The Chemistry of Water and theWater and the Nature...

The Chemistry of Water and theWater and the

Nature of Liquids

Chapter 11

11.2

CHAPTER OUTLINE

I. The Structure of Water: An Introduction to Intermolecular Forces II A Closer Look at Intermolec lar ForcesII. A Closer Look at Intermolecular Forces

A. London Dispersion Forces : Induced Dipoles, B. Permanent Dipole : Dipole Forces, C. Hydrogen Bonds

III Impact of Intermolecular Forces on the Physical Properties of Water IIII. Impact of Intermolecular Forces on the Physical Properties of Water, I IV. Phase Diagrams V. Impact of Intermolecular Forces on the Physical Properties of Water, II

A Vi it B S f T i C C ill A tiA. Viscosity, B. Surface Tension, C. Capillary Action

VI. Water: The Universal Solvent A. Why Do So Many Substances Dissolve in Water?

VII M f S l ti C t tiVII. Measures of Solution Concentration A. Measures Based on Moles, B. Measures Based on Mass

VIII. The Effect of Temperature and Pressure on Solubility A. Temperature Effects, B. Pressure Effects

IX. Colligative Properties A. Vapor Pressure Lowering, B. Boiling Point Elevation, C. Freezing Point Depression,

D Osmosis E Reverse Osmosis F Back to the Future

1 | 2Prof. Myeong Hee Moon

D. Osmosis, E. Reverse Osmosis, F. Back to the Future

11.3

What we learn from chap. 11

Our approach to this chapter is different than typical texts because although we spend much of thetexts because although we spend much of the chapter on water, we look at it as only one (albeit the key) type of liquid, and take an approach that covers y) yp q , ppliquids broadly. That is, we believe that the understanding of liquids includes water, rather than being distinct from water. The chapter is spilt into two parts, with the first part dealing with pure liquidsand the second part primarily dealing with aqueousand the second part primarily dealing with aqueous solutions. – the worldwide use of water : majority of j y

worldwide water use is for agriculture– the structure and properties of water

the need for clean water


– the need for clean water

11.4

Water Consumption (UNESCO )


11.5

11.1 Structure of Water

• Bent molecule• H-O-H bond angle of 104.5o

• Polar molecule – permanent dipole• In the liquid state, intermolecular forcesIn the liquid state, intermolecular forces

cause between 3 and 6 molecules to aggregate.

• Water vs MethaneO-H (BE 940kJ/mol) C-H (BE 1650kJ/mol)

∆Hvap=44kJ/mol ∆Hvap=9kJ/mol


11.6

Structure of Water


11.7

11.2 Intermolecular Forces

• London dispersion forces : Induced dipole

• Permanent Dipole-Dipole ForcesPermanent Dipole Dipole Forces

• Hydrogen Bonds


11.8

London Dispersion ForcesLondon Dispersion Forces

• Induced dipoles polarizabilityp

• The result of temporary dipoles.


11.9

London Dispersion Forces


11.10

Polarizability vs Distance


11.11

Permanent Dipole-Dipole Forces

• The result of the dipole in polar covalent• The result of the dipole in polar covalent molecules

A i t l 1% t l t• Approximately 1% as strong as a covalent bond


11.12

Dipole-Dipole Forces


11.13

Hydrogen Bonds

• The attraction between a hydrogen, bonded y gto F, O, or N in a molecule, and the lone electrons of F, O, or N in another molecule., ,

• The strongest of all intermolecular forces, about 10% of a covalent bondabout 10% of a covalent bond.


11.14

Hydrogen Bonds


11.15


11.16

11.3 Phase Changes

• Evaporation– the process of molecules leaving the surface of p g

the liquid phase and entering the vapor phase.

• Condensation– the process of molecules leaving the vapor

phase and entering the liquid phase.

• Sublimation– the escape of molecules from the solid phasethe escape of molecules from the solid phase

directly to the vapor phase.

: Opposite behavior -- Deposition


pp p

11.17

Water -- Vapor


11.18

Vapor Pressure


11.19

Vapor Pressure

• The pressure of a vapor over a liquid.

• Vapor pressure increases with temperature.Vapor pressure increases with temperature.

• Heavier molecules have lower vapor pressures than lighter moleculespressures than lighter molecules.

• Molecules with the strongest intermolecular forces will have the lowest vapor pressureforces will have the lowest vapor pressure.


11.20

Vapor Pressure


11.21

Heating Curves


11.22

Changes of State

• Boiling point– the pressure of a liquid’s vapor is equal to the p q p q

surrounding pressure.

• Normal boiling pointg p– the boiling point of a liquid if the surrounding

pressure is 1 atm.

• Melting point– the temperature at which a solid changes to athe temperature at which a solid changes to a

liquid.


11.23

Changes of State

• Heat of Fusion (fusH=334 J/g for ice)

– the amount of heat necessary to convert athe amount of heat necessary to convert a solid to a liquid at its melting point and constant pressure.constant pressure.

• Heat of Vaporization (vapH=2.44 kJ/g for water)water)

– the amount of heat needed to convert a liquid to a vapor at its normal boiling pointliquid to a vapor at its normal boiling point.


11.24

Sample Problem

How much heat is necessary to bring 10.0 g of ice at y g g–10.0oC to a temperature of 50.0oC? The specific heat of ice is 2.05 J/goC, the heat of fusion of water is 334 J/g, the specific heat of water is 4.184 J/goC.

: There are three steps:

1. warming the ice;

2. melting the ice; and

i i iq m SH t

3. warming the water.

warming ice ice

o o o

q m SH t

10.0 g 2.05 J/g C 0.0 C 10.0 C


205 J

11.25

Sample Problem (cont)p ( )

melting ice fusq = m H

g

10.0 g 334 J/g

3340 J

3340 J

heating water waterq m SH t

. . .

g

o o o10 0 g 4 184 J/g C 50 0 C- 0.0 C

2092 J

2092 J

total warming melting heatingq q q q

3

205 J + 3340 J + 2092 J

= 5.64 10 J


11.26

11.4 Phase Diagram

(4.6torr)

Phase diagram of CO2


Phase diagram of CO2

11.27

Phase Diagram

• Triple point– the point representing the temperature and pressure

at which the three phases coexist in equilibrium.

• Critical temperature– the temperature above which the liquid state can no p q

longer exist at any pressure.

• Critical pressurep– the vapor pressure at the critical temperature.

• Critical pointCritical point– the point defined by the critical temperature and the

critical pressure.


critical pressure.

11.28

11.5 Properties of Water

• Viscosity– the resistance of a liquid to flow.– Water : 0.890 mPa.s(25°C)→0.378 mPa.s(75°C)a e 0 890 a s( 5 C) 0 3 8 a s( 5 C)– C5H12: 0.224 mPa.s, C8H18: 0.508 mPa.s– intermolecular force, temperature (average kinetic energy)

• Surface tension• Surface tension– a measure of the energy per area on the surface of a liquid– To maximize the # of hydrogen bond (minimum energy)

W t f h ith i i fWater forms a shape with minimum surface area

• Capillary action– the upward rise of a liquid in a small diameter tube caused by the p q y

adhesion of molecules to the surface of the tube.– Hwater interact with the Oglass : adhesion– Cohesive force


11.29

Cohesive force & Adhesion

meniscus


11.30

11.6 Water : The Universal Solvent

• Solution– solvent, solute– aqueous solution– universal solvent

• Process of Dissolving – Dissolving can be described as occurring in three steps:

• Solute separation.• Solvent separation.

El t t ti i t ti b t l t d• Electrostatic interaction between solvent and solute.


11.31

Solution Process


11.32

Solution Process

Ion-dipole interaction

cf) solvation, hydration


11.33

Solution Process

• Solution formation can be exothermic or endothermic.

• Heat of solution is:

H = H + H + HHsol = Hsolute + Hsolvent + Hsolvation

• Like dissolves like

– polar molecules dissolve in polar liquids.

– nonpolar molecules dissolve in nonpolar liquids.


11.34

Solution Process


11.35

Sample Problem

Which of the following substances are soluble in water?KBr, ethanol (CH3CH2OH), hexane (C6H12)

• KBr is an ionic compound it will dissolve in water.

• ethanol is polar it will dissolve in polar water.

• hexane is nonpolar it will not dissolve in• hexane is nonpolar, it will not dissolve in water.


11.36

Solubility

solubility

g solute


solubility100 mL solvent

11.37

10.7 Molarity

• Molarity (M) is the number of moles of solutes per volume of solution in liters.

mol of soluteMolarity ( ) =

L solutionM

L solution


11.38

Sample Problem

Calculate the molarity of a solution of 24.0g of HCl made up to a volume

f

g pof 500. mL.

mol of solute=

L solutionM

1 mol HCl24.0g HCl

36.5g HCl

g

0.500 L1.32

M


11.39

MolalityMolality

• Molality (m) is the number of moles of solute per kilogram of solvent.

mol of soluteMolality ( ) =

kg solventm

g


11.40

Sample Problem

Calculate the molality of a solution of 13.5g of KF dissolved in 250. g of water.

mol of solute =

kg solventm

g g

kg solvent

1 mol KF13 5g

13.5g58.1

0.250 kg

g

0.929 m


11.41

Mole Fraction

• Mole fraction (χ) is the ratio of the number of o e ac o (χ) s e a o o e u be omoles of a substance over the total number of moles of substances in solution.

i

number of moles of i

t t l b f li

i

total number of moln

se

i

T

= n


11.42

Sample ProblemC i b t it-Conversions between units-

• ex) What is the molality of a 0.200 M aluminum

nitrate solution (d = 1 012g/mL)?nitrate solution (d = 1.012g/mL)?

– Work with 1 liter of solution. mass = 1012 g

– mass Al(NO3)3 = 0.200 mol × 213.01 g/mol = 42.6 g ;

– mass water = 1012 g -43 g = 969 g

0.2000.206 /

0 969

molMolality mol kg

k

0.969kg


11.43

Sample Problem

Calculate the mole fraction of 10.0g of NaCl dissolved in 100. g of water.

mol of NaCl

g

NaCl2

o o aC

mol of NaCl + mol H O

1 mol NaCl

2

1 mol NaCl10.0g

58.5g NaCl

1 mol H O1 mol NaCl10 0 100 H O

22

2

1 mol H O 1 mol NaCl10.0g 100.g H O

58.5g NaCl 18.0g H O

0.0299


0.0299

11.44

Conc. Based on massCo c ased o ass

% % 100

g solutemass or weight

g solution

6

10

g solution

g soluteppm

l ti

9

10

g solution

g soluteppb

12

10

10

ppbg solution

g soluteppt

10

pptg solution


11.45ppm, ppb, ppt in dilute aqueoussolutionsolution

1 1g solute mg soluteppm

610

1 1

ppmg of solution L solution

l t l t

9

1 1

10

g solute g soluteppb

g of solution L solution

12

1 1

10

g solute ng soluteppt

g of solution L solution

10 g of solution L solution


11.46

11.8 Effect of Temperature on Solubility

• The solubility of a gasdecreases with temperature.


11.47

Solubility of O2


11.48

Effect of Temperature on Solubility

• The solubility of an ionic solid generally g yincreases with temperaturetemperature.


11.49

Effect of Pressure on Solubility

Henry’s Law

Pgas = kgasCgasPgas kgasCgas

P = pressure of the gas above the solution– Pgas = pressure of the gas above the solution

– Cgas = concentration of the gas

k = Henry’s law constant– kgas = Henry s law constant

Henry’s law holds best for gases O and N doesHenry’s law holds best for gases O2 and N2, does not hold HCl


11.50

Sample Problem

A liter of water dissolves 0.0404 g of oxygen at 25oC at a pressure of 760. torr. What would be th t ti f (i /L) if ththe concentration of oxygen (in g/L) if the pressure were increased to 1880 torr at the

t t ?

P P

same temperature?

1 2

1 2

P P=

C C

C0 0404 g/L 2

2

C0.0404 g/L=

760. torr 1880 torrC = 0.0999 g/L


2 g

11.51

11.9 Colligative Properties

• Vapor pressure lowering – the vapor pressure of a solvent is lowered by the addition of a nonvolatile solute.


cf) volatile solute

11.52


11.53

Raoult’s Law

o osolution solvent solventP P

Posolvent = vapor pressure of the pure solvent


11.54

Sample Problem

What will be the vapor pressure of a solution made b di l i 6 2 f l C H O i 0 0 fby dissolving 6.25g of glucose, C6H12O6 , in 50.0g of water at 25oC? How much was the vapor pressure of the pure water lowered? The vapor pressure ofthe pure water lowered? The vapor pressure of water at 25oC is 23.8 torr

1 molmol glucose = 6.25 g 0.0347 mol glucose

180 g

180. g

1 molmol water = 50.0 g 2.78 mol water

18.0g

water

osoln water water

2.78 0.9880.0347+2.78

P P 0.988 23.8 torr 23.5 torr


soln water water

vapor pressure lowering = 23.8 torr - 23.5 torr = 0.2 torr

11.55

Colligative Properties

Boiling point elevation– the change in the boiling point is:

Tb = iKbm

i = sum of the coefficients of the ions(i = 1 for molecular compounds)

Kb = boiling point elevation constantm = molality


11.56


Freezing point depression –the change in the freezing point is:

Tf = iKfmTf iKfm

i = sum of the coefficients of the ionsi = sum of the coefficients of the ions

(i = 1 for molecular compounds)

Kf = freezing point depression constantf g p p

m = molality


11.57

Antifreeze solutiont ee e so ut o


11.58

Sample ProblemSa p e ob e

Calculate the boiling point elevation and the freezing pointCalculate the boiling point elevation and the freezing point depression of a solution made by dissolving 12.2g of KCl in 45.0g of water. Kb = 0.512oC/m and Kf = 1.86oC/m

• i = 2 for KCl K+ + Cli 2 for KCl K Cl

1 molmol KCl = 12.2g 0.164 mol

74.6g

KCl

74.6g

mol KCl 0.164 mol3.64

kg water 0.045 kg

m m

o ob b

o of f

T K 2 0.512 C/ 3.64 3.73 C

T K 2 1 86 C/ 3 64 13 5 C

i m m m

i m m m


f fT K 2 1.86 C/ 3.64 13.5 C i m m m

11.59


Osmotic pressure

П = iMRTП iMRT– i = sum of the coefficients of the ions (i

= 1 for molecular compounds)p )

– M = molarity

– R = gas constant (0.0821 L•atm/mol•K)R gas constant (0.0821 L atm/mol K)

– T = temperature in Kelvin


11.60

Osmosis


11.61

Osmosis


11.62

Osmosis


11.63


11.64

Sample Problem

What is the osmotic pressure of a 100. mL solution containing 9.50 g of glucose, C6H12O6, at 20.0oC?

1 molmol glucose = 9.50 g =0.0528 mol

180. g

0 0528 l

glucose

0.0528 molM = = 0.528 mol/L

0.100 LL atm

π = (1) 0 528 mol/L 0 08206 [20 0+273] K = 12 7 atm π = (1) 0.528 mol/L 0.08206 [20.0+273] K = 12.7 atm

K mol


11.65

Problems

• 4,8,36,47,48,66,72,84,98,106, 112


The Chemistry of Water and theWater and the Nature...

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