Chapter 6

Solutions

Denniston Topping Caret

6th Edition

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

6.1 Properties of Solutions

• Solution - homogeneous mixture• Solute - the substance in the mixture present

in lesser quantity• Solvent - the substance present in the largest

quantity• Aqueous solution - solution where the

solvent is water• Solutions can be liquids as well as solids and

gases

Examples of Solutions

• Air - oxygen and several trace gases are dissolved in the gaseous solvent, nitrogen

• Alloys - brass and other homogeneous metal mixtures in the solid state

• Focus on liquid solutions as many important chemical reactions take place in liquid solutions

6.1

Prop

ertie

s of S

olut

ions

• Clear, transparent, no visible particles• May have color• Electrolytes are formed from solutes that are

soluble ionic compounds• Nonelectrolytes do not dissociate

• Volumes of solute and solvent are not additive– 1 L ethanol + 1 L water does not give 2 L of solution

6.1

Prop

ertie

s of S

olut

ions

)(-Cl )(Na )NaCl( OH2 aqaqs +→ +

General Properties of Liquid Solutions

6.1

Prop

ertie

s of S

olut

ions Solutions and Colloids

• Colloidal suspension - contains solute particles which are not uniformly distributed– Due to larger size of particles (1nm - 200 nm)– Appears identical to solution from the

naked eye– Smaller than 1 nm, have solution– Larger than 1 nm, have a precipitate

6.1

Prop

ertie

s of S

olut

ions Tyndall Effect

• Tyndall effect - the ability of a colloidal suspension to scatter light– See a haze when shining light through the mixture

– Solutions: light passes right through without scattering

Colloidal suspension Solution-light as haze, scatters light -no haze

6.1

Prop

ertie

s of S

olut

ions Degree of Solubility

• Solubility - how much of a particular solute can dissolve in a certain solvent at a specified temperature

• Factors which affect solubility:1 Polarity of solute and solvent

• The more different they are, the lower the solubility2 Temperature

• Increase in temperature usually increases solubility3 Pressure

• Usually has no effect• If solubility is of gas in liquid, directly proportional

to applied pressure

6.1

Prop

ertie

s of S

olut

ions Saturation

• Saturated solution - a solution that contains all the solute that can be dissolved at a particular temperature

• Supersaturated solution - contains more solute than can be dissolved at the current temperature

• How is this done? • Heat solvent, saturate it with solute then cool slowly• Sometimes the excess will precipitate out• If it doesn’t precipitate, the solution will be

supersaturated

6.1

Prop

ertie

s of S

olut

ions Solubility and Equilibrium

• If excess solute is added to a solvent, some dissolves

• At first, rate of dissolution is large

• Later, reverse reaction – precipitation – occurs more quickly

• When equilibrium is reached the rates of dissolution and precipitation are equal, there is some dissolved and some undissolved solute

• A saturated solution is an example of a dynamic equilibrium

6.1

Prop

ertie

s of S

olut

ions Solubility of Gases: Henry’s Law

• Henry’s law – the number of moles of a gas dissolved in a liquid at a given temperature is proportional to the partial pressure of the gas above the liquid

• Gas solubility in a liquid is directly proportional to the pressure of the gas in the atmosphere in contact with the liquid

• Gases are most soluble at low temperatures• Solubility decreases significantly at higher

temperatures– Carbonated beverages – CO2 solubility less when warm– Respiration – facilitates O2 and CO2 exchange in lungs

6.2 Concentration Based on Mass

• Concentration - amount of solute dissolved in a given amount of solution

• Concentration of a solution has an effect on – Physical properties

• Melting and boiling points– Chemical properties

• Solution reactivity

• Amount of solute = mass of solute in grams• Amount of solution = volume in milliliters

• Express concentration as a percentage by multiplying ratio by 100% = weight/volume percent or % (W/V)

%100solution of smilliliter

solute of gramsVW% ×=

solution ofamount solute ofamount ion concentrat =

Weight/Volume Percent6.

2 C

once

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tion

Bas

ed o

n M

ass

6.2

Con

cent

ratio

n B

ased

on

Mas

sCalculating Weight/Volume

PercentCalculate the percent composition or % (W/V) of 2.00 x 102 mL containing 20.0 g sodium chloride

20.0 g NaCl, mass of solute

2.00 x 102 mL, total volume of solution

% (W/V) = 20.0g NaCl / 2.00 x 102 mL x 100%

= 10.0% (W/V) sodium chloride

6.2

Con

cent

ratio

n B

ased

on

Mas

sCalculate Weight of Solute from

Weight/Volume PercentCalculate the number of grams of glucose in 7.50 x 102 mL of a 15.0% solution

15.0% (W/V) = Xg glucose/7.50 x 102 mL x 100%

Xg glucose x 100% = (15.0% W/V)(7.50 x 102 mL) Xg glucose = 113 g glucose

%100solution of smillilitersolute of grams

VW% ×=

%100solutions grams

solute gramsWW% ×=

6.2

Con

cent

ratio

n B

ased

on

Mas

sWeight/Weight Percent

• Weight/weight percent is most useful for solutions of 2 solids whose masses are easily obtained

• Calculate % (W/W) of platinum in gold ring with 14.00 g Au and 4.500 g Pt[4.500 g Pt / (4.500 g Pt + 14.00 g Au)] x 100%= 4.500 g / 18.50 g x 100% = 24.32% Pt

6.2

Con

cent

ratio

n B

ased

on

Mas

sParts Per Thousand (ppt) and

Parts Per Million (ppm)

• As percentage is the number of parts of solute in 100 parts of solution, ppt and ppm change the calculation only by orders of magnitude– ppt = g solute / g solution x 103 ppt– ppm = g solute / g solution x 106 ppt– ppt and ppm are most often used for very

dilute solutions

6.3 Concentration of Solutions: Moles and Equivalents

• Chemical equations represent the relative number of moles of reactants producing products

• Many chemical reactions occur in solution where it is most useful to represent concentrations on a molar basis

• The most common mole-based concentration unit is molarity

• Molarity– Symbolized M– Defined as the number of moles of solute per

liter of solution

Molarity6.

3 M

oles

and

Equ

ival

ents

solution Lsolute moles=M

• Calculate the molarity of 2.0 L of solution containing 5.0 mol NaOH

• Use the equation

• Substitute into the equation:MNaOH = 5.0 mol solute

2.0 L solution= 2.5 M

Calculating Molarity from Moles6.

3 M

oles

and

Equ

ival

ents

solution Lsolute moles=M

• If 5.00 g glucose are dissolved in 1.00 x 102 mL of solution, calculate molarity, M, of the glucose solution

• Convert from g glucose to moles glucose– Molar mass of glucose = 1.80 x 102 g/mol5.00 g x 1 mol / 1.80 x 102 g = 2.78 x 10-2 mol glucose– Convert volume from mL to L1.00 x 102 mL x 1 L / 103 mL = 1.00 x 10-1 L

• Substitute into the equation:

Mglucose = 2.78 x 10-2 mol glucose1.00 x 10-1 L solution

= 2.78 x 10-1 M

Calculating Molarity From Mass6.

3 M

oles

and

Equ

ival

ents

solution Lsolute moles=M

solution Lsolute moles=M6.

3 M

oles

and

Equ

ival

ents Dilution

Dilution is required to prepare a less concentrated solution from a more concentrated one

– M1 = molarity of solution before dilution– M2 = molarity of solution after dilution– V1 = volume of solution before dilution– V2 = volume of solution after dilution

moles solute = (M)(L solution)

• In a dilution will the number of moles of solute change? – No, only fewer per unit

volume

• So,

• Knowing any three terms permits calculation of the fourth

M1V1 = M2V2

6.3

Mol

es a

nd E

quiv

alen

ts Dilution

• Calculate the molarity of a solution made by diluting 0.050 L of 0.10 M HCl solution to a volume of 1.0 L– M1 = 0.10 M molarity of solution before dilution– M2 = X M molarity of solution after dilution– V1 = 0.050 L volume of solution before dilution– V2 = 1.0 L volume of solution after dilution

• Use dilution expression

• X M = (0.10 M) (0.050 L) / (1.0 L)

0.0050 M HCl OR 5.0 x 10-3 M HCl

M1V1 = M2V2

6.3

Mol

es a

nd E

quiv

alen

ts Calculating Molarity After Dilution

6.3

Mol

es a

nd E

quiv

alen

ts Representation of Concentration of Ions in Solution

Two common ways of expressing concentration of ions in solution:

1. Moles per liter (molarity)• Molarity emphasizes the number of

individual ions

2. Equivalents per liter (eq/L)• Emphasis on charge

ionon charges ofnumber (g)ion of massmolar ionan of equivalent One =6.

3 M

oles

and

Equ

ival

ents Comparison of Molarity and

Equivalents1 M Na3PO4

• What would the concentration of PO43- ions be?

• 1 M

• Equivalent is defined by the charge

• One Equivalent of an ion is the number of grams of the ion corresponding to Avogadro’s number of electrical charges

6.3

Mol

es a

nd E

quiv

alen

ts Molarity vs. Equivalents – 1 M Na3PO4

• 1 mol Na+ = 1 equivalent Na+

• 1 mol PO43- = 3 equivalents PO4

3-

• Equivalents of Na+?– 3 mol Na+ = 3 equivalents of Na+

• Equivalents of PO43-?

– 1 mol PO43- = 3 equivalents of PO4

3-

6.3

Mol

es a

nd E

quiv

alen

ts Calculating Ion Concentration• Calculate eq/L of phosphate ion, PO4

3- in a solution with 5.0 x 10-3 M phosphate

• Need to use two conversion factors:– mol PO4

3- mol charge

– mol charge eq PO43

5.0 x 10-3 mol PO43- x 3 mol charge x 1 eq

1 L 1 mol PO43- 1mol charge

• 1.5 x 10-2 eq PO43- /L

6.4 Concentration-Dependent Solution Properties

• Colligative properties - properties of solutions that depend on the concentrationof the solute particles, rather than the identity of the solute

• Four colligative properties of solutions1. vapor pressure lowering2. boiling point elevation3. freezing point depression4. osmotic pressure

Vapor Pressure of a LiquidConsider Raoult’s law in molecular

terms• Vapor pressure of a solution

results from escape of solvent molecules from liquid to gas phase

• Partial pressure of gas phase solvent molecules increases until equilibrium vapor pressure is reached

• Presence of solute molecules hinders escape of solvent molecules, lowering equilibrium vapor pressure6.

4 C

once

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Dep

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nt

Solu

tion

Prop

ertie

s

6.4

Con

cent

ratio

n-D

epen

dent

So

lutio

n Pr

oper

ties

Vapor Pressure Lowering• Raoult’s law - when a nonvolatile solute is added

to a solvent, vapor pressure of the solvent decreases in proportion to the concentration of the solute

• Solute molecules (red below) serve as a barrier to the escape of solvent molecules resulting in a decrease in the vapor pressure

6.4

Con

cent

ratio

n-D

epen

dent

So

lutio

n Pr

oper

ties

Freezing Point Depression and Boiling Point Elevation

• Freezing point depression may be explained considering the equilibrium between solid and liquid states– Solute molecules interfere with the rate at which

liquid water molecules associate to form the solid state

• Boiling point elevation can be explained considering the definition as the temperature at which vapor pressure of the liquid equals the atmospheric pressure– If a solute is present, then the increase in boiling

temperature is necessary to raise the vapor pressure to atmospheric temperature

• Freezing point depression (DTf) - is proportional to the number of solute particles – Solute particles, not just solute

• How does an electrolyte behave?– Dissociate into ions

• An equal concentration of NaCl will affect the freezing point twice as much as glucose (a nonelectrolyte)

• Each solvent has a unique freezing point depression constant or proportionality factor

DTf=kf m6.4

Con

cent

ratio

n-D

epen

dent

So

lutio

n Pr

oper

ties

Freezing Point Depression

• Boiling point elevation (DTb) - is proportional to the number of solute particles

• An electrolyte will affect boiling point to a greater degree than a nonelectrolyte of the same concentration

• Each solvent has a unique boiling point elevation constant

DTb=kb m

6.4

Con

cent

ratio

n-D

epen

dent

So

lutio

n Pr

oper

ties

Boiling Point Elevation

Molality• Solute concentration is expressed in

mole-based units– Number of particles is critical, not the mass

of solute

• Molality (m) = moles of solute per kg of solvent– The denominator is in kg solvent, not in kg

solution

6.4

Con

cent

ratio

n-D

epen

dent

So

lutio

n Pr

oper

ties

solvent kgsolute moles Molality =

Osmotic Pressure• Some types of membranes appear impervious

to matter, but actually have a network of small holes called pores

• These pores may be large enough to permit small solvent molecules to move from one side of the membrane to the other

• Solute molecules cannot cross the membrane as they are too large

• Semipermeable membrane - allows solvent but not solute to diffuse from one side to another

6.4

Con

cent

ratio

n-D

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dent

So

lutio

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ties

6.4

Con

cent

ratio

n-D

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So

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oper

ties

Osmotic Pressure

• Osmosis - the movement of solvent from a dilute solution to a more concentratedsolution through a semipermeable membrane

• Requires pressure to stop this flow

• Osmotic pressure () - the amount of pressure required to stop the flow across a semipermeable membrane

• Osmolarity - the molarity of particles in solution– Osmol, used for osmotic pressure

calculation

=MRT

6.4

Con

cent

ratio

n-D

epen

dent

So

lutio

n Pr

oper

ties

Osmotic Pressure

6.4

Con

cent

ratio

n-D

epen

dent

So

lutio

n Pr

oper

ties

Calculating OsmolarityCalculate the osmolarity of 5.0 x 10-3 M Na3PO4

Na3PO4 is an ionic compound forming electrolytes

1 mol Na3PO4 yields 4 product ions[5.0 x 10-3 mol Na3PO4 / L] x 4 mol particles

1 mol Na3PO4

= 2.0 x 10-2 mol particles / L2.0 x 10-2 mol particles / L = 2.0 x 10–2 osmol

6.4

Con

cent

ratio

n-D

epen

dent

So

lutio

n Pr

oper

ties

Calculating Osmotic PressureCalculate the osmotic pressure of a 5.0 x 10-2 Msolution of NaCl at 25°C (298 K)

Definition of osmotic pressure, =MRTM is osmolarity =[5.0 x 10-2 mol NaCl / L] x 2 mol particles

1 mol NaCl = 1.0 x 10-1 mol particles / L

Substitute into osmotic pressure equation: = 1.0 x 10-1 mol particles x 0.0821 L x atm x 298K

Liter K x mol = 2.4 atm

6.4

Con

cent

ratio

n-D

epen

dent

So

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n Pr

oper

ties

Tonicity and the Cell• Living cells contain aqueous solution and these cells are

also surrounded by aqueous solution• Cell function requires maintenance of the same osmotic

pressure inside and outside the cell• Solute concentration of fluid surrounding cells higher

than inside results in a hypertonic solution causing water to flow into the surroundings, causing collapse = crenation

• Solute concentration of fluid surrounding cells too low, results in a hypotonic solution causing water to flow into the cell, causing rupture = hemolysis

• Isotonic solutions have identical osmotic pressures and no osmotic pressure difference across the cell membrane

IsotonicHemolysisCrenation

6.4

Con

cent

ratio

n-D

epen

dent

So

lutio

n Pr

oper

ties

Tonicity and the Cell

Pickling Cucumber in Hypertonic Brine Due to Osmosis

6.4

Con

cent

ratio

n-D

epen

dent

So

lutio

n Pr

oper

ties

6.5 Water as a Solvent• Water is often referred to as the “universal

solvent”• Excellent solvent for polar molecules • Most abundant liquid on earth• 60% of the human body is water

– transports ions, nutrients, and waste into and out of cells

– solvent for biochemical reactions in cells and digestive tract

– reactant or product in some biochemical processes

6.6 Electrolytes in Body FluidsCATIONS IN THE BLOOD and CELLS• Na+ and K+ two most important cations

Na+

K+

Blood Cells

135 meq/L 3.5 – 5.0 meq/L

10 meq/L 125 meq/L

• Active transport - the transporting of Na+ and K+ ions across the cell membrane

• Cellular energy must be expended to make concentration of ions different on each side of the cell membrane

• This is accomplished via large protein molecules embedded in cell membranes

6.6

Elec

troly

tes i

n B

ody

Flui

ds

• Danger to the body occurs when Na+

and K+ both in blood and in cells becomes too high or low

- Na+ too low: - Decrease of urine output

- Dry mouth

- Flushed skin

- Fever

- Na+ too high:- Confusion, stupor, or coma6.

6 El

ectro

lyte

s in

Bod

y Fl

uids

- K+ too high:- Death by heart failure

- K+ too low:- Death by heart failure

ANIONS IN THE BLOOD

• Cl-

- acid/base balance- maintenance of osmotic pressure- oxygen transport by hemoglobin6.

6 El

ectro

lyte

s in

Bod

y Fl

uids

• HCO3-

- Form in which most waste CO2 is carried out of the body

PROTEINS IN THE BLOOD• Blood clotting factors

• Antibodies

• Albumins (carriers of nonpolar substances which cannot dissolve in water)

• Proteins are transported as a colloidal suspension

• The blood also transports nutrients and waste products6.

6 El

ectro

lyte

s in

Bod

y Fl

uids

Chapter 7

Energy, Rate, and Equilibrium

Denniston Topping Caret

6th Edition

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

7.1 Thermodynamics• Thermodynamics – the study of energy,

work, and heat– applied to chemical change

• Calculate the quantity of heat obtained from combustions of one gallon of fuel oil

– applied to physical change• Determine the energy released by boiling water

• The laws of thermodynamics help us to understand why some chemical reactions occur and others do not

Basic Concepts – from Kinetic Molecular Theory– Molecules and atoms in a reaction mixture are

in constant, random motion

– Molecules and atoms frequently collide with each other

– Only some collisions, those with sufficient energy, will break bonds in molecules

– When reactant bonds are broken, new bonds may be formed and products result

The Chemical Reaction and Energy7.

1 Th

erm

odyn

amic

s

7.1

Ther

mod

ynam

ics

Change in Energy and Surroundings

• Absolute value for energy stored in a chemical system cannot be measured

• Can measure the change in energy during these chemical changes

• System – contains the process under study

• Surroundings – the rest of the universe

7.1

Ther

mod

ynam

ics

Changes in the System• Energy can be lost from the system to the

surroundings

• Energy may be gained by the system at the expense of the surroundings– This energy change is usually in the form of heat

– This change can be measured

• First Law of Thermodynamics – energy of the universe is constant

• This law is also called the Law of Conservation of Energy

• Where does the reaction energy come from that is released and where does the energy go when it is absorbed?7.1

Ther

mod

ynam

ics

Law of Conservation of Energy

Changes in Chemical Energy

• Consider the reaction converting AB and CD to AD and CB

• Each chemical bond is stored chemical energy

• If a reaction will occur – Bonds must break– Breaking bonds requires energy7.

1 Th

erm

odyn

amic

s

A-B + C-D A-D + C-B

If the energy required to break the bonds is less than the energy released when the bonds are formed, there is a net release of energy

– This is called an Exothermic reaction– Energy is a product in this reaction

These bonds must be broken in the

reaction, requiringenergy

These bonds are formed, releasing

energy

7.1

Ther

mod

ynam

ics

A-B + C-D A-D + C-B

Exothermic Reactions

• If the energy required to break the bonds is larger than the energy released when the bonds are formed, there will need to be an external supply of energy– This is called an Endothermic reaction

These bonds must be broken in the

reaction, requiringenergy

These bonds are formed, releasing

energy

7.1

Ther

mod

ynam

ics

A-B + C-D A-D + C-B

Endothermic Reactions

Endothermic ReactionDecomposition

22 kcal + 2NH3(g) N2(g) + 3H2(g)

7.1

Ther

mod

ynam

ics

Exothermic ReactionCombustion

CH4(g) + 2O2(g)CO2(g) + 2H2O(g) + 211 kcal

7.1

Ther

mod

ynam

ics

Enthalpy• Enthalpy – represents heat energy

• Change in Enthalpy (DHo) – energy difference between the products and reactants of a chemical reaction

• Energy released, exothermic reaction, enthalpy change is negative – In the combustion of CH4, DHo = -211 kcal

• Energy absorbed, endothermic, enthalpy change is positive.– In the decomposition of NH3, DHo=+22 kcal

7.1

Ther

mod

ynam

ics

Spontaneous and Nonspontaneous Reactions

• Spontaneous reaction - occurs without any external energy input

• Most, but not all, exothermic reactions are spontaneous

• Thermodynamics is used to help predict if a reaction will occur

• Another factor is needed, Entropy

DSo is positive

DSo is negative

7.1

Ther

mod

ynam

ics

Spontaneous and Nonspontaneous Reactions

7.1

Ther

mod

ynam

ics

Spontaneous and Nonspontaneous Reactions

Are the following processes exothermic or endothermic?

– Fuel oil is burned in a furnace

– C6H12O6(s) 2C2H5OH(l) + 2CO2(g)

DH=-16 kcal– N2O5(g) + H2O(l) 2HNO3(l) + 18.3 kcal

7.1

Ther

mod

ynam

ics

Entropy

• The second law of thermodynamics – the universe spontaneously tends toward increasing disorder or randomness

• Entropy (So) – a measure of the randomness of a chemical system• High entropy – highly disordered system, the

absence of a regular, repeating pattern

• Low entropy – well organized system such as a crystalline structure

• No such thing as negative entropy

DSo of a reaction = So(products) -So(reactants)

7.1

Ther

mod

ynam

ics

Entropy of Reactions

• A positive DSo means an increase in disorder for the reaction

• A negative DSo means a decrease in disorder for the reaction

All of these processes have a positive DSo

7.1

Ther

mod

ynam

ics

Processes Having Positive Entropy

Phase change

Melting

Vaporization

Dissolution

7.1

Ther

mod

ynam

ics

Entropy and Reaction Spontaneity

• If exothermic and positive DSo…

SPONTANEOUS

• If endothermic and negative DSo…

NONSPONTANEOUS

• For any other situations, it depends on the relative size of DHo and DSo

Greatest Entropy

• Which substance has the greatest entropy?

– He(g) or Na(s)

– H2O(l) or H2O(g)

7.1

Ther

mod

ynam

ics

• Free energy (DGo) – represents the combined contribution of the enthalpy and entropy values for a chemical reaction

• Free energy predicts spontaneity of chemical reactions– Negative DGo…Always Spontaneous

– Positive DGo…Never Spontaneous

DGo = DHo - TDSo

T in Kelvin

7.1

Ther

mod

ynam

ics

Free Energy

Free Energy and Reaction Spontaneity

• Need to know both DH and DS to predict the sign of DG, making a statement on reaction spontaneity

• Temperature also may determine direction of spontaneity DH +, DS - : DG always +, regardless of TDH -, DS + : DG always -, regardless of TDH +, DS + : DG sign depends on TDH -, DS - : DG sign depends on T7.

1 Th

erm

odyn

amic

s

7.2 Experimental Determination of Energy Change in Reactions

• Calorimetry – the measurement of heat energy changes in a chemical reaction

• Calorimeter – device which measures heat changes in calories

• The change in temperature is used to measure the loss or gain of heat

• Change in temperature of a solution, caused by a chemical reaction, can be used to calculate the gain or loss of heat energy for the reaction– Exothermic reaction – heat released is absorbed

– Endothermic reaction – reactants absorb heat from the solution

• Specific heat (SH) – the number of calories of heat needed to raise the temperature of 1 g of the substance 1 oC

sss SHT ×∆×= mQ

Heat Energy in Reactions7.

2 D

eter

min

atio

n of

Ene

rgy

Cha

nge

in R

eact

ions

• Specific heat of the solution along with the total number of g solution and the temperature change permits calculation of heat released or absorbed during the reaction

• S.H. for water is 1.0 cal/goC

• To determine heat released or absorbed, need:– specific heat– total number of grams of solution– temperature change (increase or decrease)

Heat Energy in Reactions7.

2 D

eter

min

atio

n of

Ene

rgy

Cha

nge

in R

eact

ions

• Q is the product – ms is the mass of solution in the calorimeterDTs is the change in temperature of the solution from

initial to final state– SHs is the specific heat of the solution

• Calculate with this equation

– Units are: calories = gram x ºC x calories/gram - ºC

Calculation of Heat Energy in Reactions

7.2

Det

erm

inat

ion

of E

nerg

y C

hang

e in

Rea

ctio

ns

sss SHT ×∆×= mQ

7.2

Det

erm

inat

ion

of E

nerg

y C

hang

e in

Rea

ctio

nsCalculating Energy Involved in

Calorimeter ReactionsIf 0.10 mol HCl is mixed with 0.10 mol KOH in a “coffee cup” calorimeter, the temperature of 1.50 x 102 g of the solution increases from 25.0oC to 29.4oC. If the specific heat of the solution is 1.00 cal/goC, calculate the quantity of energy evolved in the reactionDTs = 29.4oC - 25.0oC = 4.4oC

Q = ms x DTs x SHs

= 1.50 x 102 g solution x 4.4oC x 1.00 cal/goC

= 6.6 x 102 cal

7.2

Det

erm

inat

ion

of E

nerg

y C

hang

e in

Rea

ctio

nsCalculating Energy Involved in

Calorimeter ReactionsIs the reaction endothermic or exothermic

– 0.66 kcal of heat energy was released to the surroundings, the solution

– The reaction is exothermic

What would be the energy evolved for each mole of HCl reacted?– 0.10 mol HCl used in the original reaction – [6.6 x 102 cal / 0.10 mol HCl] x 10 = 6.6 kcal

7.2

Det

erm

inat

ion

of E

nerg

y C

hang

e in

Rea

ctio

nsBomb Calorimeter and

Measurement of Calories in FoodsNutritional Calorie

(large “C” Calorie) = – 1 kilocalorie (1kcal)

– 1000 calories• the fuel value of

food• Bomb Calorimeter

is used to measure nutritional Calories

7.2

Det

erm

inat

ion

of E

nerg

y C

hang

e in

Rea

ctio

nsCalculating the Fuel Value of

Foods1 g of glucose was burned in a bomb calorimeter. 1.25 x 103 g H2O was warmed from 24.5oC to 31.5oC. Calculate the fuel value of the glucose (in Kcal/g).DTs = 31.5oC - 24.5oC = 6.1oC

Surroundings of calorimeter is water with specific heat capacity = 1.00 cal/g H2O oCFuel Value =

= 1.25 x 103 g H2O x 6.1oC x 1.00 cal/g H2O oC= 7.6 x 103 cal

7.6 x 103 cal x 1 Calorie / 103 cal = 7.6 nutritional Calories

sss SHT ×∆×= mQ

7.3 Kinetics

• Thermodynamics determines if a reaction will occur spontaneously but tells us nothing about the amount of time the reaction will take

• Kinetics – the study of the rate (or speed) of chemical reactions– Also supplies an indication of the mechanism –

step-by-step description of how reactants become products

Kinetic Information• Kinetic information represents changes over

time, seen here:– disappearance of reactant, A– appearance of product, B

7.3

Kin

etic

s

7.3

Kin

etic

sAlternative Presentation of

Kinetic DataRather than the graph shown before, this figure demonstrates the change from purple reactant to green product over time from the molecular perspective

7.3

Kin

etic

sKinetic Data Assessed by Color

Change•Change in color over time can be used to monitor the progress of a chemical reaction•The rate of color change can aid in calculating the rate of the chemical reaction

7.3

Kin

etic

sThe Chemical Reaction

• C-H and O=O bonds must be broken • C=O and O-H bonds must be formed

• Energy is required to break the bonds– This energy comes from the collision of the molecules

– If sufficient energy available, bonds break and atoms recombine in a lower energy arrangement

– Effective collision is one that produces product molecules

– leads to a chemical reaction

CH4(g) + 2O2(g) CO2(g) +2H2O(g) + 211 kcal

7.3

Kin

etic

sActivation Energy and the

Activated Complex• Activation energy – the minimum amount of

energy required to initiate a chemical reaction

• Picture a chemical reaction in terms of changes in potential energy occurring during the reaction

– Activated complex – an extremely unstable, short-lived intermediate complex

– Formation of this activated complex requires energy (Ea) to overcome the energy barrier to start the reaction

• Reaction proceeds from reactants to products via the activated complex

• Activated complex – can’t be isolated from the reaction mixture

• Activation energy (Ea) is the difference between the energy of the reactants and that of the activated complex

7.3

Kin

etic

sActivation Energy and the

Activated Complex

•To be an Exothermic reaction requires a net release of energy (DHo)

7.3

Kin

etic

sActivation Energy in the Endothermic Reaction

• This figure diagrams an endothermic reaction

• Reaction takes place slowly due to the large activation energy required

• The energy of the products is greater than that of the reactants

7.3

Kin

etic

sFactors That Affect Reaction

Rate

1. Structure of the reacting species

2. Concentration of reactants

3. Temperature of reactants

4. Physical state of reactants

5. Presence of a catalyst

7.3

Kin

etic

sStructure of Reacting Species

• Oppositely charged species react more rapidly• Dissociated ions in solution whose bonds are already

broken have a very low activation energy• Ions with the same charge do not react• Bond strength plays a role

• Covalent molecules bonds must be broken with the activation energy before new bonds can be formed

• Magnitude of the activation energy is related to bond strength

• Size and shape influence the rate• Large molecules may obstruct the reactive part of the

molecule• Only molecular collisions with correct orientation lead to

product formation

7.3

Kin

etic

sThe Concentration of Reactants

• Rate is related to the concentration of one or more of the reacting substances

• Rate will generally increase as concentration increases– Higher concentration means more reactant

molecules per unit volume

– More reactant molecules means more collisions per unit time

7.3

Kin

etic

sThe Temperature of Reactants

• Rate increases as the temperature increases– Increased temperature relates directly to

increased average kinetic energy

– Greater kinetic energy increases the speed of particles

– Faster particles increases likelihood of collision

• Higher Kinetic Energy means a higher percentage of these collisions will result in product formation

7.3

Kin

etic

sThe Physical State of Reactants

Reactions occur when reactants can collide frequently with sufficient energy to react•Solid state: atoms, ions, compounds are close together but restricted in motion

•Gaseous state: particles are free to move but often are far apart causing collisions to be relatively infrequent

•Liquid state: particles are free to move and are in close proximity

•Reactions to be fastest in the liquid state and slowest in the solid state

• Liquid > Gas> Solid

7.3

Kin

etic

sThe Presence of a Catalyst

• Catalyst – a substance that increases the reaction rate– Undergoes no net change– Does not alter the final product of the reaction– Interacts with the reactants to create an alternative

pathway for product production

7.3

Kin

etic

sUse of a Solid Phase Catalyst

Haber Process is a synthesis of ammonia facilitated by a solid phase catalyst

– Diatomic gases bind to the surface– Bonds are weakened– Dissociation of diatomic gases and reformation of NH3– Newly formed NH3 leaves the solid surface with

catalyst unchanged

N2+3H2 2NH3

• Consider a decomposition reaction with the following balanced chemical equation:

• When heated N2O5 decomposes to 2 products – NO2 and O2

• When holding all factors constant, exceptconcentration, rate of reaction is proportional to the concentration

)(O)(NO4)(ON2 2252 ggg +→∆

7.3

Kin

etic

sMathematical Representation of

Reaction Rate

• Reaction rate is proportional to reactant concentration –– Concentration of N2O5 is denoted as [N2O5]– Replace proportionality symbol with = and

proportionality constant k– k is called the rate constant

)(O)(NO4)(ON2 2252 ggg +→∆

]O[Nrate 52∝

]Ok[N rate 52=

7.3

Kin

etic

sMathematical Representation of

Reaction Rate

• For a reaction Aproducts we write the equation:rate = k[A]n

• This is called the rate equation (or rate law)

• The exponent n is the order of the reaction– If n=1, first order– If n=2, second order

– n must be determined experimentally

– This exponent is not the same as the coefficient of the reactant in the balanced equation

7.3

Kin

etic

sRate Equation

• For the equation A + B productsthe rate equation is:

– rate = k[A]n[B]m

• What would be the general form of the rate equation for the reaction:

CH4+2O2CO2+2H2O

– Rate = k[CH4]n[O2]m

• Knowing the rate equation and the order helps industrial chemists determine the optimum conditions for preparing a product

7.3

Kin

etic

sRate Equation

• Write the form of the rate equation for the oxidation of ethanol (C2H5OH)

• The reaction has been experimentally determined to be first order in ethanol and third order in oxygen– Rate expression involves only the reactants– Concentrations: [C2H5OH][O2]– Raise each to exponent corresponding to its order

rate = k [C2H5OH][O2]3

– Remember that 1 as an exponent is understood and NOT written

– Knowing the rate equation and the order helps industrial chemists determine the optimum conditions for preparing a product

7.3

Kin

etic

sWriting Rate Equations

7.4 Equilibrium

Rate and Reversibility of Reactions• Equilibrium reactions – chemical reactions

that do not go to completion– Completion – all reactants have been converted

to products– Equilibrium reactions are also called Incomplete

reactions– Seen with both physical and chemical processes

• After no further observable change, measurable quantities of reactants andproducts remain

7.4

Equi

libriu

mPhysical Equilibrium

• Physical equilibria are reversible reactions– Dissolved oxygen in lake water– Stalactite and stalagmite formation– Sugar dissolved in water

• Reversible reaction – a process that can occur in both directions– Use the double arrow symbol

• Dynamic equilibrium – the rate of the forward process in a reversible reaction is exactly balanced by the rate of the reverse process

1. If add 2-3 g of sugar into 100 mL water• All will dissolve with stirring in a short time• No residual solid sugar, sugar dissolved completely

Sugar (s) Sugar (aq)

2. If add 100 g of sugar in 100 mL of water

• Not all of it will dissolve even with much stirring• Over time, you observe no further change in the

amount of dissolved sugar• Appears nothing is happening – Incorrect!

7.4

Equi

libriu

mSugar in Water

2. If add 100 g of sugar in 100 mL of water• Appears nothing is happening – Incorrect!

– Individual sugar molecules are constantly going into and out of solution

– Both happen at the same rate– Over time the amount of sugar dissolved in the

measured volume of water does not change

• An equilibrium situation has been established

• Some molecules dissolve and others return to the solid state – the rate of each process is equal

Sugar(s) Sugar(aq)

7.4

Equi

libriu

mSugar in Water

sugar(s) sugar(aq)

7.4

Equi

libriu

mDynamic Equilibrium

• The double arrow serves as an indicator of– a reversible process– an equilibrium process– the dynamic nature of the process

• Continuous change is taking place without observable change in the amount of sugar in either the solid or the dissolved form

)][sugar()][sugar(Keq s

aqkk

r

f ==

7.4

Equi

libriu

m

• ratef = forward rate rater = reverse rate

• at equilibrium: ratef = rater

• ratef = kf[sugar(s)]

• rater = kr[sugar(aq)]

• kf[sugar(s)]=kr[sugar(aq)]• Equilibrium constant (Keq) – ratio of the two rate

constants

Equilibrium Constant

7.4

Equi

libriu

mChemical Equilibrium

The Reaction of N2 and H2

N2(g) + 3H2(g) 2NH3(g)

• Mix components at elevated temperature

• Some molecules will collide with sufficient energy to break N-N and H-H bonds

• Rearrangement of the atoms will produce the product NH3

7.4

Equi

libriu

mChemical Equilibrium

• Initially the forward reaction is rapid– Reactant concentrations are

high– Product concentration

negligible• Forward reaction rate

decreases with time– Concentrations of reactants

are decreasing– Product concentration

increasing

N2(g) + 3H2(g) 2NH3(g)

Equilibrium occurs when the rate of reactant depletion is equal to the rate of product depletion Rates of forward and reverse reactions are Equal

Chemical Equilibrium

Basic equation divides into 2 parts:

• forward rxn: N2(g) + 3H2(g) 2NH3(g)

• reverse rxn: 2NH3(g)N2(g) + 3H2(g)

• ratef = kf[N2]n[H2]m

• rater = kr[NH3]p

• ratef=rater

N2(g) + 3H2(g) 2NH3(g)

m2

n2

p3

eq ]H[][N][NHK ==

r

f

kk7.

4 Eq

uilib

rium

m2

n2

p3

eq ]H[][N][NHK ==

r

f

kk

• The exponents in the rate expression are numerically equal to the coefficients

• Keq is a constant at constant temperature

322

23

eq ]H][[N][NHK =

7.4

Equi

libriu

mChemical Equilibrium

ba

dc

B][[A]D][C][Keq =

aA + bB cC + dD

7.4

Equi

libriu

mThe Generalized Equilibrium-

Constant Expression for a Chemical Reaction

• A and B are reactants• C and D are products• a, b, c, and d are the coefficients of the

balanced equation

7.4

Equi

libriu

mWriting Equilibrium-Constant

Expressions• Equilibrium constant expressions can only be

written after a correct, balanced chemical equation

• Each chemical reaction has a unique equilibrium constant value at a specified temperature

• The brackets represent molar concentration

• All equilibrium constants are shown as unitless

• Only the concentration of gases and substances in solution are shown• Concentration for pure liquids and solids are not shown

7.4

Equi

libriu

mWriting an Equilibrium-Constant

ExpressionWrite an equilibrium-constant expression for the reversible reaction:

H2(g) + F2(g) 2HF(g)•No solids or liquids are present

– All reactants and products appear in the expression– Numerator term is the product term [HF]2

– Denominator term is the reactants [H2] and [F2] – Each term contains an exponent identical to the

corresponding coefficient in the balanced equationKeq = [HF]2

[H2][F2]

7.4

Equi

libriu

mWriting an Equilibrium-Constant

ExpressionWrite an equilibrium-constant expression for the reversible reaction:

MnO2(s) + 4H+(aq) + 2Cl-(aq) Mn2+(aq) + Cl2(g) + H2O(l)

• MnO2 is a solid

• H2O(l) is a product, but negligible compared to solvent water– Numerator term is the product terms [Mn2+] and [Cl2]– Denominator term is the reactants [H+]4 and [Cl-]2

– Each term contains an exponent identical to the corresponding coefficient in the balanced equation

Keq = [Mn2+] [Cl2][H+]4 [Cl-]2

7.4

Equi

libriu

mInterpreting Equilibrium Constants

• Reversible arrow in chemical equation indicates equilibrium exists

• The numerical value of the equilibrium constant tells us the extent to which reactants have converted to products

1. Keq greater than 1 x 102

• Large value of Keq indicates numerator (product term) >>> denominator (reactant term)

• At equilibrium mostly product present

7.4

Equi

libriu

mInterpreting Equilibrium Constants

2. Keq less than 1 x 10-2

• Small value of Keq indicates numerator (product term) <<< denominator (reactant term)

• At equilibrium mostly reactant present

3. Keq between 1 x 10-2 and 1 x 102

• Equilibrium mixture contains significant concentration of both reactants and products

2NO2(g) N2O4(g)

7.4

Equi

libriu

mCalculating Equilibrium Constants

• A reversible reaction is allowed to proceed until the system reaches equilibrium

• Amount of reactants and products no longer changes

• Analyze reaction mixture to determine molar concentrations of each product and reactant

HI placed in a sealed container and comes to equilibrium; equilibrium reaction is:

2HI(g) H2(g) + I2(g)

•Equilibrium concentrations:– [HI] = 0.54 M– [H2] = 1.72 M– [I2] = 1.72 M

•Substitute concentrations:7.4

Equi

libriu

mCalculating an Equilibrium

Constant

Keq = [H2] [I2][HI]2

Keq= [1.72] [1.72] = 2.96[0.54]2

0.29= 10.1 or 1.0 x 101

2 significant fig res

7.4

Equi

libriu

mLeChateleir’s Principle

• LeChateleir’s Principle – if a stress is placed on a system at equilibrium, the system will respond by altering the equilibrium composition in such a way as to minimize the stress

• If reactants and products are present in a fixed volume and more NH3 is added into the container, the system will be stressed– Stressed = the equilibrium will be disturbed

N2(g) + 3H2(g) 2NH3(g)

7.4

Equi

libriu

mLeChateleir’s Principle

• Adding NH3 to the system causes stress– To relieve stress, remove as much of added

material as possible by converting it to reactants

• Adding N2 or H2 to the system causes stress also– To relieve stress, remove as much of added

material as possible by converting it to product

N2(g) + 3H2(g) 2NH3(g)Equilibrium shifted

Product introduced:Reactant introduced:

N2(g) + 3H2(g) 2NH3(g)

7.4

Equi

libriu

mEffect of Concentration

• Adding or removing either reactants or products at a fixed volume is saying that the concentration is changed

• Removing material decreases concentration

• System will react to this stress to return concentrations to the appropriate ratio

A B C

A: Reaction at equilibriumB: Shift to reactant with more red colorC: Shift to product with loss of red color

• Exothermic reactions: treat heat as a productN2(g) + 3H2(g) 2NH3(g) + 22 kcal

7.4

Equi

libriu

mEffect of Heat

• Addition of heat is treated as increasing the amount of product

• More product shifts equilibrium to the left

– Increases amount of reactants

– Decreases amount of product Heat favors the blue species

while cold favors the pink

• Endothermic Reaction – treat heat as a reactant

39 kcal + 2N2(g) + O2(g) 2NH3(g)

• This reaction shift will shift to the right if heat is added by increasing the temperature

7.4

Equi

libriu

mEffect of Heat

7.4

Equi

libriu

mEffect of Pressure

• Pressure affects the equilibrium only if one or more substances in the reaction are gases

• Relative number of gas moles on reactant and product side must differ

• When pressure goes up…shift to side with less moles of gas

• When pressure goes downs…shifts to side with more moles of gas

N2(g) + 3H2(g) 2NH3(g)

• If increase pressure, which way will the equilibrium shift?– Increased pressure favors decreased volume with more

product (2 moles) formed and less reactant (4 moles)

2HI(g) H2(g) + I2(g)• If increase the pressure in this reaction, which way

will the equilibrium shift?– No shift in equilibrium as both reactant and product

have 2 moles of gas

7.4

Equi

libriu

mEffect of Pressure

7.4

Equi

libriu

mEffect of a Catalyst

• A catalyst has no effect on the equilibrium composition

• It increases the rate of both the forward and reverse reaction to the same extent

• While equilibrium composition and concentration do not change, equilibrium is reached in a shorter time

Chapter 8

Acids and Bases and Oxidation-Reduction

Denniston Topping Caret

6th Edition

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

8.1 Acids and Bases

• Acids: Taste sour, dissolve some metals, cause plant dye to change color

• Bases: Taste bitter, are slippery, are corrosive

• Two theories that help us to understand the chemistry of acids and bases1. Arrhenius Theory2. Brønsted-Lowry Theory

8.1

Aci

ds a

nd B

ases

• Acid - a substance, when dissolved in water, dissociates to produce hydrogen ions– Hydrogen ion: H+ also called “protons”

HCl is an acid:HCl(aq) H+(aq) + Cl-(aq)

Arrhenius Theory of Acids and Bases

8.1

Aci

ds a

nd B

ases

Arrhenius Theory of Acids and Bases

• Base - a substance, when dissolved in water, dissociates to produce hydroxide ions

NaOH is a baseNaOH(aq) Na+(aq) + OH-(aq)

8.1

Aci

ds a

nd B

ases

Arrhenius Theory of Acids and Bases

• Where does NH3 fit? • When it dissolves in water it has basic

properties but it does not have OH- ions in it

• The next acid-base theory gives us a broader view of acids and bases

8.1

Aci

ds a

nd B

ases

Brønsted-Lowry Theory of Acids and Bases

• Acid - proton donor• Base - proton acceptor

– Notice that acid and base are not defined using water

– When writing the reactions, both accepting and donation are evident

HCl(aq) + H2O(l) Cl-(aq) + H3O+(aq)

What donated the proton? HClIs it an acid or base? Acid

What accepted the proton? H2OIs it an acid or base? Base

Brønsted-Lowry Theory of Acids and Bases

base

8.1

Aci

ds a

nd B

ases

acid

.

base acidNH3(aq) + H2O(l) NH4

+(aq) + OH-(aq)

8.1

Aci

ds a

nd B

ases

Brønsted-Lowry Theory of Acids and Bases

Now let us look at NH3 and see why it is a base

Did NH3 donate or accept a proton? Accept

Is it an acid or base? Base

What is water in this reaction? Acid

Acid-Base Properties of Water

• Water possesses both acid and base properties– Amphiprotic – a substance possessing both acid

and base properties– Water is the most commonly used solvent for

both acids and bases– Solute-solvent interactions between water and

both acids and bases promote solubility and dissociation

8.1

Aci

ds a

nd B

ases

8.1

Aci

ds a

nd B

ases

Acid and Base Strength

• Acid and base strength – degree of dissociation– Not a measure of concentration– Strong acids and bases – reaction with water is

virtually 100% (Strong electrolytes)

8.1

Aci

ds a

nd B

ases

Strong Acids and Bases

• Strong Acids:– HCl, HBr, HI Hydrochloric Acid, etc.– HNO3 Nitric Acid– H2SO4 Sulfuric Acid– HClO4 Perchloric Acid

• Strong Bases:– NaOH, KOH, Ba(OH)2

– All metal hydroxides

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

H2CO3(aq) + H2O(l) HCO3-(aq) + H3O+(aq)

8.1

Aci

ds a

nd B

ases

Weak Acids

• Weak acids and bases – only a small percent dissociates (Weak electrolytes)

• Weak acid examples:– Acetic acid:

– Carbonic Acid:

• Weak base examples:– Ammonia:

– Pyridine:

– Aniline:C6H5NH2(aq) + H2O(l) C6H5NH3

+(aq) + OH-(aq)

C5H5NH2(aq) + H2O(l) C5H5NH3+(aq) + OH-(aq)

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

8.1

Aci

ds a

nd B

ases

Weak Bases

• The acid base reaction can be written in the general form:

• Notice the reversible arrows• The products are also an acid and base

called the conjugate acid and base

acid baseHA + B A- + HB+

8.1

Aci

ds a

nd B

ases

Conjugate Acids and Bases

acid base

• Conjugate Acid – what the base becomes after it accepts a proton.

• Conjugate Base – what the acid becomes after it donates its proton

• Conjugate Acid-Base Pair – the acid and base on the opposite sides of the equation

base acid

HA + B A- + HB+

8.1

Aci

ds a

nd B

ases

HA + B A- + HB+

8.1

Aci

ds a

nd B

ases

Acid-Base Dissociation

• The reversible arrow isn’t always written– Some acids or bases essentially dissociate 100%– One way arrow is used

• HCl + H2O Cl- + H3O+

– All of the HCl is converted to Cl-

– HCl is called a strong acid – an acid that dissociates 100%

• Weak acid - one which does not dissociate 100%

8.1

Aci

ds a

nd B

ases

Conjugate Acid-Base Pairs

• Which acid is stronger:

HF or HCN? HF

• Which base is stronger:

CN- or H2O? CN -

8.1

Aci

ds a

nd B

ases

Acid-Base PracticeWrite the chemical reaction for the following acids

or bases in water. Identify the conjugate acid base pairs.

1. HF (a weak acid)

2. H2S (a weak acid)

3. HNO3 (a strong acid)

4. CH3NH2 (a weak base)

Note: The degree of dissociation also defines weak and strong bases

• Pure water is virtually 100% molecular• Very small number of molecules dissociate

– Dissociation of acids and bases is often called ionization

• Called autoionization• Very weak electrolyte

H2O(l) + H2O(l) H3O+(aq) + OH-(aq)

8.1

Aci

ds a

nd B

ases

The Dissociation of Water

• H3O+ is called the hydronium ion• In pure water at room temperature:

– [H3O+] = 1 x 10-7 M– [OH-] = 1 x 10-7 M

• What is the equilibrium expression for:

Remember, liquids are not included in equilibrium expressions

]OH][O[HK -3eq

+=H2O(l) + H2O(l) H3O+(aq) + OH-(aq)

8.1

Aci

ds a

nd B

ases

Hydronium Ion

• This constant is called the ion product for water and has the symbol Kw

• Since [H3O+] = [OH-] = 1.0 x 10-7 M, what is the value for Kw?– 1.0 x 10-14

– It is unitless

]OH][O[HK -3w

+=

8.1

Aci

ds a

nd B

ases

Ion Product of Water

8.2 pH: A Measurement Scale for Acids and Bases

• pH scale – a scale that indicates the acidity or basicity of a solution– Ranges from 0 (very acidic) to 14 (very basic)

• The pH scale is rather similar to the temperature scale assigning relative values of hot and cold

• The pH of a solution is defined as:pH = -log[H3O+]

• Use these observations to develop a concept of pH– if know one concentration, can calculate the

other– if add an acid, [H3O+] and [OH-] – if add a base, [OH-] and [H3O+] – [H3O+] = [OH-] when equal amounts of acid

and base are present• In each of these cases 1 x 10-14 = [H3O+][OH-]8.

2 pH

: A M

easu

rem

ent

Scal

e fo

r Aci

ds a

nd B

ases A Definition of pH

• pH of a solution can be:– Calculated if the concentration of either is

known• [H3O+] • [OH-]

– Approximated using indicator / pH paper that develops a color related to the solution pH

– Measured using a pH meter whose sensor measures an electrical property of the solution that is proportional to pH8.

2 pH

: A M

easu

rem

ent

Scal

e fo

r Aci

ds a

nd B

ases Measuring pH

• How do we calculate the pH of a solution when either the hydronium or hydroxide ion concentration is known?

• How do we calculate the hydronium or hydroxide ion concentration when the pH is known?

• Use two facts:

8.2

pH: A

Mea

sure

men

t Sc

ale

for A

cids

and

Bas

es Calculating pH

pH = -log[H3O+]

1 x 10-14 = [H3O+][OH-]

8.2

pH: A

Mea

sure

men

t Sc

ale

for A

cids

and

Bas

esCalculating pH from Acid

MolarityWhat is the pH of a 1.0 x 10-4 M HCl solution?

– HCl is a strong acid and dissociates in water– If 1 mol HCl is placed in 1 L of aqueous

solution it produces 1 mol [H3O+]– 1.0 x 10-4 M HCl solution has [H3O+]=1.0x10-4M

= -log [H3O+]= -log [1.0x10-4]= -[-4.00] = 4.00

pH = -log[H3O+]

8.2

pH: A

Mea

sure

men

t Sc

ale

for A

cids

and

Bas

es Calculating [H3O+] from pHWhat is the [H3O+] of a solution with pH = 6.00?

• 4.00 = -log [H3O+] • Multiply both sides of equation by –1

• -4.00 = log [H3O+]• Take the antilog of both sides

• Antilog –4.00 = [H3O+]• Antilog is the exponent of 10

• 1.0 x 10-4 M = [H3O+]

pH = -log[H3O+]

8.2

pH: A

Mea

sure

men

t Sc

ale

for A

cids

and

Bas

es Calculating the pH of a BaseWhat is the pH of a 1.0 x 10-3 M KOH solution?• KOH is a strong base (as are any metal hydroxides)• 1 mol KOH dissolved and dissociated in aqueous

solution produces 1 mol OH-

• 1.0 x 10-3 M KOH solution has [OH-] = 1.0 x 10-3 M

• Solve equation for [H3O+] = 1 x 10-14 / [OH-]• [H3O+] = 1 x 10-14 / 1.0 x 10-3 = 1 x 10-11

• pH = -log [1 x 10-11]= 11.00

1 x 10-14 = [H3O+][OH-]

pH = -log[H3O+]

8.2

pH: A

Mea

sure

men

t Sc

ale

for A

cids

and

Bas

esCalculating pH from Acid

MolarityWhat is the pH of a 2.5 x 10-4 M HNO3 solution?

• We know that as a strong acid HNO3dissociates to produce 2.5 x 10-4 M [H3O+]

• pH = -log [2.5 x 10-4]

• = 3.60

pH = -log[H3O+]

8.2

pH: A

Mea

sure

men

t Sc

ale

for A

cids

and

Bas

es Calculating [OH-] from pH

What is the [OH-] of a solution with pH = 4.95?• First find [H3O+]• 4.95 = -log [H3O+]• [H3O+] = 10-4.95

• [H3O+] = 1.12 x 10-5

• Now solve for [OH-]• [OH-] = 1 x 10-14 / 1.12 x 10-5

= 1.0 x 10-9

pH = -log[H3O+]

1 x 10-14 = [H3O+][OH-]

The pH Scale8.

2 pH

: A M

easu

rem

ent

Scal

e fo

r Aci

ds a

nd B

ases

1.0 x 100 0.001.0 x 10-1 1.001.0 x 10-2 2.001.0 x 10-3 3.001.0 x 10-4 4.001.0 x 10-5 5.001.0 x 10-6 6.001.0 x 10-7 7.00

For a strong acidHCl molarity pH

Mor

e Aci

dic

1.0 x 100 14.001.0 x 10-1 13.001.0 x 10-2 12.001.0 x 10-3 11.001.0 x 10-4 10.001.0 x 10-5 9.001.0 x 10-6 8.001.0 x 10-7 7.00

For a strong baseNaOH molarity pH

Mor

e ba

sic

Each 10 fold change in concentration changes the pH by one unit

8.2

pH: A

Mea

sure

men

t Sc

ale

for A

cids

and

Bas

es

8.3 Reactions Between Acids and Bases

• Neutralization reaction – the reaction of an acid with a base to produce a salt and water

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)Acid Base Salt Water

• Break apart into ions:H+ + Cl- + Na+ + OH-Na+ + Cl- + H2O

• Net ionic equation– Show only the changed components– Omit any ions appearing the same on both sides of

equation = Spectator Ions

H+ + OH- H2O

8.3

Rea

ctio

ns B

etw

een

Aci

ds a

nd B

ases • The net ionic neutralization reaction is more

accurately written:H3O+(aq) + OH-(aq) 2H2O(l)

• This equation applies to any strong acid / strong base neutralization reaction

• An analytical technique to determine the concentration of an acid or base is titration

• Titration involves the addition of measured amount of a standard solution to neutralize the second, unknown solution

• Standard solution – solution of known concentration

Net Ionic Neutralization Reaction

Buret – long glass tube calibrated in mL which contains the standard solution

Flask contains a solution of unknown concentration plus indicator

Indicator – a substance which changes color as pH changes

Standard solution is slowly added until the color changes

The equivalence point is when the moles of H3O+

and OH- are equal

8.3

Rea

ctio

ns B

etw

een

Aci

ds a

nd B

ases

Acid – Base Titration

8.3

Rea

ctio

ns B

etw

een

Aci

ds a

nd B

ases

Relationship Between pH and Color in Acid-Base Indicators

Determine the Concentration of a Solution of Hydrochloric Acid• Place a known volume of acid whose concentration

is not known into a flask• Add an indicator, experience guides selection, here

phenol red is good• Known concentration of NaOH is placed in a buret• Drip NaOH into the flask until the indicator

changes color

8.3

Rea

ctio

ns B

etw

een

Aci

ds a

nd B

ases

Determine the Concentration of a Solution of Hydrochloric Acid

• Indicator changes color – equivalence point is reached – Mol OH- = Mol H3O+ present in the unknown acid

• Volume dispensed from buret is determined• Calculate acid concentration:

– Volume of Hydrochloric Acid: 25.00 mL– Volume of NaOH added: 35.00 mL– Concentration of NaOH: 0.1000 M– Balanced reaction shows that HCl and NaOH react 1:1 8.

3 R

eact

ions

Bet

wee

n A

cids

and

Bas

es

Determine the Concentration of a Solution of Hydrochloric Acid

• 35.00 mL NaOH x 1L NaOH x 0.1000 mol NaOH103 mL NaOH 1L NaOH= 3.500 x 10-3 mol NaOH

• 3.500 x 10-3 mol NaOH x 1 mol HCl1 mol NaOH

= 3.500 x 10-3 mol HCl– this amount of HCl is contained in 25.00 mL

• 3.500 x 10-3 mol HCl x 103 mL HCl25.00 mL HCl 1 L HCl

• = 1.400 x 10-1 mol HCl / L HCl = 0.1400 M HCl8.3

Rea

ctio

ns B

etw

een

Aci

ds a

nd B

ases

Determine the Concentration of a Solution of Hydrochloric Acid

• Alternative strategy to solve the acid concentration(Macid)(Vacid) = (Mbase)(Vbase)

• (Macid) = (Mbase)(Vbase)(Vacid)

• (Macid) = (0.1000 M) (35.00 mL)(25.00 mL)

• = 0.1400 M HCl

8.3

Rea

ctio

ns B

etw

een

Aci

ds a

nd B

ases

8.3

Rea

ctio

ns B

etw

een

Aci

ds a

nd B

ases

Calculating the Concentration of Sodium Hydroxide

Calculate [NaOH] if 25.00 mL of this solution were required to neutralized 45.00 mL of 0.3000 M HCl•Alternative strategy to solve the acid concentration

(Macid)(Vacid) = (Mbase)(Vbase)

•(Mbase) = (Macid)(Vacid)(Vbase)

•(Mbase) = (0.3000 M) (45.00 mL)(25.00 mL)

= 0.5400 M NaOH

• The previous examples have the acid and base at a 1:1 combining ratio– Not all acid-base pairs do this

• Polyprotic substance – donates or accepts more than one proton per formula unit– Hydrochloric acid is monoprotic, producing one H+

ion for each unit of HCl

– Sulfuric acid is diprotic, each unit of H2SO4produces 2 H+ ions

H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2 H2O(l)8.3

Rea

ctio

ns B

etw

een

Aci

ds a

nd B

ases

Polyprotic Substances

Step 1.

H2SO4(aq) + H2O(l) HSO4-(aq) + H3O+(aq)

Step 2.

HSO4-(aq) + H2O(l) SO4

2-(aq) + H3O+(aq)

• In Step 1 H2SO4 behaves as a strong acid –dissociating completely

• In Step 2 HSO4-( behaves as a weak acid –

reversibly dissociating, note the double arrow8.3

Rea

ctio

ns B

etw

een

Aci

ds a

nd B

ases

Dissociation of Polyprotic Substances

8.4 Acid-Base Buffers

• Buffer solution - solution which resists large changes in pH when either acids or bases are added

• These solutions are frequently prepared in laboratories to maintain optimum conditions for chemical reactions

• Buffers are also used routinely in commercial products to maintain optimum conditions for product behavior

8.4

Aci

d-B

ase

Buf

fers • Buffers act to establish an equilibrium between a

conjugate acid – base pair• Buffers consist of either

– a weak acid and its salt (conjugate base)– a weak base and its salt (conjugate acid)

– Acetic acid (CH3COOH) with sodium acetate (CH3COONa)

• An equilibrium is established in solution between the acid and the salt anion

• A buffer is LeChatelier’s Principle in action

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

The Buffer Process

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)8.4

Aci

d-B

ase

Buf

fers

Addition of Base (OH-) to a Buffer Solution

• Adding a basic substance to a buffer causes changes– The OH- will react with the H3O+ producing water– Acid in the buffer system dissociates to replace

the H3O+ consumed by the added base– Net result is to maintain the pH close to the initial

level

• The loss of H3O+ (the stress) is compensated by the dissociation of the acid to produce more H3O+

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)8.4

Aci

d-B

ase

Buf

fers

Addition of Acid (H3O+) to a Buffer Solution

• Adding an acidic substance to a buffer causes changes– The H3O+ from the acid will increase the overall

H3O+

– Conjugate base in the buffer system reacts with the H3O+ to form more acid

– Net result is to maintain the H3O+ concentration and the pH close to the initial level

• The gain of H3O+ (the stress) is compensated by the reaction of the conjugate base to produce more acid

8.4

Aci

d-B

ase

Buf

fers

Buffer Capacity• Buffer Capacity – a measure of the

ability of a solution to resist large changes in pH when a strong acid or strong base is added

• Also described as the amount of strong acid or strong base that a buffer can neutralize without significantly changing pH

• Buffering process is an equilibrium reaction described by an equilibrium-constant expression

– In acids, this constant is Ka

• If you want to know the pH of the buffer, solve for [H3O+], then calculate pH

COOH]CH[]COOCH][OH[K

3

-33

a

+

=

8.4

Aci

d-B

ase

Buf

fers

Preparation of a Buffer Solution

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

8.4

Aci

d-B

ase

Buf

fers

Calculating the pH of a Buffer Solution

Calculate the pH of a buffer solution in which – Both the acetic acid (acid) and sodium acetate (salt)

concentrations are 2.0 x 10-2 M– Sodium acetate, the salt, is also the conjugate base

– Ka = 1.75 x 10-5

– [H3O+] = [acid]Ka[conjugate base]

= {(2.0 x 10-2 M) x (1.75 x 10-5)} / 2.0 x 10-2 M= 1.75 x 10-5

– pH = -log 1.75 x 10-5 = 4.76

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

COOH]CH[]COOCH][OH[K

3

-33

a

+

=

8.4

Aci

d-B

ase

Buf

fers

Henderson-Hasselbalch Equation• Solution of equilibrium-constant expression

and pH can be combined into one operation• Henderson-Hasselbalch Equation is this

combined expression• Using these two equations:

– pKa = -log Ka just as pH = -log[H3O+]

– pKa = pH – log ( [CH3COO-] / [CH3COOH] )

– Henderson-Hasselbalch –

– pH = pKa + log( [CH3COO-] / [CH3COOH] )

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)COOH]CH[

]COOCH][OH[K3

-33

a

+

=

8.4

Aci

d-B

ase

Buf

fers

Henderson-Hasselbalch Equation

• pH = pKa + log( [CH3COO-] / [CH3COOH] ) can be rewritten pH = pKa + log ( [conjugate base] / [weak acid])

8.5 Oxidation-Reduction Processes

• Oxidation-reduction processes are responsible for many types of chemical change

• Oxidation - defined by one of the following – loss of electrons – loss of hydrogen atoms– gain of oxygen atoms

• Example: NaNa+ + e-

– Oxidation half reaction

8.5

Oxi

datio

n-R

educ

tion

Proc

esse

s• Reduction - defined by one of the

following:– gain of electrons– gain of hydrogen– loss of oxygen

• Example: Cl + e- Cl-

– Reduction half reaction

• Cannot have oxidation without reduction.

Oxidation-Reduction Processes

Na + Cl Na+ + Cl-

Oxidizing Agent• Is reduced• Gains electrons• Causes oxidation

Reducing Agent• Is oxidized• Loses electrons• Causes reduction

8.5

Oxi

datio

n-R

educ

tion

Proc

esse

sOxidation and Reduction as Complementary Processes

Na Na+ + e-

Cl + e- Cl-

8.5

Oxi

datio

n-R

educ

tion

Proc

esse

sApplications of Oxidation and

Reduction• Corrosion - the deterioration of metals

caused by an oxidation-reduction process– Example: rust (oxidation of iron)

4Fe(s) + 3O2(g) 2Fe2O3(s)

• Combustion of Fossil Fuels– Example: natural gas furnaces

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

8.5

Oxi

datio

n-R

educ

tion

Proc

esse

s• Bleaching

• Most bleaching agents are oxidizing agents

• The oxidation of the stains produces compounds that do not have color– Example: Chlorine bleach - sodium

hypochlorite (NaOCl)

Applications of Oxidation and Reduction

8.5

Oxi

datio

n-R

educ

tion

Proc

esse

sBiological Processes

• Respiration– Electron-transport chain of aerobic

respiration uses reversible oxidation and reduction of iron atoms in cytochrome c

• Metabolism– Break down of molecules into smaller pieces

by enyzmes

• Is Zn oxidized or reduced?• Oxidized

• Copper is reduced8.5

Oxi

datio

n-R

educ

tion

Proc

esse

sVoltaic Cells

• Voltaic cell – electrochemical cell that converts stored chemical energy into electrical energy

• Let’s consider the following reaction:

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

8.5

Oxi

datio

n-R

educ

tion

Proc

esse

sVoltaic Cells

• If the two reactants are placed in the same flask they cannot produce electrical current

• A voltaic cell separates the two half reactions

• This makes the electrons flow through a wire to allow the oxidation and reduction to occur

Zn Zn2+ + 2e-

Oxidationanode – electrode

where oxidation occurs

Cu2+ + 2e- CuReduction

cathode – electrode where reduction occurs8.

5O

xida

tion-

Red

uctio

n Pr

oces

ses

Voltaic Cell Generating Electrical Current

8.5

Oxi

datio

n-R

educ

tion

Proc

esse

sVoltaic Cell Generating

Electrical Current

8.5

Oxi

datio

n-R

educ

tion

Proc

esse

sSilver Battery

• Batteries use the concept of the voltaic cell

• Modern batteries are– Smaller

– Safer

– More dependable

8.5

Oxi

datio

n-R

educ

tion

Proc

esse

sElectrolysis

• Electrolysis reactions – uses electrical energy to cause nonspontaneous oxidation-reduction reactions to occur

• These reactions are the reverse of a voltaic cell– Rechargeable battery

• When powering a device behaves as voltaic cell• With time the chemical reaction nears completion• Battery appears to “run down”• Cell reaction is reversible when battery attached

to charger

8.5

Oxi

datio

n-R

educ

tion

Proc

esse

sComparison of Voltaic and

Electrolytic Cells

Chapter 9

The Nucleus, Radioactivity and Nuclear Medicine

Denniston Topping Caret

6th Edition

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

9.1 Natural Radioactivity• Radioactivity – process by which atoms

emit energetic particles or rays• Radiation – the particles or rays emitted

– comes from the nucleus• Nuclear symbols – what we use to designate

the nucleus– Atomic symbol– Atomic number– Mass number

9.1

Nat

ural

Rad

ioac

tivity

B115

atomic symbol

atomic number number of protons

mass numbernumber of

protons and neutrons

Nuclear Symbols

B115

9.1

Nat

ural

Rad

ioac

tivity Writing Nuclear Symbols

• This defines an isotope of boron

• In nuclear chemistry often called a nuclide

• This is not the only isotope of boron– boron-10 also exists– How many protons and neutrons does

boron-10 have?• 5 protons, 5 neutrons

Three Isotopes of Carbon• Each nucleus contains the same number of protons• Only the number of neutrons is different• With different numbers of neutrons the mass of

each isotope is different

9.1

Nat

ural

Rad

ioac

tivity

9.1

Nat

ural

Rad

ioac

tivity Unstable Isotopes

• Some isotopes are stable

• The unstable isotopes are the ones that produce radioactivity

• To write nuclear equations we need to be able to write the symbols for the isotopes and the following:– alpha particle

– beta particles

– gamma rays

α α He He 42

42

2 42

+9.1

Nat

ural

Rad

ioac

tivity Alpha Particles

• Alpha particle (a) – 2 protons, 2 neutrons

• Same as He nucleus (He2+)

• Slow moving, and stopped by small barriers

• Symbolized in the following ways:

β β e 01-

01−

9.1

Nat

ural

Rad

ioac

tivity Beta Particles and Positrons

• Beta particles (b) – fast-moving electron

• Emitted from the nucleus as a neutron is converted to a proton

• Higher speed particles, more penetrating than alpha particles

• The symbol is…

g

9.1

Nat

ural

Rad

ioac

tivity Gamma Rays

• Gamma Rays (g) – pure energy (electromagnetic radiation)

• Highly energetic

• The most penetrating form of radiation

• Symbol is simply…

9.1

Nat

ural

Rad

ioac

tivity Properties of Alpha, Beta, and

Gamma Radiation

• Ionizing radiation – produces a trail of ions throughout the material that it penetrates

• The penetrating power of the radiation determines the ionizing damage that can be caused

• Alpha particle < beta particle < gamma rays

9.2 Writing a Balanced Nuclear Equation

• Nuclear equation – used to represent nuclear change

• In a nuclear equation, you do not balance the elements, instead...– the total mass on each side of the reaction

arrow must be identical– the sum of the atomic numbers on each side of

the reaction arrow must be identical

9.2

Writ

ing

a B

alan

ced

Nuc

lear

Equ

atio

ns He Th U 42

23490

23892 +→

238 = 234 + 4

92 = 90 + 2

mass number

atomic number

Alpha Decay

• Upon decomposition, nitrogen-16 produces oxygen-16 and a beta particle

• In beta decay, one neutron in nitrogen-16 is converted to a proton and the electron, the beta particle is released

e ON 01-

168

167 +→

9.2

Writ

ing

a B

alan

ced

Nuc

lear

Equ

atio

nsBeta Decay

Positron Emission• A positron has same mass as an electron, or beta

particle, BUT opposite charge (+)• Unlike beta decay, the product nuclide has the

same mass number as the parent BUT the atomic number has decreased by one

01115

116

01

115

116

B C

e B C

+→

+→

9.2

Writ

ing

a B

alan

ced

Nuc

lear

Equ

atio

ns

• Gamma radiation occurs to increase the stability of an isotope – The energetically unstable isotope is called a

metastable isotope

• The atomic mass and number do not change

• Usually gamma rays are emitted along with alpha or beta particles

Tc Tc 9943

m9943 +→9.

2 W

ritin

g a

Bal

ance

d N

ucle

ar E

quat

ions

Gamma Production

• To predict the product, simply remember that the mass number and atomic number are conserved

• What is the identity of X?

e01-

23992 XU +→

93239Np9.

2 W

ritin

g a

Bal

ance

d N

ucle

ar E

quat

ions

Predicting Products of Nuclear Decay

9.3 Properties of RadioisotopesNuclear Structure and Stability• Binding Energy – the energy that holds the

protons, neutrons, and other particles together in the nucleus

• Binding energy is very large• When isotopes decay (forming more stable

isotopes,) binding energy is released

9.3

Prop

ertie

s of

Rad

iois

otop

es Important factors for stable isotopes– Ratio of neutrons to protons– Nuclei with large number of protons (84 or more)

tend to be unstable– The “magic numbers” of 2, 8, 20, 50, 82, or 126 help

determine stability – these numbers of protons or neutrons are stable

– Even numbers of protons or neutrons are generally more stable than those with odd numbers

– All isotopes (except 1H) with more protons than neutrons are unstable

Stable Radioisotopes

9.3

Prop

ertie

s of

Rad

iois

otop

esHalf-Life

• Half-life (t1/2) – the time required for one-half of a given quantity of a substance to undergo change

• Each radioactive isotope has its own half-life – Ranges from a fraction of a second to a

billion years

– The shorter the half-life, the more unstable the isotope

Half-Lives of Selected Radioisotopes

9.3

Prop

ertie

s of

Rad

iois

otop

es

9.3

Prop

ertie

s of

Rad

iois

otop

esDecay Curve for the Medically Useful Radioisotope Tc-99m

9.3

Prop

ertie

s of

Rad

iois

otop

esPredicting the Extent of

Radioactive DecayA patient receives 10.0 ng of a radioisotope with a half-life of 12 hours. How much will remain in the body after 2.0 days, assuming radioactive decay is the only path for removal of the isotope from the body.• Calculate n, the number of half-lives elapsed using the half-life as the conversion factorn = 2.0 days x 1 half-life / 0.5 days = 4 half lives

• Calculate the amount remaining10.0 ng 5.0 ng 2.5 ng 1.3 ng 0.63 ng

1st half-life 2nd half-life 3rd half-life 4th half-life

• 0.63 ng remain after 4 half-lives

Radiocarbon Dating

• Radiocarbon dating – the estimation of the age of objects through measurement of isotopic ratios of carbon– Ratio of carbon-14 and carbon-12

• Basis for dating:– Carbon-14 (a radioactive isotope) is

constantly being produced by neutrons from the sun

H C n N 11

146

10

147 +→+

9.5

Rad

ioca

rbon

Dat

ing

• Living systems are continually taking in carbon– The ratio of carbon-14 to carbon-12 stays

constant during its lifetime

• Once the living system dies, it quits taking in the carbon-14– The amount of carbon-14 decreases

according to the reaction:

e N C 01-

147

146 +→

• The half-life of carbon-14 is 5730 years– This information is used to calculate the age

Radiocarbon Dating

9.4 Nuclear PowerEnergy Production

E = mc2

• Equation by Albert Einstein shows the connection between energy (E) and mass (m)

• c is the speed of light • The equation shows that a very large amount of

kinetic energy can be formed from a small amount of matter– Release this kinetic energy to convert liquid water into

steam– The steam drives an electrical generator producing

electricity

9.4

Nuc

lear

Pow

er • Fission (splitting) occurs when a heavy nuclear particle is split into smaller nuclei by a smaller nuclear particle

•Accompanied by a large amount of energy

•Is self-perpetuating

•Can be used to generate steam

energy n 3 Ba Kr U U n 10

14156

9236

23692

23592

10 +++→→+

Nuclear Fission

9.4

Nuc

lear

Pow

erFission of Uranium-235

• Chain reaction – the reaction sustains itself by producing more neutrons

9.4

Nuc

lear

Pow

erRepresentation of the “Energy Zones” of a Nuclear Reactor

• A nuclear power plant uses a fissionable material as fuel– Energy released by the fission heats water– produces steam– drives a generator or turbine– converts heat to electrical energy

• Fusion (to join together) – combination of two small nuclei to form a larger nucleus

• Large amounts of energy is released

• Best example is the sun

• An Example:

• No commercially successful plant exists in U.S.

energy n He H H 10

42

31

21 ++→+

9.4

Nuc

lear

Pow

erNuclear Fusion

9.4

Nuc

lear

Pow

erBreeder Reactors

• Breeder reactor – fission reactor that manufactures its own fuel

• Uranium-238 (non-fissionable) is converted to plutonium-239 (fissionable)

• Plutonium-239 undergoes fission to produce energy

9.5 Medical Applications of Radioactivity

• Modern medical care uses the following:– Radiation in the treatment of cancer– Nuclear medicine - the use of

radioisotopes in the diagnosis of medical conditions

9.5

Med

ical

App

licat

ions

of

Rad

ioac

tivity

• Based on the fact that high-energy gamma rays cause damage to biological molecules

• Tumor cells are more susceptible than normal cells

• Example: cobalt-60• Gamma radiation can cure cancer but

can also cause cancer

Cancer Therapy Using Radiation

9.5

Med

ical

App

licat

ions

of

Rad

ioac

tivity

Nuclear Medicine• The use of isotopes in diagnosis

• Tracers – small amounts of radioactive substances used as probes to study internal organs

• Nuclear imaging – medical techniques involving tracers

• Example:– Iodine concentrates in the thyroid gland.– Using radioactive 131I and 125I will allow the study

of how the thyroid gland is taking in iodine

9.5

Med

ical

App

licat

ions

of

Rad

ioac

tivity

Tracer Studies• Isotopes with short half-lives are preferred for

tracer studies. Why?– They give a more concentrated burst

– They are removed more quickly from the body

• Examples of imaging procedures:– Bone disease and injury using technetium-99m

– Cardiovascular disease using thallium-201

– Pulmonary disease using xenon-133

9.5

Med

ical

App

licat

ions

of

Rad

ioac

tivity

Making Isotopes for Medical Applications

• Artificial radioactivity – a normally stable, nonradioactive nucleus is made radioactive

• Made in two ways:

• In core of a nuclear reactor

• In particle accelerators – small nuclear particles are accelerated to speeds approaching the speed of light and slammed into another nucleus

• Tracer in the liver

• Used in the diagnosis of Hodgkin’s disease

Au n Au 19879

10

19779 →+

Ga p Zn 6731

11

6630 →+

9.5

Med

ical

App

licat

ions

of

Rad

ioac

tivity

Examples of Artificial Radioactivity

9.5

Med

ical

App

licat

ions

of

Rad

ioac

tivity

Preparation of Technetium-99m

• Some isotopes used in nuclear medicine have such a short half-life that they need to be generated on site

• 99mTc has a half-life of only 6 hours

e Tc Mo 01-

99m43

9942 +→

9.6 Biological Effects of Radiation

Radiation Exposure and SafetyThe Magnitude of the Half-Life• Isotopes with short half-lives have one major

disadvantage and one major advantage– Disadvantage: larger amount of radioactivity per

unit time– Advantage: if accident occurs, reaches

background radiation levels more rapidly

9.6

Bio

logi

cal E

ffec

ts o

f R

adia

tion

Shielding• Alpha and beta particles need low level

of shielding: lab coat and gloves• Lead, concrete or both required for

gamma rays

Distance from the Radioactive Source• Doubling the distance from the source

decreases the intensity by a factor of 4

Radiation Exposure and Safety

9.6

Bio

logi

cal E

ffect

s of

Rad

iatio

nRadiation Exposure and Safety

Time of Exposure• Effects are cumulative

Types of Radiation Emitted• Alpha and beta emitters are generally less

hazardous then gamma emitters

Waste Disposal• disposal sites are considered temporary

9.7 Measurement of Radiation

Nuclear Imaging• Isotope is administered• Isotope begins to concentrate in the organ• Photographs (nuclear images) are taken at

periodic intervals• Emission of radioactive isotope creates the

image

• Computers and television are coupled

• Gives a continuous and instantaneous record of the voyage of the isotope throughout the body– Gives increased

sensitivity– CT scanner is an

example

Computer Imaging9.

7 M

easu

rem

ent o

f R

adia

tion

9.7

Mea

sure

men

t of

Rad

iatio

nThe Geiger Counter

• Detects ionizing radiation

• Has largely been replaced by more sophisticated devices

9.7

Mea

sure

men

t of

Rad

iatio

nFilm Badges

• A piece of photographic film that is sensitive to energies corresponding to radioactive emissions

• The darker the film, when developed, the longer the worker has been exposed

9.7

Mea

sure

men

t of

Rad

iatio

nUnits of Radiation Measurement

The Curie• The amount of radioactive material

that produces 3.7 x 1010 atomic disintegrations per second

• Independent of the nature of the radiation

The Roentgen• The amount of radiation needed to

produce 2 x 109 ion pairs when passing through one cm3 of air at 0oC

• Used for very high energy ionizing radiation only

9.7

Mea

sure

men

t of

Rad

iatio

nUnits of Radiation Measurement

9.7

Mea

sure

men

t of

Rad

iatio

n Rad – Radiation absorbed dosage

• The dosage of radiation able to transfer 2.4 x 10-3 cal of energy to one kg of matter

• This takes into account the nature of the absorbing material

Units of Radiation Measurement

9.7

Mea

sure

men

t of

Rad

iatio

nThe Rem

• Roentgen Equivalent for Man

• Obtained by multiplication of the rad by a factor called the relative biological effect (RBE)• RBE = 10 for alpha particles• RBE = 1 for beta particles

• Lethal dose (LD50) - the acute dosage of radiation that would be fatal for 50% of the exposed population– LD50 = 500 rems

Units of Radiation Measurement

Chapter 10

An Introduction to Organic Chemistry:The Saturated Hydrocarbons

Denniston Topping Caret

6th Edition

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

10.1 The Chemistry of CarbonWhy are there so many organic compounds?1. Carbon forms stable, covalent bonds with other

carbon atoms• Consider three allotropic forms of elemental carbon

– Graphite in planar layers– Diamond is a three-dimensional network– Buckminsterfullerene is 60 C in a roughly spherical

shape

Why are there so many organic compounds?

2. Carbon atoms form stable bonds with other elements, such as:

– Oxygen– Nitrogen– Sulfur– Halogen

• Presence of these other elements confers many new physical and chemical properties on an organic compound

Why are there so many organic compounds?

3. Carbon atoms form double or triple bonds with:

– Other carbon atoms (double & triple)

– Oxygen (double only)– Nitrogen (double & triple)

• These combinations act to produce a variety of organic molecules with very different properties

CH2 O

CH2 NH

CH2 CH2

CH CH

CH N

double bonds

triple bonds

Why are there so many organic compounds?

4. Carbon atoms can be arranged with these other atoms; is nearly limitless

– Branched chains– Ring structures– Linear chains

• Two organic compounds may even have the same number and kinds of atoms but completely different structures and thus, different properties

– These are called isomers

CH3CH2CH2CH3 CH2 CH2

CH2

CH2

CH2

Isomers

• Many carbon compounds exist in the form of isomers• Isomers are compounds with the same molecular formula but different structures• An isomer example: both are C4H10but have different structures

– Butane– Methylpropane

Isomers

All have the same molecular formula: C4H8

Important Differences Between Organic and Inorganic Compounds

• Bond type–Organics have covalent bonds

• Electron sharing–Inorganics usually have ionic bonds

• Electron transfer• Structure

–Organics• Molecules• Nonelectrolytes

–Inorganics• Three-dimensional crystal structures• Often water-soluble, dissociating into ions -

electrolytes

Important Differences Between Organic and Inorganic Compounds

• Melting Point & Boiling Point–Organics have covalent bonds

• Intermolecular forces broken fairly easily–Inorganics usually have ionic bonds

• Ionic bonds require more energy to break• Water Solubility

–Organics• Nonpolar, water insoluble

–Inorganics• Water-soluble, readily dissociate

Comparison of Major Properties of Organic and Inorganic Compounds

Families of Organic Compounds

• Hydrocarbons contain only carbon and hydrogen

• They are nonpolar molecules– Not soluble in water – Are soluble in typical nonpolar organic

solvents • Toluene• Pentane

Families of Organic Compounds

• Hydrocarbons are constructed of chains or rings of carbon atoms with sufficient hydrogen atoms to fulfill carbon’s need for four bonds

• Substituted hydrocarbon is one in which one or more hydrogen atoms is replaced by another atom or group of atoms

Division of the Family of Hydrocarbons

Hydrocarbon Saturation• Alkanes are compounds that contain only

carbon-carbon and carbon-hydrogen single bonds– A saturated hydrocarbon has no double or

triple bonds• Alkenes and alkynes are unsaturated

because they contain at least one carbon to carbon double or triple bond

Cyclic Structure of Hydrocarbons

• Some hydrocarbons are cyclic– Form a closed ring– Aromatic hydrocarbons contain a benzene

ring or related structure

Common Functional Groups

10.2 Alkanes

• The general formula for a chain alkane is CnH2n+2– In this formula n = the number of carbon atoms

in the molecule• Alkanes are saturated hydrocarbons

– Contain only carbon and hydrogen– Bonds are carbon-hydrogen and carbon-carbon

single bonds

Formulas Used in Organic Chemistry

• Molecular formula - lists kind and number of each type of atom in a molecule, no bonding pattern

• Structural formula - shows each atom and bond in a molecule

• Condensed formula - shows all the atoms in a molecule in sequential order indicating which atoms are bonded to which

• Line formula - assume a carbon atom at any location where lines intersect– Assume a carbon at the end of any line – Each carbon in the structure is bonded to the correct

number of hydrogen atoms

The Tetrahedral Carbon Atom

(a) Lewis dot structure(b)The tetrahedral shape around the carbon atom(c) The tetrahedral carbon drawn with dashes and

wedges(d)The stick drawing of the tetrahedral carbon

atom(e) Ball and stick model of methane

Drawing Methane and Ethane

C

H

HH

CH

HH

109.5 o

Staggered form of ethane

H

HH

H

in plane

in front of plane

behind plane

Comparison of Ethane and Butane Structures

Names and Formulas of the First Ten Straight-Chain Alkanes

Comparison of Physical Properties of Five Isomers of Hexane

Compare the basic linear structure of hexane– All other isomers have one or more carbon atoms

branching from the main chain– Branched-chain forms of the molecule have a much

smaller surface area• Intermolecular forces are weaker• Boiling and melting points are lower than straight chains

Physical Properties of Organic Molecules

1. Nonpolar2. Not water soluble3. Soluble in nonpolar organic solvents4. Low melting points5. Low boiling points6. Generally less dense (lighter) than water7. As length (molecular weight) increases, melting

and boiling points increase as does the density

Properties of Alkanes

-250

-200

-150

-100

-50

0

50

100

150

200

0 1 2 3 4 5 6 7 8 9 10

Number of Carbons in Chain

Melting PointBoiling Point

Properties of Alkanes

• Most of the alkanes are hydrophobic: water hating

• Straight chain alkanes comprise a homologous series: compounds of the same functional class that differ by a –CH2- group

• Nonpolar alkanes are:– Insoluble in water (a highly polar solvent)– Less dense than water and float on it

Alkyl Groups

• An alkyl group is an alkane with one hydrogen atom removed

• It is named by replacing the -ane of the alkane name with -yl

• Methane becomes a methyl group

or CH3H CH

HH

H CH

H

Alkyl Groups

• All six hydrogens on ethane are equivalent• Removing one H generates the ethyl group• All 3 structures shown at right are the same

CH

CHH

HH

H

CH3CH2CH3CH2

C2H5

Names and Formulas of the First Five Alkyl Groups

Alkyl Group Classification

• Alkyl groups are classified according to the number of carbons attached to the carbon atom that joins the alkyl group to a molecule

• All continuous chain alkyl groups are 1º• Isopropyl and sec-butyl are 2º groups

Iso- Alkyl Groups• Propane: removal of a hydrogen generates

two different propyl groups depending on whether an end or center H is removed

n-propyl isopropyl

CH3CH CH3CH3CH2CH2

CH3 CH2 CH3

Sec- Alkyl Groups• n-butane gives two butyl groups depending

on whether an end (1º) or interior (2º) H is removed

sec-butyln-butyl

CH3 CH2 CH2 CH3

CH3 CH CH2 CH3CH3 CH2 CH2 CH2

Structures and Names of Some Branched-Chain Alkyl Groups

CH3 CH CH3

CH3

CH3 CH CH2

CH3

CH3 C CH3

CH3

More Alkyl Group Classification

• Isobutane gives two butyl groups depending on whether a 1o or 3o H is removed

isobutyl t-butyl

1o C 3o C

Nomenclature

• The IUPAC (International Union of Pure and Applied Chemistry) is responsible for chemical names

• Before learning the IUPAC rules for naming alkanes, the names and structures of eight alkyl groups must be learned

• These alkyl groups are historical names accepted by the IUPAC and integrated into modern nomenclature

Carbon Chain Length and Prefixes

IUPAC Names for Alkanes

1. The base or parent name for an alkane is determined by the longest chain of carbon atoms in the formula

– The longest chain may bend and twist, it is seldom horizontal

– Any carbon groups not part of the base chain are called branches or substituents

– These carbon groups are also called alkyl groups

IUPAC Names for Alkanes

• Rule 1 applied – Find the longest chain in each molecule

• A=7 B=8

CH3

CH2

CH2CH2CH CH2CH3

CH3

CH3CHCH2

CH3

CH2CHCH2CH2

CH2

CH3

CH3

A

B

IUPAC Names for Alkanes

2. Number the carbon atoms in the chain starting from the end with the first branch

– If both branches are equally from the ends, continue until a point of difference occurs

IUPAC Names for AlkanesNumber the carbon atoms correctly• Left: first branch is on carbon 3• Right: first branch is on carbon 3 (From

top) not carbon 4 (if number from right)

CH3

CH2

CH2CH2CH CH2CH3

CH3

CH3CHCH2

CH3

CH2CHCH2CH2

CH2

CH3

CH3

1

2

3 4 5

6 7 8

this branch would be on C-4if you started at correct C-8

123

45

6

7

IUPAC Names for Alkanes

3. Write each of the branches/substituents in alphabetical order before the base/stem name (longest chain)

– Halogens usually come first– Indicate the position of the branch on the main

chain by prefixing its name with the carbon number to which it is attached

– Separate numbers and letters with a hyphen– Separate two or more numbers with commas

IUPAC Names for Alkanes

Name : 4-ethyl-2-methylhexane

CH3CH2CH CH2CH CH3

CH3CH2

CH3

IUPAC Names for Alkanes• Hyphenated and number prefixes are

not considered when alphabetizing groups– Name the compound below

– 5-sec-butyl-4-isopropylnonane

CH CHCH3

CH3 CH2 CH2 CH3

CHCHCH3

CH2 CH3CH2 CH2 CH2 CH3

IUPAC Names for Alkanes

• When a branch/substituent occurs more than once– Prefix the name with

• di• tri• tetra

– Then list the number of the carbon branch for that substituent to the name with a separate number for each occurrence

• Separate numbers with commas

• e.g., 3,4-dimethyl or 4,4,6-triethyl

IUPAC Names for Alkanes

5-ethyl-2,3-dimethylheptaneethyl>dimethyl

CH3CHCH3 CH CH2CH

CH2CH3

CH3

CH2CH3

Name

Practice: IUPAC Name

6-ethyl-6-isobutyl-3,3-dimethyldecane

CH3CCH3

CH2

CH3

CH2CH2C CH2CH3

CH2

CH CH3CH3

CH2CH2CH2CH3

Name1

2

3 4 5 6

7 8 9 10

Structural Isomers• Constitutional/Structural Isomers differ in how

atoms are connected– Two isomers of butane have different physical

properties– The carbon atoms are connected in different

patterns

ButaneBp –0.4 oCMp –139 oC

IsobutaneBp –12 oC

Mp –145 oC

CH3 CH2 CH2 CH3 CH3 CH CH3

CH3

10.3 Cycloalkanes

• Cycloalkanes have two less hydrogens than the corresponding chain alkane– Hexane=C6H14; cyclohexane=C6H12

• To name cycloalkanes, prefix cyclo- to the name of the corresponding alkane– Place substituents in alphabetical order before

the base name as for alkanes– For multiple substituents, use the lowest

possible set of numbers; a single substituent requires no number

Cycloalkane Structures

Cyclopropane

Cyclobutane

Cyclohexane

Type of Formula: Structural Condensed Line

Naming a Substituted Cycloalkane

Name the two cycloalkanes shown below• Parent chain 6 carbon ring 5 carbon ring

cyclohexane cyclopentane• Substituent 1 chlorine atom a methyl group

chloro methyl• Name Chlorocyclohexane Methylcyclopentane

cis-trans Isomers in Cycloalkanes• Atoms of an alkane can rotate freely around the

carbon-carbon single bond having an unlimited number of arrangements

• Rotation around the bonds in a cyclic structure is limited by the fact that all carbons in the ring are interlocked– Formation of cis-trans isomers, geometric isomers, is a

consequence of the lack of free rotation• Stereoisomers are molecules that have the same

structural formulas and bonding patterns, but different arrangements of atoms in space– cis-trans isomers of cycloalkanes are stereoisomers

whose substituents differ in spatial arrangement

cis-trans Isomers in Cycloalkanes• Two groups may be on the same side (cis) of the imagined

plane of the cycloring or they may be on the opposite side (trans)

• Geometric isomers do not readily interconvert, only by breaking carbon-carbon bonds can they interconvert

10.4 Conformations of Alkanes• Conformations differ only in rotation about carbon-

carbon single bonds • Two conformations of ethane and butane are shown

– The first (staggered form) is more stable because it allows hydrogens to be farther apart and thus, the atoms are less crowded

Two Conformations of Cyclohexane

Chair form (more stable) Boat formA

EE

A

AE

E A

E

E

A

A

H HH H

H H

HH H

H

HH

E=equitorial A=axial

10.5 Reactions of Alkanes

• Alkanes, cycloalkanes, and other hydrocarbons can be: – Oxidized (by burning) in the presence of excess

molecular oxygen, in a process called combustion

– Reacted with a halogen (usually chlorine or bromine) in a halogenation reaction

Alkane Reactions

The majority of the reaction of alkanes are combustion reactions

– Complete CH4 + 2O2 CO2 + 2H2O Complete combustion produces

– Carbon dioxide and water

– Incomplete 2CH4 + 3O2 2CO + 4H2O• Incomplete combustion produces

– Carbon monoxide and water– Carbon monoxide is a poison that binds

irreversibly to red blood cells

HalogenationHalogenation is a type of substitution reaction, a reaction that results in a replacement of one group for another

– Products of this reaction are: • Alkyl halide or haloalkane• Hydrogen halide

– This reaction is important in converting unreactive alkanes into many starting materials for other products

– Halogenation of alkanes ONLY occurs in the presence of heat and/or light (UV)

HH +Br2

heat orlight +HBr

BrH

Petroleum Processing

Fraction Boiling Pt Range ºC

Carbon size

Typical uses

Gas -164-30 C1-C4 Heating, cooking

Gasoline 30-200 C5-C12 Motor fuel

Kerosene 175-275 C12-C16 Fuel for stoves, diesel and jet engines

Heating oil Up to 375 C15-C18 Furnace oil

Lubricating oil

350 and up C16-C20 Lubrication, mineral oil

Greases Semisolid C18-up Lubrication, petroleum jelly

Paraffin (wax)

Melts at 52-57 C20-up Candles, toiletries

Pitch / tar Residue in boiler High Roofing, asphalt