dissolve well in water

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• Calculating Osmotic pressure = nRT =MRTn= V V RT

n = moles = M (M is molarity)V L

In a problem solving for molecular weight, solve for n (moles), then divide a given mass by n to get g/mol. This type of problem gives the mass, volume, temperature and osmotic pressure.

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• Example- Molecular weight from osmotic pressureA sample of 2.05 g pf polystyrene was dissolved in enough toluene to form 0.100 L of solution. The osmotic pressure of the solution was found to be 1.21 kPa at 25 C. Find molar mass of polystyrene.1.21 kPa |1 atm = 0.0119 atm 101.325 kPan =V=0.0119 atm(0.100 L) = 4.84 x 10-5 mol RT (0.0821 L-atm/mol-K)(298K)Molar mass = 2.05 g = 4.24 104 g__ 4.84 x 10-5 mol mol

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• Vapor Pressure: Raoults LawVapor = gas formed by the boiling or evaporation of a liquid or a solidVapor pressure of a liquid is the pressure exerted by the vapor in equilibrium with its liquid.A solution containing a nonvolatile solute has a lower vapor pressure than the pure solvent.

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• Raoults LawPsolvent = XsolventPsolventPsolvent: Vapor pressure of solvent in the solutionXsolvent: Mole fraction of solvent in solutionPsolvent: Vapor pressure of the pure solvent As the mole fraction (%solvent) goes up, its vapor pressure also goes up proportionally.If the solute is nonvolatile, then Psolvent = PsolutionLowering of vapor pressure is defined:Psolvent=Psolvent Psolvent OR Psolvent = XsolutePsolvent

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• Example 1: Raoults LawAt 25 C, determine the vapor pressure lowering of 1.25 m sucrose solution that has been made with 50.0 g C12H22O11 and 117 g H2O. The vapor pressure of pure water is 23.8 torr at 25 C.117 g H2O = 6.50 mol 50.0 g C12H22O11 = 1.46 mol Psolvent = XsolutePsolventsucrose = 1.46 mole/(1.46 + 6.50)moles = 0.0220 Psolvent =0.0220(23.8 torr) = 0.524 torr

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• Example 2: Raoults LawAt 40 C, the v.p. of pure hexane is 92.0 torr and the v.p. of pure octane is 31.0 torr. In a solution containing 1.00 mole heptane and 4.00 moles octane, calculate the vapor pressure of each component and the v.p. above the solution.Psolvent = XsolventPsolventheptane = 1.00 mole/(1.00 + 4.00)moles = 0.200octane = 1- heptane = 0.800Poctane=XsolventPsolvent =0.800(31.0 torr)=24.8 torrPheptane=XsolutePsolute=0.200(92.0 torr)=18.4 torrPtotal = 18.4 + 24.8 torr = 43.2 torr

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• Henrys LawSolubility of gas in liquidThe solubility of a gas is directly proportional to the partial pressure of that gas on the surface of the liquid.Soda bottle: High pressure at the surface while the bottle is closed, so lots of CO2 in the liquidOpen bottle, pressure on surface lowers to room atmosphere and CO2 leaves the liquidHigh pressure = High gas concentrationLow pressure = low gas concentration

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