Bài soạn Chapter 12 Solutions

Solutions • solute is the dissolved substance seems to “disappear” “takes on the state” of the solvent • solvent is the substance solute dissolves in does not appear to change state •

Solution

• homogeneous mixtures

 composition may vary from one sample to another

 appears to be one substance, though really contains

multiple materials

• most homogeneous materials we encounter are

actually solutions

 e.g., air and sea water

• nature has a tendency toward spontaneous mixing

 generally, uniform mixing is more energetically

favorable

Chemistry, Julia Burdge, 2 nd e., McGraw Hill.

Solutions animation

Solutions

• solute is the dissolved substance

seems to “disappear”

“takes on the state” of the solvent

• solvent is the substance solute

dissolves in

does 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

Chemistry, Julia Burdge, 2 nd e., McGraw Hill.

• drinking seawater will dehydrate you and give you diarrhea

• 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 Phase Solute 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

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

Tro, Chemistry: A Molecular Approach 8

Solubility

• when one substance (solute) dissolves in another

(solvent) it is said to be soluble

salt is soluble in water

bromine is soluble in methylene chloride

• when one substance does not dissolve in another it is said to be insoluble

oil 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

Spontaneous Mixing

Tro, Chemistry: A Molecular Approach 10

Solubility

• there is usually a limit to the solubility of one

substance in another

gases are always soluble in each other

two liquids that are mutually soluble are said to be

miscible

 alcohol and water are miscible

 oil 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 Process

Entropy

• formation of a solution does not necessarily

lower the potential energy of the system

 the difference in attractive forces between atoms of

two separate ideal gases vs two mixed ideal gases is

negligible

 yet the gases mix spontaneously

• the gases mix because the energy of the system

is lowered through the release of entropy

• entropy is the measure of energy dispersal

throughout the system

• energy has a spontaneous drive to spread out

over as large a volume as it is allowed

Tro, Chemistry: A Molecular Approach 12

Intermolecular Forces and the Solution Process

• at least some of the energy to do this comes from

making new solute-solvent attractions

 exothermic

Intermolecular Attractions

Dissolution animation 1

Dissolution animation 2

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

Tro, Chemistry: A Molecular Approach 16

Solution Interactions

Will It Dissolve?

• Chemist’s Rule of Thumb –

Like Dissolves Like

• a chemical will dissolve in a solvent if it has a similar structure to the solvent

• when the solvent and solute structures are similar,

the solvent molecules will attract the solute particles

at least as well as the solute particles to each other

Tro, Chemistry: A Molecular Approach 18

Classifying Solvents

Solvent Class

Structural Feature

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

C H

C

C C

O

O

OH

O H

C H C

H2

OH OH

Vitamin C

The 4 OH groups make

the molecule highly

polar and it will also

H-bond to water.

Vitamin C is water

soluble

Tro, Chemistry: A Molecular Approach 20

C H

C H

C H

C H

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 –

Tro, Chemistry: A Molecular Approach 22

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

Energy of Solution animation

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

Tro, Chemistry: A Molecular Approach 26

Heat of Hydration

Tro, Chemistry: A Molecular Approach 28

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

Tro, Chemistry: A Molecular Approach 30

Solubility Limit

• a solution that has the maximum amount of solute

dissolved in it is said to be saturated

 depends on the amount of solvent

 depends 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 disturbed

 e.g., shaking a carbonated beverage

Tro, Chemistry: A Molecular Approach 32

Adding Solute to a Supersaturated

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 increases

 when 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

Tro, Chemistry: A Molecular Approach 34

Solubility Curves

Temp KCl NaCl NH 4 Cl Li 2 SO 4 Ca(OH) 2 Ce 2 (SO 4 ) 3

Tro, Chemistry: A Molecular Approach 36

Solubility of Some Salts in Water

Li2SO4

Ce2(SO4)3•9H2O

KNO3

Temperature Dependence of Solubility

of Gases in Water

• solubility is generally given in moles of solute that will dissolve in 1 Liter of solution

• generally lower solubility than ionic or polar

covalent solids because most are nonpolar

molecules

• for all gases, the solubility of the gas decreases

as the temperature increases

the Hsolution is exothermic because you do not need

to overcome solute-solute attractions

Tro, Chemistry: A Molecular Approach 38

Solubility of Gases in Water at Various

Temperatures

0 0.05

Tro, Chemistry: A Molecular Approach 40

Solubility of Gases in Water at Various Temperatures

0 0.001

Pressure Dependence of Solubility of

Gases in Water

• the larger the partial pressure of a gas in contact with a liquid, the more soluble the gas is in the liquid

Tro, Chemistry: A Molecular Approach 42

Henry’s Law

• the solubility of a gas (Sgas) is

directly proportional to its

partial pressure, (Pgas)

Relationship between Partial Pressure

and Solubility of a Gas

Tro, Chemistry: A Molecular Approach 44

Solubility of Gases in Water at Various

persrst

Tro, Chemistry: A Molecular Approach 46

atm5

3atm

M10

14.3

M.12

0

1 -

2 H

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

• 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

Tro, Chemistry: A Molecular Approach 48

Solution Concentration

Molarity

• 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

liters of solution

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-1(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-1 =

2.0 M Cl

Tro, Chemistry: A Molecular Approach 50

Solution Concentration

Molality, m

• moles of solute per 1 kilogram of solvent

defined in terms of amount of solvent, not solution

 like the others

• does not vary with temperature

because based on masses, not volumes

solvent of

kg

solute of

moles m

molality, 

• parts of solute in every 100 parts solution

• mass percent = mass of solute in 100 parts

solution by mass

if a solution is 0.9% by mass, then there are 0.9

grams of solute in every 100 grams of solution

 or 0.9 kg solute in every 100 kg solution

Solution of

Mass Solvent

of Mass Solute

of Mass

%

100 g

Solution, of

Mass

g Solute, of

Mass Percent

Tro, Chemistry: A Molecular Approach 52

Percent Concentration

Solution of

Mass Solvent

of Mass Solute

of Mass

%

100 g

Solution, of

Mass

g Solute, of

Mass Percent

Volume Solvent

of Volume Solute

of Mass

%

100 mL

Solution, of

Volume

g Solute, of

Mass e

Mass/Volum Percent

Volume Solvent

of Volume Solute

of Volume

%

100 mL

Solution, of

Volume

mL Solute,

of

Volume Percent

Whole

(solute) Part

Using Concentrations as

Conversion Factors

• concentrations show the relationship between the amount of solute and the amount of solvent

12%(m/m) sugar(aq) means 12 g sugar  100 g solution

 or 12 kg sugar  100 kg solution; or 12 lbs  100 lbs solution

 5.5%(m/v) Ag in Hg means 5.5 g Ag  100 mL solution

22%(v/v) alcohol(aq) means 22 mL EtOH  100 mL solution

• The concentration can then be used to convert the amount

of solute into the amount of solution, or vice versa

Tro, Chemistry: A Molecular Approach 54

Preparing a Solution

• need to know amount of solution and

concentration of solution

• calculate the mass of solute needed

start with amount of solution

use 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.”

Example - How would you prepare 250.0 g of 5.00% by mass glucose solution (normal glucose)?

100

Glucose g

5.00

Apply Solution Map:

Answer:

glucoseg

12.5solution

g100

glucoseg

.00

5solution

g0

Solution Concentration

PPM

• 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 water

for water solutions we often approximate the kg of the solution as the kg or L of water

• 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 = X = moles of components A

Tro, Chemistry: A Molecular Approach 59

mol 1

7 27

0 O

H C g 62.07

O H C mol

1 O

H C g 2

17

2 6 2

2 6

2 2

0 mL

1

L

0.001 mL

15

5 0.538  M 

M

L 0.515

O H C mol 1

mol 1

7 27

0 O

H C g 62.07

O H C mol

1 O

H C g 2

17

2 6 2

2 6

2 2

O H kg 0.500

O H C mol 1

7

0.27 m

2

2 6 2





Tro, Chemistry: A Molecular Approach 61

O H g 10 00

.

5 O

H kg 1

O H g

1000 O

H kg 500

sol' g

solute g

%  

n sol' g 2 7 51 O

H g 0 50 O

H C g 2

17 3.33%2 6 2  2 

%

%

100 n

sol' g

.2 7 51

O H C g

O H mol 5

7

27.

O H g 18.02

O H mol

1 kg

1

g

1000 O

H kg 500

7 27

0 O

H C g 62.07

O H C mol

1 O

H C g 2

17

2 6 2

2 6

2 2

2 2

6 2

2 6 2

10 9.89

O H mol 5

7 7 2 O

H C mol 1

7 0.27

O H C mol 1

7 0.27

Tro, Chemistry: A Molecular Approach 63

% 0.989

%

%

100 O

H mol 5

7 7 2 O

H C mol 1

7 0.27

O H C mol 1

7

0.27

%

2 2

6 2

2 6 2

7

27.

O H g 18.02

O H mol

1 kg

1

g

1000 O

H kg 500

7 27

0 O

H C g 62.07

O H C mol

1 O

H C g 2

17

2 6 2

2 6

2 2

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