Properties of SolutionsProperties of Solutions

Chapter 11Chapter 11

TopicsTopics

Concentration of solutions using different Concentration of solutions using different termsterms

Solution process and solubilitySolution process and solubility Factors affecting the solubilityFactors affecting the solubility Vapor pressure of solutionVapor pressure of solution Collegative properties of solutionCollegative properties of solution

Boiling point elevationBoiling point elevation Freezing point depressionFreezing point depression Osmotic pressureOsmotic pressure

Section 11-8: Colloids-Self studySection 11-8: Colloids-Self study

A solution is a homogenous mixture of 2 or more substances

The solute is(are) the substance(s) present in the smaller amount(s)

The solvent is the substance present in the larger amount

11.1 Solution composition

Various types of solutions

Concentration UnitsThe concentration of a solution is the amount of solute present in a given quantity of solvent or solvent or solution.solution.

Percent by Mass

% by mass = x 100%mass of solutemass of solute + mass of solvent

= x 100%mass of solutemass of solution

Mole Fraction (X)

XA = moles of A

sum of moles of all components

BA

AA nn

nAoffractionMole

M =

Molarity (M)

Molality (m)

m =moles of solute

mass of solvent (kg)

moles of solute

volume of solution (liters)

ExampleExample

1.00 g C1.00 g C22HH55OH is added to 100.0 g of water to OH is added to 100.0 g of water to

make 101 mL of solution. Find the molarity, make 101 mL of solution. Find the molarity, mass %, mole fraction and molality of mass %, mole fraction and molality of ethanol.ethanol.

MolarityMolarity

Number of moles of solute per L orNumber of moles of solute per L or

solutionsolution

solution of liters

solute of molesM

M 0.215L 0.101

mol 102.17

mL 1000L 1

mL 101

g 46.07mol 1

OHHC g 1.00M

252

Mass PercentMass Percent

Also called weight percentAlso called weight percent Percent by mass of the solute in the solutionPercent by mass of the solute in the solution

100solution of mass

solute of mass% Mass

% 0.990100solution g) 100.0g (1.00

OHHC g 1.00% Mass 52

Mole FractionMole Fraction

ratio of number of moles of a part ofratio of number of moles of a part of

solution to total number of moles of solutionsolution to total number of moles of solution

BA

AA nn

n

00389.0

1017.202.18

10.100

1017.2

22

522

52

molg

molOgH

OHHmolCOHHC

MolalityMolality

Number of moles of solute per kg of solventNumber of moles of solute per kg of solvent

solvent of kg

solute of molesm

0.217m

solution1000g

1kg100.0g

OHHmolC102.17m 52

2

ExampleExample

An aqueous antifreeze solution is 40% An aqueous antifreeze solution is 40% ethylene glycol (Cethylene glycol (C22HH66OO22) by mass. The ) by mass. The

density of the solution is 1.05 g/cmdensity of the solution is 1.05 g/cm33. . Calculate the molality, molarity and Calculate the molality, molarity and mole fraction of ethylene glycolmole fraction of ethylene glycol

mol/kg 07.11000

0.6007.62

mol10.40

2

kg

g

OHggEG

EGEGg

Molality

Lmol

mL

L

solutiong

mLsolutiong

EGg

EGmolEGg

Molarity /77.6

1000

1

05.1

10.100

07.62

10.40

162.0644.033.3

644.0EG

where EG = ethylene glycol (C2H6O2)

# mol water = 60.0/18.0 = 3.33 mol

# mol EG = 0.644 mol EG

Mass of water (solvent) = 100-40 = 60.0 g

What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL?

m =moles of solute

mass of solvent (kg)M =

moles of solute

liters of solution

Assume 1 L of solution:Mass of solute = mass of 5.86 moles ethanol = 270 g ethanolMass of solution= mass of 1 L solution= (1000 mL x 0.927 g/mL) = 927 g of solution

mass of solvent = mass of solution – mass of solute

= 927 g – 270 g = 657 g = 0.657 kg

m =moles of solute

mass of solvent (kg)=

5.86 moles C2H5OH

0.657 kg solvent= 8.92 m

“like dissolves like”

Two substances with similar intermolecular forces are likely to be soluble in each other.

• Non-polar molecules are soluble in non-polar solvents CCl4 in C6H6

• Polar molecules are soluble in polar solvents C2H5OH in H2O

• Ionic compounds are more soluble in polar solvents NaCl in H2O or NH3 (l)

11.2 Energies of Solution FormationDissolution of a solute in liquids

• Any explanation for this behavior?

Solubility ProcessSolubility Process

The formation of aThe formation of aliquid solution takesliquid solution takesplace in 3 stepsplace in 3 steps

1. Expand solute1. Expand solutemoleculesmolecules

2. Expand solvent2. Expand solventmoleculesmolecules

3. Mixing solute and3. Mixing solute andsolventsolvent

Endothermic

Endothermic

Often Exothermic

Energy of Solubility ProcessEnergy of Solubility Process

Steps 1 and 2 require energy to overcome Steps 1 and 2 require energy to overcome intermolecular forces:intermolecular forces: endothermicendothermic

Step 3 usually releases energyStep 3 usually releases energy often exothermicoften exothermic

enthalpy of solutionenthalpy of solution sum of ∆H valuessum of ∆H values can be – or +can be – or + ∆∆HHsoln soln = ∆H= ∆H11 + ∆H + ∆H22 + ∆H + ∆H33

Energy of Solubility ProcessEnergy of Solubility Process

soln is negative

Case 1: oil and waterCase 1: oil and water

Oil is nonpolar (London forces)Oil is nonpolar (London forces) Water is polar (H-bonding)Water is polar (H-bonding) ∆∆HH11 will be small for typical molecular size will be small for typical molecular size ∆∆HH22 will be large will be large ∆∆HH33 will be small since there won’t be much will be small since there won’t be much

interaction between the twointeraction between the two ∆∆HHsolnsoln will be will be large and +velarge and +ve because because

energy required by steps 1 and 2 is larger energy required by steps 1 and 2 is larger than the amount released by 3than the amount released by 3

Case 2: NaCl and waterCase 2: NaCl and water

NaCl is ionic NaCl is ionic water is polar (H-bonding)water is polar (H-bonding) ∆∆HH11 will be large will be large

∆∆HH22 will be large will be large

∆∆HH33 will be large and –ve because of the will be large and –ve because of the

strong interaction between ions and waterstrong interaction between ions and water ∆∆HHsolnsoln will be close to zero- small but +ve will be close to zero- small but +ve

Enthalpy of hydrationEnthalpy of hydration - ∆H- ∆Hhydhyd

enthalpy change associated with dispersal enthalpy change associated with dispersal of gaseous solute species in waterof gaseous solute species in water

It combinesIt combines ∆H∆H22 (for expanding solvent) and ∆H (for expanding solvent) and ∆H3 3 (for (for

solvent-solute interaction)solvent-solute interaction)

NaCl(s) NaCl(s) Na Na++(g) + Cl(g) + Cl--(g)(g)

∆∆HH1 1 =786 kJ/mol=786 kJ/mol

HH22O(l) + NaO(l) + Na++(g) + Cl(g) + Cl--(g) (g) Na Na++(aq) + Cl(aq) + Cl--(aq)(aq)

∆∆HHhydhyd=∆H=∆H22 + ∆H + ∆H33=-783 kJ/mol=-783 kJ/mol

∆∆HHsolnsoln= ∆H= ∆H1 1 + ∆H+ ∆Hhydhyd = = 3 kJ/mol3 kJ/mol

Energy

Reactants

Solution

H1

H2

H3

Solvent

Solute and Solvent

Size of Size of HH33 determines whether a solution will form determines whether a solution will form

H3

Solution

∆∆HHsolnsoln is small but NaCl is highly soluble, Whyis small but NaCl is highly soluble, Why??

wo factors explains the solubility: wo factors explains the solubility:

1.1. n increase in the probability of mixing n increase in the probability of mixing favors the processfavors the process

2.2. Processes that require large amounts of Processes that require large amounts of energy tend not to occurenergy tend not to occur

If If HHsolnsoln is small and positive, a solution will still is small and positive, a solution will still

form because of high probability (entropy) for form because of high probability (entropy) for mixingmixing

There are many more ways for them to become There are many more ways for them to become mixed than there is for them to stay separate.mixed than there is for them to stay separate.

Energy of Solubility ProcessEnergy of Solubility Process

11 . .Structure EffectsStructure Effects

Water soluble molecules must have dipole moments Water soluble molecules must have dipole moments -polar bonds.-polar bonds.

To be soluble in non polar solvents the molecules To be soluble in non polar solvents the molecules must be non polar.must be non polar.

11.311.3 Factors Affecting SolubilityFactors Affecting Solubility

• Pentane C5H12 is misciblemiscible with hexane C6H14 and immiscibleimmiscible with water• Solubility of alcohols decreases with the molar mass? Cl3OH CH3(CH2)3OH CH3 (CH2)6OH Soluble Insoluble (Hydrophilic) (Hydrophobic)

Polarity decreases

OH is smaller portionHydrocarbon is larger

Structure effects: Vitamins and the bodyStructure effects: Vitamins and the body

Hydrophobic soluteHydrophobic solute: : water- fearingwater- fearing: : nonpolarnonpolar Insoluble in waterInsoluble in water

Hydrophilic soluteHydrophilic solute: : water-lovingwater-loving:: polarpolar Soluble in waterSoluble in water

Hydrophobic, accumulatesin the body. The body cantolerate a diet deficientin vitamin A

Hydrophilic, excreted bythe body and must be consumed regularly. The body cannot tolerate a diet deficientin vitamin C

22 . .Pressure effectsPressure effects

Changing the pressure Changing the pressure doesn’t effectdoesn’t effect the amount the amount of solid or liquid that dissolves; of solid or liquid that dissolves; they are incompressible.

Changing the pressure Changing the pressure effectseffects gases. gases. Pressure effects the amount of gas that can Pressure effects the amount of gas that can

dissolve in a liquid.dissolve in a liquid. The dissolved gas is at equilibrium with the gas The dissolved gas is at equilibrium with the gas

above the liquid.above the liquid.

The gas molecules above the The gas molecules above the liquid are at equilibrium with liquid are at equilibrium with the gas molecules dissolved the gas molecules dissolved in this solution.in this solution.

The equilibrium is dynamicThe equilibrium is dynamic..

If the pressure is increased If the pressure is increased the gas molecules dissolve the gas molecules dissolve faster.faster.

The equilibrium is disturbed.The equilibrium is disturbed.

The system reaches a new The system reaches a new equilibrium with more gas equilibrium with more gas dissolved.dissolved.

Henry’s Law.Henry’s Law. C= kPC= kPConcentration = Pressure X constantConcentration = Pressure X constant

Henry’s LawHenry’s Law

C C P P C = kPC = kP C is concentration and P is partial pressure of C is concentration and P is partial pressure of

gaseous solute gaseous solute The law is obeyed best by The law is obeyed best by dilute solutionsdilute solutions of of

gases that don’t dissociate or react with solventgases that don’t dissociate or react with solvent

•Amount of gas dissolved is directly Amount of gas dissolved is directly proportional to P of gas above solutionproportional to P of gas above solution

2

1

2

1

P

P

C

C

ExampleExample

AA soft drink bottled at 25°C contains soft drink bottled at 25°C contains COCO22 at pressure of 5.0 atm over liquid. at pressure of 5.0 atm over liquid.

Assume that PAssume that PCO2CO2 in atmosphere is 4.0 x in atmosphere is 4.0 x

1010-4-4 atm. Find the equilibrium atm. Find the equilibrium concentration in soda before and after concentration in soda before and after opening. k=3.1 X 10opening. k=3.1 X 10-2-2 mol/L.atm at 25°C mol/L.atm at 25°C

ExampleExample

before opening:before opening:

after opening:after opening:

LmolXCPkCCOCOCO

/16.0)atm 5.0mol/L.atm(101.3 2

222

LmolatmxPkCCOCOCO

/102.1)104(mol/L.atm) (3.1x10 54-2

222

ExampleExample

Solubility of pure NSolubility of pure N22 in blood at body temp, in blood at body temp,

3737ooC and 1 atm is 6.2X10C and 1 atm is 6.2X10-4-4 M. If a diver M. If a diver breaths air ( = 0.78) at a depth where the breaths air ( = 0.78) at a depth where the total pressure is 2.5 atm, calculate the total pressure is 2.5 atm, calculate the concentration of Nconcentration of N22 in his body. in his body.

2N

.......22

ONtotalPPP

22 NtotNPP

atmatmPN

0.278.05.22

2

1

2

1

P

P

C

C

atm

atm

C

M

0.2

0.1102.6

2

4

M10 02.1blood) sdiver' in the N concof( -3

22C

33 . .Temperature Effects (for aqueous solutions)Temperature Effects (for aqueous solutions)

Solids dissolve more rapidly at higher Solids dissolve more rapidly at higher temperaturestemperatures

But it is not possible to predict whether the But it is not possible to predict whether the solubility (amount dissolved) increases or solubility (amount dissolved) increases or decreasesdecreases

Solubility of most solids increases with Solubility of most solids increases with temperature but some solids showed a decrease temperature but some solids showed a decrease in solubility with temperaturein solubility with temperature

Temperature effect cannot be predicted; only by Temperature effect cannot be predicted; only by experiment experiment

Temperature effectsTemperature effects

dissolving a solid dissolving a solid occurs faster at occurs faster at higher Thigher T

but the amount to but the amount to be dissolved does be dissolved does not changenot change

Temperature effectsTemperature effects

Solubility of gas in Solubility of gas in water decreases water decreases

with Twith T

Solubility and EnvironmentSolubility and Environment

Thermal pollutionThermal pollution Water used as a coolant when pumped again into Water used as a coolant when pumped again into

the source (lakes and rivers) floats on the cold the source (lakes and rivers) floats on the cold water causing a decrease in solubility of Owater causing a decrease in solubility of O22 and and consequently affecting the aquatic life.consequently affecting the aquatic life.

COCO22 dissolves in water that contains CO dissolves in water that contains CO332-2- causing causing

formation of HCOformation of HCO33-- that is soluble in water. that is soluble in water.

COCO332-2- (aq) + CO (aq) + CO22(aq) 2HCO(aq) 2HCO33

--

When temp increases COWhen temp increases CO22 will be driven off the water will be driven off the water causing precipitation of COcausing precipitation of CO33

2-2- again forming scales again forming scales on the wools that blocks the pipes and reduce the on the wools that blocks the pipes and reduce the heating efficiency heating efficiency

11.411.4 The vapor pressures of The vapor pressures of solutionssolutions

A A nonvolatilenonvolatile solvent lowers the vapor pressure of solvent lowers the vapor pressure of the solution.the solution.

Nonvolatile solute Nonvolatile solute decreasesdecreases # of solvent molecules # of solvent molecules per unit volume. per unit volume.

Thus, # of solute molecules escaping will be loweredThus, # of solute molecules escaping will be lowered. .

Aqueous Solution

Pure water

Water has a higher vapor pressure than a solutionWater has a higher vapor pressure than a solution

Aqueous solution and pure ware in closed environment

P1 Po1

Aqueous Solution

Pure water

Water molecules evaporate faster from pure water than Water molecules evaporate faster from pure water than from the solutionfrom the solution

P1Po

1<

Water molecules condense faster in the solution so it Water molecules condense faster in the solution so it should all end up there.should all end up there.

Aqueous Solution

Pure water

empty

Vapor pressures of solutions containing nonvolatile solvents were studied by Raoult

RaoultRaoult’’s Laws Law

The presence of a nonvolatile solute lowers the vapor pressure of the solvent.

Psolution = Observed Vapor pressure of the solution (vapor pressure of solvent in solution)

P0solvent = Vapor pressure of the pure solvent

solvent = Mole fraction of the solvent

0

solventsolventsolutionPP

• This law applies only to an ideal solution where the solute doesn’t contribute to the vapor pressure, i.e., • solute and solvent are alike: solute-solute, solvent-solvent and solute-solvent interactions are very similar ).

0)1(solventsolutesolution

PP Only one solute

solvento

solutesosolvent

o PPP ln

solvento

solutePVPL

Vapor pressure lowering

Vapor pressure lowering when ionic Vapor pressure lowering when ionic compounds are dissolvedcompounds are dissolved

When NaCl is dissolved in water, the VPL is twice as When NaCl is dissolved in water, the VPL is twice as much as expected.much as expected.

1 mol NaCl dissociates into 1 mol Na1 mol NaCl dissociates into 1 mol Na++ and 1 mol Cl- and 1 mol Cl- # mols of solute = 2 X # mols NaCl# mols of solute = 2 X # mols NaCl Thus, vapor pressure measurement can give Thus, vapor pressure measurement can give

information about the nature of the soluteinformation about the nature of the solute When NaWhen Na22SOSO44 is dissolved VPL is is dissolved VPL is 3 x expected

ExampleExample

Find the vapor pressure at 25°C for Find the vapor pressure at 25°C for solution of 158.0 g of sucrose (Csolution of 158.0 g of sucrose (C1212HH2222OO1111) )

in 643.5 mL of water. in 643.5 mL of water.

The density of water at 25°C is 0.9971 The density of water at 25°C is 0.9971 g/mL and the partial pressure of water g/mL and the partial pressure of water vapor at 25°C is 23.76 torr.vapor at 25°C is 23.76 torr.

ExampleExample

solventsolventso PP ln

9872.0

3.3421

0.15802.18

119971.0

5.643

02.181

19971.0

5.643

gmol

gg

molmL

gmL

gmol

mLg

mL

water

torrtorrPso 46.23)76.23)(9872.0(ln

Liquid-liquid solutions in which both Liquid-liquid solutions in which both components are volatilecomponents are volatile

Modified Raoult's Law:Modified Raoult's Law:

00BBAABATOTAL PPPPP

P0 is the vapor pressure of the pure solvent PA and PB are the partial pressures resulting from A

& B in the vapor above the solution

Nonideal solutions

RaoultRaoult’’s Law s Law –– Ideal Solution Ideal Solution

A solution of two liquids that obeys Raoult’s Law is called an ideal solution

Negative Deviations from RaoultNegative Deviations from Raoult’’s Laws Law

Strong solute-solvent interaction results in a vapor pressure lower than predicted

Exothermic mixing = Negative deviation

Positive Deviations from RaoultPositive Deviations from Raoult’’s Laws Law

Weak solute-solvent interaction results in a vapor pressure higher than predicted

Endothermic mixing = Positive deviation

Colligative Properties of solutionsColligative Properties of solutions

Colligative properties are properties that depend only on the number of solute particles (molecules or ions)(molecules or ions) in solution and not on the nature (identity) of the solute particles.

11.5 Boiling point elevation and freezing Point depression11.5 Boiling point elevation and freezing Point depression

• Colligative properties include: 1. Vapor pressure lowering2. Boiling point elevation and freezing point

depression3. Osmotic pressure

• Each of these properties is a consequence of a decrease in the escaping tendency of solvent molecules brought by the presence of solute particles.

Boiling Point ElevationBoiling Point Elevation

A nonvolatile solute A nonvolatile solute lowers the vaporlowers the vapor pressurepressure

A higher TA higher T is required to reach the 1 atm of is required to reach the 1 atm of pressure which defines boiling pointpressure which defines boiling point

A nonvolatile solute A nonvolatile solute elevates the boilingelevates the boiling point of the solventpoint of the solvent

The amount of the elevation depends on The amount of the elevation depends on concentration of the soluteconcentration of the solute

solutebmKT

Freezing Point DepressionFreezing Point Depression

Vapor pressure of solid and liquid are equal at Vapor pressure of solid and liquid are equal at freezing pointfreezing point

nonvolatile solute lowers the vapor pressure so a nonvolatile solute lowers the vapor pressure so a lower T is needed to decrease the vapor pressure to lower T is needed to decrease the vapor pressure to that of the solidthat of the solid

a nonvolatile solute depresses the freezing point of a nonvolatile solute depresses the freezing point of the solventthe solvent

the amount of the depression depends on the amount of the depression depends on concentration of the soluteconcentration of the solute

solutef mKT

1 atm

Vapor Pressure of solution

Vapor Pressure of pure water

1 atm

Freezing and boiling points of water

1 atm

Freezing and boiling points of solution

1 atm

TfTb

Example 1Example 1

18.00 g of glucose are added to 150.0 g of 18.00 g of glucose are added to 150.0 g of water. The boiling point of the solution is water. The boiling point of the solution is 100.34 C. The boiling point constant is 0.51 100.34 C. The boiling point constant is 0.51 C*kg/mol. C*kg/mol.

Find the molar mass of glucose.Find the molar mass of glucose.

Example 1Example 1

soluteb mKT

soluteif mmol

kgCTTT

)51.0(00.10034.100

kg

mol

molkgC

Cmsolute 67.0

/51.0

34.0

Example 1Example 1

waterkg 0.1500670 glucosen

kg

mol.

molkgkg

moln eglu 10.0)1500.0)(67.0(cos

molgmol

g

moles

mass/180

10.0

00.18massmolar

Example 2Example 2

What mass of CWhat mass of C22HH66OO22 (M=62.1 g/mol) needs (M=62.1 g/mol) needs

to be added to 10.0 L Hto be added to 10.0 L H22O to make a solution O to make a solution

that freezes at -23.3°C? density is 1.00 that freezes at -23.3°C? density is 1.00 g/mL; boiling point constant is 1.86°C*kg/molg/mL; boiling point constant is 1.86°C*kg/mol

solutef mKT

soluteif mmol

kgCTTT

)86.1(0.03.23

kg

mol

molkgC

Cmsolute 5.12

/86.1

3.23

2623

262

262

2

2622 OHC g 107.76

OHC mol 1

OHC g 62.1

OH kg 1

OHC mol 12.5OH kg 10.0

kg 10.0g 1000

kg 1

mL 1.00

g 1

L 1

mL 1000OH L 10.0 2

What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g.

Tf = Kf m

m =moles of solute

mass of solvent (kg)= 2.41 m=

3.202 kg solvent

478 g x 1 mol62.01 g

Kf water = 1.86 0C/m

Tf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C

Tf = T f – Tf0

Tf = T f – Tf0 = 0.00 0C – 4.48 0C = -4.48 0C

Experimental approachExperimental approach

Semipermeable membraneSemipermeable membrane It is a partition with pores that allow small solvent particles to pass It is a partition with pores that allow small solvent particles to pass

through but not bigger solute particles. Thus, it separates a solution through but not bigger solute particles. Thus, it separates a solution and a pure solventand a pure solvent

11.6 Osmoses and Osmotic Pressure11.6 Osmoses and Osmotic Pressure

Initial Final

OsmosisOsmosis

The flow of solvent molecules from The flow of solvent molecules from a pure solvent through a semi-a pure solvent through a semi-permeable membrane into a permeable membrane into a solutionsolution

when the system has reached when the system has reached equilibrium, the water levels are equilibrium, the water levels are differentdifferent

Because the liquid levels are Because the liquid levels are different, there is a greater different, there is a greater hydrostatic pressurehydrostatic pressure on the on the solutionsolution than on than on pure solventpure solvent

After

Before Osmotic pressure,

Osmotic pressure of a solutionOsmotic pressure of a solution: The minimum pressure that stops the osmosis

Osmotic PressureOsmotic Pressure

The minimum pressure that stops the osmosis is equal to the osmotic pressure of the solution

Osmotic pressure, Osmotic pressure, ,, and concentration of and concentration of nonelectrolytesnonelectrolytes

molarity of soultionmolarity of soultion PV = nRT (Ideal gas equation)PV = nRT (Ideal gas equation) Relation between Relation between and M is the and M is the

same: same: V = nRTV = nRT

where where M is the molarity of the solutionM is the molarity of the solution R is the gas law constantR is the gas law constant T is the temperature in KelvinT is the temperature in Kelvin

MRT RTV

n

ExampleExample When 1.00x10When 1.00x10-3-3 g of a protein is mixed with g of a protein is mixed with

water to make 1.00 mL of solution, the water to make 1.00 mL of solution, the osmotic pressure is 1.12 torr at 25.0°C.osmotic pressure is 1.12 torr at 25.0°C.

Find the molar mass of this protein. Find the molar mass of this protein.

TRM

)298(8206.0760

112.1 K

Kmol

atmLM

torr

atmtorr

solution L 1

protein mol 106.01 5M

protein mol 106.01L 1

mol106.01

mL 1000

L 1soln mL 1.00 8

5

molgmol

g

moles

mass/1066.1

1001.6

1000.1massmolar 4

8

3

Osmosis and blood cells

hypotonicsolution

hypertonicsolution

isotonicsolution

Cell remains stable

0.95% NaCl & 0.31M glucose

0.95% NaCl & 0.31 M glucose

0.95% NaCl & 0.31M glucose

<0.95% NaCl <& 0.31 M glucose

<0.95% NaCl <& 0.31 M glucose

Cell swells and burst

Cell shrinks

HemolysesCrenation

DialysisDialysis

Separation of small species (molecules of solute and Separation of small species (molecules of solute and solvent and ions) from big species in a solution by solvent and ions) from big species in a solution by means of a semi permeable membranemeans of a semi permeable membrane

In the artificial kidney, the blood is circulated from In the artificial kidney, the blood is circulated from the patient through cellophane tubes immersed in the patient through cellophane tubes immersed in solution containing all essential ions and small solution containing all essential ions and small molecules in blood at the appropriate molecules in blood at the appropriate concentrations.concentrations.

Only waste products dialyze from blood through the Only waste products dialyze from blood through the membranemembrane. .

Functioning of an artificial kidneyFunctioning of an artificial kidney

Reverse osmosisReverse osmosis

.

Reverse osmosis is the process of reversing the normal net flow of solvent molecules through a semipermeable membrane by applying to the solution a pressure exceeding the osmotic pressure

Reverse osmosis is the process of reversing the normal net flow of solvent molecules through a semipermeable membrane by applying to the solution a pressure exceeding the osmotic pressure

Reverse OsmosisReverse OsmosisA net flow of Solvent molecules(blue) from solution to the solvent

Desalination (Water purification) is an application of reverse osmosis

11.7 Collegative properties of electrolytes solutions11.7 Collegative properties of electrolytes solutions

Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

Since colligative properties only depend on the Since colligative properties only depend on the number of solute particlesnumber of solute particles

Ionic compounds (salts) should have a bigger Ionic compounds (salts) should have a bigger effect.effect.

When they dissolve they dissociate.When they dissolve they dissociate. NaCl dissociate into NaNaCl dissociate into Na++ and Cl and Cl-- ions ions 1 mole of NaCl makes 2 moles of ions.1 mole of NaCl makes 2 moles of ions. 1mole Al(NO1mole Al(NO33))33 makes 4 moles ions. makes 4 moles ions.

Electrolytes have a bigger impact on melting and Electrolytes have a bigger impact on melting and freezing points per mole because they make more freezing points per mole because they make more species.species.

Relationship is expressedRelationship is expressed using theusing the

van’t Hoff factorvan’t Hoff factor i i

The expected value can be determined from the formula The expected value can be determined from the formula of the salt.of the salt.

dissolved particles solute of moles

solutionin particles of molesi

van’t Hoff Factorvan’t Hoff Factor

Observed Observed ii value is smaller than expected value is smaller than expected Ion pairing most affects Ion pairing most affects ii value for highly charged value for highly charged

ionsions affects colligative propertiesaffects colligative properties

The actual value (effect on colligative The actual value (effect on colligative properties) is usually less becauseproperties) is usually less because

At any given instant some of the ions in solution At any given instant some of the ions in solution will be paired.will be paired.

Ion pairing tends to be higher for highly charged Ion pairing tends to be higher for highly charged ionsions

Ion pairing increases with concentration.Ion pairing increases with concentration. ii decreases with concentration. decreases with concentration.

Thus, van’t Hoff factor should be added to Thus, van’t Hoff factor should be added to the equations of collegativethe equations of collegative properties: properties:

H = H = iiKKmm

Dissociation Equations and theDissociation Equations and the Determination of Determination of ii

NaCl(s)

AgNO3(s) MgCl2(s)

Na2SO4(s)

AlCl3(s)

Na+(aq) + Cl-(aq)

Ag+(aq) + NO3-(aq)

Mg2+(aq) + 2 Cl-(aq)

2 Na+(aq) + SO42-

(aq)Al3+(aq) + 3 Cl-(aq)

i = 2

i = 2

i = 3

i = 3

i = 4

ExampleExample Osmotic pressure for 0.10 M solution of Fe(NHOsmotic pressure for 0.10 M solution of Fe(NH22))22(SO(SO44))22

at 25°C was 10.8 atm. Compare the van’t Hoff Factor at 25°C was 10.8 atm. Compare the van’t Hoff Factor observed and expected.observed and expected.

iiexpexp= 1+2+2=5= 1+2+2=5

iiobsobs < < iiexpexp because of high ion pairing because of high ion pairing

4.4)298(08206.0)/10.0(

8.10

K

KmolatmL

Lmol

atm

MRTiobs

iMRT