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

Mullis2 Colligative Properties Depend on NUMBER of solute particles, not the kind of particles in a solution. FOUR types 1.Boiling point elevation 2.Freezing point depression 3.Osmotic pressure 4.Vapor pressure lowering

Mullis3 Boiling Point Elevation Δ T = K b mi K b is a proportionality constant that is given or retrieved from a table. It is unique for a particular solvent. (Usually °C/m) Δ T is the change in temp, final – initial (Usually °C) i is the van’t 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.

Mullis4 Freezing Point Depression Δ T = K f mi K f is a proportionality constant that is given or retrieved from a table. It is unique for a particular solvent. (Usually °C/m) Δ T is the change in temp, final – initial (Usually °C) i is the van’t 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.

Mullis5 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 K K Cl K

Mullis6 Calculating i for weak electrolytes i = ΔT f (actual) = K f m effective = m effective ΔT f (if nonelectrolyte) K f m stated m stated A problem that asks for the calculation of i will provide an actual freezing or boiling point at the given concentration. First find Δ T assuming that the solvent is pure water.

Mullis7 Example: Calculating i i = ΔT f (actual) = K f m effective = m effective ΔT f (if nonelectrolyte) K f m stated m stated The freezing point of a 0.100m Na 2 SO 4 (aq) solution is °C. Calculate i. Use the ratio above. m effective =.0349°C = m 1.86°C/m m effective = m = 1.88 m stated m

Mullis8 But I thought i = 3 for Na 2 SO 4 ! 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 Na 2 SO 4 (aq)). The freezing point of a 1.00m Na 2 SO 4 (aq) solution is -3.29°C. Calculate i. m effective = 3.29°C = 1.77 m 1.86°C/m m effective = m = 1.77 Value of i is smaller since m stated 1.00 m concentration increased.

Mullis9 Osmotic Pressure Click 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.

Mullis10 Usefulness of Osmotic Pressure This 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: 1.Only a small amount of solute or solution is available. 2.The solute is expensive 3.The solute does not dissolve well in water

Mullis11 Calculating Osmotic pressure Π = nRT Π =MRTn= ΠV 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.

Mullis12 Example- Molecular weight from osmotic pressure A sample of 2.05 g pf polystyrene was dissolved in enough toluene to form L of solution. The osmotic pressure of the solution was found to be 1.21 kPa at 25 ° C. Find molar mass of polystyrene kPa |1 atm = atm kPa n = ΠV = atm(0.100 L) = 4.84 x mol RT ( L-atm/mol-K )(298K) Molar mass = 2.05 g = g__ 4.84 x mol mol

Mullis13 Vapor Pressure: Raoult’s Law Vapor = gas formed by the boiling or evaporation of a liquid or a solid Vapor 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.

Mullis14 Raoult’s Law P solvent = X solvent P° solvent P solvent : Vapor pressure of solvent in the solution X solvent : Mole fraction of solvent in solution P° solvent : 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 P solvent = P solution Lowering of vapor pressure is defined: ΔP solvent =P° solvent – P solvent OR Δ P solvent = X solute P° solvent

Mullis15 Example 1: Raoult’s Law At 25° C, determine the vapor pressure lowering of 1.25 m sucrose solution that has been made with 50.0 g C 12 H 22 O 11 and 117 g H 2 O. The vapor pressure of pure water is 23.8 torr at 25 ° C. 117 g H 2 O = 6.50 mol 50.0 g C 12 H 22 O 11 = 1.46 mol Δ P solvent = X solute P° solvent χ sucrose = 1.46 mole/( )moles = Δ P solvent =0.0220(23.8 torr) = torr

Mullis16 Example 2: Raoult’s Law At 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. P solvent = X solvent P° solvent χ heptane = 1.00 mole/( )moles = χ octane = 1- χ heptane = P octane =X solvent P° solvent =0.800(31.0 torr)=24.8 torr P heptane =X solute P° solute =0.200(92.0 torr)=18.4 torr P total = torr = 43.2 torr

Mullis17 Henry’s Law Solubility of gas in liquid The 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 CO 2 in the liquid –Open bottle, pressure on surface lowers to room atmosphere and CO 2 leaves the liquid High pressure = High gas concentration Low pressure = low gas concentration