1 Chapter 13 Physical Properties of Solutions Insert picture from First page of chapter.

1

Chapter 13

Physical Properties of Solutions

Insert picture fromFirst page of chapter

Copyright McGraw-Hill 2009 2

13.1 Types of Solutions


13.1 Types of Solutions

Solution classifications based on amount of solute dissolved relative to the maximum:

• Saturated – maximum amount at a given temperature (This amount is termed the solubility of the solute.)

• Unsaturated – less than the maximum

• Supersaturated – more than a saturated solution but is an unstable condition


unsaturated

supersaturated

saturated

heat


Conversion of a Supersaturated Solution to a Saturated Solution


13.2 A Molecular View of the Solution Process

Factors that determine solubility

• Intermolecular forces present in the formation of a solution– Solute-solute interactions– Solvent-solvent interactions– Solute-solvent interactions


321soln HHHH


321soln HHHH

0soln321 HHHH

0soln321 HHHH

endothermic

exothermic


Solubility can be predicted based on “like dissolves like” in terms of intermolecular forces.For example: water and methanol are both polar

and dissolve in each other

• Two liquids that are soluble in each other in all proportions are term miscible.

• Ions readily dissolve in polar solvent due to solvation by the solvent molecules.

O

HH


Predict whether Vitamin B6 is water soluble

or fat soluble.


Water soluble due to the presence of polar

groups.

polar groups


• Energy and entropy – Exothermic processes are generally more

favorable than endothermic processes– Solutions do form when the overall process is

endothermic– Entropy (randomness or disorder) contributes

to the solution process

• Entropy tends to increase for all process

• A solution is more disordered than the isolated solute and solvent


13.3 Concentration Units

• Molality (m) – number of moles of solute dissolved in 1 kg (1000 g) of solvent

• Percent by Mass – ratio of mass of solute to the mass of the solution times 100

kg)(insolutionofmass

soluteofmolesmolality m

x100solventofmasssoluteofmass

soluteofmassmassby percent


Units analogous to percent by mass (part per hundred) to express very small concentrations

• Parts per million (ppm)

• Parts per billion (ppb)

*masses must be expressed in the same units

6x10solutionof*mass

soluteof*massppm

9x10solutionof*mass

soluteof*massppb


Choice of units depends on the purpose of the experiment

• Mole fraction – used for gases and vapor pressures of solutions

• Molarity – commonly used since volumes of solutions are easy to measure

• Molality – temperature independent• Percent by Mass – temperature independent

and need not know molar masses• Conversion between units requires the use of

density if any mass to volume or volume to mass conversion in needed.


Determine the a) molality, b) percent by

mass and c) the ppm concentrations of a

solution prepared by dissolving 15.56 g of

glucose (C6H12O6) in 255 g of water.


a) molality

Molar mass of glucose = 180.2 g/mol

mol0.086349g180.2

molxg15.56glucosemol

mm 0.339kg

g10x

g 255

1 x mol0.086349

3


b) percent by mass

c) ppm

glucose5.751% x100g 270.56

g 15.56

glucose5.751x10 x10g 270.56

g 15.56 46 ppm

(original slide typo!)


13.4 Factors that Affect Solubility

• Temperature– The solubility of solids may increase,

decrease or remain relatively constant with increasing temperature

– The solubility of gases decreases with increasing temperature• Thermal pollution – a consequence of

the relation between gas solubility and temperature


Temperature Dependence of the Solubility of Selected Solids


• Pressure – significantly affects only the solubility of gases– Henry’s law – the solubility of a gas, c, is

directly proportional to the pressure of the gas, P, over the solution

where c, is in mol/L, k is Henry’s law

constant with units of mol/L. atm and P is in atm.

Pc kPc


Molecular View at Different Pressures

P1 P2

P1< P2


13.5 Colligative Properties

Colligative properties depend only on the number of solute particles in solution and not the nature of the solute• Vapor-Pressure lowering

– Nonvolatile solute (no appreciable pressure)• Vapor pressure of the solvent is decreased

– Volatile solute (exhibit appreciable pressure)• Vapor pressure is the sum of the individual

pressures


– Raoult’s law – quantitative expression of the solution vapor pressure

• Nonvolatile solute (P1 is the solution vapor pressure, P0 is the vapor pressure of the pure substance, and is the mole fraction.)

• Volatile solute (PT is the solution vapor pressure, and are vapor pressures of pure solution components)

0111 PP

0BB

0AAT PPP

0AP 0

BP Volatile solute: not exam material


Calculate the vapor pressure of a solutionmade by dissolving 115 g of urea, a nonvolatilesolute, [(NH2)2CO; molar mass = 60.06 g/mol]in 485 g of water at 25°C. (At 25°C, .)mmHg23.8OH2

P


0OHOHOH 222

PP

0.9336mol1.915mol26.91

mol26.91OH2

mol26.91g18.02

molgx485mol OH2

mol1.915g60.06

mol xg115molurea

mmHg22.2mmHg23.8 x 0.9392OH2P


• Boiling-Point Elevation (Tb)

– The boiling point of a solution, Tb, of a nonvolatile solute will be higher than that of the pure solvent, .

– Elevation is directly proportional to the molal concentration.

– Kb is the molal boiling-point elevation constant.

0bT

mKT bb

0bbb TTT


Effect of Vapor Pressure Lowering

Effect on Boiling Point



• Freezing-Point Depression (Tf)

– The freezing point of a solution, Tf, will be lower than that of the pure solvent, .

– Depression is directly proportional to the molal concentration.

– Kf is the molal freezing-point elevation constant.

0fT

mKT ff

f0

ff TTT


Effect of Vapor Pressure Lowering

Effect on Freezing Point


Calculate a) the freezing point and b) the

boiling point of a solution containing 268 g of

ethylene glycol and 1015 g of water. (The

molar mass of ethylene glycol (C2H6O2) is

62.07 g/mol. Kb and Kf for water are 0.512°C/m

and 1.86°C/m, respectively.)


a) freezing point

C7.914.254C1.86 o

o

f mxm

T

mol4.318g62.07

molxg268glycolethylenemol

mm 4.254kg

g10x

g 1015

1 x mol3184

3

.

foo C0.00C7.91 T

C7.91of T


b) boiling point

C100.00C2.18 ob

o T

C2.184.254C0.512 o

o

b mxm

T

C102.18ob T


Osmosis - Selective passage of solvent molecules through a semipermeable membrane

solvent solution solvent solution


• Osmotic Pressure () – Pressure required to stop osmosis

– Directly proportional to molar concentration, M

= MRT

where R is 0.08206 L.atm/mol . K and T is in

kelvins


• Solutions of Electrolytes

– Dissociation of strong and weak electrolytes affects the number of particles in a solution

– van’t Hoft factor (i) – accounts for the effect of dissociation

solutionindissolvedinitiallyunitsformulaofnumber

ondissociatiaftersolutioninparticlesofnumberactuali


Modified Equations for Colligative Properties

miKT ff

miKT bb

iRTM



Formation of ion pairs affects colligative properties



The freezing-point depression of a 0.100 m

MgSO4 solution is 0.225°C. Determine the

experimental van’t Hoff factor of MgSO4 at

this concentration.


miKT ff

One approach:

mxm

i 0.100C1.86

C0.225o

o

1.21i

Note, at this concentration the dissociation of MgSO4 is not complete.


Ideal freezing point depression

C0.1860.100x C1.86 o

o

f mm

T

Compare ideal and real freezing point depression

211C0.186

C0.225o

o

.i

Another approach to the same problem:


A solution made by dissolving 25.0 mg ofinsulin in 5.00 mL of water has an osmoticpressure of 15.5 mmHg at 25°C. Calculatethe molar mass of insulin. (Assume that there is no change in volume when theinsulin is added to the water and that insulinis a nondissociating solute.)


K298

1

atmL0.08206

Kmolxatm2.039x10 2 x

RTM

Calculate the M of the solution

atm2.039x10mmHg760

atmxmmHg15.5 2

L

mol10x8.33810x8.338

44

MM


Calculate the moles of insulin

mol x104.169mL

L10xmL5.00x

L

mol10x8.340mol 6

34

Molar mass is ratio of grams to moles

mol x104.169

1x

mg

g10xmg25.0

6

3

M

mol

g10x6.00 3

M


13.7 ColloidsA colloid is a dispersion of particles of onesubstance throughout another substance.• Intermediate between a homogenous and heterogeneous

mixture• Range of particle size: 103 – 106 pm• Categories

– Aerosols – Foams– Emulsions– Sols– Gels


Aerogel

• Lowest density solids known

• Good thermal, electric, sound insulators



A 2.5-kg brick supported by a 2-g piece of aerogel

(source: Wikipedia)


• Exhibit the Tyndall effect

colloid solution


Fog – A Familiar Manifestation of the Tyndall Effect


• Many important colloids are aqueous and can be further classified.

– Hydrophilic (water loving)

– Hydrophobic (water fearing)

Hydrophilic groups on the surface of a protein stabilize the molecule in water


Stabilization of Hydrophobic Colloidal Particles by Ion Adsorption


Removal of Grease by Soap (Sodium Stearate)


Emulsification – • stabilization of an unstable colloid• accomplished by the addition of an emulsifier or emulsifying agent

Sodium glycoholate – a bile salt or biological emulsifying agent for ingested fats


Lecithin is a mixture of substances found in natural sources such as egg yolk. It contains molecules such asphosphatidylcholine (shown right) that have both hydrophobic (nonpolar) and hydrophilic (polar; charged) regions. Lecithin acts as an emulsifier.

This is why egg yolk is used to make mayonnaise. The egg yolk is typically whisked with a small amount of vinegar or lemon juice (the aqueous phase). Vegetable oil is then added gradually while whisking. Mayonnaise is the resulting emulsion of vegetable oil and aqueous acid.

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1 Chapter 13 Physical Properties of Solutions Insert picture from First page of chapter.

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