Download - Chapter 12 Solutions 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro
Chapter 12Solutions
2008, Prentice Hall
Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro
Tro, Chemistry: A Molecular Approach 2
Solution• homogeneous mixtures
composition may vary from one sample to anotherappears to be one substance, though really contains
multiple materials
• most homogeneous materials we encounter are actually solutionse.g., air and sea water
• nature has a tendency toward spontaneous mixinggenerally, uniform mixing is more energetically
favorable
Tro, Chemistry: A Molecular Approach 3
Solutions• solute is the dissolved substance
seems to “disappear”“takes on the state” of the solvent
• solvent is the substance solute dissolves indoes not appear to change state
• when both solute and solvent have the same state, the solvent is the component present in the highest percentage
• solutions in which the solvent is water are called aqueous solutions
Tro, Chemistry: A Molecular Approach 4
Spontaneous Mixing
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Seawater• drinking seawater will dehydrate you and give you diarrhea
• Seawater draws water to itself – spontaneous mixing
• the cell wall acts as a barrier to solute moving
• the only way for the seawater and the cell solution to have uniform mixing is for water to flow out of the cells of your intestine and into your digestive tract
Tro, Chemistry: A Molecular Approach 6
Common Types of Solution
Solution PhaseSolute Phase
Solvent Phase Example
gaseous solutions gas gas air (mostly N2 & O2)
liquid solutions
gas
liquid
solid
liquid
liquid
liquid
soda (CO2 in H2O)
vodka (C2H5OH in H2O)
seawater (NaCl in H2O)
solid solutions solid solid brass (Zn in Cu)
• solutions that contain Hg and some other metal are called amalgams
• solutions that contain metal solutes and a metal solvent are called alloys
7
BrassType Color % Cu % Zn Density
g/cm3
MP
°C
Tensile
Strength
psi
Uses
Gilding redish 95 5 8.86 1066 50K pre-83 pennies,munitions, plaques
Commercial bronze 90 10 8.80 1043 61K door knobs,grillwork
Jewelry bronze 87.5 12.5 8.78 1035 66K costume jewelry
Red golden 85 15 8.75 1027 70K electrical sockets,fasteners & eyelets
Low deep yellow
80 20 8.67 999 74K musical instruments,clock dials
Cartridge yellow 70 30 8.47 954 76K car radiator cores
Common yellow 67 33 8.42 940 70K lamp fixtures,bead chain
Muntz metal yellow 60 40 8.39 904 70K nuts & bolts,brazing rods
Tro, Chemistry: A Molecular Approach 8
Solubility• when one substance (solute) dissolves in another (solvent) it
is said to be solublesalt is soluble in waterbromine is soluble in methylene chloride
• when one substance does not dissolve in another it is said to be insolubleoil is insoluble in water
• the solubility of one substance in another depends on two factors – nature’s tendency towards mixing, and the types of intermolecular attractive forces
Tro, Chemistry: A Molecular Approach 9
Solubility• there is usually a limit to the solubility of one
substance in anothergases are always soluble in each othertwo liquids that are mutually soluble are said to be
miscible alcohol and water are miscibleoil and water are immiscible
• the maximum amount of solute that can be dissolved in a given amount of solvent is called the solubility
• the solubility of one substance in another varies with temperature and pressure
Tro, Chemistry: A Molecular Approach 10
Mixing and the Solution ProcessEntropy
• formation of a solution does not necessarily lower the potential energy of the system the difference in attractive forces between atoms of
two separate ideal gases vs. two mixed ideal gases is negligible
yet the gases mix spontaneously
• the gases mix because the energy of the system is lowered through the release of entropy
• entropy is the measure of energy dispersal throughout the system
• energy has a spontaneous drive to spread out over as large a volume as it is allowed
Tro, Chemistry: A Molecular Approach 11
Intermolecular Forces and the Solution ProcessEnthalpy of Solution
• energy changes in the formation of most solutions also involve differences in intermolecular forces
solvent-solute interactionssolute-solute interactionssolvent-solvent interactions
Tro, Chemistry: A Molecular Approach 12
Intermolecular Attractions
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Solution Interactions
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Relative Interactions and Solution Formation
• when the solute-to-solvent attractions are weaker than the sum of the solute-to-solute and solvent-to-solvent attractions, the solution will only form if the energy difference is small enough to be overcome by the entropy
Solute-to-Solvent >Solute-to-Solute +
Solvent-to-SolventSolution Forms
Solute-to-Solvent =Solute-to-Solute +
Solvent-to-SolventSolution Forms
Solute-to-Solvent <Solute-to-Solute +
Solvent-to-SolventSolution May or May Not Form
Tro, Chemistry: A Molecular Approach 15
Will It Dissolve?• Chemist’s Rule of Thumb –
Like Dissolves Like• a chemical will dissolve in a solvent if it has a
similar structure to the solvent
• when the solvent and solute structures are similar, the solvent molecules will attract the solute particles at least as well as the solute particles to each other
Tro, Chemistry: A Molecular Approach 16
Classifying Solvents
Solvent Class
Structural
Feature
Water, H2O polar O-H
Methyl Alcohol, CH3OH polar O-H
Ethyl Alcohol, C2H5OH polar O-H
Acetone, C3H6O polar C=O
Toluene, C7H8 nonpolar C-C & C-H
Hexane, C6H14 nonpolar C-C & C-H
Diethyl Ether, C4H10O nonpolar C-C, C-H & C-O,
(nonpolar > polar)
Carbon Tetrachloride nonpolar C-Cl, but symmetrical
Tro, Chemistry: A Molecular Approach 17
Example 12.1a predict whether the following vitamin is soluble in fat or water
CH
CC
C
O
O
OH
OH
CHCH2
OHOH
Vitamin C
The 4 OH groups make the molecule highly polar and it will also H-bond to water.
Vitamin C is water soluble
Tro, Chemistry: A Molecular Approach 18
CHCH
CH
CH
C
CC
CH
CC
O
O
CH3
Example 12.1b predict whether the following vitamin is soluble in fat or water
Vitamin K3
The 2 C=O groups are polar, but their geometric symmetry suggests their pulls will cancel and the molecule will be nonpolar.
Vitamin K3 is fat soluble
Tro, Chemistry: A Molecular Approach 19
Tro, Chemistry: A Molecular Approach 20
Energetics of Solution Formation
• overcome attractions between the solute particles – endothermic
• overcome some attractions between solvent molecules – endothermic
• form new attractions between solute particles and solvent molecules – exothermic
• the overall H depends on the relative sizes of the H for these 3 processes
Hsoln = Hsolute + Hsolvent + Hmix
Tro, Chemistry: A Molecular Approach 21
Solution Process
1. add energy in to overcome solute-solute attractions
2. add energy in to overcome some solvent-solvent attractions3. form new solute-solvent attractions, releasing energy
Tro, Chemistry: A Molecular Approach 22
Energetics of Solution Formation
if the total energy cost for breaking attractions between particles in the pure solute and pure solvent is less than the energy released in making the new attractions between the solute and solvent, the overall process will be exothermic
if the total energy cost for breaking attractions between particles in the pure solute and pure solvent is greater than the energy released in making the new attractions between the solute and solvent, the overall process will be endothermic
Tro, Chemistry: A Molecular Approach 23
Heats of Hydration• for aqueous ionic solutions, the energy added to overcome the attractions between water
molecules and the energy released in forming attractions between the water molecules and ions is combined into a term called the heat of hydration attractive forces in water = H-bonds attractive forces between ion and water = ion-dipole
• .Hhydration = heat released when 1 mole of gaseous ions dissolves in water
always largely –ve because ion-dipole interaction >>>> H-Bonds
Hsoln = Hsolute + Hhydration (+ve) (-ve)
Heats of Hydration
24
Tro, Chemistry: A Molecular Approach 25
Heat of Hydration
Hsol’n = Hsolute (+ve) + Hhydration (-ve)
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Solution Equilibrium
• the dissolution of a solute in a solvent is an equilibrium process
• initially, when there is no dissolved solute, the only process possible is dissolution
• shortly, solute particles can start to recombine to reform solute molecules – but the rate of dissolution >> rate of deposition and the solute continues to dissolve
• eventually, the rate of dissolution = the rate of deposition – the solution is saturated with solute and no more solute will dissolve
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Solution Equilibrium
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Solubility Limit
• a solution that has the maximum amount of solute dissolved in it is said to be saturateddepends on the amount of solventdepends on the temperature
and pressure of gases
• a solution that has less solute than saturation is said to be unsaturated
• a solution that has more solute than saturation is said to be supersaturated
Tro, Chemistry: A Molecular Approach 29
How Can You Make a Solvent Hold More Solute Than It Is Able To?
• solutions can be made saturated at non-room conditions – then allowed to come to room conditions slowly
• for some solutes, instead of coming out of solution when the conditions change, they get stuck in-between the solvent molecules and the solution becomes supersaturated
• supersaturated solutions are unstable and lose all the solute above saturation when disturbede.g., shaking a carbonated beverage
Tro, Chemistry: A Molecular Approach 30
Adding Solute to a Supersaturated Solution of NaC2H3O2
http://www.youtube.com/watch?v=uy6eKm8IRdI
Tro, Chemistry: A Molecular Approach 31
Temperature Dependence of Solubility of Solids in Water
• solubility is generally given in grams of solute that will dissolve in 100 g of water
• for most solids, the solubility of the solid increases as the temperature increaseswhen Hsolution is endothermic
• solubility curves can be used to predict whether a solution with a particular amount of solute dissolved in water is saturated (on the line), unsaturated (below the line), or supersaturated (above the line)
Tro, Chemistry: A Molecular Approach 32
Solubility Curves
Tro, Chemistry: A Molecular Approach 35
Temperature Dependence of Solubility of Gases in Water
• solubility is generally given in moles of solute that will dissolve in 1 Liter of solution
• generally lower solubility than ionic or polar covalent solids because most are nonpolar molecules
• for all gases, the solubility of the gas decreases as the temperature increases the Hsolution is exothermic because you do not need to
overcome solute-solute attractions
Tro, Chemistry: A Molecular Approach 38
Solubility of Gases in Water at Various Temperatures
0
0.001
0.002
0.003
0.004
0.005
0.006
0.007
0.008
0 10 20 30 40 50 60 70 80 90 100 110
Temperature, °C
So
lub
ilit
y, g
in
100
g w
ater
N2
O2
39
Pressure Dependence of Solubility of Gases in Water
• the larger the partial pressure of a gas in contact with a liquid, the more soluble the gas is in the liquid
Tro, Chemistry: A Molecular Approach 40
Henry’s Law
• the solubility of a gas (Sgas) is directly proportional to its partial pressure, (Pgas)
Sgas = kHPgas
• kH is called Henry’s Law Constant
Tro, Chemistry: A Molecular Approach 43
persrst
Tro, Chemistry: A Molecular Approach 44
atm 5.3atmM 1014.3
M .1201-2
H
k
SP
Ex 12.2 – What pressure of CO2 is required to keep the [CO2] = 0.12 M at 25°C?
the unit is correct, the pressure higher than 1 atm meets our expectation from general experience
Check:
Solve:
S = kHP, kH = 3.4 x 10-2 M/atm
Concept Plan:
Relationships:
S = [CO2] = 0.12 M,
P of CO2, atm
Given:
Find:
[CO2] P
H
S
kP
Tro, Chemistry: A Molecular Approach 45
Tro, Chemistry: A Molecular Approach 46
Concentrations
• solutions have variable composition• to describe a solution, need to describe components
and relative amounts• the terms dilute and concentrated can be used as
qualitative descriptions of the amount of solute in solution
• concentration = amount of solute in a given amount of solution
Tro, Chemistry: A Molecular Approach 47
Solution ConcentrationMolarity, M
• moles of solute per 1 liter of solution
• used because it describes how many molecules of solute in each liter of solution
• if a sugar solution concentration is 2.0 M, 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar
molarity = moles of soluteliters of solution
Preparing 1 L of a 1.00 M NaCl Solution
Dilution
• often, solutions are stored as concentrated stock solutions
• to make solutions of lower concentrations from these stock solutions, more solvent is added the amount of solute doesn’t change, just the volume of
solutionmoles solute in solution 1 = moles solute in solution 2
• the concentrations and volumes of the stock and new solutions are inversely proportional
M1∙V1 = M2∙V2
Dilution
E.g. find the volume of 12 M HCl required to make 1 L of 1.0 M HCl
M1∙V1 = M2∙V2
12 x V1 = 1 M x 1 LV1 = 1 / 12 = 0.083 L = 83 mL
Add 83 mL to a 1 L volumetic flask containing a few 100 mL of DI water (never add water to acid!) and then make up to the mark with DI water
Tro, Chemistry: A Molecular Approach 51
Molarity and Dissociation• the molarity of the ionic compound allows you to determine the molarity
of the dissolved ions
CaCl2(aq) = Ca+2(aq) + 2 Cl-(aq)
• A 1.0 M CaCl2(aq) solution contains 1.0 moles of CaCl2 in each liter of solution 1 L = 1.0 moles CaCl2, 2 L = 2.0 moles CaCl2
• Because each CaCl2 dissociates to give one Ca+2 = 1.0 M Ca+2
1 L = 1.0 moles Ca+2, 2 L = 2.0 moles Ca+2
• Because each CaCl2 dissociates to give 2 Cl- = 2.0 M Cl-
1 L = 2.0 moles Cl-, 2 L = 4.0 moles Cl-
Tro, Chemistry: A Molecular Approach 52
Solution ConcentrationMolality, m
• moles of solute per 1 kilogram of solventdefined in terms of amount of solvent, not solution
like the others
• does not vary with temperaturebecause based on masses, not volumes
Molality, (m) = moles of solute kg of solvent
Tro, Chemistry: A Molecular Approach 53
Percent• parts of solute in every 100 parts solution• mass percent = mass of solute in 100 parts
solution by massif a solution is 0.9% by mass, then there are 0.9
grams of solute in every 100 grams of solutionor 0.9 kg solute in every 100 kg solution
Solution of Mass Solvent of Mass Solute of Mass
%100g Solution, of Mass
g Solute, of Mass Percent Mass
Tro, Chemistry: A Molecular Approach 54
Percent Concentration
Solution of Mass Solvent of Mass Solute of Mass
%100g Solution, of Mass
g Solute, of Mass Percent Mass
Solution of Volume Solvent of Volume Solute of Mass
%100mL Solution, of Volume
g Solute, of Mass eMass/VolumPercent
Solution of Volume Solvent of Volume Solute of Volume
%100mL Solution, of Volume
mL Solute, of Volume Percent Volume
%100(solution) Whole
(solute)Part Percent
Tro, Chemistry: A Molecular Approach 55
Using Concentrations asConversion Factors
• concentrations show the relationship between the amount of solute and the amount of solvent 12%(m/m) sugar(aq) means 12 g sugar 100 g solution
or 12 kg sugar 100 kg solution; or 12 lbs. 100 lbs. solution 5.5%(m/v) Ag in Hg means 5.5 g Ag 100 mL solution 22%(v/v) alcohol(aq) means 22 mL EtOH 100 mL solution
• The concentration can then be used to convert the amount of solute into the amount of solution, or vice versa
Tro, Chemistry: A Molecular Approach 56
Preparing a Solution
• need to know amount of solution and concentration of solution• calculate the mass of solute needed
start with amount of solutionuse concentration as a conversion factor
5% by mass 5 g solute 100 g solution“Dissolve the grams of solute in enough solvent to total the
total amount of solution.”
Tro, Chemistry: A Molecular Approach 57
Example - How would you prepare 250.0 g of 5.00% by mass glucose solution (normal glucose)?
Given: 250.0 g solution
Find: g Glucose
Equivalence: 5.00 g Glucose 100 g solution
Solution Map:g solutiong solution g Glucoseg Glucose
solution g 100Glucose g 5.00
Apply Solution Map:
Answer: Dissolve 12.5 g of glucose in enough water to total 250.0 g
glucose g 12.5 solution g 100glucose g .005
solution g 0250 .
58
Solution ConcentrationPPM
• grams of solute per 1,000,000 g of solution
• mg of solute per 1 kg of solution
• 1 liter of water = 1 kg of waterfor water solutions we often approximate the kg of
the solution as the kg or L of water
grams solutegrams solution
x 106
mg solutekg solution
mg soluteL solution
Tro, Chemistry: A Molecular Approach 59
sucrose g 5.42soln g 100
sucrose g 5.11
soln mL
soln g .041soln 55mL3
Ex 12.3 – How much sugar is in a 355 mL can of pop that is 11.5 % sucrose by mass?
the unit is correctCheck:
Solve:
1 g soln. = 1.04 g sugar, 100 g soln = 11.5 g sucro
Concept Plan:
Relationships:
355 mL pop, density = 1.04 g/mL
sucrose, g
Given:
Find:
sugar g 1.04
soln mL 1mL soln. g soln. g sugar
sucrose g 11.5
soln g 100
Tro, Chemistry: A Molecular Approach 60
Solution ConcentrationsMole Fraction, XA
• the mole fraction is the fraction of the moles of one component in the total moles of all the components of the solution
• total of all the mole fractions in a solution = 1• unitless • the mole percentage is the percentage of the moles of one
component in the total moles of all the components of the solution= mole fraction x 100%
mole fraction of A = XA = moles of components A total moles in the solution
Tro, Chemistry: A Molecular Approach 61
mol 1727.0OHC g 62.07
OHC mol 1OHC g 2.17
262
262262
Ex 12.4a – What is the molarity of a solution prepared by mixing 17.2 g of C2H6O2 with 0.500 kg of H2O to
make 515 mL of solution?
the unit is correct, the magnitude is reasonableCheck:
Solve:
M = mol/L, 1 mol C2H6O2 = 62.07 g, 1 mL = 0.001 L
Concept Plan:
Relationships:
17.2 g C2H6O2, 0.500 kg H2O, 515 mL soln.
M
Given:
Find:
g C2H6O2 mol C2H6O2
mL soln. L soln. ML
molM
L 515.0mL 1
L 0.001 mL 155 M 0.538 M
L 0.515
OHC mol 170.27M 262
Tro, Chemistry: A Molecular Approach 62
mol 1727.0OHC g 62.07
OHC mol 1OHC g 2.17
262
262262
Ex 12.4b – What is the molality of a solution prepared by mixing 17.2 g of C2H6O2 with 0.500 kg of H2O to
make 515 mL of solution?
the unit is correct, the magnitude is reasonableCheck:
Solve:
m = mol/kg, 1 mol C2H6O2 = 62.07 g
Concept Plan:
Relationships:
17.2 g C2H6O2, 0.500 kg H2O, 515 mL soln
m
Given:
Find:
g C2H6O2 mol C2H6O2
kg H2Omkg
molm
mol/kg 0.554 molality
OH kg 0.500
OHC mol 170.27molality
2
262
Tro, Chemistry: A Molecular Approach 63
OH g1000.5OH kg 1
OH g 1000OH kg .5000 2
2
2
22
Ex 12.4c – What is the percent by mass of a solution prepared by mixing 17.2 g of C2H6O2 with 0.500 kg of
H2O to make 515 mL of solution?
the unit is correct, the magnitude is reasonableCheck:
Solve:
1 kg = 1000 g
Concept Plan:
Relationships:
17.2 g C2H6O2, 0.500 kg H2O, 515 mL sol’n
%(m/m)
Given:
Find:
g C2H6O2
g solvent g sol’n %
%100nsol' g
solute g%
nsol' g .2751 OH g 050OHC g 2.17 2262 3.33% %
%100nsol' g .2751
OHC g 17.2% 262
Tro, Chemistry: A Molecular Approach 64
OH mol 5727.OH g 18.02
OH mol 1
kg 1
g 1000 OH kg .5000 2
2
22
the unit is correct, the magnitude is reasonable
mol 1727.0OHC g 62.07
OHC mol 1OHC g 2.17
262
262262
Ex 12.4d – What is the mole fraction of a solution prepared by mixing 17.2 g of C2H6O2 with 0.500 kg of
H2O to make 515 mL of solution?
Check:
Solve:
= molA/moltot, 1 mol C2H6O2 = 62.07 g, 1 mol H2O = 18.02 g
Concept Plan:
Relationships:
17.2 g C2H6O2, 0.500 kg H2O, 515 mL sol’n
Given:
Find:
g C2H6O2 mol C2H6O2
g H2O mol H2Ototal
A
mol
mol
3-
2262
262
109.89
OH mol 57.72OHC mol 170.27
OHC mol 170.27
Tro, Chemistry: A Molecular Approach 65
%0.989 %
%100OH mol 57.72OHC mol 170.27
OHC mol 170.27%
2262
262
OH mol 5727.OH g 18.02
OH mol 1
kg 1
g 1000 OH kg .5000 2
2
22
the unit is correct, the magnitude is reasonable
mol 1727.0OHC g 62.07
OHC mol 1OHC g 2.17
262
262262
Ex 12.4d – What is the mole percent of a solution prepared by mixing 17.2 g of C2H6O2 with 0.500 kg of
H2O to make 515 mL of solution?
Check:
Solve:
= molA/moltot, 1 mol C2H6O2 = 62.07 g, 1 mol H2O = 18.02 g
Concept Plan:
Relationships:
17.2 g C2H6O2, 0.500 kg H2O, 515 mL sol’n
Given:
Find:
g C2H6O2 mol C2H6O2
g H2O mol H2O
%100mol
mol%
total
A
Tro, Chemistry: A Molecular Approach 66
Converting Concentration Units• assume a convenient amount of solution
given %(m/m), assume 100 g solution given %(m/v), assume 100 mL solution given ppm, assume 1,000,000 g solution given M, assume 1 liter of solution given m, assume 1 kg of solvent given X, assume you have a total of 1 mole of solutes in the solution
• determine amount of solution in non-given unit(s) if assume amount of solution in grams, use density to convert to mL and then to L if assume amount of solution in L or mL, use density to convert to grams
• determine the amount of solute in this amount of solution, in grams and moles• determine the amount of solvent in this amount of solution, in grams and moles• use definitions to calculate other units
67
Tro, Chemistry: A Molecular Approach 68
Tro, Chemistry: A Molecular Approach 69
Vapor Pressure of Solutions• the vapor pressure of a liquid is the pressure of the
gas above the liquid when in dynamic equilibrium
evaporation = condensation
The pure solvent establishes an liquid vapor equilibrium
Tro, Chemistry: A Molecular Approach 70
Vapor Pressure of Solutions• What is the effect of a nonvolatile solute on the VP
of the liquid?
Tro, Chemistry: A Molecular Approach 71
Vapor Pressure of Solutions• the VP of a solution is lower than the VP of
the pure solventthe solute particles replace some of the solvent
molecules at the surface
The pure solvent establishes an liquid vapor equilibrium
Addition of a nonvolatile solute reduces the rate of vaporization, decreasing the amount of vapor
Eventually, equilibrium is re-established, but a smaller number of vapor molecules – therefore the
vapor pressure will be lower
Tro, Chemistry: A Molecular Approach 72
Thirsty Solutions
• a concentrated solution will draw solvent molecules toward it due to the natural drive for materials in nature to mix
• similarly, a concentrated solution will draw pure solvent vapor into it due to this tendency to mix
• the result is reduction in vapor pressure
Tro, Chemistry: A Molecular Approach 73
Thirsty Solutions
Beakers with equal liquid levels of pure solvent and a solution are place in a bell jar. Solvent molecules evaporate from each one and fill the bell jar, establishing an equilibrium with the liquids in the beakers.
When equilibrium is established, the liquid level in the solution beaker is higher than the solution level in the pure solvent beaker – the thirsty solution grabs and holds solvent vapor more effectively
VP solution <<< VP pure solvent
Tro, Chemistry: A Molecular Approach 74
Raoult’s Law
• the VP of a volatile solvent above a solution is equal to its mole fraction of its normal vapor pressure, P°
Psolvent in solution = solvent∙P°
since the mole fraction of the solvent is always less than 1, the vapor pressure of the solvent in solution will always be less than the vapor pressure of the pure solvent
Tro, Chemistry: A Molecular Approach 75
Raoult’s Law
• the VP of a volatile solvent above a solution is equal to its mole fraction of its normal vapor pressure, P°
Psolvent in solution = solvent∙P°
e.g. water at 25 °C contains 0.90 mol water and 0.10 mol sucrose
Pure water has VP 23.8 torr,
P soln = H2O∙P°H2O = 0.90 x 23.8 torr = 21.4 torr
VP of solution is lowered in proportion to mole composition of the solvent, VP is 90% of the VP of pure solvent
Tro, Chemistry: A Molecular Approach 76
OH mol 56.61OH g 18.02
OH mol 1
mL 1
g 1.00 OH mL 00.03 2
2
22
112212112212
112212112212 OHC mol 7029.0
OHC g 342.30
OHC mol 1OHC g 9.59
Ex 12.5 – Calculate the vapor pressure of water in a solution prepared by mixing 99.5 g of C12H22O11 with
300.0 mL of H2O
Solve:
P°H2O = 23.8 torr, 1 mol C12H22O11 = 342.30 g, 1 mol H2O = 18.02 g
Concept Plan:
Relationships:
99.5 g C12H22O11, 300.0 mL H2O
PH2O
Given:
Find:
82.980 OH mol 56.16OHC mol 700.29
OH mol 56.16
2112212
2
g C12H22O11 mol C12H22O11
g H2O mol H2OmL H2O
PH2O
PP
torr23.4
torr8.238298.0
OH
OHOHOH
2
222
P
PP
total
A
mol
mol
Tro, Chemistry: A Molecular Approach 77
Ionic Solutes and Vapor Pressure
• according to Raoult’s Law, the effect of solute on the vapor pressure simply depends on the number of solute particles
• when ionic compounds dissolve in water, they dissociate – so the number of solute particles is a multiple of the number of moles of formula units
• the effect of ionic compounds on the vapor pressure of water is magnified by the dissociation since NaCl dissociates into 2 ions, Na+ and Cl, one mole of NaCl
lowers the vapor pressure of water twice as much as 1 mole of C12H22O11 molecules would
Tro, Chemistry: A Molecular Approach 78
Effect of Dissociation
Tro, Chemistry: A Molecular Approach 79
Ex 12.6 – What is the vapor pressure of H2O when 0.102 mol Ca(NO3)2 is mixed with 0.927 mol H2O @ 55°C?
the unit is correct, the pressure lower than the normal vapor pressure makes sense
Check:
Solve:
P = ∙P°
Concept Plan:
Relationships:
0.102 mol Ca(NO3)2, 0.927 mol H2O, P° = 118.1 torr
P of H2O, torr
Given:
Find:
H2O, P°H2O PH2OPP
Ca(NO3)2 Ca2+ + 2 NO3 3(0.102 mol solute)
8175.0
OH mol 927.0solute mol 102.03
OH mol 927.0
OH
2
2OH
2
2
torr88.8
torr18.118175.0
OH
OHOHOH
2
222
P
PP
Tro, Chemistry: A Molecular Approach 80
Raoult’s Law for Volatile Solute• when both the solvent and the solute can evaporate, both
molecules will be found in the vapor phase
• the total VP above the solution will be the sum of the VPs s of the solute and solvent for an ideal solution
Ptotal = Psolute + Psolvent
• the solvent decreases the solute VP in the same way the solute decreased the solvent’s
Psolute = solute∙P°solute and Psolvent = solvent∙P°solvent
Tro, Chemistry: A Molecular Approach 84
Freezing Point Depression• the melting/freezing point of a solution is lower than the
melting/freezing point of the pure solvent
Allows salt to melt ice even if ambient temp is below freezing
Tro, Chemistry: A Molecular Approach 85
Freezing Point Depression
VP of soln. is shifted downward
Net effect: lower mpt, higher bpt
Depends on amount of solute not type = colligative property
Tro, Chemistry: A Molecular Approach 86
Freezing Point Depression• the difference between the freezing point of the solution and freezing point of
the pure solvent is directly proportional to the molal concentration of solute particles
FPsolvent – FPsolution) = Tf = m∙Kf
• m = molality of solution (mol solute/kg solvent)• the proportionality constant is called the Freezing Point Depression Constant,
Kf
the value of Kf depends on the solvent
the units of Kf are °C/m
Tro, Chemistry: A Molecular Approach 87
Kf
Tro, Chemistry: A Molecular Approach 88
C 3.2
86.1 7.1
OH, 2
f
m
C
ff
T
m
KmT
Ex 12.8 – What is the freezing point of a 1.7 m aqueous ethylene glycol (Antifreeze) solution, C2H6O2?
the unit is correct, the freezing point lower than the normal freezing point makes sense
Check:
Solve:
Tf = m ∙Kf, Kf for H2O = 1.86 °C/m, FPH2O = 0.00°C
Concept Plan:
Relationships:
1.7 m C2H6O2(aq)
Tf, °C
Given:
Find:
m Tf
ff KmT
CFP
CFPC
TFPFP f
2.3
2.300.0
nsol'
nsol'
nsol'OH2
Tro, Chemistry: A Molecular Approach 89
Boiling Point Elevation• the boiling point of a solution is higher than the boiling
point of the pure solvent
Tro, Chemistry: A Molecular Approach 90
Boiling Point Elevation• the difference between the bpt. of the solution and bpt. of the pure solvent
is directly proportional to the molal concentration of solute particles
BPsolution – BPsolvent) = Tb = m∙Kb
• the proportionality constant is called the Boiling Point Elevation Constant, Kb
the value of Kb depends on the solvent
the units of Kb are °C/m
For water Kb = 0.512 C/m
Tro, Chemistry: A Molecular Approach 91
mm
mCm
C
77.9
0.1000.105 512.0
Ex 12.9 – How many g of ethylene glycol, C2H6O2, must be added to 1.0 kg H2O to give a solution that boils at 105°C?
Solve:
Tb = m ∙Kb, Kb H2O = 0.512 °C/m, BPH2O = 100.0°CMMC2H6O2 = 62.07 g/mol, 1 kg = 1000 g
Concept Plan:
Relationships:
1.0 kg H2O, Tb = 105°C
mass C2H6O2, g
Given:
Find:
Tb mbb KmT
kg H2O mol C2H6O2 g C2H6O2
solvent kg
solute molm
mol 1
g 62.07
bb KmT
2622
262
262
2
2622 OHC g 101.6
OHC mol 1
OHC g 62.07
OH kg 0.1
OHC mol 9.77OH kg 0.1
Tro, Chemistry: A Molecular Approach 92
Osmosis
• osmosis is the flow of solvent through a semi-permeable membrane from solution of low concentration to solution of high concentration
• the amount of pressure needed to keep osmotic flow from taking place is called the osmotic pressure
• the osmotic pressure, , is directly proportional to the molarity of the solute particlesR = 0.08206 (atm∙L)/(mol∙K)
= MRT
Tro, Chemistry: A Molecular Approach 93
Tro, Chemistry: A Molecular Approach 94
protein mol 10813.1L 1
protein mol 811.3
mL 001.0
L 1 mL 0.10 6
M 10318.1
K 29808206.0 torr/atm760
torr45.2
4
Kmol
Latm
M
M
Ex 12.10 – What is the molar mass of a protein if 5.87 mg per 10 mL gives an osmotic pressure of 2.45 torr at 25°C?
Solve:
MRT, T(K) = T(°C)+273.15, R = 0.08206atm∙L/mol∙KM = mol/L, 1 mL = 0.001 L, MM = g/mol, 1 atm = 760 torr
Concept Plan:
Relationships:
5.87 mg/10 mL, P = 2.45 torr, T = 25°C
molar mass, g/mol
Given:
Find:
T MMRT
mL L mol proteinmL 1
L 0.001LM mol
MRTK 298273.1525
273.15C)T( T(K)
g/mol 1045.4mol 10318.1
g1087.5 36
3
MM
Tro, Chemistry: A Molecular Approach 95
Colligative Properties• colligative properties are properties whose value
depends only on the number of solute particles, and not on what they areVapor Pressure Depression, Freezing Point Depression,
Boiling Point Elevation, Osmotic Pressure
• the van’t Hoff factor, i, is the ratio of moles of solute particles to moles of formula units dissolved
• measured van’t Hoff factors are often lower than you might expect due to ion pairing in solution
Tro, Chemistry: A Molecular Approach 96
Dissociation is not ideally complete
Tro, Chemistry: A Molecular Approach 97
Colligative Properties
• Ionic solutions:
Tf = i m∙Kf
Tb = i m∙Kb
= i MRT
Use the van’t Hoff factor, i
e.g. which has the highest bpt?
1 M sugar solution, 1 M NaCl solution or 1 M MgCl2 solution?
Tro, Chemistry: A Molecular Approach 98
An isosmotic solution has the same osmotic pressure as the solution inside the cell – as a result there is no net the flow of water into or out of the cell.
A hyperosmotic solution has a higher osmotic pressure than the solution inside the cell – as a result there is a net flow of water out of the cell, causing it to shrivel
A hyposmotic solution has a lower osmotic pressure than the solution inside the cell – as a result there is a net flow of water into the cell, causing it to swell
Tro, Chemistry: A Molecular Approach 99
Colloids
• a colloidal suspension is a heterogeneous mixture in which one substance is dispersed through another
most colloids are made of finely divided particles suspended in a medium
Tro, Chemistry: A Molecular Approach 100
Properties of Colloids
• the particles in a colloid exhibit Brownian motion
Tro, Chemistry: A Molecular Approach 101
Properties of Colloids
• colloids exhibit the Tyndall Effectscattering of light as it passes through a suspensioncolloids scatter short wavelength (blue) light more
effectively than long wavelength (red) light
Tro, Chemistry: A Molecular Approach 102
When a light beam passes through a colloidal suspension it is visible due to scattering, it is not visible in pure water
Tro, Chemistry: A Molecular Approach 103
Soap and Micelles• soap molecules - hydrophilic (polar/ionic)
“head” and a hydrophobic (nonpolar) “tail”
• part of the molecule is attracted to water, but the majority of it isn’t
• the nonpolar tail tends to coagulate together to form a spherical structure called a micelle
Tro, Chemistry: A Molecular Approach 104
Soap Action
• most dirt and grease is made of nonpolar molecules – making it hard for water to remove it from the surface
• soap molecules form micelles around the small oil particles with the polar/ionic heads pointing out
• this allows the micelle to be attracted to water and stay suspended
Tro, Chemistry: A Molecular Approach 105
The polar heads of the micelles attract them to the water, and simultaneously repel other micelles so they will not coalesce and settle out.
Colloidal suspensions are kept stable by electrostatic repulsions