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Chapte r 11Chapte r 11

Solutions and Their PropertiesSolutions and Their Properties

Hemolysis(bursts) shrinks(crenates)

Chapter 11 Slide 2

SolutionsSolutionsSolution: A homogeneous mixture ( uniform in composition ).

Solvent: The major component.

Solute: A minor component.

Chapter 11 Slide 3

• Saturated: Contains the maximum amount of solute that will dissolve in a given solvent.

• Unsaturated: Contains less solute than a solvent has the capacity to dissolve.

• Supersaturated: Contains more solute than would be present in a saturated solution.

• Crystallization: The process in which dissolved solute comes out of the solution and forms crystals.

Solution Formation 01Solution Formation 01

• Saturated: Contains the maximum amount of solute that will dissolve in a given solvent.

• Unsaturated: Contains less solute than a solvent has the capacity to dissolve.

• Supersaturated: Contains more solute than would be present in a saturated solution.

• Crystallization: The process in which dissolved solute comes out of the solution and forms crystals.

Chapter 11 Slide 4


Chapter 11 Slide 5


Chapter 11 Slide 6

How Does a Solution Form?How Does a Solution Form?

As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them.

Chapter 11 Slide 7

How Does a Solution FormHow Does a Solution Form

If an ionic salt is soluble in water, it is because the ion-dipole interactions are strong enough to overcome the lattice energy of the salt crystal.

Chapter 11 Slide 8

Entropy of Solution Formation 03Entropy of Solution Formation 03

∆S <0

Energy Changes and the Solution ProcessEnergy Changes and the Solution Process

ΔG = ΔH - TΔS

-ΔG+ΔG

spontaneous:nonspontaneous:

+ΔH-ΔH

endothermic:exothermic:

Chapter 11 Slide 10


• Exothermic ∆Hsoln:

• The solute–solvent interactions are stronger than solute–solute or solvent–solvent.

Chapter 11 Slide 11


• Endothermic ∆Hsoln:

• The solute–solvent interactions are weaker than solute–solute or solvent–solvent.

Chapter 11 Slide 12

Why Do Endothermic Processes Occur?Why Do Endothermic Processes Occur?

Things do not tend to occur spontaneously (i.e., without outside intervention) unless the energy of the system is lowered.

ΔG<0

Chapter 11 Slide 13

Why Do Endothermic Processes Occur?Why Do Endothermic Processes Occur?

Yet we know that in some processes, like the dissolution of NH4NO3 in water, heat is absorbed, not released.

Chapter 11 Slide 14

Enthalpy Is Only Part of the PictureEnthalpy Is Only Part of the Picture

The reason is that increasing the disorder or randomness (known as entropy) of a system tends to lower the energy of the system.

Chapter 11 Slide 15

Enthalpy Is Only Part of the PictureEnthalpy Is Only Part of the Picture

So even though enthalpy may increase, the overall energy of the system can still decrease if the system becomes more disordered.

Chapter 11 Slide 16


• Solubility: A measure of how much solute will dissolve in a solvent at a specific temperature.

• Miscible: Two (or more) liquids that are completely soluble in each other in all proportions.

• Solvation: The process in which an ion or a molecule is surrounded by solvent molecules arranged in a specific manner.

Chapter 11 Slide 17

Solution FormationSolution Formation

Predict whether each of the following substances is more likely to dissolve in carbon tetrachloride (CCl4) or in water: C7H16, Na2SO4, HCl, and I2.

Chapter 11 Slide 18

Solution FormationSolution Formation

C7H16 is a hydrocarbon, so it is molecular and nonpolar. Na2SO4, a compound containing a metal and nonmetals, is ionic; HCl, a diatomic molecule containing two nonmetals that differ in electronegativity, is polar; and I2, a diatomic molecule with atoms of equal electronegativity, is nonpolar. We would therefore predict that C7H16and I2 would be more soluble in the nonpolar CCl4than in polar H2O, whereas water would be the better solvent for Na2SO4 and HCl.

Chapter 11 Slide 19

C5H12 < C5H11 Cl < C5H11 OH < C5H10(OH)2(in order of increasing polarity and hydrogen-bonding ability)

Solution Formation

Arrange the following substances in order of increasing solubility in water:

Chapter 11 Slide 20


1. Is iodine (I2) more soluble in water or in carbon disulfide (CS2)?

2. Which would have the largest (most negative) hydration energy and which should have the smallest? Al3+, Mg2+, Na+

Chapter 11 Slide 21

Size and Charge of the ionSize and Charge of the ion

• The smaller the ion and the greater its charge, it will be the more hydrated, therefore more hydration energy

Chapter 11 Slide 22

Concentration Units 01Concentration Units 01

Chapter 11 Slide 23

• Concentration: The amount of solute present in a given amount of solution.

• Percent by Mass (weight percent): The ratio of the mass of a solute to the mass of a solution, multiplied by 100%.

% bymassof solute =mass of solute

mass of solution × 100%

mass of solution =mass of solute +mass of solvent


Chapter 11 Slide 24


• Parts per Million:

• Parts per million (ppm) =

= % mass x 104

• One ppm gives 1 gram of solute per 1,000,000 g or one mg per kg of solution. For dilute aqueous solutions this is about 1 mg per liter of solution.

610xsolutionofmassTotal

componentofMass

Chapter 11 Slide 25

A solution is made by dissolving 13.5 g of glucose (C6H12O6) in 0.100 kg of water.a) What is the mass percentage of solute in this

solution?b) What is the ppm of solute in this solution?

Problem:

Chapter 11 Slide 26

SolutionSolution

Mass ppm Glucose = 11.9 x 104

Chapter 11 Slide 27


• Mole Fraction (X):

• Molarity (M):

• Molality (m):

molesofnumber TotalAofMoles

=AX

SOLUTIONof LiterssoluteofMolesMolarity =

SOLVENT of KilogramssoluteofMoles=Molality

Chapter 11 Slide 28

Calculating ConcentrationsCalculating ConcentrationsDissolve 62.1 g (1.00 mol) of ethylene glycol in Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H250. g of H22O. Calculate mol fraction, O. Calculate mol fraction, molalitymolality, , and weight % of glycol.and weight % of glycol.

Chapter 11 Slide 29

Calculating ConcentrationsCalculating Concentrations

250. g H250. g H22O = 13.9 molO = 13.9 mol

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate X, m, and % of glycol.Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of Hof H22O. Calculate X, m, and % of glycol.O. Calculate X, m, and % of glycol.

Xglycol = 1.00 mol glycol1.00 mol glycol + 13.9 mol H2O

X X glycolglycol = 0.0672= 0.0672

Chapter 11 Slide 30


Calculate molalityCalculate Calculate molalitymolality


conc (molality) = 1.00 mol glycol0.250 kg H2O

= 4.00 molal

%glycol = 62.1 g62.1 g + 250. g

x 100% = 19.9%

Calculate weight %Calculate weight %

Chapter 11 Slide 31

Effect of Temperatureon Solubility of Solids in water 01Effect of Temperatureon Solubility of Solids in water 01

heat + Glucose + water → dissolved Glucose

Chapter 11 Slide 32

Effect of Temperature on Solubility of Gases in Water 02Effect of Temperature on Solubility of Gases in Water 02

gaseous O2 + water → saturated O2 solution + heat

Chapter 11 Slide 33

The Effect of Pressure on theSolubility of Gases, Henry's Law 01The Effect of Pressure on theSolubility of Gases, Henry's Law 01

• Henry’s Law: • The solubility of a gas is proportional to the pressure of the gas over the solution.

c ∝ Pc = k·P

gg kPS =

Chapter 11 Slide 34

Flash Animation - Click to ContinueFlash Animation Flash Animation -- Click to ContinueClick to Continue

The Effect of Pressure on theSolubility of Gases 02The Effect of Pressure on theSolubility of Gases 02

Chapter 11 Slide 35

Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over the liquid at 25°C. The Henry’s law constant for CO2 in water at this temperature is 3.1 × 10–2 mol/L-atm.

PRACTICE EXERCISECalculate the concentration of CO2 in a soft drink after the bottle is opened and equilibrates at 25°C under a CO2 partial pressure of 3.0 × 10–4 atm.

2

Chapter 11 Slide 36

Colligative Propertiesof Nonvolatile Solutes 01Colligative Propertiesof Nonvolatile Solutes 01

• Colligative Properties: Depend only on the number of

solute particles in solution. Solutions of electrolytes (which dissociate upon dissolving) should show greater changes than those of nonelectrolytes. These changes are :

1. Vapor pressure lowering2. Freezing point depression3. Boiling point elevation4. Osmotic pressure

Chapter 11 Slide 37

Vapor PressureVapor Pressure

Because of solute-solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase.

Chapter 11 Slide 38

Vapor PressureVapor Pressure

Therefore, the vapor pressure of a solution is lower than that of the pure solvent.

Chapter 11 Slide 39

Calculating Change of Vapor Pressure when the solute is

Nonvolatile . 03Calculating Change of Vapor Pressure when the solute is

Nonvolatile . 03

• Raoult’s Law: P1 = x1P°1 where x1 is the solvent molefraction.

• For a single solute solution, x1= 1 – x2 , where x2 is the solute mole fraction.

• We can obtain an expression for the change in vapor pressure of the solvent (the vapor pressure lowering).

ΔP = P°1 – P1 = P°

1 – x1 P°1 = P°

1 – (1 – x2 ) P°1

∆P = x2 P°1

Chapter 11 Slide 40

Assume the solution containing 62.1 g of glycol in 250. g of Assume the solution containing 62.1 g of glycol in 250. g of water . What is the vapor pressure of water over the solution atwater . What is the vapor pressure of water over the solution at30 30 ooCC? ? (The VP of pure H(The VP of pure H22O is 31.8 mm Hg at 30 O is 31.8 mm Hg at 30 °°C)C)

250. g H250. g H22O = 13.9 molO = 13.9 mol


Xglycol = 1.00 mol glycol1.00 mol glycol + 13.9 mol H2O

X X glycolglycol = 0.0672= 0.0672

Chapter 11 Slide 41


Calculate molalityCalculate Calculate molalitymolality


conc (molality) = 1.00 mol glycol0.250 kg H2O

= 4.00 molal

%glycol = 62.1 g62.1 g + 250. g

x 100% = 19.9%

Calculate weight %Calculate weight %

Chapter 11 Slide 42

RaoultRaoult’’ss LawLawAssume the solution containing 62.1 g of glycol in 250. g of watAssume the solution containing 62.1 g of glycol in 250. g of water . er . What is the vapor pressure of water over the solution at 30 What is the vapor pressure of water over the solution at 30 ooCC? ? (The VP of pure H(The VP of pure H22O is 31.8 mm Hg)O is 31.8 mm Hg)

SolutionSolutionXXglycolglycol = 0.0672 = 0.0672 ( see previous slides)( see previous slides) and so and so XXwaterwater = ?= ?Because Because XXglycolglycol + + XXwaterwater = 1= 1XXwaterwater = 1.000 = 1.000 -- 0.0672 = 0.93280.0672 = 0.9328PPwaterwater = = XXwaterwater •• PPoo

waterwater = (0.9328)(31.8 mm Hg)= (0.9328)(31.8 mm Hg)PPwaterwater = 29.7 mm Hg= 29.7 mm Hg

Chapter 11 Slide 43

Calculating Change of Vapor Pressure when the solute is

Volatile . 01Calculating Change of Vapor Pressure when the solute is

Volatile . 01

• What happens if both components are volatile(have measurable vapor pressures)?

• The vapor pressure has a value intermediate between the vapor pressures of the two pure liquids.

PT = PA + PB = XAP°A + XBP°

B

Chapter 11 Slide 44

Vapor Pressure of Two Volatile Liquids 02Vapor Pressure of Two Volatile Liquids 02

Chapter 11 Slide 45

Boiling-Point Elevation and Freezing-Point DepressionBoiling-Point Elevation and Freezing-Point Depression

Chapter 11 Slide 46

Boiling-Point Elevation and Freezing-Point Depression 01Boiling-Point Elevation and Freezing-Point Depression 01

• Boiling-Point Elevation (∆Tb): The boiling point of the solution (Tb) minus the boiling point of the pure solvent (T°

b):

∆Tb = Tb – T°b

∆Tb is proportional to concentration in molality:

∆Tb = Kb mKb = molal boiling-point elevation constant.

Chapter 11 Slide 47

Change in Boiling Point Change in Boiling Point Dissolve 62.1 g of glycol (1.00 mol) in 250. g of Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the BP of the solution?water. What is the BP of the solution?

KKBPBP = +0.512 = +0.512 ooC.kgC.kg/mol for water /mol for water (see Table (see Table 11.4).11.4).

SolutionSolution1.1. Calculate solution Calculate solution molalitymolality = 4.00 m = 4.00 m (see (see

previous slides).previous slides).2.2. ∆∆TTBPBP = K= KBPBP •• mm

∆∆TTBPBP = +0.512 = +0.512 ooC.kgC.kg/mol/mol (4.00 (4.00 molalmolal))∆∆TTBPBP = +2.05 = +2.05 ooCCBP = 102.05 BP = 102.05 ooCC

Chapter 11 Slide 48


• Freezing-Point Depression (∆Tf): The freezing point of the pure solvent (T°

f) minus the freezing point of the solution (Tf).

∆Tf = T°f – Tf

∆Tf is proportional to concentration:

∆Tf = Kf mKf = molal freezing-point depression constant.

Chapter 11 Slide 49


Chapter 11 Slide 50

Calculate the FP of a 4.00 Calculate the FP of a 4.00 molalmolalglycol/water solution.glycol/water solution.

KKFPFP = 1.86 = 1.86 ooC/molalC/molal (Table 11.4)(Table 11.4)SolutionSolution∆∆TTFPFP = K= KFPFP •• mm

= (= (--1.86 1.86 ooC/molal)(4.00 m)C/molal)(4.00 m)∆∆TTFP FP = = -- 7.44 7.44 ooCCRecall that Recall that ∆∆TTBPBP = +2.05 = +2.05 ˚̊C for this solution.C for this solution.

Freezing Point DepressionFreezing Point Depression

Chapter 11 Slide 51

Colligative Properties of ElectrolytesColligative Properties of Electrolytes

Colligative Properties: Depend only on the number of solute particles in solution However, a 1 M solution of NaCldoes not show twice the change in freezing point that a 1 Msolution of methanol does.

Chapter 11 Slide 52

van’t Hoff Factorvan’t Hoff Factor

One mole of NaCl in water does not really give rise to two moles of ions.

Chapter 11 Slide 53

van’t Hoff Factorvan’t Hoff Factor

Some Na+ and Cl−reassociate for a short time, so the true concentration of particles is somewhat less than two times the concentration of NaCl.

Chapter 11 Slide 54

The van’t Hoff FactorThe van’t Hoff Factor• Reassociation is more likely

at higher concentration.• Therefore, the number of

particles present is concentration dependent.

Chapter 11 Slide 55

The van’t Hoff FactorThe van’t Hoff Factor

We modify the previous equations by multiplying by the van’t Hoff factor, i

ΔTf = Kf m i

Chapter 11 Slide 56

Approximation of Van’t Hoff Factor, iApproximation of Van’t Hoff Factor, i

• Van’t Hoff Factor, i:• At very low concentrations, This factor is approximately the number of ions produced from each molecule of a compound upon dissolving.i = 1 for CH3OH i = 3 for CaCl2i = 2 for NaCl i = 5 for Ca3(PO4)2

• For compounds that dissociate on dissolving, use:

∆Tb = i⋅Kb m ∆Tf = i⋅Kf m ∆P = i⋅x2 P°

1 (X2 solute mole fraction)

Chapter 11 Slide 57

How much How much NaClNaCl must be dissolved in 4.00 kg of water must be dissolved in 4.00 kg of water to lower FP to to lower FP to --10.00 10.00 ooCC?.?.

SolutionSolution

Calc. required Calc. required molalitymolality

∆Tf = i⋅Kf m 10.00 10.00 ooCC = (1.86 = (1.86 ooC/molalC/molal) ) •• (( i.mi.m))

(( i.mi.m)) = 5.38 = 5.38 molalmolal , , i = 2i = 2

m = 5.38 m = 5.38 molalmolal / 2 = / 2 = 2.69 mol 2.69 mol NaClNaCl / kg ( 58.5g/mol)/ kg ( 58.5g/mol)157 g 157 g NaClNaCl / kg/ kg(157 g (157 g NaClNaCl / kg)/ kg)••(4.00 kg) = (4.00 kg) = 629 g 629 g NaClNaCl

Freezing Point DepressionFreezing Point Depression

Chapter 11 Slide 58

List the following aqueous solutions in order of their expected freezing point: 0.050 m CaCl2, 0.15 m NaCl, 0.10 m HCl, 0.050 m HC2H3O2, 0.10 m C12H22O11.

Because the freezing points depend on the total molality of particles in solution, the expected ordering is 0.15 m NaCl (lowest freezing point), 0.10 mHCl, 0.050 m CaCl2, 0.10 m C12H22O11, and 0.050 m HC2H3O2, (highest freezing point).

Chapter 11 Slide 59

Osmosis and Osmotic Pressure 01Osmosis and Osmotic Pressure 01

Chapter 11 Slide 60

• During osmosis, water flows across the semipermeable membrane from the 4% starch solution into the 10% solution.

semipermeablemembrane

4% starch

10% starch

H2O

OsmosisOsmosis

Chapter 11 Slide 61

• Eventually, the flow of water across the semipermeable membrane becomes equal in both directions.

7% starch

7% starch

H2O

EquilibriumEquilibrium

Chapter 11 Slide 62

Chapter 11 Slide 63Fig. 7-15b, p.229

Chapter 11 Slide 64

Osmosis in Blood CellsOsmosis in Blood Cells

• If the solute concentration outside the cell is greater than that inside the cell, the solution is hypertonic.

• Water will flow out of the cell, and crenation results.

Chapter 11 Slide 65

Osmosis in CellsOsmosis in Cells

• If the solute concentration outside the cell is less than that inside the cell, the solution is hypotonic.

• Water will flow into the cell, and hemolysis results.

Chapter 11 Slide 66

• Osmosis: The selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one.

• Osmotic pressure (π or ∏): The pressure required to stop osmosis.

π = i⋅MRTR = 0.08206 (L⋅atm)/(mol⋅K)M = M = MolarityMolarity mol/Lmol/L


Chapter 11 Slide 67


Chapter 11 Slide 68


Chapter 11 Slide 69


• Isotonic: Solutions have equal concentration of solute, and so equal osmotic pressure.

• Hypertonic: Solution with higher concentration of solute.

• Hypotonic: Solution with lower concentration of solute.

Chapter 11 Slide 70

Uses of Colligative Properties 01Uses of Colligative Properties 01

• Desalination:

Chapter 11 Slide 71

Osmosis Osmosis Calculating a Molar MassCalculating a Molar Mass

Dissolve 35.0 g of hemoglobin in enough Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. water to make 1.00 L of solution. ∏∏measured to be 10.0 mm Hg at 25 measured to be 10.0 mm Hg at 25 ˚̊C. C. Calc. molar mass of hemoglobin.Calc. molar mass of hemoglobin.

Solution : Solution : ∏∏ = = iMRTiMRT(a)(a) Calc. Calc. ∏∏ in atmospheresin atmospheres

∏∏ = 10.0 mmHg = 10.0 mmHg •• (1 (1 atmatm / 760 / 760 mmHg)mmHg)

= 0.0132 = 0.0132 atmatm(b)(b) Calc. concentrationCalc. concentration

Chapter 11 Slide 72

Osmosis Osmosis Calculating a Molar MassCalculating a Molar Mass

M = 5.39 x 10M = 5.39 x 10--44 mol/Lmol/L(c)(c) Calc. molar massCalc. molar mass

Molar mass = 35.0 g / 5.39 x 10Molar mass = 35.0 g / 5.39 x 10--44 mol/Lmol/LMolar mass = 65,100 g/molMolar mass = 65,100 g/mol

Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. of solution. ∏∏ measured to be 10.0 mm Hg at 25 measured to be 10.0 mm Hg at 25 ˚̊C. Calc. C. Calc. molar mass of hemoglobin.molar mass of hemoglobin.

SolutionSolution(b)(b) Calc. concentration from Calc. concentration from ∏∏ = = iMRTiMRT

Conc = 0.0132 atm(0.0821 L • atm/K • mol)(298K)

Chapter 11 Slide 73


• Fractional Distillation is the separation of volatile liquid mixtures into fractions of different composition.

Fractional Distillation of Liquid MixturesFractional Distillation of Liquid Mixtures

Chapter 11 Slide 75

Fractional Distillation of Liquid MixturesFractional Distillation of Liquid Mixtures

Chapter 11 Slide 76

Fractional DistillationFractional Distillation

http://www.indiana.edu/~issgraph/crandall.html

http://www.indiana.edu/~issgraph/crandall.html

Chapter 11 Slide 77


• Fractional distillation can be represented on a phase diagram by plotting temperature against composition.

http://www2.wwnorton.com/college/chemistry/gilbert/tutorials/ch12.htm

Chapter 11 Slide 78

Molar heat of vaporization (ΔHvap) is the energy required to vaporize 1 mole of a liquid.

ln P = -ΔHvap

RT+ C

Clausius-Clapeyron EquationP = (equilibrium) vapor pressure

T = temperature (K)

R = gas constant (8.314 J/K•mol)

Chapter 11 Slide 7979

• By taking measurements at two temps, we get:

⎟⎠

⎞⎜⎝

⎛−

Δ=

12

vap

21 11ln

TTRH

PP

ln P = -ΔHvap

RT+ C

Clausius-Clapeyron Equation

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