Brdy-6Ed-Ch12-Solution

Brady/Jespersen/Hyslop, Chemistry7E, Copyright 2015 John Wiley & Sons, Inc. All Rights Reserved.

Chapter 12Mixtures at the Molecular

Level: Properties of Solutions

Chemistry, 7th Edition International Student Version

Brady/Jespersen/Hyslop

. Personal Used Only - Use Wisely .


Chapter 12: SolutionsSolution

Homogeneous mixture Composed of solvent and solute(s)

Solvent More abundant component of mixture

Solute(s) Less abundant or other component(s) of mixture

e.g., Lactated Ringers solution NaCl, KCl, CaCl2, NaC3H5O3 in water

2



Why do Solutions Form?

Two driving forces behind formation of solution

1. Disordering/Randomizing*2. Intermolecular ForcesWhether or not a solution forms depends on

both opposing forces

*In the thermodynamics chapter this is called entropy and is defined more precisely

3



Spontaneous Mixing Two gases mix

spontaneously Due to random motions Mix without outside

work Never separate

spontaneously Tendency of system

left to itself, to become increasingly disordered

4

Gas A Gas B

separate mixed



Intermolecular Forces Attractive forces between molecules in pure

substances (solvents and solutes) were considered in the last chapter

Now we consider the attractive forces between a solvent molecule and a solute molecule

Strength of intermolecular attractive forces in a mixture, compared to the pure substances, will indicate if a mixture can form or if the substances will remain separate

5



Formation of Solutions Depends on Intermolecular Forces

When mixed, for a solution to form, Solvent-to-solute attractions must be similar to

attractions between solute alone and solvent alone

Initially solute and solvent separate to make room for each other Solute molecules held together by intermolecular

forces: energy added Solvent molecules held together by

intermolecular forces: energy added

6



Formation of Solutions Depends on Intermolecular Forces

When the separated solute and solvent are mixed to make a solution Solute molecules are attracted to solvent

molecules: energy released Solvent molecules are attracted to solute

molecules: energy released If there is a net release of energy, the

solution will form.

7



Miscible LiquidsMiscible liquids

Two liquids that dissolve in one another in all proportions

Form solution Strengths of intermolecular attractions are similar in

solute and solvent Similar polaritye.g., Ethanol and water

8

Ethanol; polar bondHOCC

H

H

H

H

H

+



Immiscible Liquids Two insoluble liquids Do not mix Get two separate phases Strengths of intermolecular forces are

different in solute and solvent Different polaritye.g., Benzene and water

9

Benzene;no polar bond

C

CC

C

CC

H

H

H

H

H

H



Rule of Thumb Like dissolves like Use polar solvent for polar solute Use nonpolar solvent for nonpolar solute

When strengths of intermolecular attractions are similar in solute and solvent, solutions form because net energy exchange is about the same

10



Learning CheckWhich of the following are miscible in water?

11



Your Turn!Which of the following molecules is most soluble in hexane, C6H14?A. NH3B. CH3NH2C. CH3OHD. CH3CH3E. H2O

12



Your Turn!Which of the following molecules is most soluble in methanol CH3OH?A. N(CH3)3B. CH3NH2C. CH3COCH3D. CH3CH2CH2CH2CH2CH3E. HOCH2CH2OH

13



Solutions of Solids in Liquids Basic principles remain the same when

solutes are solids Sodium chloride (NaCl)

Ionic bonding Strong intermolecular forces Ions dissolve in water because ion-dipole forces

of water with ions strong enough to overcome ion-ion attractions

14



Hydration of Solid Solute Ions at edges, fewer

ion-ion attractions Multiple ion-dipole

attractions formed with water

These overcome stronger ion-ion electrostatic attractions

New ion at surface Process continues until

all ions in solution Hydration of ions

Completely surrounded by solvent

15



Hydration vs. Solvation

Hydration Ions surrounded by water molecules

Solvation General term for surrounding solute particle by

solvent molecules

16



Polar Molecules Dissolve in Water

17

H2O reorients so Positive H atoms are near negative ends of solute Negative O atoms are near positive ends of solute



Solvation in Nonpolar Solvents?

Wax is a mixture of long chain alkanes and is nonpolar

Benzene is a nonpolar solvent Benzene can dissolve waxes

Weak London dispersion forces in both solute and solvent. Weak London forces will form between solute and solvent.

Wax molecules Easily slip from solid Slide between benzene molecules Form new London forces between solvent and solute

18



Heat of Solution Energy change associated with formation of

solution Difference in intermolecular forces between

isolated solute and solvent and mixture Cost of mixing Energy exchanged between system and

surroundingsMolar Enthalpy of Solution (Hsoln)

Amount of enthalpy exchanged when one mole of solute dissolves in a solvent at constant pressure to make a solution

19



Heat of Solution

Hsoln > 0 (positive) Costs energy to make solution Endothermic Increase in potential energy of system

Hsoln < 0 (negative) Energy given off when solution is made Exothermic Decrease in potential energy of system

Which occurs depends on your system

20



Modeling Formation of Solution

Formation of solution from solid and liquid can be modeled as two-step process

Step 1: Separate solute and solvent molecules Break intermolecular forces Endothermic, increase in potential energy of

system, cost Step 2: Mix solute and solvent

Come together Form new intermolecular forces Exothermic, decrease in potential energy of

system, profit21



Two-Step Process Application of Hess Law

Way to take two things we can measure and use to calculate something we cant directly measure Hsoln is path independent

Method works because enthalpy is state function Hsoln = Hsoln + Hsolute + Hsolvent

Overall, steps take us from solid solute + liquid solvent final solution

These steps are not the way solution would actually be made in lab

22



Enthalpy Diagram 1. Break up solid lattice

Hlattice = lattice enthalpy Increase in potential energy

2. Dissolve gas in solvent HSolvation = solvation enthalpy Decrease in potential energy

Hsolution = Hlattice + HSolvation

Whether Hsolution is positive or negative depends on two steps

In lab, solution formed directly

23



Figure 12.5 Dissolving KI in H2OHlattice (KI) = 632 kJ mol1 Hhydration (KI) = 619 kJ mol1

Hsolution = Hlatt + HhydrHsoln = 632 kJ mol1 619 kJ mol1 Hsoln = +13 kJ mol1 Formation of KI(aq) is endothermic

24



Dissolving NaBr in H2OHlattice (NaBr) = 728 kJ mol1 Hhydration (NaBr) = 741 kJ mol1

Hsolution = Hlatt + Hhydr

25

Hsoln = 728 kJ mol1 741 kJ mol1 Hsoln = 13 kJ mol1 Formation of NaBr(aq) is exothermic



Your Turn!Of the following ions which would have the most exothermic enthalpy of hydration?A. Ga3+

B. Ca2+

C. Li+

D. Na+

26



Your Turn!When 2.50 g of solid sodium hydroxide is added to 100.0 g of water in a calorimeter, the temperature of the mixture increases by 6.50 C. Determine the molar enthalpy of solution for sodium hydroxide. Assume the specific heat of the mixture is 4.184 J g1 K1.A. +44.6 kJ/molB. +43.5 kJ/molC. 44.6 kJ/molD. 43.5 kJ/mol

27

mol NaOH = 2.5 g 1 mol NaOH40.0 g NaOH

= 0.0625 mol NaOH

Hsoln

= 102.5 g( ) 4.184 J g1 K 1( ) 6.5 K( ) 1 kJ1000 J

0.0625 mol NaOH= 44.6 kJ/mol NaOH



Solutions Containing Liquid SoluteSimilar treatment but with three step path1. Solute expanded to gas H is positive Increase in potential

energy2. Solvent expanded to

gas H is positive Increase in potential

energy3. Solvation occurs H is negative Decrease in potential

energy28

Hsoln = Hsolute + Hsolvent + Hsolvation



Ideal Solution One in which inter-

molecular attractive forces are identical Hsoln = 0

e.g., Benzene in CCl4 All London forces Hsoln ~ 0

Step 1 + Step 2 = Step 3 or

Hsolute + Hsolvent = Hsolvation 29



Gaseous Solutes in Liquid Solution Only very weak attractions exist between

gas molecules There are no intermolecular attractions in ideal

gases When making solution with gas solute

Energy required to expand solute is negligible

Heat absorbed or released when gas dissolves in liquid has two contributions:1.Expansion of solvent Hsolvent 2.Solvation of gas Hsolvation

30



Gas dissolves in organic solvent

Generally endothermic Hsolvation > 0

Gas dissolves in H2O Generally exothermic Hsolvation < 0

31

Gaseous Solutes in Liquid Solution



Your Turn!

The solubility of a substance increases with increased temperature if:A. Hsolution > 0B. Hsolution < 0C. Hsolution = 0

32



Your Turn!

In which of the following situations would the hsolution be exothermic:A. Hhydration > HlatticeB. Hhydration< HlatticeC. Hhydration = Hlattice

33



Solubility Mass of solute that forms saturated solution with

given mass of solvent at specified temperature

If extra solute added to saturated solution, extra solute will remain as separate phase

34

solubility = g solute100 g solvent

soluteundissolved

solutedissolved



Effect of Temperature on Solubility

Generally substances become more soluble at elevated temperatures Increased disorder in the solution is often the reason

for this

35



Solubility of Most Substances Increases with Temperature

Most substances become more soluble as T increases

Amount solubility increases Varies considerably Depends on

substance

36



Effect of T on Gas Solubility in Liquids Solubility of gases usually decreases as T increases

Solubilities of Common Gases in Water

37



Case Study: Dead Zones

During the industrial revolution, factories were built on rivers so that the river water could be used as a coolant for the machinery. The hot water was dumped back into the river and cool water recirculated. After some time, the rivers began to darken and many fish died. The water was not found to be contaminated by the machinery. What was the cause of the mysterious fish kills?

38

Increased temperature, lowered the amounts of dissolved oxygen



Henrys Law Pressure-Solubility Law Concentration of gas in liquid at any given

temperature is directly proportional to partial pressure of gas over solution

Cgas = kHPgas (T is constant)

Cgas = concentration of gas Pgas = partial pressure of gas kH = Henrys Law constant

Unique to each gas Tabulated

39



Henrys Law True only at low concentrations and

pressures where gases do NOT react with solvent

Alternate form

C1 and P1 refer to an initial set of conditions C2 and P2 refer to a final set of conditions

40

2

2

1

1PC

PC

=. Personal Used Only - Use Wisely .


Effect of Pressure on Gas Solubility

Solubility increases as P increases

41



Effect of Pressure on Gas SolubilityA. At some P, equilibrium exists between vapor phase and

solution ratein = rateout

B. Increase in P puts stress on equilibrium Increase in frequency of collisions so ratein > rateout

C. More gas dissolved rateout will increase until rateout = ratein again

42



Ex. Using Henrys LawCalculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 5 atm over the liquid at 25 C. The Henrys Law constant for CO2 in water at this temperature is 3.12 102 mol L1 atm1.

43

= (3.12 102 mol/L atm) (5.0 atm)= 0.156 mol/L 0.16 mol/L

CCO2

= kH(CO

2)P

CO2



Ex. Using Henrys LawCalculate the concentration of CO2 in a soft drink after the bottle is opened and equilibrates at 25 C under a partial pressure of CO2 of 4.0 104 atm.

44

C2 = 1.2 104 mol CO2/LCompared to the concentration of CO2 at

5 atm: 0.16 mol CO2/L

C2 =

0.156 mol/L( ) 4.0 104 atm( )5.0atm

C1

P1

=C

2

P2

C2=

P2C

1

P1. Personal Used Only - Use Wisely .


Learning Check

What is the concentration of dissolved nitrogen in a solution that is saturated in N2 at 2.0 atm? kH= 8.42 107 mol L1 atm1

45

Cg=kHPg Cg= (8.42 107 mol L1 atm1) 2.0 atm Cg=1.7 106 mol L1



Your Turn!How many grams of oxygen gas at 1.0 atm will dissolve in 10.0 L of water at 25 C if Henrys constant is 1.3 103 mol L1 atm1 at this temperature?A. 0.42 gB. 0.013 gC. 0.042 gD. 0.21 gE. 2.4 g

46

1.0 atm O2 .0013 mol O2

L atm 32.00 g

1.00 mol10 L

1 =0.416 g



Your Turn! The solubility of O2 in water is approximately

0.00380 g L-1 when the temperature is 25.0 C and the partial pressure of gaseous oxygen is 760 torr. What will the solubility of oxygen be if the pressure is adjusted to 1000 torr?

A. 0.00289 g L-1

B. 0.00500 g L-1

C. 1.49 g L-1

D. 2.89 x 103 g L-1

E. 3.46 x 103 g L-1

47

0.00380 g/L760 torr

= X g/L1000 torr

X = 3.80760

= 0.00500 g/L



Solubility of Polar vs. Nonpolar Gases Gas molecules with polar bonds are much more

soluble in water than nonpolar molecules like oxygen and nitrogen CO2, SO2, NH3 >> O2, N2, Ar

Form H-bonds with H2O Some gases have increased solubility because they

react with H2O to some extente.g., CO2(aq) + H2O H2CO3(aq) H+(aq) + HCO3(aq)

SO2(aq) + H2O H2SO3(aq) H+(aq) + HSO3(aq)

NH3(aq) + H2O NH4+(aq) + OH(aq)

48



Case Study

When you open a bottle of seltzer, it fizzes. How should you store it to increase the time before it goes flat?

49

Gases are more soluble at low temperature and high pressure.



Concentration What units we use depends on situation Stoichiometric calculations Molarity

Units: mol/LProblem: M varies with temperature

Volume varies with temperature Solutions expand and contract when heated

and cooled If temperature independent concentration is

needed, must use other units50

solution of Lsolute of mol

=M



Temperature Independent Concentration

1. Percent Concentrations Also called percent by mass or percent by

weight

This is sometimes indicated %(w/w) where w stands for weight or mass

The (w/w) is often omitted

51

%100solution of masssolute of mass massby percent =. Personal Used Only - Use Wisely .


Ex. Percent by Mass What is the percent by mass of NaCl in a solution consisting of 12.5 g of NaCl and 75.0 g water?

52

percent by massNaCl

= 14.3% NaCl

percent by massNaCl

= 12.5 g12.5 g + 75.0 g

100%

%100solution of masssolute of mass

massby percent =. Personal Used Only - Use Wisely .


Learning CheckSeawater is typically 3.5% sea salt and has a density of 1.03 g/mL. How many grams of sea salt would be needed to prepare enough seawater solution to fill a 62.5 L aquarium?What do we need to find?

62.5 L ? g sea saltWhat do we know?

3.5 g sea salt 100 g solution 1.03 g solution 1.00 mL solution 1000 mL 1.00 L

53

= 2.2 103 g sea saltsoln g 100

salt sea g 5.3solnmL 1.00

soln g 1.03L 1mL 1000

L 5.62



More Temperature Independent Concentration Units

Molality (m) Number of moles of solute per kilogram solvent

Also molal concentration Independent of temperature m vs. M Similar when d = 1.00 g/mL Different when d >> 1.00 g/mL or

d


Ex. Concentration CalculationIf you prepare a solution by dissolving 25.38 g of I2 in 500.0 g of water, what is the molality (m) of the solution? What do we need to find? 25.38 g ? m

What do we know? 253.8 g I2 1 mol I2 m = mol solute/kg solvent 500.0 g 0.5000 kg

Solve it

55

= 0.2000 mol I2/kg water = 0.2000 mwater kg .50000

1I g 253.8

I mol 1 I g 5.382

2

22



Ex. Concentration Calcn. (cont)What is the molarity (M) of this solution? The density of this solution is 1.00 g/mL. What do we need to find? 25.38 g ? M

What do we know? 253.8 g I2 1 mol I2 M = mol solute/L soln 1.00 g soln 1 mL soln

Solve it

56

= 0.1903 mol I2/ L soln = 0.1903 M

25.38 g I2

1 mol I2

253.8 g I2

1525.38 g soln

1.00 g1.00 mL

1000 mL1 L



Ex. M and m in CCl4What is the molality (m) and molarity (M) of a solution prepared by dissolving 25.38 g of I2 in 500.0 g of CCl4? The density of this solution is 1.59 g/mL. What do we need to find? 25.38 g ? m

What do we know? 253.8 g I2 1 mol I2 m = mol solute/kg solvent 500.0 g soln 0.5000 kg soln

Solve it

57

= 0.2000 mol I2/kg CCl4= 0.2000 m

25.38 g I2

1 mol I2

253.8 g I2

1500.0 g CCl

4

1000 g1 kg



Ex. M and m in CCl4 (cont)What is the molarity (M) of this solution? The density of this solution is 1.59 g/mL.What do we need to find? 25.38 g ? M

What do we know? 253.8 g I2 1 mol I2 M = mol solute/L solution 1.59 g solution 1 mL solution g of solution = g I2 + g CCl4 = 500.0 g + 25.38 g

Solve it

58

= 0.3030 mol I2/L soln = 0.303 M

25.38 g I2

1 mol I2

253.8 g I2

1525.38 g soln

1.59 g1.00 mL

1000 mL1 L



Converting between ConcentrationsCalculate the molarity and the molality of a 40.0% HBr solution. The density of this solution is 1.38 g/mL.

If we assume 100.0 g of solution, then 40.0 g of HBr.

If 100 g solution, then

mass H2O = 100.0 g solution 40.0 g HBr = 60.0 g H2O

59

mol HBr= 40.0 g HBr80.91 g HBr/mol

=0.494 mol HBr

40.0% HBr = wt%= 40.0 g HBr100 g solution

100%

m= mol HBrkg of H

2O

m = molHBrkgH

2O

= 0.494molHBr0.0600kgH

2O=8.24m



Converting between Concentrations (cont.)

Now calculate molarity of 40% HBr

60

mol HBr = 0.494 mol

= 72.46 mL

= 6.82 M

Volume solution= massdensity

= 100 g1.38 g/mL

M = 0.494 mol HBr72.46 mL solution

1000 mL1 L

M = mol HBrL solution



Your Turn!What is the molality of 50.0% (w/w) sodium hydroxide solution? A. 0.500 mB. 1.25 mC. 0.025 mD. 25 mE. 50 m

61

Molar Masses H2O: 18.02 g/mol ; NaOH: 40.00 g/mol

100.0 g soln = 50.0 g NaOH + 50.0 g water

= 25 m

50.0 g NaOH50.0 g water

1 mol NaOH40.00 g NaOH

1000 g water1 kg water



Your Turn!What is the molarity of the 50%(w/w) solution if its density is 1.529 g/mL?A. 19 MB. 1.25 MC. 1.9 MD. 0.76 M

62



Your Turn! - Solution

63

?M = 50 g NaOH 100 g solution

= 50 g NaOH 100 g solution

1 mol NaOH40.0 g NaOH

1.529 g solutionmL solution

1000 mLL

= 19 M



Your Turn!What mass of NH4Cl must be dissolved in 100 mL of H2O to make a 0.250 M solution?

A. 1340 gB. 21.40 gC. 13.4 gD. 1.34 g

64



Your Turn! - Solution

65

=0.250 moles NH

4Cl

1 L solution0.100 L

1

53.49 g NH4Cl

1 Mole

= 1.34 g



More Temperature-Independent Concentration Units

Mole Fraction

Mole %

66

XA= # mol A

Total moles of all components

mol %A = XA 100%



Colligative Properties Physical properties of solutions Depend mostly on relative populations of particles

in mixtures Dont depend on their chemical identities

Effects of solute on vapor pressure of solvents Solutes that cant evaporate from solution are

called nonvolatile solutes

Fact: All solutions of nonvolatile solutes have lower vapor pressures than their pure solvents

67



Raoults Law Vapor pressure of solution, Psolution, equals

product of mole fraction of solvent, Xsolvent and its vapor pressure when pure, Psolvent

Applies for dilute solutions

68

solvent pure of pressurevapor

solvent the of fraction mole solution the of pressure vapor

=

=

=

solvent

solvent

solution

P

XP

solventsolventsolution PXP =



Alternate form of Raoults Law Plot of Psoln vs. Xsolvent

should be linear

Slope =

Intercept = 0 Change in vapor

pressure can be expressed as

Usually more interested in how solutes mole fraction changes the vapor pressure of solvent

69

P = change in P = (Psolvent P

solution)

P = Xsolute

Psolvent

Xsolvent

Psolvent



Ex. Glycerin (using Raoults Law)Glycerin, C3H8O3, is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25 C. Calculate vapor pressure at 25 C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25 C is 23.8 torr.To solve use:

First we need Xsolvent, so we need to determine the moles of glycerin and moles of water.

70

Psolution

= Xsolvent

Psolvent



Ex. Glycerin (cont.)

71

Mole glycerin

Mole water

Mole fraction glycerin

50.0 mL C3H

8O

3 1.26 g

mL

1 mol C3H

8O

3

92.1 g C3H

8O

3

= 0.684 mol C3H

8O

3

500.0 mL H2O 1.00 g

mL

1 mol H2O

18.02 g H2O

= 27.75 mol H2O

XC3H8O3

= 27.75 mol27.75 mol + 0.684 mol

= 0.976

Psolution

= csolvent

Psolvent

= 0.9759( ) 23.8 torr = 23.2 torr



Learning CheckThe vapor pressure of 2-methylhexane is 37.986 torr at 15C. What would be the pressure of the mixture of 78.0 g 2-methylhexane and 15 g naphthalene, which is nearly non-volatile at this temperature?

72

Psolution = solvent Posolvent

= 33.02 torr = 33 torrP = 0.869 37.986 torr( )

mole 2-methylhexane = 78.0 g100.2 g/mol

= 0.7784 mol

mole naphthalene = 15 g128.17 g/mol

= 0.117 mol

c2methylhexane

= 0.7784 mol 0.7784 mol + 0.117 mol

= 0.869

naphthaleneC10H8

MM 128.17

2-methylhexaneC7H16MM 100.2



Why Nonvolatile Solute Lowers Vapor Pressure

A. Lots of solvent molecules in liquid phase Rate of evaporation and

condensation high and equal

B. Fewer solvent molecules in liquid Rate of evaporation is

lower so rate of condensation must be lower which is achieved by fewer gas particles

At equilibrium, fewer gas molecules give lower pressure

73



Solutions That Contain Two or More Volatile Components

Vapor contains two components Partial pressure of each component A and B is

given by Raoults Law

Total pressure of solution of components A and B given by Daltons Law of Partial Pressures

Since XA + XB = 1 then 1 XA = XB and we can write

74

PA= X

AP

A and P

B= X

BP

B

Ptotal

= PA+P

B= X

AP

A + X

BP

B

Ptotal

= XAP

A + (1 X

A)P

B



For Ideal, Two Component Solution of Volatile Components

75

PA= X

AP

A

PB= X

BP

Bo

Ptotal

= PA+P

B= X

AP

A + X

BP

B



Ex. Benzene and Toluene

Consider a mixture of benzene, C6H6, and toluene, C7H8, containing 1.0 mol benzene and 2.0 mol toluene. At 20 C, the vapor pressures of the pure substances are:P benzene = 75 torr and P toluene = 22 torr

Assuming the mixture obeys Raoults law, what is the total pressure above this solution?

76



Ex. Benzene and Toluene (cont.)1. Calculate mole fractions of A and B

2. Calculate partial pressures of A and B

3. Calculate total pressure

77

Pbenzene

= Xbenzene

Pbenzeneo = 0.33 75 torr = 25 torr

Ptoluene

= Xtoluene

Ptolueneo = 0.67 22 torr = 15 torr

Xbenzene

= 1.0 mol benzene1.0 mol + 2.0 mol

= 0.33 benzene

Xtoluene

= 2.0 mol1.0 + 2.0( ) mol

= 0.67 toluene

Ptotal

= Pbenzene

+Ptoluene

= (25 +15) torr = 40 torr



Learning CheckThe vapor pressure of 2-methylheptane is 233.95 torr at 55 C. At the same temperature, 3-ethylpentane has a vapor pressure of 207.68 torr. What would be the pressure of the mixture of 78.0 g 2-methylheptane and 15 g 3-ethylpentane?

78

Psolution = XAP oA + XBP oB

2-methylheptaneC8H18

MM 114.23 g/mol

3-ethylpentaneC7H16

MM 100.2 g/mol



Learning Check

79

P = 230 torr

P = 0.827 233.95 torr( )+ 0.173207.68 torr( )

mole 2-methylheptane = 78.0 g114.23 g/mol

= 0.683 mol

mole 3-ethylpentane = 15.0 g100.2 g/mol

= 0.150 mol

X3-ethylpentane =0.150 mol

(0.683 mol + 0.150 mol) = 0.173

X2-methylpentane =0.683 mol

(0.683 mol + 0.150 mol) = 0.827. Personal Used Only - Use Wisely .


Your Turn!n-Hexane and n-heptane are miscible. If the vapor pressure of pure hexane is 151.28 mm Hg, and heptane is 45.67 at 25 C, which equation can be used to determine the mole fraction of hexane in the mixture if the mixtures vapor pressure is 145.5 mm Hg?

A. X (151.28 mmHg) = 145.5 mmHgB. X (151.28 mmHg) + (X )(45.67 mm Hg)

= 145.5 mmHgC. X (151.28 mmHg) + (1 X )(45.67 mm Hg) =

145.5 mm HgD. None of these

80



Solutes also Affect Freezing and Boiling Points of Solutions

Facts: Freezing point of solution always lower than

pure solvent Boiling point of solution always higher than

pure solventWhy? Consider the phase diagram of H2O

Solid, liquid, gas phases in equilibrium Blue lines P vs. T

81



SolutionEffect of Solute Solute molecules stay

in solution only None in vapor None in solid

Crystal structure prevents from entering

Liquid/vapor Decrease number solvent

molecules entering vapor Need higher T to get all

liquid to gas Line at higher T along phase

border (red)

82



SolutionEffect of Solute Triple point moves

lower and to the left Solid/liquid

Solid/liquid line to left (red)

Lower T all along phase boundary

Solute keeps solvent in solution longer

Must go to lower T to form crystal

83



Freezing Point Depression and Boiling Point Elevation

Solution Observe increase in boiling point and decrease in

freezing point compared to pure solvent Both Tf and Tb depend on relative amounts of

solvent and solute

Colligative properties Boiling Point Elevation (Tb)

Increase in boiling point of solution vs. pure solvent

Freezing Point Depression (Tf ) Decrease in freezing point of solution vs. pure solvent

84



Freezing Point Depression (Tf)

Tf = Kf mwhere Tf = (Tf Tsoln)m = concentration in MolalityKf = molal freezing point depression constant

Units of C/molalDepend on solvent, see Table 12.4

85



Boiling Point Elevation (Tb)

Tb = Kb mwhere Tb = (Tsoln Tb)m = concentration in MolalityKb = molal boiling point elevation constant

Units of C/mDepend on solvent, see Table 12.4

86



Table 12.3 Kf and Kb

87



Ex. Freezing Point DepressionEstimate the freezing point of a permanent type of antifreeze solution made up of 100.0 g ethylene glycol, C2H6O2, (MM = 62.07 g/mol) and 100.0 g H2O (MM = 18.02 g/mol).

88

Tf = Kfm = (1.86 C/m) 16.11 m Tf = (Tfp Tsoln)30.0 C = 0.0 C Tsoln Tsoln = 0.0 C 30.0 C

= 1.611 mol C2H6O2

= 16.11 m C2H6O2

= 30.0 C

= 30.0 C

100.0 g C2H

6O

2

1 mol C2H

6O

2

62.07 g C2H

6O

2

m = mol solutekg solvent

=1.611 mol C

2H

6O

2

0.100 kg water



Your Turn!When 0.25 g of an unknown organic compound is added to 25.0 g of cyclohexane, the freezing point of cyclohexane is lowered by 1.6 C. Kf for the solvent is 20.2 C m1. Determine the molar mass of the unknown.A. 505 g/molB. 32 g/molC. 315 g/molD. 126 g/mol

89

Tf= K

fm

1.6 oC = 20.2oCm

0.250 gmolar mass

0.025 kgMW = 126 g/mol



Ex. Boiling Point ElevationA 2.00 g sample of a substance was dissolved in 15.0 g of CCl4. The boiling point of this solution was determined to be 77.85 C. Calculate the molar mass of the compound. [For CCl4, the Kb = 5.07 C/m and BP = 76.50 C]

90

Tb= K

bm m = Tb

Kb=

77.85 76.50( ) C5.07 C /m

= 0.2684 m

m = mol solutekg solvent

mol solute = 0.2684 m0.0150 kg CCl4= 4.026103 mol

molar mass = 2.00 g compound4.026103 mole

= 497 g/mol



Learning CheckAccording to the Sierra Antifreeze literature, the freezing point of a 40/60 solution of Sierra antifreeze and water is 4 F. What is the molality of the solution?

91

m = 11 m

tF = 1.8 tC + 32

4 F = 1.8 tC + 32

tC = 20. C

T = Tf Tsoln

Tsolution = m Kfp

0 (20.)( ) C = m 1.86 C m 1



Learning CheckIn the previous sample of a Sierra antifreeze mixture, 100 mL is known to contain 42 g of the antifreeze and 60. g of water. What is the molar mass of the compound found in this antifreeze if it has a freezing point of 4 F?

92

From before:m = 11 m

MM solute = 64 g/mol

= 0.66 mol solutemol solute = 11 m 0.060 kg solvent

solute mol 0.66solute g 42

MM =

solvent kgsolute mol

=m. Personal Used Only - Use Wisely .


Your Turn!If all of the following generic compounds have the same Kb, which will cause a greater change in the boiling point of a 0.25 m nonelectrolyte solution

(Assume all dissociate completely in solution)

A. MXB. MX2C. M2X3D. MX3E. Not enough information given

93



Membranes and PermeabilityMembranes

Separators Example: cell walls Keep mixtures separated

Permeability Ability to pass substances

through membrane

Semipermeable Some substances pass,

others dont Membranes are

semipermeable Selective

94



Membranes and Permeability Degree of permeability depends on type of

membrane Some pass water only Some pass water and small ions only

Membranes separating two solutions of different concentration Two similar phenomena occur Depends on membrane

Dialysis When semipermeable membrane lets both H2O and

small solute particles through Membrane called dialyzing membrane Keeps out large molecules such as proteins

95



OsmosisOsmotic Membrane

Semipermeable membrane that lets only solvent molecules through

Osmosis Net shift of solvent molecules (usually water)

through an osmotic membrane Direction of flow in osmosis,

Solvent flows from dilute to more concentrated side Flow of solvent molecules across osmotic membrane

Increase in concentration of solute on dilute side Decrease in concentration of solute on more

concentrated side

96



Osmosis and Osmotic Pressure

A. Initially, solution B separated from pure water, A, by osmotic membrane. No osmosis occurred yet

B. After a while, volume of fluid in tube higher. Osmosis has occurred.

C. Need back pressure to prevent osmosis = osmotic pressure.

97



Osmotic Pressure Exact back pressure needed to prevent osmotic

flow when one liquid is pure solvent.Why does osmosis eventually stop? Extra weight of solvent as rises in column

generates this opposing pressure When enough solvent transfers to solution so that

when osmotic pressure is reached, flow stops If osmotic pressure is exceeded, then reverse

process occurssolvent leaves solution Reverse osmosisused to purify sea water

98



Equation for Osmotic Pressure Similar to ideal gas law

V = nRT or = MRT = osmotic pressure V = volume n = moles M = molarity of solution (n/V) T = Temperature in Kelvins R = Ideal gas constant

= 0.082057 L atm mol1 K1

99



Osmometer Instrument to measure osmotic pressure Very important in solutions used for

biological samplesIsotonic solution

Same salt concentration as cells Same osmotic pressure as cells

Hypertonic solution Higher salt concentration than cells Higher osmotic pressure than cells Will cause cells to shrink and dehydrate

Hypotonic solution Lower salt concentration than cells Lower osmotic pressure than cells Will cause cells to swell and burst

100



Ex. Osmotic PressureEye drops must be at the same osmotic pressure as the human eye to prevent water from moving into or out of the eye. A commercial eye drop solution is 0.327 M in electrolyte particles. What is the osmotic pressure in the human eye at 25 C?

101

= MRT T = 25 C + 273.15

= 0.327 M( ) 0.08206 L atmK mol

298 K( ) = 8.00 atm



Ex. Using to determine MMThe osmotic pressure of an aqueous solution of certain protein was measured to determine its molar mass. The solution contained 3.50 mg of protein in sufficient H2O to form 5.00 mL of solution. The measured osmotic pressure of this solution was 1.54 torr at 25 C. Calculate the molar mass of the protein.

102

M = PRT

= 1.54 torr 1 atm

760 torr

0.08206 L atm mol1 K 1( )298 K = 8.28 105 mol

L

mol = 8.28 105 M( ) 5.00 103 L( ) = 4.14 107 mol

Molar mass = gmol

= 3.50 103 g

4.14 107 mol = 8.45 103 g/mol



Learning Check: OsmosisA solution of 5% dextrose (C6H1206) in water is placed into the osmometer shown at right. It has a density of 1.0 g/mL. The surroundings are filled with distilled water. What is the expected osmotic pressure at 25 C?

103

= 7 atm

= 0.277 molL

0.082057 L atmmol K

298 K( )

5 g C6H

12O

6

100 g solution

mol C6H

12O

6

180.16 g

1.0 g solnmL soln

1000 mLL

= M

= MRT. Personal Used Only - Use Wisely .


Learning CheckFor a typical blood plasma, the osmotic pressure at body temperature (37 C) is 5409 mm Hg. If the dominant solute is serum protein, what is the concentration of serum protein?

104

Molarity = 0.280 M

7.117 atm = ? molL

0.082057 L atmmol K

310. K( )

5409 mm Hg( ) 1 atm760 mm Hg

=

= MRT. Personal Used Only - Use Wisely .


Your Turn!Which aqueous solution has the highest osmotic pressure. A)0.100 molar Al(NO3)3B) 0.150 molar Ba(NO3)2C) 0.100 molar CaCl2D) 0.150 molar NaClE) 0.200 molar NH3

105

Molar Mass H2S: 34.076 g/mol

All solutions have the same temperature so the only influence is number of ions.



Colligative Properties of Electrolyte Solutions Differ

Kf (H2O) = 1.86 C/m Expect 1.00 m solution of NaCl to freeze at

1.86 C Actual freezing point = 3.37 C About twice expected T

Why? Colligative properties depend on

concentration (number) of particles One NaCl dissociates to form two particles

NaCl(s) Na+(aq) + Cl(aq)

106



Colligative Properties of Electrolyte Solutions Depend on Number of Ions Actual concentration of ions = 2.00 m

(Started with 1.00 m NaCl) Now use this to calculate T Tf = 1.86 C/m 2.00 m = 3.72 COr Tfinal = Tinitial Tf = 0.00 3.72 C

= 3.72 C Not exactly = to actual Tf = 3.37 C This method for ions gives rough estimate if

you assume that all ions dissociate 100%.107



Why isnt this Exact for Electrolytes? Assumes 100% dissociation of ions

Electrolytes dont dissociate 100%, especially in concentrated solutions

Some ions exist in ion pairs Closely associated pairs of oppositely charged ions that

behave as a single particle in solution So, fewer particles than predicted Result: freezing point depression and boiling point

elevation not as great as expected As you go to more dilute solutions, electrolytes

more fully dissociated and observe freezing point and boiling point closer to calculated value. Model works better at dilute concentrations

108



vant Hoff Factor = i Scales solute molality to correct number of

particles Measure of dissociation of electrolytes vant Hoff factor is equivalent to percent

ionization In general, it varies with concentration (see

Table 12.5, page 613)

109

i =(T

f)

measured

(Tf)

calcd as nonelectrolyte



Table 12.5 vant Hoff Factors vs. Concentration

110

Note: 1. As concentration decreases, iobserved iexpected 2. MgSO4 much less dissociated than NaCl or KCl



Your Turn!What is the vant Hoff factor for FeCl3 if a solution of 81.1 g dissolved in 500 g H2O has a boiling point of 101.74 C?A. 4B. 1C. 3.4D. 2.7E. Not enough information given

111

MM: FeCl3 = 162.2 g/mol; Kb=0.512 C/m

i = tKbm

= 1.74C0.512 C/m * 1.00 m

= 3.4. Personal Used Only - Use Wisely .


Nonelectrolytes Some molecular solutes produce weaker

colligative effects than predicted by their molal concentrations

Evidence of solute molecule clustering or associating

Result: only half the number of particles expected based on molality of solution, so Tf only half of what expected

112

C6H5 C

O

O H

2 C6H5 C

O

O H

C6H5C

O

OH. Personal Used Only - Use Wisely .


Nonelectrolytes

Result: calculated molar mass double what is expected Concentration of particles is about half of

what is expected and Tf only half of what expected

So size of solute particles is important Common with organic acids and alcohols

113

C6H5 C

O

O H

2 C6H5 C

O

O H

C6H5C

O

OH



Learning CheckIn preparing pasta, 2 L of water at 25 C are combined with about 15 g salt (NaCl, MM= 58.44 g/mol) and the solution brought to a boil. What is the change in the boiling point of the water?

114

T = i m Kbpmass of water = volume density = 2000 mL 1.0 g/mL

= 2000 g water = 2 kg

mNaCl = 0.26 mol / 2 kg = 0.1 m

T = 0.1 CT = 2 ionmol

0.1 m1

0.51 Cm

mol NaCl = 15 g NaCl58.44 g mol1

= 0.257 mol



Case StudySuppose you run out of salt. What mass of sugar (C12H22O11, MM = 342.30 g/mol) added to 2 L of water would raise the temperature of water by 0.10 C?

115

T=i m Kbp

mass of water = volume density = 2000 mL 1.0 g/mL

= 2000 g water = 2 kg 0.196 m = y mol / 2 kg

molality = 0.196 m

y = 0.392 mol

mass of sucrose =130 g

0.10 oC = 1 ionmol

x m( ) 0.51

oCm

0.392 mol = mass sucrose342.30 g mol1



Colligative Properties SummaryColligative properties depend on number

of particles i = mapp/mmolecular Raoults Law Freezing point depression Boiling point elevation Osmotic pressure

Must look at solute and see if molecular or ionic If molecular, i = 1 If ionic, must include i > 1 in equations

116



Colligative Properties

Raoults law

Freezing point depression Tf = iKf m

Boiling point elevation Tb = iKb m

Osmotic pressure = iMRT

117

Psolution

= csolvent

Psolvento



Heterogeneous Mixtures

118

Dispersions: 2 or more phases that are not homogeneous Many consumer products are of this type

Two major types Suspensions

Large particles mixed in a solvent Two phases

Colloidal dispersions Very small particles Do not settle or have separate phases Look like true solutions but scatter visible light



Colloidal Dispersions

119

Tyndall effect Scattering of light observed in colloidal

dispersions


Slide 1Chapter 12: SolutionsWhy do Solutions Form?Spontaneous MixingIntermolecular ForcesFormation of Solutions Depends on Intermolecular ForcesFormation of Solutions Depends on Intermolecular ForcesMiscible LiquidsImmiscible LiquidsRule of ThumbLearning CheckYour Turn!Your Turn!Solutions of Solids in LiquidsHydration of Solid SoluteHydration vs. SolvationPolar Molecules Dissolve in WaterSolvation in Nonpolar Solvents?Heat of SolutionHeat of SolutionModeling Formation of SolutionTwo-Step ProcessEnthalpy DiagramFigure 12.5 Dissolving KI in H2ODissolving NaBr in H2OYour Turn!Your Turn!Solutions Containing Liquid SoluteIdeal SolutionGaseous Solutes in Liquid SolutionSlide 31Your Turn!Your Turn!SolubilityEffect of Temperature on SolubilitySolubility of Most Substances Increases with TemperatureEffect of T on Gas Solubility in LiquidsCase Study: Dead ZonesHenrys LawHenrys LawEffect of Pressure on Gas SolubilityEffect of Pressure on Gas SolubilityEx. Using Henrys LawEx. Using Henrys LawLearning CheckYour Turn!Your Turn!Solubility of Polar vs. Nonpolar GasesCase StudyConcentrationTemperature Independent ConcentrationEx. Percent by MassLearning CheckMore Temperature Independent Concentration UnitsEx. Concentration CalculationEx. Concentration Calcn. (cont)Ex. M and m in CCl4Ex. M and m in CCl4 (cont)Converting between ConcentrationsConverting between Concentrations (cont.)Your Turn!Your Turn!Your Turn! - SolutionYour Turn!Your Turn! - SolutionMore Temperature-Independent Concentration UnitsColligative PropertiesRaoults LawAlternate form of Raoults LawEx. Glycerin (using Raoults Law)Ex. Glycerin (cont.)Learning CheckWhy Nonvolatile Solute Lowers Vapor PressureSolutions That Contain Two or More Volatile ComponentsFor Ideal, Two Component Solution of Volatile ComponentsEx. Benzene and TolueneEx. Benzene and Toluene (cont.)Learning CheckLearning CheckYour Turn!Solutes also Affect Freezing and Boiling Points of SolutionsSolutionEffect of SoluteSolutionEffect of SoluteFreezing Point Depression and Boiling Point ElevationFreezing Point Depression (Tf)Boiling Point Elevation (Tb)Table 12.3 Kf and KbEx. Freezing Point DepressionYour Turn!Ex. Boiling Point ElevationLearning CheckLearning CheckYour Turn!Membranes and PermeabilityMembranes and PermeabilityOsmosisOsmosis and Osmotic PressureOsmotic PressureEquation for Osmotic PressureOsmometerEx. Osmotic PressureEx. Using to determine MMLearning Check: OsmosisLearning CheckYour Turn!Colligative Properties of Electrolyte Solutions DifferSlide 107Why isnt this Exact for Electrolytes?vant Hoff Factor = iTable 12.5 vant Hoff Factors vs. ConcentrationYour Turn!NonelectrolytesNonelectrolytesLearning CheckCase StudyColligative Properties SummaryColligative PropertiesHeterogeneous MixturesColloidal Dispersions

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