Properties of Solutions Chapter 11. A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller.

Properties of Solutions

Chapter 11

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 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

M = moles of solute

volume of solution (liters)

Molarity (M)

Molality (m)

m =moles of solute

mass of solvent (kg)

BA

AA nn

nAoffractionMole

Example

1.00 g C2H5OH is added to 100.0 g of water to make 101 mL of solution. Find the molarity, mass %, mole fraction and molality of ethanol.

Molarity Number of moles of solute per L or

solution

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 Percent also called weight percent percent 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 Fraction ratio of number of moles of a part of

solution 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

Molality

Number of moles of solute per kg of solvent

solvent of kg

solute of molesm

0.217m

solution1000g

1kg100.0g

OHHmolC102.17m 52

2

Example

An aqueous antifreeze solution is 40% ethylene glycol (C2H6O2) by mass. The density of the solution is 1.05 g/cm3. Calculate the molality, molarity and mole 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 solvent = 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

• Why this behavior occur?

Solubility Process

The formation of a liquidsolution takes place in 3 steps

1. Expand solutemolecules

2. Expand solventmolecules

3. Mixing solute andsolvent

Endothermic

Endothermic Often Exothermic

Energy of Solubility Process Steps 1 and 2 require energy to overcome

IMFs

• endothermic Step 3 usually releases energy

• exothermic enthalpy of solution

• sum of ∆H values• can be – or +• ∆Hsoln = ∆H1 + ∆H2 + ∆H3

Energy of Solubility Process

Case 1: oil and water Oil is nonpolar (LD forces) Water is polar (H bonding) ∆H1 will be small for typical size ∆H2 will be large ∆H3 will be small since there won’t be

much interaction between the two ∆Hsoln will be large and +ve because

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

Case 2: NaCl and water NaCl is ionic water is polar (H bonding) ∆H1 will be large

∆H2 will be large

∆H3 will be large and –ve because of the strong interaction between ions and water

∆Hsoln will be close to zero- small but +ve

Energy of solubility process Enthalpy of hydration - ∆Hhyd

enthalpy change associated with dispersal

of gaseous solute in water

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

∆H1=786 kJ/mol

H2O(l) + Na+(g) + Cl-(g) Na+(aq) + Cl-(aq)

∆Hhyd=∆H2 + ∆H3=-783 kJ/mol

∆Hsoln=3 kJ/mol

Energy

Reactants

Solution

H1

H2

H3

Solvent

Solute and Solvent

Size of H3 determines whether a solution will form

H3

Solution

Mixing solvent and solute

wo factors explains the solubility:

1. n increase in the probability of mixing favors the process

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

If Hsoln is small and positive, a solution will still form because of entropy.

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

Energy of Solubility Process

1 .Structure and Solubility

Water soluble molecules must have dipole moments -polar bonds.

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

11.311.3 Factors Affecting SolubilityFactors Affecting Solubility

• Pentane C5H12 is miscible with hexane C6H14

and immiscible 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 body

hydrophobic: water- fearing: nonpolar• Insoluble in water

hydrophilic: water-loving: polar• Soluble in water

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

Hydrophilic, excreted bythe body and must be consumed regularly

2 .Pressure effects Changing the pressure doesn’t effect the amount

of solid or liquid that dissolves They are incompressible. It does effect gases.

Pressure effects the amount of gas that can dissolve in a liquid.

The dissolved gas is at equilibrium with the gas above the liquid.

The gas is at equilibrium with the dissolved gas in this solution.

The equilibrium is dynamic.

If the pressure is increased the gas molecules dissolve faster.

The equilibrium is disturbed.

The system reaches a new equilibrium with more gas dissolved.

Henry’s Law.

P= kC

Pressure = constant x Concentration

of gas

Henry’s LawHenry’s Law

C C P C = kP P C = kP• C is concentration and P is partial pressure C is concentration and P is partial pressure

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

gases that don’t dissociate or react with gases that don’t dissociate or react with solventsolvent

•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

A soft drink bottled at 25°C contains A soft drink bottled at 25°C contains COCO22 at pressure of 5.0 atm over at pressure of 5.0 atm over

liquid. Assume that Pliquid. Assume that PCO2CO2 in in

atmosphere is 4.0 x 10atmosphere is 4.0 x 10-4-4 atm. Find the atm. Find the equilibrium concentration in soda equilibrium concentration in soda before and after opening. k=32 before and after opening. k=32 L*atm/mol at 25°CL*atm/mol at 25°C

ExampleExample before opening:before opening:

after opening:after opening:

LmolmolatmL

atm

k

PCCkP CO

COCOCO /16.032

0.52

222

LmolmolatmL

atm

k

PC CO

CO /102.132

104 54

2

2

Example Solubility of pure N2 in blood at body

temp, 37oC and 1 atm is 6.2X10-4 M. If a diver breaths air ( = 0.78) at a depth where the total pressure is 2.5 atm, calculate the concentration of N2 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

3 .Temperature Effects (for aqueous solutions)

Increased temperature increases the rate at which a solid dissolves.

But it is not possible to predict whether it will increase the amount of solid that dissolves.

Solubility can be predicted only from a graph of experimental data.

20 40 60 80 100

Temperature effectsTemperature effects

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

but the amount able but the amount able to be dissolved to be dissolved does not changedoes not change

Temperature effectsTemperature effects

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

with Twith T

Temperature and Solubility

Solid solubility and temperature

solubility increases with increasing temperature

solubility decreases with increasing temperature

12.4

Solubility and environment

Thermal pollution• Water used as a coolant when pumped again

into the source (lakes and rivers) floats on the cold water causing a decrease in solubility of O2 and consequently affecting the aquatic life.

CO2 dissolves in water that contains CO32-

causing formation of HCO3- that is soluble in

water. When temp increases CO2 will be driven off the water causing precipitation of CO3

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

CO32- (aq) + CO2(aq) 2HCO3

-

11.411.4 The vapor pressures of solutionsThe vapor pressures of solutions A nonvolatile solvent lowers the vapor pressure

of the solution. For the molecules of the solvent

to escape, they must overcome the forces of both the other solvent

molecules and the solute molecules.

Nonvolatile solute decreases # of solvent molecules per unit volume. Thus, # of solute molecules escaping will be lowered.

Aqueous Solution

Pure water

Water has a higher vapor pressure than a solution

How a nonvolatile solute affects a solvent?

P1 Po1

Aqueous Solution

Pure water

Water molecules evaporate faster from pure water than from the solution

P1Po

1<

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

Aqueous Solution

Pure water

empty

Raoult’s LawRaoult’s LawThe 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

Applies only to an ideal solution where the solute doesn’t contribute to the vapor pressure (solute and solvent are alike: solute-solute, solvent-solvent and solute-solvent interactionsAre very similar ). Solute acts to dilute the solvent

0)1(solventsolutesolution

PP Only one solute

solvento

solutesosolvent

o PPP ln

solvento

solutePVPL

Vapor pressure lowering

Deviations If solvent has a strong affinity for solute (H bonding). Lowers solvents ability to escape. Lower vapor pressure than expected. Negative deviation from Raoult’s law. If Hsoln is large and negative (exothermic): strong

interaction exists between the solute and the solvent; thus a negtive deviation from Rault’s law because both components will have lower escaping tendency in the solution than in pure liquids.

If the two liquids mix endothermically (solute-solvent interactions are weaker than interactions among molecules in the pure liquids. More energy is required to expand liquids than is released when liquids are mixed). In this case molecules in the solution have higher tendency to escape than expected and positive deviations from Raul’s law are observed.

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

Nonideal solutions

Raoult’s Law – Ideal SolutionRaoult’s Law – Ideal Solution

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

Negative Deviations from Raoult’s LawNegative Deviations from Raoult’s LawStrong solute-solvent interaction results in a vapor pressure lower than predicted

Exothermic mixing = Negative deviation

Positive Deviations from Raoult’s LawPositive Deviations from Raoult’s Law

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

Endothermic mixing = Positive deviation

PA = XA P A0

PB = XB P B0

PT = PA + PB

PT = XA P A0 + XB P B

0

Ideal Solution

An ______________

is one that obeys Raoult’s Law

PT is ______________ predicted by Raoults’s law

PT is ______________predicted by Raoults’s law

ForceA-B

ForceA-A

ForceB-B< &

ForceA-B

ForceA-A

ForceB-B> &

Vapor pressure lowering when ionic compounds are dissolved

VPL, when NaCl is dissolved in water, is twise as much as expected.

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

about the nature of the solute When Na2SO4 is dissolved VPL is 3Xexpected

ExampleExample

Find the vapor pressure at 25°C for Find the vapor pressure at 25°C for solution of 158.0 g of sucrose solution of 158.0 g of sucrose (C(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 g/mL and the partial pressure of water vapor at 25°C is 23.76 torr.water 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

Colligative Properties of solutions

Colligative properties depend only on the number of solute particles (molecules or ions) in a solution

They do not depend on the identity of particles

11.5 Boiling point elevation and 11.5 Boiling point elevation and

freezing Point depressionfreezing 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 Elevation

A nonvolatile solute lowers the vapor pressure

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

A nonvolatile solute elevates the boiling point of the solvent

The amount of the elevation depends on concentration of the solute

solutebmKT

Freezing Point Depression Vapor pressure of solid and liquid are equal

at freezing point nonvolatile solute lowers the vapor pressure

so a lower T is needed to decrease the vapor pressure to that of the solid

a nonvolatile solute depresses the freezing point of the solvent

the amount of the depression depends on concentration 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

Boiling-Point Elevation

Tb = Tb – T b0

Tb > T b0 Tb > 0

T b is the boiling point of the pure solvent

0

T b is the boiling point of the solution

Tb = Kb m

m is the molality of the solution

Kb is the molal boiling-point elevation constant (0C/m)

Freezing-Point Depression

Tf = T f – Tf0

T f > Tf0 Tf > 0

T f is the freezing point of the pure solvent

0

T f is the freezing point of the solution

Tf = Kf m

m is the molality of the solution

Kf is the molal freezing-point depression constant (0C/m)

Molal Boiling-Point Elevation and Freezing-Point Depression Constants of Several Common Liquids

Example 1 18.00 g of glucose are added to 150.0 g of

water. The boiling point of the solution is 100.34 C. The boiling point constant is 0.51 C*kg/mol.

Find the molar mass of glucose.

Example 1

soluteb mKT

soluteif mmol

kgCTTT

)51.0(00.10034.100

kg

mol

molkgC

Cmsolute 67.0

/51.0

34.0

Example 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 2

What mass of C2H6O2 (M=62.1 g/mol) needs to be added to 10.0 L H2O to make a solution that freezes at -23.3°C? density is 1.00 g/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

Understanding osmoses: Experimental approach

Semipermeable membrane is a partition with pores that allow small solvent particles to pass through but not bigger solute particles. Thus, it separates a solution and a pure solvent

11.6 Osmoses and Osmotic Pressure11.6 Osmoses and Osmotic Pressure

Initial Final

Osmosis The flow of solvent molecules from

a pure solvent through a semi-permeable membrane into a solution

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

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

BEFORE

AFTER

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, , and concentration of nonelectrolytes

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

same: V = nRT

where • M is the molarity of the

solution• R is the gas law constant• T is the temperature in Kelvin

MRT RTV

n

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

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

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

Dialysis Separation of small species (molecules of solute and

solvent and ions) from big species in a solution by means of a semi permeable membrane

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

Only waste products dialyze from blood through the membrane.

Functioning of an artificial kidney

Reverse 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

A net flow of Solvent molecules(blue) from solution to the solvent

Desalination (Water purification) is an application of reverse osmosis

Solutions of Electrolytes

The van’t Hoff factor (i) is used to modify the equations for colligative properties

FPD: Tf = –i Kfm BPE: Tb = i Kbm OP: = i M RT

EOS

i is dependent on solution molality

11.7 Collegative properties of electrolytes solutions

Since colligative properties only depend on the number of solute particles

Ionic compounds (salts) should have a bigger effect.

When they dissolve they dissociate. Individual Na+ and Cl- ions fall apart. 1 mole of NaCl makes 2 moles of ions. 1mole Al(NO3)3 makes 4 moles ions.

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

Relationship is expressed using the

van’t Hoff factor i

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

dissolved particles solute of moles

solutionin particles of molesi

van’t Hoff Factor Observed i value is smaller than expected Ion pairing most affects i value for highly

charged ions affects colligative properties

The actual value (effect on colligative properties) is usually less because• At any given instant some of the ions in

solution will be paired.• Ion pairing tends to be higher for highly

charged ions• Ion pairing increases with concentration.• i decreases with concentration.

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

H = iKm

Colligative Properties of Electrolyte Solutions

0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions

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

0.1 m NaCl solution 0.2 m ions in solution

van’t Hoff factor (i) = actual number of particles in soln after dissociation

number of formula units initially dissolved in soln

nonelectrolytesNaCl

CaCl2

i should be

12

3

Dissociation Equations and Dissociation Equations and the Determination of the 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

Boiling-Point Elevation Tb = i Kb m

Freezing-Point Depression Tf = i Kf m

Osmotic Pressure () = iMRT

Colligative Properties of Electrolyte Solutions

Vapor pressure lowering, VPL ) ( solventpureo

solutePiVPL

Ideal vs. Real van’t Hoff FactorIdeal vs. Real van’t Hoff Factor

The ideal van’t Hoff Factor is only achieved in The ideal van’t Hoff Factor is only achieved in VERY DILUTEVERY DILUTE solution. solution.

Example 1 Osmotic pressure for 0.10 M solution of

Fe(NH2)2(SO4)2 at 25°C was 10.8 atm. Compare the van’t Hoff Factor observed and expected.

iexp= 1+2+2=5

iobs < iexp because of high ion pairing

4.4)298(08206.0)/10.0(

8.10

K

KmolatmL

Lmol

atm

MRTiobs

iMRT

Ch. 11 SolutionsCh. 11 Solutions

11.8 Colloids11.8 Colloids

ColloidsColloids

Tyndall EffectTyndall Effect scattering of light particles used to scattering of light particles used to

determine whether something is a determine whether something is a solution or suspensionsolution or suspension

ColloidColloid also called colloidal dispersionalso called colloidal dispersion suspension of tiny but visible particles in suspension of tiny but visible particles in

a mediuma medium

Types of Colloids Types of Colloids

ColloidsColloids

particles in a colloid are 1-1000nm in particles in a colloid are 1-1000nm in diameter and overall neutraldiameter and overall neutral

they have layer of negative ions on they have layer of negative ions on outsideoutside

so repel each otherso repel each other CoagulationCoagulation

when a colloid is destroyed through when a colloid is destroyed through heating or addition of electrolyteheating or addition of electrolyte