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Chapter 13. The Properties of Mixtures: Solutions and Colloids. Terms Associated with Solutions. solute the material dissolved; the lesser amount solvent the material the solute is dissolved in the larger amount; usually water. Terms Associated with Solutions. miscible.

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Chapter 13


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slide1

Chapter 13

The Properties of Mixtures:

Solutions and Colloids

terms associated with solutions
Terms Associated with Solutions

solute

the material dissolved; the lesser amount

solvent

the material the solute is dissolved in

the larger amount;

usually water

solubility s
Solubility (S)

Solubility:

# moles solute dissolved / L solution

or…

# grams solute dissolved / 100 ml water

“Like Dissolves Like”

factors affecting solubility
Factors Affecting Solubility

Given the structures of Vitamin A and Vitamin C, which vitamin must be consumed regularly?

Which vitamin will be stored in the fatty tissues of the body?

Vitamin C

slide6

Figure 13.3

Like dissolves like: solubility of methanol in water

demos
Demos
  • Salting Out
  • Water, Iodine, Hexane, Ethanol
slide8

Figure 13.1

Ion-Dipole Forces

Explains solubility of ionic compounds in water

Example:

Any ion in water

slide9

Figure 13.1

Hydrogen Bonding

Chemistry is FON!

Example:

Organic compounds (sugars/alcohols/amino acids) in

cellular fluid

slide10

Dipole-Dipole Forces

Solubility of polar molecules in each other

Figure 13.1

Example:

Polar organic molecules in polar non-aqueous solvents

slide11

Ion-Induced Dipole

An ion’s charge distorts the electron cloud

of a nearby polar molecule

Example:

Fe2+ binding to hemoglobin

Solubility of salts in ethanol

slide12

Dipole-Induced Dipole Forces

Polar molecule distorts electron cloud in nonpolar substance

Example:

Solubility of gases in water

slide13

Dispersion Forces

Contribute to the solubility of ALL solutes in ALL solvents

The principal type of IMF in solutions of nonpolar substances

Example:

Petroleum and gasoline

slide14

LIKE DISSOLVES LIKE

Substances with similar types of intermolecular forces dissolve in each other.

When a solute dissolves in a solvent, solute-solute interactions and solvent-solvent interactions are being replaced with solute-solvent interactions. The forces must be comparable in strength in order to have a solution occur.

liquid solutions and the role of molecular polarity
Liquid Solutions and the role of Molecular Polarity
  • Liquid – Liquid solutions

When a solute dissolves in a solvent, solute-solute interactions and solvent-solvent interactions are being replaced with solute-solvent interactions. The forces must be comparable in strength in order to have a solution occur.

slide17

PROBLEM:

Predict which solvent will dissolve more of the given solute:

(a) Methanol - NaCl is ionic and will form ion-dipoles with the -OH groups of both methanol and propanol. However, propanol is subject to the dispersion forces to a greater extent.

(b) Water - Hexane has no dipoles to interact with the -OH groups in ethylene glycol. Water can H bond to the ethylene glycol.

(c) Ethanol - Diethyl ether can interact through a dipole and dispersion forces. Ethanol can provide both while water would like to H bond.

SAMPLE PROBLEM 13.1

Predicting Relative Solubilities of Substances

(a) Sodium chloride in methanol (CH3OH) or in propanol (CH3CH2CH2OH)

(b) Ethylene glycol (HOCH2CH2OH) in hexane (CH3CH2CH2CH2CH2CH3)

or in water.

(c) Diethyl ether (CH3CH2OCH2CH3) in water or in ethanol (CH3CH2OH)

SOLUTION:

slide18

Gas – Liquid Solutions

Why aren’t gases very soluble in water?

Table 13.3

Correlation Between Boiling Point and Solubility in Water

Gas

Solubility (M)*

bp (K)

He

4.2 x 10-4

4.2

Ne

6.6 x 10-4

27.1

N2

10.4 x 10-4

77.4

CO

15.6 x 10-4

81.6

O2

21.8 x 10-4

90.2

NO

32.7 x 10-4

121.4

* At 273K and 1 atm

gas solutions and solid solutions
Gas Solutions and Solid Solutions
  • Gas – Gas Solutions
    • Infinitely soluble in each other
  • Gas – Solid Solutions
    • Gas molecules occupy spaces between the solid particles
  • Solid – Solid solutions
    • Alloys
slide24

solute (aggregated) + heat solute (separated) DHsolute > 0

solvent (aggregated) + heat solvent (separated) DHsolvent > 0

solute (separated) + solvent (separated) solution + heat DHmix < 0

Heats of solution and solution cycles

1. Solute particles separate from each other - endothermic

2. Solvent particles separate from each other - endothermic

3. Solute and solvent particles mix - exothermic

DHsoln = DHsolute + DHsolvent + DHmix

slide26

H2O

M+ (g) [or X-(g)] M+(aq) [or X-(aq)] DHhydr of the ion < 0

M+ (g) + X-(g) MX(s) DHlattice is always (-)

Heats of Hydration

solvation of ions by water - always exothermic

DHhydr is related to the charge density of the ion, that is, coulombic charge and size matter.

Lattice energy is the DH involved in the formation of an ionic solid from its gaseous ions.

heats of hydration ionic solids in water
Heats of Hydration: Ionic Solids in Water

Solvation/Hydration:

The process of surrounding a solute particle with solvent particles

DHsoln = DHsolute + DHsolvent + DHmix

DHsoln = DHlattice + DHhydr of the ions

slide28

Figure 13.2

Hydration shells around an aqueous ion

slide29

Table 13.4 Trends in Ionic Heats of Hydration

Ion

Ionic Radius (pm)

DHhydr (kJ/mol)

Group 1A(1)

Li+

76

-510

Na+

102

-410

K+

138

-336

Rb+

152

-315

Cs+

167

-282

Group 2A(2)

Mg2+

72

-1903

Ca2+

100

-1591

Sr2+

118

-1424

Ba2+

Group 7A(17)

F-

133

-431

Cl-

181

-313

Br-

196

-284

I-

220

-247

slide30

Figure 13.6

Dissolving ionic compounds in water

NaCl

NH4NO3

NaOH

the solution process and the change in entropy
The Solution Process and the Change in Entropy
  • The natural tendency of a system to disperse matter or energy
  • Which has highest entropy: solid, liquid, or gas?
  • Which has highest entropy: pure solute, pure solvent, solution?
slide33

Enthalpy diagrams for dissolving NaCl and octane in hexane

Figure 13.7

Entropy drives the reaction (DH negligible)

Entropy not big enough to overcome DH

solubility as an equilibrium process
Solubility as an Equilibrium Process
  • Saturated
  • Unsaturated
  • Supersaturated
  • Equilibrium
slide35

solute (undissolved) solute (dissolved)

Figure 13.8

Equilibrium in a saturated solution

slide36

Figure 13.9

Sodium acetate crystallizing from a supersaturated solution

slide37

Figure 13.10

The relation between solubility and temperature for several ionic compounds

factors affecting solubility39
Factors Affecting Solubility

equilibrium

Increase

pressure

New

equilibrium

William Henry

Henry’s Law:

As Pressure increases, concentration of dissolved gases ____

slide40

Henry’s Law

Sgas = kH X Pgas

The solubility of a gas (Sgas) is directly proportional to the partial pressure of the gas (Pgas) above the solution.

slide42

PROBLEM:

The partial pressure of carbon dioxide gas inside a bottle of cola is 4 atm at 250C. What is the solubility of CO2? The Henry’s law constant for CO2 dissolved in water is 3.3x10-2 mol/L*atm at 250C.

S = (3.3x10-2 mol/L*atm)(4 atm) =

CO2

SAMPLE PROBLEM 13.2

Using Henry’s Law to Calculate Gas Solubility

SOLUTION:

0.1 mol/L

slide43

amount (mol) of solute

Molarity (M)

volume (L) of solution

amount (mol) of solute

Molality (m)

mass (kg) of solvent

mass of solute

Parts by mass

mass of solution

volume of solute

Parts by volume

volume of solution

amount (mol) of solute

Mole fraction 

amount (mol) of solute + amount (mol) of solvent

Table 13.5 Concentration Definitions

Concentration Term

Ratio

slide44

PROBLEM:

Hydrogen peroxide is a powerful oxidizing agent used in concentrated solution in rocket fuels and in dilute solution a a hair bleach. An aqueous solution H2O2 is 30.0% by mass and has a density of 1.11 g/mL. Calculate its

0.882 mol H2O2

SAMPLE PROBLEM 13.5

Converting Concentration Units

(a) Molality

(b) Mole fraction of H2O2

(c) Molarity

PLAN:

(a) To find the mass of solvent we assume the % is per 100 g of solution. Take the difference in the mass of the solute and solution for the mass of peroxide.

(b) Convert g of solute and solvent to moles before finding c.

(c) Use the density to find the volume of the solution.

70.0 g H2O

SOLUTION:

(a)

g of H2O = 100. g solution - 30.0 g H2O2 =

30.0 g H2O2

mol H2O2

34.02 g H2O2

molality =

= 12.6 m H2O2

kg H2O

70.0 g H2O

103 g

slide45

mol H2O

18.02 g H2O

mL

1.11 g

L

103 mL

SAMPLE PROBLEM 13.5

Converting Concentration Units

continued

(b)

70.0 g H2O

= 3.88 mol H2O

0.882 mol H2O2

= 0.185 c of H2O2

0.882 mol H2O2 + 3.88 mol H2O

(c)

= 90.1 mL solution

100.0 g solution

0.882 mol H2O2

= 9.79 M H2O2

90.1 mL solution

a little harder
A little harder…
  • The electrolyte in car batteries is a 3.75 M sulfuric acid solution that has a density of 1.230 g/ml. Calculate the molality of the sulfuric acid.
colligative properties of sol ns
Colligative Properties of Sol’ns
  • The presence of solute particles changes the physical properties of the solution
  • Depends on NUMBER of solute particles
  • Colligative (“collective”) Properties:
    • Vapor pressure lowering, BP elevation, FP depression, osmotic pressure
vapor pressure of solutions
Vapor Pressure of Solutions

Which has higher vapor pressure: water or aqueous solution? Why?

collig properties of nonvolatile nonelectrolyte sol ns
Collig Properties of Nonvolatile Nonelectrolyte Sol’ns
  • i.e. sugar

Effect on Vapor Pressure:

Less surface area—

less evaporation

slide50

Figure 13.15

In terms of entropy?

l -> g increases entropy

Mixture already has higher entropy

Than the pure solvent, so

Less evaporation is necessary

predict what will happen
Predict What Will Happen

Higher

vapor

pressure

Lower

vapor

pressure

raoult s law
Raoult’s Law:

Vapor pressure of a solution is directly proportional to the mole fraction of solvent present.

Psoln = X solvent ∙ Posolvent

Vapor

pressure

of pure solvent

Observed

vapor

pressure

of solution

Mole fraction

of nonvolatile

solvent

slide53
Ideal Solution

one that follows Raoult’s Law at any concentration

Raoult’s Law works better for dilute solutions

slide54

PROBLEM:

Calculate the vapor pressure lowering, DP, when 10.0 mL of glycerol (C3H8O3) is added to 500. mL of water at 50.0C. At this temperature, the vapor pressure of pure water is 92.5 torr and its density is 0.988 g/mL. The density of glycerol is 1.26 g/mL.

0.988 g H2O

1.26 g C3H8O3

mol H2O

mol C3H8O3

mL H2O

mL C3H8O3

18.02 g H2O

92.09 g C3H8O3

c = 0.00498

SAMPLE PROBLEM 13.6

Using Raoult’s Law to Find the Vapor Pressure

Lowering

SOLUTION:

10.0 mL C3H8O3

= 0.137 mol C3H8O3

x

= 27.4 mol H2O

500.0 mL H2O

x

0.137 mol C3H8O3

DP =

x

92.5 torr

= 0.461 torr

0.137 mol C3H8O3 + 27.4 mol H2O

usefulness of raoult s law
Usefulness of Raoult’s Law
  • Calculate MW of solute

Psoln = X solvent ∙ P0solvent

Measure this

Measure this

slide56

Colligative Properties--Quantitative

Raoult’s Law (vapor pressure of a solvent above a solution, Psolvent)

Psolvent = csolvent X P0solvent

where P0solvent is the vapor pressure of the pure solvent

P0solvent - Psolvent = DP = csolute x P0solvent

what if the solute is volatile
What if the solute is volatile?

The solute and solvent contribute to Pvap

Ptotal = PA + PB = X A ∙ P0A + XB ∙ P0B

ideal vs nonideal solutions
Ideal vs Nonideal Solutions

Positive

Deviation

observed

pressure

higher than

predicted

Negative

Deviation

observed

pressure

lower than

predicted

Follows

Raoult’s

Law—

when solute

and solvent

are similar in

polarity

colligative properties

P atm

Colligative Properties

Pvap

Patm

1) Boiling-Point Elevation

  Normal Boiling Point: when ___ = ____

A non-volatile solute (increases, decreases) the vapor pressure of solvent

So, to boil the solution,

increase or decrease temp?

Pvap

colligative properties60
Colligative Properties

2) Freezing-Point Depression (link)

 Normal Freezing Point: when ___ =____

A non-volatile solute (increases, decreases) the vapor pressure of water

So, to freeze the solution,

increase or decrease temp?

Pvap

liquid

Pvap

solid

slide61

0oC

100oC

equations
Equations

Boiling Point Elevation Freezing Point Depression

DTb = Kb msolute -DTf = Kfmsolute

Kb = molal boiling point elevation constant

Kb for water = 0.51 o C/m

Kf = molal freezing point depression constant

Kf for water = 1.86 o C/m

sample problem
Sample Problem:

The following data were obtained in an experiment designed to find the molecular mass of a solute by freezing point depression.

Solvent: para-dichlorobenzene Kf : 7.0 oC/m

Freezing point of pure solvent: 53.02 oC Mass pdb: 24.80 g

Mass of unknown substance: 2.04 g FP of sol’n: 50.78 oC

Calculate the molecular mass of the solute.

slide64

PROBLEM:

You add 1.00 kg of ethylene glycol antifreeze (C2H6O2) to your car radiator, which contains 4450 g of water. What are the boiling and freezing points of the solution?

PLAN:

Find the # mols of ethylene glycol; m of the solution; multiply by the boiling or freezing point constant; add or subtract, respectively, the changes from the boiling point and freezing point of water.

mol C2H6O2

62.07 g C2H6O2

SAMPLE PROBLEM 13.7

Determining the Boiling Point Elevation and Freezing Point Depression of a Solution

SOLUTION:

1.00x103 g C2H6O2

= 16.1 mol C2H6O2

16.1 mol C2H6O2

= 3.62 m C2H6O2

4.450 kg H2O

DTbp =

0.512 0C/m

x

3.62m

= 1.850C

DTfp =

1.86 0C/m

x

3.62m

BP = 101.85 0C

FP = -6.73 0C

3 osmotic pressure
3) Osmotic Pressure :

Osmosis:

flow of solvent from high to low concentration

osmotic pressure:

 pressure needed to prevent osmosis

  Apply pressure to which side?

semi-permeable

membrane; only

lets small

particles through

osmosis equation
Osmosis Equation

How does osmotic pressure depend on concentration of solute?

pa M (directly proportional)

p = RT x M

p = MRT

constant

slide67

semipermeable membrane

The development of osmotic pressure

Applied pressure needed to prevent volume increase

Figure 13.17

osmotic pressure

pure solvent

solution

net movement of solvent

solute molecules

solvent molecules

interesting
Interesting…

p = MRT

p = RT

pV = nRT

PV = nRT

colligative properties of electrolyte solutions
Colligative Properties of Electrolyte solutions

C6H12O6 (s)  C6H12O6 (aq)

NaCl (s)  Na+1 (aq) + Cl-1

K2SO4 (s)  2K+1 (aq) + SO42- (aq)

Trend: DTf increases with more ions present

van t hoff factor i
Van’t Hoff factor (i)

2

1.9

3

2.7

1.3

2

ION PAIRING

1

1

Why is i (observed) < i (expected)?

slide71

Figure 13.19

Nonideal behavior of electrolyte solutions

slide72

Figure 13.20

An ionic atmosphere model for nonideal behavior of electrolyte solutions.

modified equations
  Modified equations:

 Boiling Point Elevation: DTb = i Kb msolute

 Freezing Point Depression: -DTf = i Kfmsolute

 Osmotic Pressure: p = iMRT

slide74

PROBLEM:

Biochemists have discovered more than 400 mutant varieties of hemoglobin, the blood protein that carries oxygen throughout the body. A physician studing a variety associated with a fatal disease first finds its molar mass (M). She dissolves 21.5 mg of the protein in water at 5.00C to make 1.50 mL of solution and measures an osmotic pressure of 3.61 torr. What is the molar mass of this hemoglobin variant?

PLAN:

We know p as well as R and T. Convert p to atm and T to degrees K. Use the p equation to find M and then the amount and volume of the sample to get to M.

atm

p

760 torr

RT

L

2.08x10-4 mol

L

103 mL

g

1

103 mg

3.12x10-8 mol

SAMPLE PROBLEM 13.8

Determining Molar Mass from Osmotic Pressure

3.61 torr

SOLUTION:

M =

=

= 2.08x10-4M

(0.0821 L*atm/mol*K)(278.1K)

(1.50 mL)

= 3.12x10-8 mol

21.5 mg

= 6.89x104 g/mol