matter pure impure one substance at least two substances Elementscompounds molecular ionic consist of ions NaCl consist of molecules H 2 O mono-atomic:

Solution properties

matter

pure impure

one substance at least two substances

Elements compounds

molecularionicconsist of ions

NaClconsist of molecules

H2O

mono-atomic: HeDiatomic: O2

Polyatomic: Fe

mixture

matter

pure impure

one substance at least two substances

mixture

homogeneous heterogeneousone phaseTab water

two phasesoil/water

milk

solution

Phaseregion of physical uniformity

all parts of that space have the same physical properties

color, density, conductivity, magnetism, …

mixture

homogeneous heterogeneousone phaseTab water

two phasesoil/water

milk

solution

solvent Solute(s)more abundant

component of mixturewater in tab water

nitrogen in air

Less abundant or other component(s) of mixture

salts in tab water

Water always solvent even in 98% H2SO4

concentartionamount of solute to solventamount of solute in solution

Molarity

Molality

Mole fraction

mass%

NaCl

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?

NaCl %3.14% NaClwt

7

10075.0)g (12.5g 12.5

%

NaClwt

%100solution of masssolute of mass

massby percent

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

solventkgmol

m of

solute of molality

8

If 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 is the molarity (M) of this solution? The density of this solution is 1.00 g/mL.

molmolg

g

M

mn

OHHC

OHHC

OHHC

2

11017.2

.07.46

1

52

52

52

ML

mol

L

mol

V

nM

solution

OHHC

OHHC 215.0215.0101.0

1017.2 2

52

52

%99.0%1001001

1%

%100%100%

52

52

52

5252

gg

gOHHC

mm

m

m

mOHHC

waterOHHC

OHHC

solution

OHHC

molmolg

g

M

mn

OHHC

OHHC

OHHC

2

11017.2

.07.46

1

52

52

52

molmolg

g

M

mn

water

waterwater 56.5

.0.18

1001

waterOHHC

OHHC

solution

OHHC

OHHC nn

n

n

nx

52

5252

52

00389.056.51017.2

1017.22

2

52

52

52

molmol

mol

nn

nx

waterOHHC

OHHC

OHHC

mkg

mol

kg

mol

wt

nm

waterkgsolvent

OHHC

OHHC 217.0217.0100.0

1017.2 2

/

52

52

3.75 M sulfuric acid solution has a density of 1.230 g/mL. Calculate the mass percent and molality of H2SO4 in solution.

368 g H2SO4; 862 g H2O; 29.9% H2SO4; 4.35 m

Why do Solutions Form?

1. Energetic effects:

Intermolecular Interaction forces

2. Entropy effects:

Tendency of nature towards disorder

14

Hsoln = H1solute + H2solvent + H3solvation

• Hsoln > 0 (positive)– Costs energy to make

solution– Endothermic– PE of system

• Hsoln < 0 (negative)– Energy given off when

solution is made– Exothermic– PE of system

Ideal Solution

• Step 1 + Step 2 = –Step 3Hsolute + Hsolvent = –Hsolvation

16

Ex. Benzene/CCl4

– All London forces

– Hsoln ~ 0

– Hsoln = 0

Hsoln = Hsolute + Hsolvent + Hsolvation

Why oil doesn’t dissolve in water?

Hsoln = H1 + H2 + Hsolvation

Hsolute > 0 expanding oil (bringing oil molecules apart)

but small (small London forces, nonpolar)

Hwater> 0 Expanding water (bringing water molecules apart)

very large (hydrogen bond!)

Attraction oil molecules-water molecules

Attraction: Ep decrease

small (polar-nonpolar)

Hsolv< 0

Hsoln strongly positive: strongly endo

Large amounts of energy needed to form

solution: usually can not be afforded

Immiscible Liquids • Two insoluble liquids• Do not mix• Get two separate phases• Strengths of IMFs are different in solute

and solvent• Different polarityEx. Benzene and water

18

C

CC

C

CC

H

H

H

H

H

HBenzene;no polar bond

Rule of Thumb

• “Like dissolves Like”– Use polar solvent for polar solute– Use Nonpolar solvent for nonpolar solute

19

Hydration of Solid Solute • At edges, fewer

oppositely charged ions around– H2O can come in

– Ion-dipole forces– Remove ion

• New ion at surface– Process continues until

all ions in solution• Hydration of ions

– Completely surrounded by solvent

20

Hsoln = Hsolute + Hsolvent + Hsolvation

Hsolute > 0 NaBr(s) → Na+(g) + Br-

(g) Lattice energy

DHlattice

Hhydration = Hsolvent + Hsolvation

Na+(g) + Br-

(g) → Na+(aq) + Br-

(aq) H2O

Hsoln = Hlattice + Hhydration

NaBr(s) → Na+(aq) + Br-

(aq) H2O

Dissolving NaBr in H2O

Dissolving NaBr in H2O

Hlattice (NaBr) = 728 kJ mol–1

Hhydration (NaBr) = –741 kJ mol–1

Hsolution = Hlatt + Hhydr

22

Hsoln = 728 kJ mol–1

– 741 kJ mol–1

Hsoln = – 13 kJ mol–1

Formation of NaBr(aq) is

exothermic

Dissolving KI in H2OHlattice (KI) = 632 kJ mol–1

Hhydration (KI) = – 619 kJ mol–1

Hsolution = Hlatt + Hhydr

Hsoln = 632 kJ mol–1

– 619 kJ mol–1 Hsoln = +13 kJ mol–1 • Formation of KI(aq) is endothermic

23

Still: KI dissolves in

water!!

Why?

Spontaneous Mixing• 2 gases mix

spontaneously – Due to random motions– Mix without outside work– Never separate

spontaneously• Tendency of system left

to itself, to become increasingly disordered – Entropy effect

24

Gas A Gas B

separate mixed

Maximum amount of NaCl that can be dissolved in 100 g water at 25oC is 36 g.

If 50 g added to water: only 36 g can dissolve, 14 g settle down.

If only 20 g NaCl are dissolved in 100 g H2O at 25oC: unsaturated: can take more.

solution

saturatedContains the maximum

amount of solute that can be disssolved

unsaturatedContains less than the

maximum amount of solute that can be disssolved

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

•

solvent 100

solute solubility

g

g

26

Solubility (NaCl)=36 g/100 g H2O at 25oC

Solubility of Most Substances Increases with Temperature

• Most substances become more soluble as T

• Amount solubility – Varies considerably– Depends on

substance

27

Effect of T on Gas Solubility in Liquids• Solubility of gases usually as T

28

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?

29

Increased temperature, lowered amounts of dissolved oxygen

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

and solution– ratein = rateout

B. in P– frequency of collisions so ratein > rateout

– More gas molecules dissolve than are leaving solution

C. More gas dissolved– Rateout will until Rateout = Ratein and equilibrium

restored

30

Effect of Pressure on Gas Solubility

• Solubility as P solubility as P – Soda in can

31

Henry’s Law“Concentration of gas in liquid at any given temperature is

directly proportional to partial pressure of gas over solution”

Cgas = kH Pgas (T is constant)

Cgas = concentration of gas in the liquid phase

Pgas = partial pressure of gas above solution

kH = Henry's Law constant

»Unique to each gas»Tabulated

32

Henry’s 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

2

2

1

1

PC

PC

33

Ex. 1 Using Henry’s 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 Henry’s Law constant for CO2 in water at this temperature is 3.12 102 mol/L·atm.

22)(CO2 COHCO PkC

34

= 3.12 102 mol/L·atm * 5.0 atm= 0.156 mol/L 0.16 mol/L

When under 5.0 atm pressure

Ex. 1 Using Henry’s 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.

35

C2 = 1.2 104 · mol/L

When open to air

atm0.5

atm 100.4mol/L 156.0 C

4

2

2

2

1

1

PC

PC

1

122 P

CPC

L

V

L

Voiii pxp

Raoult’s Law

Calculate the expected vapor pressure at 25oC for a solution prepared by dissolving 158.0 g common table sugar (sucrose, molar mass 342.3 g/mol) in 643.5 cm3 of water. At 25oC, the density of water is 0.9971 g/cm3 and the vapor pressure is 23.76 torr.

oiii pxp

o

waterwaterwatersolution pxpp

sugarwater

waterwater

BA

AA

nn

nx

nn

nx

molmolg

cm

M

dV

M

mn cm

g

water

waterwater

water

waterwater 6.35

.0.18

9771.05.6431

33

molmolg

g

M

mn

sucrose

sucrosesucrose 4616.0

.3.342

0.1581

9873.04616.06.35

6.35

sugarwater

waterwater nn

nx

torrp

torrpxpp

solution

o

waterwaterwatersolution

46.23

76.239873.0

xbay

xpppp

pxpxpp

pxpxp

pxpxp

ppp

BoA

oB

oAsolution

oBB

oAB

oAsolution

oBB

oABsolution

oBB

oAAsolution

BAsolution

1

Binary SystemSolute volatile

A+BL

V

BA pppsolution

BBAABATotal PXPXPPP

AAA PXP

BBB PXP

xbay

xpppp B

o

A

o

B

o

Asolution

positive

deviation

negative

deviation

P > Pideal P < Pideal

A solution is prepared by mixing 5.81 g acetone (C3H6O, molar mass

58.1 g/mol) and 11.9 g chloroform (HCCl3, molar mass 119.4 g/mol). At

35oC, this solution has a total vapor pressure of 260 torr. Is this an ideal solution? The vapor pressures of pure acetone and pure chloroform at 35oC are 345 and 293 torr, respectively.

Does it obey Raoult’s

Law?

o

chloroformchloroform

o

acetoneacetoneidealsolution

chloroformacetoneidealsolution

pxpxp

ppp

,

,

? P = Pideal ?

molmolg

g

M

mn

acetone

acetoneacetone 100.0

.1.58

81.51

molmolg

g

M

mn

HCCl

HCClHCCl 100.0

.4.119

9.111

3

33

500.0500.01

500.0100.0100.0

100.0

3

3

HCCl

HCClacetone

acetoneacetone

x

nn

nx

torrp

pxpxp

idealsolution

o

chloroformchloroform

o

acetoneacetoneidealsolution

3192935.03455.0,

,

P < Pideal

260 <

319

negative

deviation

Phase

diagram

pure water

Phase

diagram aq

solution

Colligative

Properties

solute solute

solute

Non-volatile solutes

Kf and Kb

47

A solution was prepared by dissolving 18.00 g glucose in 150.0 g water. The resulting solution was found to have a boiling point of 100.34oC. Calculate the molar mass of glucose. Glucose is a molecular solid that is present as individual molecules in solution.

solutebb mKT

molg

solute

solutesolute

solute

solutesolute

kgmol

solute

solventsolutesolute

solvent

solutesolute

kgmol

o

o

b

bsolute

mol

g

n

wtMwt

Mwt

wtn

molkgn

kgwtmnkgwt

nm

molkgC

C

K

Tm

18010.0

00.18

10.0150.067.0

//

67.0/.51.0

34.0

CCCTTT oooo

b 34.010034.100

What mass of ethylene glycol (C2H6O2, molar mass 62.1 g/mol), the

main component of antifreeze, must be added to 10.0 L water to produce a solution for use in a car’s radiator that freezes at -23.3oC? Assume the density of water is exactly 1 g/mL.

soluteff mKT

CTTT o

f

o

ff 3.23

gMwtnwtMwt

wtn

molkgn

kgwtmnkgwt

nm

molkgC

C

K

Tm

solutesolutesolute

solute

solutesolute

kgmol

solute

solventsolutesolute

solvent

solutesolute

kgmol

o

o

f

f

solute

77601.62125

1250.105.12

//

5.12/.86.1

3.23

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)

and 100.0 g H2O (MM = 18.02).

50CCCTTT

TTT

Ckgwt

nKT

kgwt

nmm

mKT

ooof

off

foff

o

EG

EGff

EG

EGEGsolute

soluteff

30300

3010.0

611.186.1

/

/

molmolg

g

Mwt

mn

EG

EGEG 611.1

/07.62

100

Osmosis and Osmotic Pressure

A. Initially, Soln B separated from pure water, A, by osmotic membrane (permeable to water). No osmosis occurred yet

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

51

Flow of water molecules

Net flow

Column risesPressure increases

Increase of flow from right to leftFinally:

Equilibrium established

Flow of water molecules

Net flow = 0

Osmotic pressure (p): Pressure needed to stop the flow.

Equation for Osmotic Pressure

• Assumes dilute solutions

p = i M R T– p = osmotic pressure– i = number of ions per formula unit = 1 for molecules– M = molarity of solution

• Molality, m, would be better, but M simplifies• Especially for dilute solutions, where m M

– T = Kelvin Temperature– R = Ideal Gas constant

= 0.082057 L·atm·mol1K1 53

Eye 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?

atmKmolK

atmLM 00.829808206.0327.0

p

p = MRT T(K) = 25°C + 273.15

Using p 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.

L

mol

KmolKatmL

torratm

torr

RTM 51028.8

29808206.0

7601

54.1

p

molLMVMn 735 1014.41000.51028.8

molgmol

g

n

massMwt /1045.8

1014.4

1050.3 37

3

solute solute p = i M R T

The observed osmotic pressure for a 0.10 M solution

of Fe(NH4)2(SO4)2 at 25oC is 10.8 atm. Compare the

expected and experimental values for i.

i=5