S olva tion

S truc ture /Inte rm olecula r F orces

Henry's L aw

B. pt. E leva tion F r. p t. D epress ion O som otic Pressure

C oll iga tive Properties

V apor Pressures Raoult's L aw

T em pera ture Pressure

Hea ts of S olution

C om pos itions S olubi l i ty Rules

T erm inology

Properties of SolutionsProperties of Solutions

TerminologyTerminology

• Solution = A homogeneous mixture.• Solute = a substance dissolved in a solvent to

form a solution; usually the smaller portion.• Solvent = The dissolving medium of a

solution; usually the greater portion.• Solubility = Amount of substance dissolved.• Dilute , Concentrated , Saturated .

The Solution ProcessThe Solution Process

Figure 13.2: The Solution ProcessFigure 13.2: The Solution ProcessHydration or SolvationHydration or Solvation

Solution Formation, Spontaneity, and Disorder

• If the process leads to a greater state of disorder, then the process is spontaneous.

• Example: a mixture of CCl4 and C6H14 is less ordered than the two separate liquids. Therefore, they spontaneously mix even though Hsoln is very close to zero.

• There are solutions that form by physical processes and those by chemical processes.

The Solution ProcessThe Solution Process

Solute-Solvent Interaction• Miscible liquids: mix in any proportions.• Immiscible liquids: do not mix. [ Oil and Vinegar ]• Generalization: “like dissolves (in) like”.

• Polar liquids tend to dissolve in polar solvents.

Factors Affecting SolubilityFactors Affecting Solubility

Mass Percentage• All methods involve quantifying amount of solute per

amount of solvent (or solution).• Generally amounts or measures are masses, moles or

liters.

Ways of Expressing ConcentrationWays of Expressing Concentration

100solution of mass total

solution incomponent of masscomponent of % mass

• Mole Fraction, Molarity, and Molality

Ways of Expressing ConcentrationWays of Expressing Concentration

solution of moles totalsolutionin component of moles

component offraction Mole

solution of literssolute moles

Molarity

solvent of kgsolute moles

Molality, m

Solution CompositionsSolution Compositions

• s = solute ; A = solvent; V = Tot. Vol. of solution.• Weight %:

• Mole Fraction:

• Molarity:

• Molality:

100% xww

ww

As

ss

As

ss nn

n

V

nM s

s

Akg

nm s

s

Example of Solution CompositionsExample of Solution Compositions

• A solution is prepared by mixing 78.9 g of ethanol (C2H5OH) with 100.0 g of water to give 190.5 mL of solution. Calculate the solution compositions.

• The electrolyte in automobile lead storage batteries is a 3.75 M H2SO4 solution that has a density of 1.230 g/mL. Calculate mass %, molality, and mole fraction in terms of H2SO4 .

• [Hint: Assume exactly one liter of solution.]• [Answers: 29.9% , 4.35 molal, 0.0727 ]

%1.441009.178

9.78% x

g

gws

s = ethanol (solute); A = water (solvent);

236.04549.5371.1

371.1

02.180.100

07.469.78

07.469.78

As

ss nn

n

199.85.190

371.1 LmolmL

mol

V

nM s

s

11.171000.0

371.1 kgmolwaterkg

mol

Akg

nm s

s

In the Dilution process of a more concentrated solution:

• The number of moles are the same in diluted and concentrated solutions.

• So:MdiluteVdilute = moles = MconcentratedVconcentrated

Concentrations of SolutionsConcentrations of Solutions

Electrolytic Properties• Three types:

• Strong electrolytes,• Weak electrolytes, and• Nonelectrolytes.

General Properties of Aqueous SolutionsGeneral Properties of Aqueous Solutions

Strong and Weak Electrolytes• Strong electrolytes: completely dissociated in solution.

e.g.

• Weak electrolytes: produce small concentration of ions when dissolved.

• e.g.

General Properties of Aqueous SolutionsGeneral Properties of Aqueous Solutions

HCl(aq) H+(aq) + Cl-(aq)

HC2H3O2(aq) H+(aq) + C2H3O2-(aq)

Precipitation ReactionsPrecipitation Reactions

Exchange (Metathesis) Reactions• Metathesis reactions involve swapping ions in solution:

AX + BY AY + BX.

Precipitation ReactionsPrecipitation Reactions

Ionic Equations• Ionic equation: used to highlight reaction between ions.

• Molecular equation: all species listed as molecules:AgNO3(aq) + NaI(aq) AgI(s) + NaNO3(aq)

• Complete ionic equation(CIE): lists all ions:

Ag+(aq) + NO3-(aq) + Na+(aq) + I-(aq) AgI(s) + Na+(aq) + NO3

-(aq)

• Net ionic equation: cancel spectator ions from CIEAg+(aq) + I-(aq) AgI(s)

Precipitation ReactionsPrecipitation Reactions

Figure 13.14: Factors Affecting SolubilityFigure 13.14: Factors Affecting Solubility

Intermolecular Forces

Pressure

Temperature

Pressure Effects

Figure 13.14: Factors Affecting SolubilityFigure 13.14: Factors Affecting Solubility

Pressure Effects

• If Sg is the solubility of a gas, k is a constant, and Pg is the partial pressure of a gas, then Henry’s Law gives:

• Carbonated beverages are bottled with a partial pressure of CO2 >1 atm, ( pressure inside can ~4 atm above liq).

• As bottle is opened, partial pressure of CO2 decreases and solubility of CO2 decreases.

• Therefore, bubbles of CO2 escape from solution.

Factors Affecting SolubilityFactors Affecting Solubility

gg PkS

CyberChem Diving Gases

Temperature Effects: Solids in Liquids• Generally, as temperature increases, solubility of solids

generally increases, BUT• Sometimes, solubility decreases as temperature increases

(e.g. Ce2(SO4)3).

Factors Affecting SolubilityFactors Affecting Solubility

Figure 13.18

Temperature Effects: Gases in Liquids• Gases get less soluble as temperature increases.

• Thermal pollution: if lakes get too warm, CO2 and O2 become less soluble and are not available for plants or animals.

Factors Affecting SolubilityFactors Affecting Solubility

Figure 13.18

• Colligative properties depend on quantity of solute molecules. (e.g. freezing point depression and boiling point elevation.)

Lowering Vapor Pressure• Non-volatile solutes reduce the ability of the surface solvent

molecules to escape the liquid.• Therefore, vapor pressure is lowered.• The amount of vapor pressure lowering depends on the

amount of solute.

Colligative PropertiesColligative Properties

Vapor Pressure on the Molecular Level

Figure 11.22: Vapor PressureFigure 11.22: Vapor Pressure

Figure 11.24

Figure 11.26: Phase DiagramsFigure 11.26: Phase Diagrams

The Phase Diagrams of H2O and CO2

Figure 11.27: Phase DiagramsFigure 11.27: Phase Diagrams

Lowering Vapor Pressure

Figure 13.20: Colligative PropertiesFigure 13.20: Colligative Properties

Lowering Vapor Pressure• Raoult’s Law:

• Where: PA = vapor pressure with solute,

• PA = vapor pressure without solute (pure solvent), and

A = mole fraction of A.

Colligative PropertiesColligative Properties

AAA PP

Vapor Pressure ExamplesVapor Pressure Examples

• Calculate the expected vapor pressure at 25oC for a solution prepared by dissolving 158.0 g of common table sugar (sucrose – MW=342.3) in 643.5 mL of water. At 25oC, the density of water is 0.9971 g/mL and the vapor pressure is 23.76 torr.

• A solution was prepared by adding 20.0 g of urea to 125 g of water at 25oC, a temperature at which pure water has a vapor pressure of 23.76 torr. The observed vapor pressure of the solution was found to be 22.67 torr. Calculate the molecular weight of urea. [Answer: 60. g/mol ]

AAA PP

Calculate the expected vapor pressure at 25oC for a solution prepared by dissolving 158.0 g of common table sugar (sucrose – MW=342.3) in 643.5 mL of water. At 25oC, the density of water is 0.9971 g/mL and the vapor pressure is 23.76 torr.

Example 1: Sugar in Water

s = sugar A = water ws 158.0gm MWs 342.3gm mol1

VA 643.5cm3 dA 0.9971gm cm

3 PoA 23.76torr MWA 18.015gm mol1

nA

VA dA

MWA nA 35.62mol nA 643.5cm

30.9971gm

cm3

1 mol

18.015gm

ns

ws

MWs ns 0.4616mol

ns 158.0gm1 mol

342.3gm

A

nA

ns nA A 0.9872 A

35.62

0.4616 35.6235.62

36.08

PA A PoA PA 23.46torr PA 0.9872 23.76torr( )

Example 2: Urea Molecular Weight

• A solution was prepared by adding 20.0 g of urea to 125 g of water at 25oC, a temperature

at which pure water has a vapor pressure of 23.76 torr. The observed vapor pressure of thesolution was found to be 22.67 torr. Calculate the molecular weight of urea.[Answer: 60. g/mol ]

s = urea A = water

ws 20.0gm wA 125gm PoA 23.76torr PA 22.67torr

PA A PoA A

PA

PoA A 0.9541 MWA 18.015gm mol

1

A

nA

ns nAnA

wA

MWA nA 6.939mol

solving for ns has solution of:

A

nA

ns nAnA

A 1

A

ns

nA 1 A

A ns 0.334mol

ns

ws

MWsMWs

ws

ns MWs 60gm mol

1

Figure 13.22

Boiling-Point Elevation

• Molal boiling-point-elevation constant, Kb, expresses how much Tb changes with molality, mS :

• Decrease in freezing point (Tf) is directly proportional to molality (Kf is the molal freezing-point-depression constant):

Colligative PropertiesColligative Properties

Sff mKT

Sbb mKT

Colligative PropertiesColligative Properties

Applications of Colligative PropertiesApplications of Colligative Properties

• A solution was made by dissolving 18.00 g of glucose in 150.0 g of water. The resulting solution was found to have a boiling point of 100.34oC. Calculate the molecular weight of glucose.

• How many kg of ethylene glycol (C2H6O2, MW=62.1), antifreeze, must be added to 10.0 L of water to produce a solution for use in a car’s radiator that freezes at –23.3oC? Assume that the density of water is 1.00 g/mL. [Answer: 7.78 kg (with d = 1.18 g/mL => 6.59 L)]

Sbb mKT

•A solution was made by dissolving 18.00 g of glucose in 150.0 g of water. The resulting solution was found to have a boiling point of 100.34oC. Calculate the molecular weight of glucose.

• A solution was made by dissolving 18.00 g of glucose in 150.0 g of water. The resulting solution was found to have a boiling point of 100.34oC. Calculate the molecular weight of glucose.

Sbb mKT

How many kg of ethylene glycol (C2H6O2, MW=62.1), antifreeze, must be added to 10.0 L of water to produce a solution for use in a car’s radiator that freezes at –23.3oC? Assume that the density of water is 1.00 g/mL.

Sff mKT

Freezing Point Depression Example

• How many kg of ethylene glycol (C2H6O2, MW=62.1), antifreeze, must be added to 10.0 L

of water to produce a solution for use in a car’s radiator that freezes at –23.3 oC? Assumethat the density of water is 1.00 g/mL.[Answer: 7.78 kg (with d = 1.18 g/mL => 6.59 L)]

Let: s = ethylene glycol A=water

Kf 1.86K kg mol1 Tf 23.3K kg_A 10.0L

1 gm1 mL

1 mL

103

L

10

3kg

1 gm

kg_A 10.0kg MWs 62.1gm mol1

Tf Kf ms Kf

ns

kg_A

Kf

ws

MWs kg_A

ws

Tf MWs kg_A

Kf

ws23.3K 62.1 gm mol

1 10.0 kg

1.86K kg mol1

ws 7.78kg

ds 1.18gm mL1 Vs

ws

ds Vs 6.59L

Colligative Properties: Homework due for Friday Dec. ?, 201?.Colligative Properties: Homework due for Friday Dec. ?, 201?.

What is the molecular weight (g/mol) of a non-volatile solute if 7.32 kg of this solute dissolved in 10.0 L of water produced a solution with a freezing point of -23.3C? Assume density of water as 1.00 g/mL.

molgMWS /4.58

Osmosis• movement of a solvent from low solute concentration to

high solute concentration across a semipermeable membrane.

Colligative PropertiesColligative Properties

Figure 13.23

Osmosis• Osmotic pressure, , is the pressure required to stop

osmosis:

Colligative PropertiesColligative Properties

TRV

n

TRnV

TRM S TRM S

Application of Osmotic PressureApplication of Osmotic Pressure

• To determine the molecular weight of a certain protein, 1.00 mg of it was dissolved in enough water to make 1.00 mL of solution. The osmotic pressure of this solution was found to be 1.12 torr at 25.0oC. Calculate the molecular weight of the protein.

[Answer: 1.66x104 g/mol]

TRM S TRM S

To determine the molecular weight of a certain protein, 1.00 mg of it was dissolved in enough water to make 1.00 mL of solution. The osmotic pressure of this solution was found to be 1.12 torr at 25.0 oC.

Calculate the molecular weight of the protein. TRM S TRM S

Colligative Properties: Osmotic Pressure Example

• To determine the molecular weight of a certain protein, 1.00 mg of it was dissolved in

enough water to make 1.00 mL of solution. The osmotic pressure of this solution was found

to be 1.12 torr at 25.0oC. Calculate the molecular weight of the protein.[Answer: 1.66x104 g/mol]

Define: s = protein A = water

ws 1.00103 gm V 1.0010

3 L 1.12torr R 0.08206L atm mol1 K

1

T 25.0 273.15( ) K 1.12 torr1atm

760torr

1.474 103 atm

Ms R Tns

V

R Tws

MWs V

R T

The only unknown is the molecular weight of the protein ( MWs ).

MWs

ws R T

V MWs

1.00103 gm 0.08206L atm mol

1 K1 298.15 K

1.474103 atm 1.0010

3 L

MWs 1.66 104 gm mol

1

Osmosis• Isotonic solutions: two solutions with the same

separated by a semipermeable membrane.

• Osmosis is spontaneous.• Red blood cells are surrounded by semipermeable

membranes.

Colligative PropertiesColligative Properties

Osmosis

Figure 13.25: Colligative PropertiesFigure 13.25: Colligative Properties

Crenation: hypertonic solution Hemolysis: hypotonic solution

IV(intravenous) fluids must be Isotonic.CyberChem: Desalination

• Colloids are suspensions in which the suspended particles are larger than molecules but too small to drop out of the suspension due to gravity.

• Particle size: 10 to 2000 Å.• Tyndall effect: ability of a Colloid to scatter light. The

beam of light can be seen through the colloid.

ColloidsColloids

ColloidsColloids

Figure 13.31: ColloidsFigure 13.31: Colloids

Solvation

Structure/Interm olecular Forces

Henry's Law

B. pt. E levation Fr. pt. Depression Osom otic Pressure

Colligative Properties

Vapor Pressures Raoult's Law

T em perature Pressure

Heats of Solution

Com positions Solubility Rules

T erm inology

Properties of SolutionsProperties of Solutions

o

AAA PP

Sfbfb mKT ,, TRM S

AS

AA nn

n

Akg

nm S

S

V

nM S

S