Chapter 11 Properties of Solutions - PowerPoint Presentation

Slide1

Chapter 11Properties of Solutions

DE Chemistry

Dr. Walker

Slide2

Solutions

In a solution, the excess material is the

solvent

, the smaller amount is the solute

Slide3

Measuring Solution Composition

Molarity is most common, covered previously in Chapter 4

Slide4

Measuring Solution Composition

Mass Percent

Slide5

Mass Percent Example

What is the weight percent of ethanol in a solution made by dissolving 5.3 g of ethanol

(C

2H

5

OH) in

85.0

mL

of water

?

Slide6

Mass Percent Example

What is the weight percent of ethanol in a solution made by dissolving 5.3 g of ethanol (C

2

H

5

OH) in 85.0 mL of water?

Density of water = 1.0 g/mL, therefore 85 mL of water = 85 g

Mass of solute/mass of solution x 100 = mass %

5.3 g ethanol/(5.3 g ethanol + 85.0 g water) =

5.9%

Slide7

Measuring Solution Composition

Mole Fraction

Slide8

Mole Fraction Example

What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C

2

H5

OH) in 85.0 mL of water?

Note: If the density of water is 1.0 g/mL, 85 mL = 85 g

Slide9

Mole Fraction Example

What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C

2

H

5

OH) in 85.0 mL of water?

Note: If the density of water is 1.0 g/mL, 85 mL = 85 g

Calculate moles of ethanol:

5.30

g

C

2

H

5

OH

1 mole

C

2

H

5

OH

46.08

g

C

2

H5OH

=

0.115

moles

C

2

H

5

OH

Slide10

Mole Fraction Example

What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C

2

H

5

OH) in 85.0 mL of water?

Note: If the density of water is 1.0 g/mL, 85 mL = 85 g

Calculate moles of water:

85.0

g H

2

O

1 mole H

2

O

=

4.72 moles

H

2

O

18.02 g H

2

O

Slide11

Mole Fraction Example

What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C

2

H

5

OH) in 85.0 mL of water?

Note: If the density of water is 1.0 g/mL, 85 mL = 85 g

Mole Fraction of ethanol

Mole Fraction of Water

0.115

moles

C

2

H

5

OH + 4.72 moles

H

2

O

0.115

moles

C

2H5OH

= 0.024

0.115

moles

C

2

H

5

OH

+

4.72

moles H

2

O

4.72

moles H

2O

=

0.976

Slide12

Measuring Solution Composition

Molality

Slide13

Example

What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C

2

H

5

OH) in 85.0 mL of water?

Note: If the density of water is 1.0 g/mL, 85 mL = 85 g

Slide14

Example

What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C

2

H

5

OH) in 85.0 mL of water?

Note: If the density of water is 1.0 g/mL, 85 mL = 85 g

Molality = moles solute/kg solvent

5.30

g

C

2

H

5

OH

1 mole

C

2

H

5

OH

46.08

g

C

2

H

5

OH

=

0.115

moles

C

2

H

5

OH

Slide15

Measuring Solution Composition

Normality

Equivalents of acids and bases

Mass

that donates or accepts a mole of

protons (usually a subscript of H

+

or OH

-

)

Equivalents

of oxidizing and reducing agents

Mass

that provides or accepts a mole of

electrons (usually a subscript)

Slide16

Normality

Find molarity first

Normality = Molarity x number of acidic protons

1.5 M

HCl

= 1.5 N

HCl

(1 acidic proton)

1.5 M H

2

SO

4

= 3.0 N H

2

SO

4

(2 acidic protons)

1.5 M H

3

PO

4

= 4.5 N H

3

PO4 (3 acidic protons)

Slide17

Energy of Solution Formation

“Like Dissolves Like”

Polar molecules and ionic compounds tend to dissolve in polar solvents

Nonpolar molecules dissolve in nonpolar compounds

Water - polar

Butane - nonpolar

Slide18

Heat of Solution

The

Heat of Solution

is the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent.

Substance

Heat of Solution

(kJ/mol)

NaOH

-44.51

NH

4

NO

3

+25.69

KNO

3

+34.89

HCl

-74.84

Slide19

Steps In Solution Formation

Slide20

Steps in Solution Formation

H

1

Step 1 -

Expanding the solute

Separating the solute into individual components – usually endothermic

Slide21

Steps in Solution Formation

H

2

Step 2 -

Expanding the solvent

Overcoming intermolecular forces of the solvent molecules – usually endothermic

Slide22

Steps in Solution Formation

H

3

Step 3 -

Interaction of solute and solvent to form the solution – usually exothermic

Slide23

Enthalpy of Solution

May be positive or negative

Negative is favorable (exothermic)

Positive values are endothermic, does not dissolve spontaneously

Solution formation increases entropy (favorable)

Slide24

Factors Affecting Solubility

Structure Effects

Polar (hydrophilic) dissolves in polar

Water soluble vitamins

Nonpolar (hydrophobic) in nonpolar

Fat soluble vitamins

Slide25

Pressure Effects – Henry’s Law

The concentration of a dissolved gas in a solution is directly proportional to the pressure of the gas above the solution

Applies most accurately for dilute solutions of gases that do not dissociate or react with the solvent

Yes



CO

2

, N

2

, O

2

No



HCl

, HI

 remember, these dissociate in solution!

Slide26

Henry’s Law Example

Slide27

Henry’s Law Example

Unopened bottle:

C =

k

P C = (3.1 x 10

-2

mol/L

.

Atm)(5.0 atm) =

0.16 mol/L

Opened bottle:

C =

k

P C = (3.1 x 10

-2

mol/L

.

Atm)(4.0 x 10

-4

atm) =

1.2 x 10

-5

mol/L

Slide28

Solubility – Temperature Effects

Solids

Increases in temperature always cause dissolving to occur more rapidly

Increases in temperature usually increases solubility (the amount that can be dissolved)

Gases

Solubility of gases always decreases with increasing temperatures

Slide29

Vapor Pressure – Nonvolatile solutes

Nonvolatile electrolytes lower the vapor pressure of a solute

Nonvolatile molecules do not enter the vapor phase

Fewer molecules are available to enter the vapor phase

Dissociation of ionic compounds has nearly two, three or more times the vapor pressure lowering of nonionic (nonelectrolyte) solutes.

Slide30

Slide31

Raoult’s Law

The presence of a nonvolatile solute lowers the vapor pressure of the solvent.

P

solution

= Observed Vapor pressure of

the solution

P

0

solvent

= Vapor pressure of the pure solvent



solvent

= Mole fraction of the solvent

This should make sense, since you’ve learned previously that dissolved

Solutes

raise

boiling points. Lower vapor pressure =

higher

boiling point

Slide32

Raoult’s Law Example

1.5 moles of cherry Kool-Aid are added to a pitcher containing 2 liters of water on a nice day at 25

o

 C. The vapor pressure of water alone is 23.8 mm Hg at 25

o

 C. What is the new vapor pressure of Kool-Aid?

Slide33

Raoult’s Law Example

1.5 moles of cherry Kool-Aid are added to a pitcher containing 2 liters of water on a nice day at 25

o

 C. The vapor pressure of water alone is 23.8 mm Hg at 25

o

 C. What is the new vapor pressure of Kool-Aid?

Usually, you must solve for the mole fraction of the solvent first:

2000 g H

2

O

1 mole H

2

O

18.02 g H

2

O

= 110.9 moles H

2

O

Mole Fraction of water (remember 1 L = 1000 g)

1.5 moles kool aid + 110.99 moles H

2

O

110.99 moles H

2

O

= 0.987

Slide34

Raoult’s Law Example

1.5 moles of cherry Kool-Aid are added to a pitcher containing 2 liters of water on a nice day at 25

o

 C. The vapor pressure of water alone is 23.8 mm Hg at 25

o

 C. What is the new vapor pressure of Kool-Aid?

Using the mole fraction from the previous page

P

Kool-Aid

 =

c

H2O

 P

pure H2O 

= (.987)(23.8 mm Hg) = 23.5 mm Hg

Slide35

Slide36

Liquid-liquid solutions in which both components are

volatile (non-ideal solutions)

Modified Raoult's Law:

P

0

is the vapor pressure of the pure solvent

P

A

and

P

B

are the partial pressures

This is a variation on Dalton’s Law of Partial Pressures that accounts for

Raoult’s

Law

Slide37

Modified Raoult’s Law Example

What is the vapor pressure

of the solution?

Slide38

Modified Raoult’s Law Example

Slide39

Modified Raoult’s Law Example

What is the vapor pressure

of the solution?

Slide40

Ideal Solutions

Liquid-liquid solution that obeys Raoult’s Law

Like gases, none are ideal, but some are close

Negative Deviations

Lower than predicted vapor pressure

Solute and solvent are similar, strong forces of attraction

In a solution of acetone and

ethanol, there is a negative

deviation due to the hydrogen

bonding interactions shown

Slide41

Ideal Solutions

Positive Deviations

Higher than predicted vapor pressure

Particles easily escape attractions in solution to enter the vapor phase

Ethanol and hexane are not attracted to each other due to differences

In polarity. As a result, its solution is a positive deviation (higher

D

H)

From Raoult’s Law.

Slide42

Colligative Properties

Properties dependent on the number of solute particles but not on their identity

Boiling-Point elevation

Freezing-Point depression

Osmotic Pressure

Slide43

Boiling Point Elevation

Each mole of solute particles raises the boiling point of 1 kilogram of water by 0.51 degrees Celsius.

K

b

= 0.51

C  kilogram/mol

m

=

molality

of the solution

i

=

van’t

Hoff

factor

Slide44

Freezing Point Depression

Each mole of solute particles lowers the freezing point of 1 kilogram of water by 1.86 degrees Celsius.

K

f

= 1.86

C  kilogram/mol

m

=

molality

of the solution

i

=

van’t

Hoff

factor

Slide45

Boiling Point Elevation

A radiator in a car has a capacity of 6 L. If a 50/50 (v/v) solution of ethylene glycol (C

2

H6

O

2

, d=1.11 g/mL) and water is used, what will be the new boiling point?

Slide46

Boiling Point Elevation

A radiator in a car has a capacity of 6 L. If a 50/50 (v/v) solution of ethylene glycol (C

2

H6

O

2

, d=1.11 g/mL) and water is used, what will be the new boiling point?

Van’t Hoff factor = 1 (not an ionic salt)

K

b

= 0.512 kg/mol

Molality = moles/kg

3000 mL C

2

H

6

O

2

1.11 g C

2

H

6

O

2

1 mL C

2

H

6

O

2

1 mole C

2

H

6

O

2

62.08 g C

2

H

6

O

2

=

53.15

moles

Slide47

Boiling Point Elevation

A radiator in a car has a capacity of 6 L. If a 50/50 (v/v) solution of ethylene glycol (C

2

H6

O

2

, d=1.11 g/mL) and water is used, what will be the new boiling point?

Van’t Hoff factor = 1 (not an ionic salt)

K

b

= 0.512 kg/mol

Molality = moles/kg

3 L water = 3 kg water

m = moles solute/kg solvent

m =

53.16

moles C

2

H

6

O

2

/3 kg water =

17.72

m

DT = i . Kb. m= 1 x 0.512 x 17.72 = 9.15 C + 100 = 109.07 degrees C

Slide48

Freezing Point Depression

Give the freezing point depression from adding 10 g ethylene glycol to 100 g water.

DT = (1) (1.86) (m)

10 g C2H6O2 / 620.8g g/mol = 0.16 moles

m = 0.16 moles/0.1 kg = 1.6 m

DT = (1)(1.86)(1.6) = 2.88 degrees

Slide49

Determination of Molar Mass by Freezing Point Depression

Slide50

Determination of Molar Mass by Freezing Point Depression

Rearrange equation

M

solute

=

D

T/K

f

M

solute

= 0.240 C / 5.12 C

.

kg/mol = 4.69 x 10

-2

mol/kg

Slide51

Determination of Molar Mass by Freezing Point Depression

Rearrange equation

M

solute

=

D

T/K

f

M

solute

= 0.240 C / 5.12 C

.

kg/mol = 4.69 x 10

-2

mol/kg

0.015 kg benzene

Moles hormone

=

4.69 x 10

-2

mol/kg

Slide52

Determination of Molar Mass by Freezing Point Depression

0.015 kg benzene

Moles hormone

=

4.69 x 10

-2

mol/kg

Molality

Moles hormone = 7.04 x 10

-4

moles

Determine molar mass

Slide53

Van’t Hoff Factor, i

For ionic compounds, the expected value of

i

is an integer greater than 1

NaCl,

i

= 2

BaCl

2

,

i

= 3

Al

2

(SO

4

)

3

,

i

= 5

Total number of ions in solution

Slide54

Dissociation Equations and the Determination of

i

NaCl(s)



AgNO

3

(s)



MgCl

2

(s)



Na

2

SO

4

(s)



AlCl

3

(s)



Na

+

(aq) + Cl

-

(aq)

Ag

+

(aq) + NO

3

-

(aq)

Mg

2+

(aq) + 2 Cl

-

(aq)

2 Na

+

(aq) + SO

4

2-

(aq)

Al

3+

(aq) + 3 Cl

-

(aq)

i

= 2

i

= 2

i

= 3

i

= 3

i

= 4

Slide55

Freezing Point Depression and Boiling Point Elevation Constants,

C/

m

Solvent

K

f

K

b

Acetic acid

3.90

3.07

Benzene

5.12

2.53

Nitrobenzene

8.1

5.24

Phenol

7.27

3.56

Water

1.86

0.512

Slide56

Osmotic Pressure

Semipermeable Membrane

Membrane which allows

solvent but not solute

molecules to pass through.

As time

passes…

One side of the membrane is mostly solvent

The other side

(which didn’t pass through) is more concentrated since the solute can’t go through the membrane

Osmosis

The flow of solvent into the solution through the semipermeable membrane

Slide57

Osmotic Pressure

The minimum pressure that stops the osmosis is equal to the osmotic pressure of the solution

http://chemwiki.ucdavis.edu/Textbook_Maps/General_Chemistry_Textbook_Maps/Map%3A_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Physical_Properties/13.07%3A_Osmotic_Pressure

Slide58

Osmotic Pressure Calculations



= Osmotic pressure

M

=

Molarity

of the solution

R

= Gas Constant = 0.08206

Latm

/

molK

i

=

van’t

Hoff Factor

See page 509 in text for example

Slide59

Example

Slide60

Dialysis

Transfer of solvent molecules as well as small solute molecules and ions

Remember in osmosis, only solute molecules are transferred

This description fits the “dialysis” used to filter the blood when the kidneys do not work properly.

Slide61

Kidney Dialysis

Dialyser contains ions and small molecules in blood, but none of the waste products.

Slide62

Osmotic Pressure and Living Cells

Crenation

Cells placed in a hypertonic (higher osmotic pressure) solution lose water to the solution, and shrink

Hemolysis

Cells placed in a hypotonic (lower osmotic pressure) solution gain water from the solution and swell, possibly bursting

Slide63

Reverse Osmosis

External pressure applied to a solution can cause water to leave the solution

Concentrates impurities (such as salt) in the remaining solution

Pure solvent (such as water) is recovered on the other side of the semipermeable membrane

Applicable to desalination plants which can make drinking water from ocean water.

Slide64

Colloids

Tiny particles suspended in some medium

Particles range in size from 1 to 1000 nm

Noticeable by shining light through the mediumParticles are large enough that they scatter light

Slide65

Examples of Colloids

Slide66

Tyndall Effect

Scattering of light by particles

Light passes through a solution

Light is scattered in a colloid

http://www.dorthonion.com/drcmcm/CHEMISTRY/Lessons/Lectures/images/14TyndallEffect.jpg