Solutions: AP NotesUse Pre-AP Notes for

background solution info

Colligative Properties

Mullis 2

Colligative PropertiesDepend on NUMBER of solute

particles, not the kind of particles in a solution.

FOUR types1. Boiling point elevation2. Freezing point depression3. Osmotic pressure4. Vapor pressure lowering

Mullis 3

Boiling Point Elevation Δ T = Kb mi

• Kb 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.

Mullis 4

Freezing Point Depression Δ T = Kf mi

• Kf 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.

Mullis 5

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

KCl

-+ +

- ++

Cl

-++

-++

-+

+

-+

+

-++

-++

K

-+ +

- ++

Cl

-++

-++

-+

+

-++

Mullis 6

Calculating i for weak electrolytes

i = ΔTf (actual) = Kfmeffective = meffective

ΔTf (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 Δ T assuming that the solvent is pure water.

Mullis 7

Example: Calculating i

i = ΔTf (actual) = Kfmeffective = meffective

ΔTf (if nonelectrolyte) Kfmstated mstated

The freezing point of a 0.100m Na2SO4 (aq) solution is -0.0349°C. Calculate i.

Use the ratio above.

meffective = .0349°C = 0.188 m1.86°C/m

meffective = 0.0188 m = 1.88

mstated 0.100 m

Mullis 8

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 -3.29°C. Calculate i.

meffective = 3.29°C = 1.77 m 1.86°C/m

meffective = 0.0188 m = 1.77 Value of i is smaller since mstated 1.00 m concentration increased.

Mullis 9

Osmotic Pressure• http://www.wasanlab.ubc.ca/assets/flash/osmosis.swf

• 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.

Mullis 10

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 expensive3. The solute does not dissolve well in

water

Mullis 11

Calculating Osmotic pressure

Π= nRT Π =MRT n= 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.

Mullis 12

Example- Molecular weight from osmotic pressure

A 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 kPa

n =ΠV=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

Mullis 13

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.

Mullis 14

Raoult’s LawPsolvent = XsolventP°solvent

• Psolvent: Vapor pressure of solvent in the solution

• Xsolvent: 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 Psolvent = Psolution

Lowering of vapor pressure is defined:ΔPsolvent=P°solvent– Psolvent OR Δ Psolvent = XsoluteP°solvent

Mullis 15

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 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

Δ Psolvent = XsoluteP°solvent

χsucrose = 1.46 mole/(1.46 + 6.50)moles = 0.0220

Δ Psolvent =0.0220(23.8 torr) = 0.524 torr

Mullis 16

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.

Psolvent = XsolventP°solvent

χheptane = 1.00 mole/(1.00 + 4.00)moles = 0.200

χoctane = 1- χheptane = 0.800

Poctane=XsolventP°solvent =0.800(31.0 torr)=24.8 torr

Pheptane=XsoluteP°solute=0.200(92.0 torr)=18.4 torr

Ptotal = 18.4 + 24.8 torr = 43.2 torr

Mullis 17

Henry’s LawSolubility 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 CO2 in the liquid– Open bottle, pressure on surface lowers to

room atmosphere and CO2 leaves the liquid

• High pressure = High gas concentration• Low pressure = low gas concentration