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


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

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Spontaneous Mixing


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


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


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


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


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


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


Intermolecular Attractions


Solution Interactions


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


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


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


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


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



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


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


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


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


Heat of Hydration

Hsol’n = Hsolute (+ve) + Hhydration (-ve)


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


Solution Equilibrium


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


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


Adding Solute to a Supersaturated Solution of NaC2H3O2

http://www.youtube.com/watch?v=uy6eKm8IRdI

http://www.youtube.com/watch?v=uy6eKm8IRdI


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)


Solubility Curves


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


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


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


persrst


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



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


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


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-


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


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


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


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


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.”

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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


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


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


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


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?


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


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?


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


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


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


%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


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


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



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


Vapor Pressure of Solutions• What is the effect of a nonvolatile solute on the VP

of the liquid?


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


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


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


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


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


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


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


Effect of Dissociation


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


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


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


Freezing Point Depression

VP of soln. is shifted downward

Net effect: lower mpt, higher bpt

Depends on amount of solute not type = colligative property


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


Kf


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


Boiling Point Elevation• the boiling point of a solution is higher than the boiling

point of the pure solvent


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


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


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



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


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


Dissociation is not ideally complete


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?


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


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


Properties of Colloids

• the particles in a colloid exhibit Brownian motion


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


When a light beam passes through a colloidal suspension it is visible due to scattering, it is not visible in pure water


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


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


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

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Chapter 12 Solutions 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro.

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