13-1 Chapter 13 Properties of Mixtures: Solutions.

13-1

Chapter 13

Properties of Mixtures: Solutions

13-2

Properties of Mixtures: Solutions

13.1 Types of solutions: Intermolecular forces and predicting solubility

13.2 Energy changes in the solution process

13.3 Solubility as an equilibrium process

13.4 Quantitative ways of expressing concentration

13.5 Colligative properties of solutions

13-3

13-4

Figure 13.1

The major types of

intermolecular forces in solutions

(from Chapter 12)

(energies in parenthesis)

13-5

“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 partly replaced with solute-solvent interactions.

The new forces created between solute and solvent must be comparable in strength to the forces destroyed within the solute and the solvent.

13-6

A major factor that determines whether a solution forms:

The relative strengths of the intermolecular forces within andbetween solute and solvent molecules

13-7

Some Definitions

Solvent: the most abundant component of a given solutionSolute: component dissolved in the solvent

Solubility (S): the maximum amount of solute that dissolves in a fixed quantity of solvent at a given temperature (in the presenceof excess solute)

Dilute and concentrated solutions: qualitative terms

13-8

Figure 13.2

Hydration shells

around an aqueous ion

Formation of ion-dipoleforces when a salt dissolves

in water

13-9

Liquid Solutions

Liquid-Liquid Gas-Liquid

Gas and Solid Solutions

Gas-Gas Gas-Solid Solid-Solid

13-10

hexane =CH3(CH2)4CH3

Competitionbetween H-bonding

and dispersion forces

13-11

Figure 13.3

Molecular Basis for the Solubility of CH3OH in H2O

H-bonding: CH3OH can serve as a donor and acceptor(maximum number of three H-bonds / molecule)

13-12

(c) Diethyl ether can interact through dipole and dispersion forces. Ethanolcan provide both while water can only H-bond.

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

SAMPLE PROBLEM 13.1 Predicting relative solubilities of substances

SOLUTION:

PROBLEM: Predict which solvent will dissolve more of the given solute:

(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)

PLAN: Consider the intermolecular forces that exist between solute molecules and consider whether the new solvent-solute interactions can substitute for them.

(a) NaCl is ionic and forms ion-dipoles with the OH groups of bothmethanol and propanol. However, propanol is subject to greaterdispersion forces (more CH bonds than methanol).

13-13

Figure B13.1

Structure-Function Correlations: A Soap

Soap: the salt form of a long-chain fatty acid; is amphipathicin character (has polar and non-polar components)

13-14

Figure B13.2

The mode of action of the

antibiotic, Gramicidin A

Destroys the Na+/K+ ionconcentration gradients

in the cell

13-15

QuickTime™ and aPhoto - JPEG decompressor

are needed to see this picture.

Gas-Liquid Solutions

Non-polar gas solubility in water is directly related to the boiling point of the gas.

important toaquatic life

13-16

Gas-gas solutions: All gases are infinitely soluble in one another.

Gas-solid solutions: The gas molecules occupy the spacesbetween the closely packed particles of the solid.

Solid-solid solutions: alloys (substitutional or interstitial)

13-17

Figure 13.4

The arrangement of atoms in two types of alloys

13-18

Heats of solution and solution cycles

1. Solute particles separate from each other - endothermic

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

2. Solvent particles separate from each other - endothermic

3. Separate solute and solvent particles mix - exothermic

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

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

Dissolution of a solid: breaking down the process into three steps

13-19

Hsoln = Hsolute + Hsolvent + Hmix

Calculating the heat of solution, Hsoln

The total enthalpy change that occurs when a solution forms by dissolving a solute into a solvent.

A thermochemical solution cycle

13-20

Figure 13.5

Solution cycles and the enthalpy components of the heat of solution

13-21

Heats of Hydration

The solvation of ions by water is always exothermic.

M+ (g) [or X- (g)] M+ (aq) [or X- (aq)] Hhydr of the ion < 0H2O

Hhydr is related to the charge density of the ion, that is, both coulombic charge and ion size are important.

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

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

Thus, Hsoln = -Hlattice + Hhydr

(for 1 mole of gaseous ions)

13-22

Heats of Hydration and Ionic Character

For a given size, greater charge leads to a more (-) Hhydr

For a given charge, smaller size leads to a more (-) Hhydr

13-23

Table 13.4 Trends in Ionic Heats of Hydration

ion ionic radius (pm) Hhydr (kJ/mol)

Group 1A

Group 2A

Group 7A

Li+

Na+

K+

Rb+

Cs+

Mg2+

Ca2+

Sr2+

Ba2+

F-

Cl-

Br-

I-

76102138152167

-510-410-336-315-282

72100118

133

-1903-1591-1424

-431181 -313196 -284220 -247

135 -1317

13-24

Figure 13.6

Enthalpy Diagrams for Dissolving Three Different Ionic Compounds in Water

NaOH

NH4NO3NaCl

13-25

Entropy Considerations

The natural tendency of most systems is to become moredisordered; entropy increases.

Dissolution: involves a change in enthalpy and a change inentropy.

Entropy always favors the formation of solutions.

13-26

Figure 13.7

Enthalpy diagrams for dissolving NaCl and octane in hexane

In this case, dissolution isentropy-driven!

NaCl in insoluble in hexane!

13-27

When excess undissolved solute is in equilibrium withthe dissolved solute: a saturated solution

An unsaturated solution: more solute can be dissolved,ultimately producing a saturated solution

A supersaturated solution: a solution that contains morethan the equilibrium amount of dissolved solute

More Definitions

13-28

Figure 13.8

Equilibrium in a saturated solution

solute (undissolved) solute (dissolved)

13-29

Figure 13.9

Sodium acetate crystallizing from a supersaturated solution

nucleation a saturated solutionresults

13-30

Solubility and Temperature

Most solids are more soluble at higher temperatures.

The sign of the heat of solution, however, does not predict reliably the effectof temperature on solubility; e.g., NaOH and NH4NO3 have

Hsoln of opposite signs, yet their solubility in H2O increaseswith temperature.

13-31

Figure 13.10

The relation between

solubility and temperature for several

ionic compounds

13-32

Gas Solubility in Water: Temperature Effects

For all gases, Hsolute = 0, Hhydr < 0; thus, Hsoln < 0

Implications: gas solubility in water decreases with increasing temperature

solute(g) + water(l) saturated solution(aq) + heat

13-33

Figure 13.11

Thermal Pollution

Leads to O2

deprivation in aquaticsystems

13-34

Pressure Effects on Solubility

Essentially zero for solids and liquids, but substantial for gases!

gas + solvent saturated solution

13-35

Figure 13.12

The effect of pressure on gas solubility

gas volume is reduced;pressure (concentration!)increases; more collisionsoccur with liquid surface

13-36

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.

A quantitative relationshipbetween gas solubility and

pressure

kH = Henry’s law constant

for a gas; units of mol/L.atm

Implications for scuba diving!

13-37

SAMPLE PROBLEM 13.2 Using Henry’s Law to calculate gas solubility

PLAN:

SOLUTION:

PROBLEM: The partial pressure of carbon dioxide gas inside a bottle of cola is 4 atm at 25 oC. What is the solubility of CO2? The

Henry’s law constant for CO2 dissolved in water is 3.3 x 10-2

mol/L.atm at 25 oC.Knowing kH and Pgas, we can substitute into the Henry’s Law equation.

0.1 mol / LS = (3.3 x 10-2 mol/L.atm)(4 atm) =CO2

13-38

Table 13.5 Concentration Definitions

concentration term ratio

molarity (M)amount (mol) of solutevolume (L) of solution

molality (m)amount (mol) of solute

mass (kg) of solvent

parts by massmass of solute

mass of solution

parts by volumevolume of solute

volume of solution

mole fraction amount (mol) of solute

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

13-39

SAMPLE PROBLEM 13.3 Calculating molality

PLAN:

SOLUTION:

PROBLEM: What is the molality of a solution prepared by dissolving 32.0 g of CaCl2 in 271 g of water?

Convert grams of CaCl2 into moles and grams of water to kg. Then substitute into the equation for molality.

molality =

= 0.288 mole CaCl2

271 g H2O

0.288 mole CaCl2

kg

103 gx

= 1.06 m CaCl2

32.0 g CaCl2mole CaCl2

110.98 g CaCl2x

13-40

Figure 13.13

The Sex Attractant of the Gypsy Moth Potent at Extremely Low Concentrations!

100-300 molecules/mL air

100 parts per quadrillion by volume!

Practical Implications: a strategy used to target andtrap specific insects (Japanese beetles)

13-41

Other Expressions of Concentration

mass percent (% w/w) = mass solute / mass of solution x 100 (related to parts per million (ppm) or parts per billion (ppb))

volume percent (% (v/v) = volume solute / volume of solution x 100

% (w/v) = solute mass / solution volume x 100

mole percent (mol%) = mole fraction x 100

13-42

SAMPLE PROBLEM 13.4 Expressing concentration in parts by mass, parts by volume, and mole fraction

PLAN:

PROBLEM: (a) Find the concentration of calcium (in ppm) in a 3.50 g pill thatcontains 40.5 mg of Ca.

(b) The label on a 0.750 liter bottle of Italian chianti indicates“11.5% alcohol by volume”. How many liters of alcoholdoes the wine contain?

(c) A sample of rubbing alcohol contains 142 g of isopropylalcohol (C3H7OH) and 58.0 g of water. What are the

mole fractions of alcohol and water?

(a) Convert mg to g of Ca, find the ratio of g Ca to g pill, and multiplyby 106.

(b) Knowing the % alcohol and the total volume, the volume ofalcohol can be calculated.

(c) Convert g of solute and solvent to moles, and find the ratios ofeach part to the total.

13-43

SAMPLE PROBLEM 13.4

SOLUTION:

(continued)

(a)

3.5 g

103 mg

g40.5 mg Ca x

106x = 1.16 x 104 ppm Ca

(b) 11.5 L alcohol

100. L chianti0.750 L chianti x = 0.0862 L alcohol

(c) moles isopropyl alcohol = 142 gmole

60.09 g = 2.36 mol C3H7OH

moles water = 58.0 gmole

18.02 g = 3.22 mol H2O

2.36 mol C3H7OH

2.36 mol C3H7OH + 3.22 mol H2O

3.22 mol H2O

2.36 mol C3H7OH + 3.22 mol H2O

= 0.423 C3H7

OH

= 0.577 H2O

x

x

13-44

SAMPLE PROBLEM 13.5 Converting concentration units

PLAN:

SOLUTION:

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

(a) molality (b) mole fraction (c) molarity

(a) To find the mass of solvent, assume the % is per 100 g of solution. Take the difference in the mass of the solute and solution to determine the mass of solvent.

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

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

(a) g of H2O = 100. g solution - 30.0 g H2O2 = 70.0 g H2O

molality =

30.0 g H2O234.02 g H2O2

mol H2O2

70.0 g H2Okg H2O103 g

= 12.6 m H2O2

x

x

13-45

SAMPLE PROBLEM 13.5 (continued)

(b)

0.882 mol H2O2

70.0 g H2Omol H2O

18.02 g H2O= 3.88 mol H2O

0.882 mol H2O2 + 3.88 mol H2O= 0.185 = of H2O2

(c)100.0 g solution

mL

1.11 g= 90.1 mL solution

0.882 mol H2O2

90.1 mL solution L

103 mL

= 9.79 M H2O2

x

x

x

13-46

Colligative Properties

Physical properties of solutions dictated by the number ofsolute particles present. Their chemical structures are

not factors in determining these properties!

vapor pressure lowering boiling point elevation freezing point depression osmotic pressure

13-47

Figure 13.14

Three types of electrolytes

strong

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

weak

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

non-electrolyte

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

13-48

Vapor Pressure Lowering

The vapor pressure of a solution of a nonvolatile nonelectrolyteis always lower than the vapor pressure of the pure solvent.

Figure 13.15

An entropy argument!

13-49

Quantitative Treatment of VP Lowering

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

Psolvent = solvent x Posolvent

where Posolvent = vapor pressure of the pure solvent

Posolvent - Psolvent = P = solute x Po

solvent

How does the amount of solute affect the magnitude of the VP lowering?( substitute 1- solute for solvent in the above equation and rearrange)

(change in VP is proportional to the mole fraction of solute)

13-50

SAMPLE PROBLEM 13.6 Using Raoult’s Law to find the vapor pressure lowering

SOLUTION:

PROBLEM: Calculate the vapor pressure lowering, P, when 10.0 mL of glycerol (C3H8O3) is added to 500. mL of water at 50. oC. 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.

PLAN: Find the mol fraction, , of glycerol in solution and multiply by the vapor pressure of water.

10.0 mL C3H8O3

1.26 g C3H8O3

mL C3H8O3

mol C3H8O3

92.09 g C3H8O3= 0.137 mol C3H8O3

500.0 mL H2O0.988 g H2O

mL H2O

mol H2O

18.02 g H2O= 27.4 mol H2O

P = 0.137 mol C3H8O3

0.137 mol C3H8O3 + 27.4 mol H2O92.5 torrx

x

x

= 0.461 torr

= 0.00498

x

x

13-51

Boiling Point Elevation

A solution boils at a higher temperature than the pure solvent.

This effect is explained by differences between the VP ofthe solution and VP of the pure solvent at a given temperature.

13-52

Figure 13.16

Superimposed phase diagrams of solvent and solution

aqueous solution:dashed lines

pure water:solid lines

13-53

(m = solution molality, Kb = molal BP elevation constant, Tb = BP elevation)

Quantitative Treatment of BP Elevation

Tb = Kbm

The magnitude of the effect is proportional to solute concentration.

Tb = Tb (solution) - Tb (solvent)

13-54

(m = solution molality, Kf = molal FP depression constant, Tf = FP depression)

Quantitative Treatment of FP Depression

Tf = Kfm

The magnitude of the effect is proportional to solute concentration.

Tf = Tf (solvent) - Tf (solution)

13-55

Table 13.6 Molal Boiling Point Elevation and Freezing Point Depression Constants of Several Solvents

solventboiling

point (oC)* Kb (oC/m) Kf (

oC/m)

melting

point (oC)

acetic acid

benzene

carbon disulfide

carbon tetrachloride

chloroform

diethyl ether

ethanol

water

117.9

80.1

46.2

76.5

61.7

34.5

78.5

100.0

3.07 16.6 3.90

2.53 5.5 4.90

2.34 -111.5 3.83

5.03 -23 30.

3.63 -63.5 4.70

2.02 -116.2 1.79

1.22 -117.3 1.99

0.512 0.0 1.86

*at 1 atm.

13-56

SAMPLE PROBLEM 13.7 Determining the boiling point elevation and freezing point depression of a solution

SOLUTION:

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 resulting solution?

PLAN: Find the number of mols of ethylene glycol and 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.

1.00 x 103 g C2H6O2

mol C2H6O2

62.07 g C2H6O2

= 16.1 mol C2H6O2

Tb = 0.512 oC/m

16.1 mol C2H6O2

4.450 kg H2O= 3.62 m C2H6O2

3.62 mx = 1.85 oC

BP = 101.85 oC

Tf = 1.86 oC/m 3.62 mx

FP = -6.73 oC

x

13-57

Osmotic Pressure

Applies only to aqueous solutions!

Two solutions of different concentrations are separated by a semi-permeable membrane (allows water but notsolute to pass through)

13-58 Figure 13.17

The development of osmotic pressure

pure solvent

solution

applied pressure needed to prevent volume

increase; equal to

the osmotic pressure

osmotic pressure

semipermeablemembrane

13-59

T is the Kelvin temperature

Quantitative Treatment of Osmotic Pressure ()

OP is proportional to the number of solute particles in a givenvolume of solution (to M).

nsolute/Vsoln or M

The constant of proportionality = RT, so = M x R x T

13-60

Underlying Principle of Colligative Properties

Each property stems from an inability of solute particles tocross between two phases.

13-61

Determination of Solute Molar Mass by Exploiting Colligative Properties

In principle, any colligative property can be used, but OP gives the most accurate results (better dynamic range).

13-62

SAMPLE PROBLEM 13.8 Determining molar mass from osmotic pressure

SOLUTION:

PROBLEM: Biochemists have discovered more than 400 mutant varieties of hemoglobin (Hb), the blood protein that carries oxygen throughout the body. A physician studying a form of Hb associated with a fatal disease first finds its molar mass (M). She dissolves 21.5 mg of the protein in water at 5.0 oC to make 1.50 mL of solution and measures an osmotic pressure of 3.61 torr. What is the molar mass of this Hb mutant?

PLAN: We know as well as R and T. Convert to atm and T to Kelvin. Use the equation to find the molarity M and then the amount and volume of the sample to calculate M.

M =

RT= 3.61 torr

atm

760 torr

(0.0821 L . atm/mol . K)(278.15 K)

= 2.08 x 10-4 M

2.08 x 10-4 molL 1.50 mL

103 mL

L= 3.12 x 10-7 mol

21.5 mgg

103 mg

1

3.12 x 10-7 mol= 6.89 x 104 g/mol

x

x x

xx

# mol = g/M

13-63

Fractional Distillation of VolatileNonelectrolytes

The presence of each volatile component lowers the vapor pressure of the other.

partial pressure = mole fraction x vapor pressure of pure gas

For vapor: mole fraction = partial pressure / total pressure(thus, the vapor has a higher mole fraction of the more volatile

solution component)

13-64

Figure 13.18

The process of fractional distillation

Gas

Gasoline 38 oCKerosene 150 oC

Heating oil 260 oC

Lubricating oil 315 oC - 370 oC

Crude oil vapors from heater

Steam

Residue (asphalt, tar)

Condenser

Gasoline vapors

13-65

Colligative Properties of Electrolyte Solutions

Must consider the full dissociation into ions!

van’t Hoff factor (i) = measured value for electrolyte solution

expected value for nonelectrolyte solution

This factor is multiplied into the appropriate equations; for example, = i (MRT).

For ideal behavior, i = mol particles in solution / mol dissolved solute

But solutions are not ideal; for example, for BP elevation of NaClsolutions, i = 1.9, not 2!

Data suggest that the ions are not behaving as independent particles!

13-66

Figure 13.19

Non-ideal behavior of electrolyte solutions

Observed values of i are lessthan the predicted (expected)

values.

13-67

Figure 13.20

An ionic atmosphere model for non-ideal

behavior of electrolyte solutions

ionic atmospheres

Concept of effectiveconcentration

13-68

Some Practical Applications

ion-exchange (water softeners)

water purification

13-69

Figure B13.4

Ion exchange for removal of

hard-water cations

Use of ion-exchangeresins

13-70

Figure B13.5

Reverse osmosis for the removal of ions

Desalination Process

13-71

End of Assigned Material

13-72

Figure 13.21

Light scattering and the Tyndall effect

Photo by C.A.Bailey, CalPoly SLO (Myanmar)

13-73 Figure 13.22

A Cottrell precipitator for removing particulates from industrial smokestack gases

13-74

Figure B13.3

The steps in a typical municipal water treatment plant