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Chapter 1 Summary Notes Main Concepts Explanations

Elements: substances that cannot

be decomposed into simpler

substances

Compounds: substances composed

of two or more elements

Law of Constant

Composition or law of

definite proportions: the

relative masses of elements

are fixed in a given

chemical substance.

Law of Multiple

Proportions: Applies

ONLY when two elements

combine to form two or

more compounds. The

masses of one element

which combine with a

fixed mass of the second

element are in a ratio of

whole numbers

Mixtures: combinations of two or

more substances

Techniques for separating

mixtures: filtration,

distillation,

chromatography

Properties:

Physical vs. chemical: Did

the sample (really) change?

Intensive vs. extensive:

Does the measurement

depend on quantity of

sample?

-Law of Constant Composition: Ex. In pure H2O, H and O

combine in a 1:8 mass ratio. Does law of constant composition

hold good for CuSO4.5H2O? Why or why not?

-Law of Multiple Proportions: Explain the following in terms of

multiple proportions:

-Separation Techniques:

Hand Separation- for mixtures that can be visually differentiated

based on mass, color, shape etc.

Filtration: Fitrate, precipitate, heterogeneous mixtures

Separating Funnel: For immiscible liquids, layers separate with lesser density layer on top.

Centrifugation: Separates particles of different masses based

on centrifugal force. Heavier particles at the bottom and the

lighter particles on top. Distillation- uses differences in the boiling points to separate a homogeneous mixture.

Chromatography-separates homogenous mixtures (mostly inks) based on the differences in solubility of the mixture in a

solvent. There is a stationary and a mobile phase

Summary of the Chapter and Important things to remember:



Naming Compounds Review: Before

naming a compound, it is important to

know its type because naming depends

upon the type. For naming purposes, we

classify compounds as ionic compounds,

molecular compounds, and acids.

Ionic Compounds can be identified

by the presence of a metal in it.

(generally solids) Ex. NaCl, K2SO4,

PbSO4

Molecular compounds are made up

of all non metals. (generally liquids

and gases) Ex. H2O, N2O5

Acids begin with H (generally present

as aq solutions or gases) Ex. HCl,

H2SO4, HClO3

Coordination compound: compound that

contains a complex ion or ions.

Ex. [Cu(NH3)4]Cl2 1. Name cation before anion; one or

both may be a complex. (Follow

standard nomenclature for non-

complexes.)

2. Within each complex (neutral or ion),

name all ligands before the metal.

-Name ligands in alphabetical order

-If more than one of the same ligand is

present, use a numerical prefix: di, tri,

tetra, penta, hexa, …

-Ignore numerical prefixes when

alphabetizing.

In any (uncharged) atom: The #f

protons= atomic number (Z)

# of e= # of p

Mass # (A)= atomic #- # of neutrons

Atomic Symbol: 6C12

Isotopes are atoms of the same

element containing different numbers

of neutrons and therefore having

different masses.

Naming Compounds Flow ChartDoes the formula start with H?

NO YES

Does it begin with a metal that has more than one

oxidation number? (e.g. Fe, Ni, Cu, Sn, Hg)

NO YES

Does the formula contain a polyatomic ion?

NO YES

Are both elements nonmetals?

NO YES

Name the first element,

Then the second element

with an –ide ending.

Name the first element using the proper prefix (never

mono–). Name the second element with the proper

prefix (including mono–) and –ide ending.

1 = mono– 4 = tetra– 7 = hepta– 10 = deca–

2 = di– 5 = penta– 8 = octa–

3 = tri– 6 = hexa– 9 = nona– (not nano–)

Name the first element, then

the polyatomic ion. If two

elements are present, name

both, then the polyatomic ion

(e.g. NaHCO3 is sodium

hydrogen carbonate).

It is an acid (must be aqueous).

Does the acid contain a polyatomic ion?

NO YES

Does the acid end with a

polyatomic ion?

–ite –ate

Name the polyatomic

ion, replacing the –ate

ending with –ic. Add

the word acid.

Name the polyatomic ion,

replacing the –ite ending

with –ous. Add the word

acid.

Write the prefix hydro–, then the

name of the second element with –ic

ending. Add the word acid.

Name the first element followed by its

oxidation number (Roman Numeral) or

“old school” –ic or –ous endings.

Practice: 1. Sodium Sulfide (Na2S), Potassium Nitrate (KNO3),

Ferrous Sulfate Fe (SO4), Ammonium Chloride (NH4Cl),

Phosphoric Acid (H3PO4)

2. What is the O.N. of P in PO43- ion?

3. What is the O.N. of Fe in Fe(NO3)3?




Atomic mass units: 1 amu = 1.66054 x 10

-24 g

or 1 g = 6.02214 x 1023

amu

1 atom of 12

C isotope defined as weighing exactly 12 amu

Percent Composition

Moles: Avogadro’s Number = 6.022 x 1023

= atoms in exactly

12.000 g of 12

C = 1 mol

-Moles important because they can be used for ratios. Mass

cannot be used for ratios. Converting: grams → moles →

molecules

Determining Empirical Formula

Start with the number of grams of each element, given in the

problem.

-If percentages are given, assume that the total mass is 100

grams so that the mass of each element = the percent given.

-Convert the mass of each element to moles using the molar

mass

-Divide each mole value by the smallest number of moles

calculated.

-Round to the nearest whole number. This is the mole ratio of

the elements and is represented by subscripts in the empirical

formula.

-If the number is too far to round (x.1 ~ x.9), then multiply

each solution by the same factor to get the lowest whole

number multiple. If one solution is 1.5, then multiply each

solution in the problem by 2 to get 3.If one solution is 1.25,

then multiply each solution in the problem by 4 to get 5.

Ex. 12

C: 98.892% x 12 amu =11.867 13

C: 1.108% x 13.00335 amu =+ 0.1441

AW = 12.011 amu

Therefore 12.011 g C = 1 mol C

Ex. What is the percent of O in

CuSO4.5H2O? 57.7%

Ex. How many molecules of H2O in 100.0

g H2O? 3.343 x 1024

molecules H2O

Ex. What is the empirical formula of a

compound that is composed of 80.%

Carbon and 20.% Hydrogen? If the molar

mass is found to be 30 g/mol, what is the

molecular formula? CH3, C2H6

Summary of the page and Important things to remember:


Chapter 3 Summary Notes Contd. Main Concepts Explanations

EF from Combustion Data: To find out the EF of a

compound, the compound is burned in the air. The

equation for combustion of these compounds is all

follows.

For Hydrocarbons: CxHy + O2 CO2 + H2O

For Compounds Containing CHN:

CxHyNz+ O2 CO2 + H2O+ NO2

For Compounds containing C, H and O

CxHyOz + O2 CO2 + H2O

OR

Limiting Reactant: The reactant that runs out first. LR can be

determined by comparing the mole ratios of the reactants. LR

determines the amount of the products formed.

100yield lTheoretica

yield Actual Yield %

100 valueAccepted

result Measured - valueAccepted Error %

Determining the Formula of a Hydrate: To determine the

formula of a hydrate, a certain mass of hydrate is heated to

drive off the water. Then the mass of water driven-off is

calculated. Now the mole ratio of water to the compound

is calculated.

Ex. A 0.6349 g sample of the unknown

produced 1.603 g of CO2 and 0.2810 g of H2O.

Determine the empirical formula of the

compound. Ans. C7H6O2

Ex. When carbon-containing compounds are burned

in a limited amount of air, some CO(g) is produced

as well as CO2 (g). A gaseous product mixture is

35.0 mass % CO and 65.0 mass % CO2. What is the

mass % C in the mixture? 32.7% C

Ex. If 6.0 g hydrogen gas reacts with 40.0 g oxygen

gas, what mass of water will be produced?

2H2 (g) + O2 (g) → 2 H2O (g)

Ex. A mixture of morphine (C17H19NO3) and an inert

solid is analyzed by combustion with O2. The

unbalanced equation for the reaction of morphine

with O2 is

C17H19NO3 + O2 → CO2 + H2O + NO2

The inert solid does not react with O2. If 4.000 g of

the mixture yields 8.72 g of CO2, calculate the

percent morphine by mass in the mixture. 83.1%

Ex. If in the previous example, only 40.0 g

water were formed, what is the percent yield and

percent error?

Ex. When 2.000 gram of Na2CO3.xH2O was

heated, 0.914 gram of anhydrous residue

remained. What is the formula of this

compound? Ans: Na2CO3.7H2O Sodium

carbonate heptahydrate




Reactions: To be able to successfully write reactions, you will need to know the

following: Solubility Rules , Nomenclature Types of Reactions (explained later in

this worksheet), How to write net ionic equations MUST KNOW For AP

Chemistry reaction prediction:

1. Always write balanced net ionic equations (meaning dissociate soluble

compounds (based on solubility rules),

2. Metal are insoluble and are atomic, written as (s) in these equations. Ex. Mg(s)

3. Molecular compounds such as gases (CO2, H2S Etc.) are written as (g) and will not dissociate into ions.

4. Water is written as (l) and does not dissociate.

5. Ionic compounds may or may not dissociate depending on solubility rules. Ex. PbSO4 insoluble and NaNO3 soluble.

6. Even a soluble ionic compound may NOT dissociate if it is in solid form

(meaning no water present to actually dissociate the ions.) 7. Weak acids and bases partly dissociate or ionize and are written with a

reversible arrow.

8. Remember PSHOFBrINCl. Phosphorus occurs as P4, Sulfer as S8 and rest as

diatomic. 9. While we are reviewing, remember the difference between Zn and Zn

2+ and Cl2

and 2 Cl-

10. Strong acids (HCl, HBr, HI, HNO3, H2 SO4 (first dissociation only!), HClO4 and HClO3) and strong bases (Group 1 alkali metal hydroxide and Ca, Ba, Sr

hydroxides from group 2) dissociate in aq. Solutions. Weak acids and bases are

not dissociated in net ionic equations.

Solubility Rules Always soluble: alkalies, NH4

+, NO3

-, C2H3O2

-

Types of Reactions: Double displacement. Precipitation, neutralization, gas forming.

H2CO3 in water = H2O & CO2

Single displacement or redox replacement: (metals displace metals and nonmetals displace nonmetals)

Combination or synthesis = two reactants result in a single product

• Metal oxide + water metallic hydroxide (base)

• Nonmetal oxide + water nonbinary acid

• Metal oxide + nonmetal oxide salt

Decomposition = one reactant becomes several products

• Metallic hydroxide metal oxide + water

• Acid nonmetal oxide + water

• Salt metal oxide + nonmetal oxide

• Metallic chlorates metallic chlorides + oxygen

• Electrolysis decompose compound into elements (water in dilute acids or solutions of

dilute acids)

• Hydrogen peroxide water + oxygen

• Metallic carbonates --> metal oxides + carbon dioxide

• Ammonium carbonate ammonia, water and carbon dioxide.

Hydrolysis = compound reacting with water.

• Watch for soluble salts that contain anions of weak acid the anion is a conjugate base and

cations of weak bases that are conjugate acids.

Reactions of coordinate compounds and complex

Synthesis Reactions:

Decomposition

Single Replacement

Combustion Reactions:


• Complex formation by adding excess source of ligand to transitional metal of highly

charged metal ion such

as Al3+ Al =4 ligands others 2X ox #

• Breakup of complex by adding an acid metal ion and the species formed when

hydrogen from the acid

reacts with the ligand

Redox = change in oxidation state= a reaction between an oxidizer and a reducer.

1. Familiarization with important oxidizers and reducers

2. “added acid” or “acidified”

3. an oxidizer reacts with a reducer of the same element to produce the element at

intermediate oxidation state

Molarity (M) = moles solute = mol

volume of solution L

Titration is a method to determine the molarity of unknown acid or base. In

titration, an acid or base of unknown molarity is titrated against a standard solution

(whose M is known) of acid or base.The end point in a titration is indicated by a

color change by the indicator. Indicators are weak acids or bases and are added in

small quantity (1-3 drops) to indicate the end point. At equivalence point (which

should be close to end point),

moles of H+= moles of OH-

M1V1= M2 V2 (sometimes used to get moles , M= moles/L , so moles= M XV)

-What other ways can you get the moles- for a solid acid or base? For a gas?

Electrolyte: substance which, in aqueous solution, ionizes and thus conducts

electricity. Ex: salt in water.

Non-electrolyte: substance which, in aqueous solution, does not dissociate and thus

does not conduct electricity

Strong & weak electrolytes: conductivity depends on degree of dissociation and

equilibrium position:

HA (aq) ↔ H+ (aq) + A- (aq)

Strong = nearly completely dissociated

Weak = partially dissociated

Molecular equation: shows complete chemical equation with states of matter,

undissociated

BaCl2 (aq) + Na2SO4 (aq) → 2 NaCl (aq) + BaSO4 (s)

Complete ionic equation: shows complete chemical equation with states of matter,

dissociated if appropriate

Ba2+

(aq) + 2 Cl-(aq) + 2 Na

+(aq) + SO4

2-(aq) →

2 Cl-(aq) + 2 Na

+(aq) + BaSO4 (s)Spectator ions:

present in reaction but do not “participate”; depend on solubility rules

Cl- (aq) and Na

+ (aq)

3. Net ionic equation: shows chemical equation without spectator ions

Ba2+

(aq) + SO42-

(aq) → BaSO4 (s)

Precipitation

Ex. How many mL of a 3M

NaOH solution are required to

completely neutralize 20.0 mL of

1.5M H2SO4? (Start by writing a

balanced equation!) Ans. 20.0

mL

Ex. How many g of NaOH is

required to completely react with

100. mL of 1M HCl?

Chapter 5 Summary Notes



Main Concepts Explanations

0th

and 1st Laws of Thermodynamics:

0. 2 systems are in thermal equilibrium when they are at

the same T

1. Energy can be neither created nor destroyed, or, energy

is conserved

Internal Energy

Includes translational, rotational, vibrational energy

Change (ΔE = Efinal - Einitial) is often measured

o ΔE > 0: Energy of system increases (gained from surr.)

o ΔE < 0: Energy of system decreases (lost to surr.)

o ΔE = q + w

q = heat added/liberated from system

q > 0 : heat added to system

q < 0: heat removed from system

w = work done on or by the system

w > 0 : work done to system

w < 0 : system does work on surr.

Calorimetry: Measurement of heat flow, experimental

technique used to measure the heat transferred in a physical or

chemical process

o Calorimeter: the apparatus used in this procedure; two

types: constant pressure (coffee cup) and constant

volume (bomb calorimeter)

o Coffee Cup Calorimeter: system in this case is the

“contents” of the calorimeter and the surroundings are

cup and the immediate surroundings

o qrxn + q solution = 0

qrxn: heat gained/ lost in the chemical reaction

qsolution: the heat gained/lost by solution

Heat Capacity, C: Amount of heat required to raise T of an

object by 1 K

o q = CΔT

Specific Heat (or Specific Heat Capacity), c: heat capacity of

1 g of substance

o q = mcΔT

Ex. Octane and Oxygen gases combust

within a closed cylinder in an engine. The

cylinder gives off 1150J of heat and a

piston is pushed down by 480J during the

reaction. What is the change in internal

energy of the system? (Ans: ΔE = -1630J)

Ex. How much energy is required to heat

40.0 g of iron (c = 0.45J/gK) from 0.0oC to

100.0oC? (Ans: q = 1800J)

Ex. 0.500g of Mg chips are placed in a

coffee-cup calorimeter and 100.0mL of

1.00M HCl is added to it. The reaction is:

Mg(s) + 2HCl(aq) H2(g) + MgCl2(aq)

The temp. of the solution increases from

22.2C to 44.8C. What’s the enthalpy

change for the reaction, per mole of Mg?

Assume specific heat capacity of solution is

4.20J/gK and density of the HCl solution is

1.00 g/mL. (Ans: ΔH= -4.64*105 J/molMg)


Chapter 5 Summary Notes Contd. Main Concepts Explanations

Enthalpy, H: change in heat content of a reaction at

constant P

o H = E + PV ΔH = ΔE + PΔV ΔH = (qp+w) +

(-w) ΔH = qp

qp = heat content

ΔH > 0: heat gained from surr. + ΔH in

endothermic reaction

ΔH < 0: heat released to surr. + ΔH in

exothermic reaction

Enthalpy of Reaction, ΔHrxn: heat of reaction, extensive

property, depends on states of reactions and products

o ΔHrxn = -ΔHreverse rxn

Hess’s Law: If a rxn is carried out in a series of steps:

ΔHrxn = Σ(ΔHsteps) = ΔH1 + ΔH2 + ΔH3 + · · ·

Enthalpy of Formation, ΔHf:

heat needed to form substance

from its elements

o Standard Enthalpy of

Formation, ΔHfo: forms

1 mole of compound

from its elements in their

standard state (at 298K)

o ΔHf of pure element (C,

O2, H2, etc) is 0.

ΔHrxno = Σ [n*ΔHf

o(products)] - Σ[n*ΔHf

o(reactants)]

Bond Enthalpy: Amount of energy required to break a

particular bond between two elements in gaseous state

o Indicates the “strength” of a bond

ΔHrxn ≈ Σ [ΔHbonds broken] - Σ[ΔHbonds formed]

o NOTE: this is the “opposite” of Hess’s Law

Ex. What is the ΔH of combustion of 100g CH4

if ΔHrxno = -890kJ? (Ans: -5550kJ)

Ex. What is ΔHrxno of the combustion of

propane?

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

Givens:

3C(s) + 4H2(g) C3H8(g) ΔH1 = -103.85kJ

C(s) + O2(g) CO2(g) ΔH2 = -393.5kJ

H2(g) + ½O2(g) H2O(l) ΔH3 = -285.8kJ

(Ans: ΔHrxno = -2219.8kJ)

Ex. What is ΔHrxno for the combustion of liquid

benzene?

C6H6(l) + 15/2O2(g) 6CO2(g) + 3H2O(l)

Givens:

ΔHfo(C6H6(l)) = +49 kJ/mol

ΔHfo(CO2(g)) = -394 kJ/mol

ΔHfo(H2O(l)) = -286 kJ/mol

(Ans: ΔHrxno = -3268 kJ/mol)

Ex. What is ΔHrxno for the following reaction?

CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)

Given:

ΔH/mol of C-H bond: +413 kJ/mol

ΔH/mol of H-Cl bond: +431 kJ/mol

ΔH/mol of C-C bond: +348 kJ/mol

ΔH/mol of Cl-Cl bond: +242 kJ/mol

ΔH/mol of C-Cl bond: +328 kJ/mol

ΔH/mol of C=C bond: +614 kJ/mol

(Ans: ΔHrxno ≈ -104 kJ/mol)




Electromagnetic Spectrum: radiant energy can travel without

matter

λν c = speed of light = 3.0 x 108 m/s

λ = wavelength (m) ν = frequency (Hz)

Planck’s Theory: Blackbody radiation can be explained if

energy can be released or absorbed in packets of a standard

size called quanta

h = Planck’s constant = 6.63 x 10-34 J-s

Photoelectric Effect: As first explained by Einstein in 1905,

the photoelectric effect is the spontaneous emission of an

electron from metal struck by light if the energy is sufficient

Atomic Emission Spectra: spectrum for specific wavelengths

of light emitted from pure substances

Bohr’s Model of the H Atom: Bohr applied idea of

quantization of energy transfer to atomic model, theorizing that

electrons travel in certain “orbits” around the nucleus

Allowed orbital energies are defined by:

2

18

2

Hn

n

10178.2

n

RE

n = principal quantum number = 1, 2, 3 ...

RH = Rydberg’s constant = 2.178 x 10-18 J

Line Series: transitions from one level to another

Heisenberg’s uncertainty principle: The position and

momentum of a particle cannot be simultaneously measured

with accuracy.

Schrödinger’s wave function: Relates probability ( 2) of

predicting position of e- to its energy.

Matter as a Wave:

m = h / c Particles (with mass) have an associated wavelength

h / mc Waves (with a wavelength) have an associated mass

and velocity

Atomic Emission Spectra for

H

Line

Series

Transition down

to (emitted)

or up from

(absorbed)…

Type of EMR

Lyman 1 UV

Balmer 2 Visible

Paschen 3 IR

Brackett 4 Far IR

Probability Plots for 1s, 2s, and 3s

Orbitals

ch h E

dt

dihU

dx

d

m

hE

2

22

2


Pauli Exclusion Principle: no two charges in an atom

can have the same set of four quantum numbers n, l, m1,

ms.

Effective Nuclear Charge: the net positive charge acting

on the outermost electron.

Shielding Effect: inner electrons shielding the outer

electron from the full charge of the nucleus.

Electron Configuration: the way the electrons are

distributed among the various orbitals of an atom.

The most stable, or ground, electron configuration is

one in which the electrons are in the lowest possible

energy states.

Hund’s Rule: for degenerate orbitals (orbitals with the

same energy), the lowest energy is attained when the

number of electrons with the same spin is maximized.

The periodic table is your best guide to the order in

which orbitals are filled.

s-block and p-block contain the representative (main

group) elements.

The ten columns in the middle that contain transition

metals, elements in which d-orbitals are being filled.

f-block metals are the ones in which the f-orbitals are

being filled.

Diamagnetic: paired electrons

Paramagnetic: unpaired electrons

Mass Spectroscopy: Helps identify # and abundance of

isotopes and structures of different compounds. Chlorine

has two isotopes, 35

Cl and 37

Cl, in the approximate ratio

of 3 atoms of 35

Cl to 1 atom of 37

Cl. You might suppose

that the mass spectrum would look like this but that is

not the case because Chlorine consists molecules that

fragment (chemguide.co.uk)

You could have the following mass fragments--35 + 35 = 70,

35 + 37 = 72, 37 + 37 = 74. So the actual mass spectrograph

will look like the one on the right.

Ex: Give ground state electron configurations for

the following: Ni2+

and Ni3+

Ans: Ni2+

= [Ar]3d8, Ni3+

= [Ar]3d7

E

x: Is Cupric ion dia or paramagnetic? Why?

Paramagnetic, due to an unpaired d e.

Mass Spectrograph for Chlorine (Cl2)

(chemguide.co.uk)



Photoemission Spectroscopy (PES) In a photoelectron spectroscopy experiment any electron can

be ionized when the atom is excited. Unlike the first

ionization, in this experiment any electron can be removed, not just the electron that requires the least amount of energy.

PES gives insight into the structure of atom. Each peak in PES

indicates the number of electrons and the position of the peak indicates the amount of energy required to remove those

electrons. Note that s electrons will require more energy than

p electrons due to higher ENC hence s electrons will be farther

out on the energy axis.

http://www.chem.arizona.edu/chemt/Flash/photoelectron.

html


http://www.chem.arizona.edu/chemt/Flash/photoelectron.html

http://www.chem.arizona.edu/chemt/Flash/photoelectron.html



Periodic Trends Key Words

Principal Energy Level: more energy

levels means bigger atoms

Nuclear Charge (# of p in an atom):

attraction of electrons to nucleus;

increased nuclear charge causes

atomic radius to decrease

Shielding Effect: inner electrons

increase atomic size by reducing the

attractive force on outermost electrons

Effective Nuclear Charge: force of

attraction felt valence e- from nucleus;

a high ENC means smaller ionic

radius (greater attraction to outermost

electrons)

Atomic Radius: half distance between

two covalently-bonded atoms

Periodic Trends

Atomic Radius: increase going down,

decrease going right (more energy

levels and shielding going down,

higher ENC going right)

Size of Ions: cations smaller than

anions (fewer e-, less e

- repulsion),

bigger going down,smaller going right

Isoelectronic series have same # of

electrons, different # of protons

Ionization Energy: the amount of

energy required to remove an electron

from the ground state of a gaseous

atom or ion.

A(g) A+ + e

-

Bigger going right, smaller going

down

Exceptions: between Group 2&13,

Group 5&6

Atomic Radius

Isoelectronic Series

Ionization Energy

Zeff = Z − S Zeff = ENC Z = atomic number S = screening constant (# of inner electrons)



Periodic Trends (cont’d)

Electron Affinity: amount of energy

required to add an electron to a

gaseous atom:

Cl(g) + e− Cl

−

Larger going right, smaller going

down

Exceptions: between Group 1&2,

Group 14&15

Electronegativity: tendency to attract

electrons in a covalent bond

Increases going up and right

Metals and Nonmetals

Metals: tend to form cations, metal

oxides are basic

Nonmetals: tend to form anions,

nonmetal oxides are acidic, poor

conductors of electricity

Metallic character increases down a

group, decreases across a period

Alkali metals (1A)—The most reactive metal

family, these must be stored under oil

because they react violently with water!

They dissolve and create an alkaline, or

basic, solution, hence their name.

Alkaline earth metals (2A)—These also are

reactive metals, but they don’t explode in

water; pastes of these are used in batteries.

Halogens (7A)—Known as the “salt

formers,” they are used in modern lighting

and always exist as diatomic molecules in

their elemental form.

Noble gases (8A)—Known for their

extremely slow reactivity, these were once

thought to never react; neon, one of the

noble gases, is used to make bright signs. http://www.sparknotes.com/testprep/books/sat2/chemistry/chapt

er4section6.rhtml

Electron Affinity

Electronegativity




Intramolecular Bonding o Ionic: electrostatic attrction between oppositely charged

ions. Generally solids. Ex: NaCl, K2SO4

o Covalent: sharing of e- between two atoms (typically

nonmetals). Generally gases/liquids. Ex: CO2, SO2

o Metallic: “sea of e-”; bonding e

- relatively free to move

throughout 3D structure. Generally solid. Ex: Fe, Al

o Covalent Network: large number of atoms/molecules

bonded in network through covalent bonding. Ex: SiO2,

Si, Ge, Diamond, Graphite

Ionic Bonding o Results as atoms lose or gain electrons to achieve a

nobel gas electron configuration. Typically exothermic.

Bonded state lower in energy (more stable)

Opposite charges create electrostatic attraction,

which determines the strength of the ionic bond

o Occurs when difference in electronegativy is > 1.7

o Use brackets when writing Lewis symbols of ions.

Lattice Energy: Measurement of the strength of the ionic bond

o Hlattice = energy required to completely separate 1

mole of solid ionic compound into its gaseous ions

o Electrostatic attraction (and thus lattice energy)

increases as ionic charges increase and as ionic radii

decrease.

DHlattice µQ+Q-

r+ + r-

Covalent Bonding: Atoms share electrons to achieve noble gas

configuration that is lower in energy (therefore more stable)

o Occurs when difference of electronegativity is ≤ 1.7

Polar covalent: 0.3 < diff in EN ≤ 1.7

Nonpolar covalent: 0 ≤ diff in EN ≤ 0.3

Coordinate covalent: shared pair contributed by

only one of the two sharing species. Ex: Lewis

acids and bases

Ex. Draw the Lewis symbol for fluoride.

Ex. In the following pairs, which has a

greater lattice energy and why?

NaCl or KCl

NaCl or MgS

Covalent bond strength is measured by

bond energy. Bond energy is calculated by

= Energy of reactant bonds- Energy of

product bonds


Metallic Bonding: Forms between metal atoms because of

the movement of valence electrons from atom to atom to

atom in a “sea of electrons”. The metal consists of cations

held together by negatively charged electron “glue”.

o Results in excellent thermal and electrical

conductivity, ductility, and malleability

o A combination of 2 metals is called an alloy

The Octet Rule: Atoms tend to gain, lose, or share

electrons until they are surrounded by 8 electrons in their

outermost energy level (filled s&p shells), and are thus

energetically stable

Lewis symbols (electron-dot symbol)

o Shows a dot for valence electrons of an atom/ion

o Places dots at top, bottom, right, and left sides and

in pairs only when necessary (Hund’s rule).

o Primarily used for representative elements only

(Groups 1A – 8A)

o Transition metals typically form +1, +2, & +3 ions

Transition metal atoms first lose both “s”

electrons, even though it is a higher energy

subshell. Ex. Cr2+

, Cr3+

Most lose electrons to end up with a filled

or half-filled subshell. Ex. Cu+ ion

Lewis structures: used to depict bonding pairs and lone

pairs of electrons in the molecule

1. Total the # of valence electrons in the system. Add

the total negative charge if you have an ion,

subtract the charge if you have a cation.

2. Number the electrons if each atom is to be “happy”

(8 electrons for octet rule, or 2 for hydrogen)

3. Calculate the number of bonds in the system.

Covalent bonds are made by the sharing of

electrons. # of bonds = (electrons in step 2 –

electrons in step 1) / 2

4. Draw the structure. The central atom is usually the

atom with the least electronegativity.

5. Double check your answer by counting total

number of electrons

Ex. Draw the Lewis structures for the following

using steps. Show work!

a) Cl2

b) CH2Cl2

c) NH3

d) NaCl

e) HCN(SO4)

f) H2O2CNS


Exceptions to the Octet Rule

o Odd-electron molecules. Ex. NO or NO2

o Incomplete octet. Ex. H2, He, BeF2, BF3

o Expanded octet (occurs in molecules when the

central atom is beyond the third period. The empty

3d subshell is used in hybridization). Ex. PCl5, SF6

Formal Charge: the numerical difference between # of

valence electrons in the isolated atom and # of electrons

assigned to that atom in the Lewis structure. Doesn’t

represent real charges, just a useful tool for selecting most

stable Lewis structure

1. Assign unshared electrons (usually in pairs) to the

atom on which they are found

2. Assign one electron from each bonding pair to each

atom in the bond (split the electrons in a bond)

3. Subtract the electrons assigned from the original

number of valence electrons

o Used to select the most stable (and therefore most

likely) structure when more than one structure is

reasonable according to rules

o Most stable:

Has FC on all atoms closest to zero

Has all negative FC on most EN atoms

Resonance Structures: Equivalent Lewis structures that

describe a molecule with more than one likely

arrangement of electrons

o Notation: use double-headed arrow between all

resonance structures

o “Real” electron structure of the molecule is an

“average” of all resonance structures

Bond Order: Indication of bond strength and bond length

Bond Enthalpy: Amount of energy required to break a

particular bond between two elements in gaseous state.

Indicates “strength” of a bond.

Hrxn ≈ (Hbonds broken) - (Hbonds formed)

# Valence electrons in free atom

– # Non-bonding electrons

– ½ (# Bonding electrons)

Formal Charge

Ex. Draw at least 2 Lewis structures for each,

then calculate the formal charge for each atom

in each structure. a) SCN

-

b) N2OBF3

Ex. Draw the Lewis structure and determine the

bond S-O, C-C, and C-H bond orders.

c) SO3

d) C6H6

Ex: CH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g)

DHrxn = ?

Bond Ave DH/mol Bond Ave DH/mol

C-H 413 Cl-Cl 242

H-Cl 431 C-Cl 328

C-C 348 C=C 614

Ans: Hrxn ≈ -104 kJ/mol




Valence Shell Electron Pair Repulsion Theory: helps construct molecular 3-D shape from 2-D

Lewis Structures

Electron Domains: areas of valence electron

density around the central atom

Includes bonding electron pairs and lone

electron pairs

A single, double, or triple bond counts as one

domain

Basis for VSEPR: each group of valence electrons

(electron domains) around a central atom tend to be

as far as possible from each other to minimize

repulsion and this determines molecular geometry

of molecule

lonepair-lonepair repulsion > lonepair-bondpair

repulsion > bondpair-bondpair repulsion

Molecular Dipole Moment: molecules with polar

covalent bonds might have a net dipole moment

depending on the compound's 3-D geometry and

symmetry

Check if individual bonds are polar and if

individual dipole moments cancel out due to

symmetry

Polarity: polar substances are soluble in water and

non-polar substances are soluble in non-polar

solvents like benzene and oil

Valence Bond Theory: predicts bond strengths

based upon orbital overlap for covalent bond

formation

Basis for VB: covalent bond forms when orbitals of

two atoms overlap

An orbital can have a max of two electrons with

opposing signs

The bond strength depends on the attraction of

nuclei for the shared electron, so greater the


overlap, the stronger the bond

Sigma Bond: end to end overlap of orbital

which allows free rotation of parts of molecule

(single bonds)

Pi Bond: side to side overlap of orbital which

restricts rotation (2nd

and 3rd

bonds in double

and triple bonds)

Draw hybrid orbital diagram for C2H4.



Hybrid Orbital Theory: an extension of VB theory, where

atomic orbitals "hybridize" to form new hybrid orbitals;

explains the bonding in terms of quantum mechanical

model of atom (s, p, d, f orbitals)

Basis for HO: valence atomic orbitals in the molecule are

very different from those in isolated atoms

The number of hybrid orbitals obtained equals the

number of atomic orbitals mixed

The type of hybrid orbitals obtained varies with types

of atomic orbitals mixed

ns and np give two sp hybrids

ns and two np give three sp2 hybrid orbitals

ns and three np give four sp3 hybrid orbitals

Delocalized Bonds are present in compounds with

resonance structures

Molecular Orbital Theory: is based on the wave nature of

the electrons and is a better model to explain

paramagnetism of oxygen

Basis for MO: Molecular orbitals form through the

combination of atomic orbitals

Bonding MO: stable orbital that forms between nuclei

Antibonding MO: less stable orbital that forms behind

nuclei

Sigma MO: orbital forming from a combination of

two 1s or 2s orbitals form different atoms or two 2pz

orbitals from different atoms

Pi MO: orbital forming from a combination of two 2px

or 2py from different atoms (do not appear until B2)

Diamagnetism: all electrons paired; no magnetic

properties

Paramagnetic: at least 1 unpaired electron; drawn into

exterior magnetic field since spins of atoms become

aligned; unlikely to retain alignment when field is

removed

Ferromagnetism: occurs primarily in Fe, Co, and Ni;

drawn into exterior magnetic field since spins of atoms

become aligned; likely to retain alignment when field is

removed




Characteristics of Gases

Particles in a gas are very far apart, and

have almost no interaction.

Gases expand spontaneously to fill their

container (have indefinite volume and

shape.)

Pressure

Pressure = a force that acts on a given

area

A

FP

Atmospheric pressure: the result of the

bombardment of air molecules upon all

surfaces

1 atm = 760 mm Hg = 760 torr = 101.3

kPa = 14.7 PSI

Barometer: measures atmospheric P

compared to a vacuum

* Invented by Torricelli in 1643

* Liquid Hg is pushed up the closed

glass tube by air pressure

Manometers: measure P of a gas

*Closed-end: difference in Hg levels

(Dh) shows P of gas in container

compared to a vacuum

* Open-end: Difference in Hg levels

(Dh) shows P of gas in container

compared to Patm

Gas Laws

Boyle’s Law: the volume (V) of a fixed

quantity (n) of a gas is inversely

proportional to the pressure at constant

temperature (T).

Closed

2211 VPVP P

1constantV



Charles’ Law: V of a fixed

quantity of a gas is directly

proportional to its absolute T at

constant P.

Gay-Lussac’s law: P of a fixed

quantity of a gas is directly

proportional to its absolute T at

constant V.

Seen as derivative of C’s and B’s

laws

Avogadro’s hypothesis: Equal

volumes of gases at the same T &

P contain equal numbers of

molecules

Combined Gas Law

2

2

1

1

T

V

T

VT constantV

T constantP2

2

1

1

T

P

T

P

constantPV

T 2

22

1

11

T

VP

T

VP




INTRAmolecular Forces: the forces holding

atoms together to form moleculs

INTERmolecular Forces: Forces between

molecules between ions, or between molecules

and ions

Intermolecular Forces (IMF)

-IMF < intramolecular forces (covalent,

metallic, ionic bonds)

-IMF strength: solids>liquids>gases

-Types of IMFs Ion-Ion Forces, Ion-Dipole

Forces, Dipole-Dipole Forces, H-bonds

extreme dipole-dipole, LDFs

Types of IMF

Electrostatic Forces: act over larger distances

in accordance with Coulomb’s Law

Ion-Dipole: between an ion and a dipole (a

neutral, polar molecule has separate partial

charges)

-Increasing with increasing polarity of

molecule and increasing ion charge

Ion-Permanent Dipole

-Water is highly polar and can interact with

positive ions to give hydrated ions in water

-Attraction between ions and dipole depends

on ion charge and ion-dipole distance

Dipole-Dipole

-Weakest electrostatic force (not all IMFs,

LDFs weaker than dipole-dipole); exist

between neutral polar molecules

-Increase with increasing polarity (dipole

moment) of molecule

Hydrogen Bonds (H-bonds)

-H is unique among elements because it has a

single e- that is also a valence e

-

-When e- is “hogged” by a highly

electronegative atom (very polar covalent

bond), the H nucleus is partially exposed and

becomes attracted to e- rich atom nearby

-Explains why ice floats on water, has lattice-

like structure, explains why molecules with H-

bonds have higher boiling points, H-bonding

in water, O -H bond is very polar

H-bonding in biology

-DNA bases bind to each other due to specific

hydrogen bonding between Lewis Bases

Boiling points and melting points are good indicators of relative

IMF strength

2d

QQF

Summary of the page:


Inductive Forces arise from induced distortion of

e- cloud

London Dispersion: between polar or

nonpolar molecules or atoms, but is generally

mentioned for non polar molecules when other

forces are absent. Very weak, motion of e-

creates an instantaneous dipole moment which

induces a dipole in an adjacent atom

Nonpolar molecules can dissolve in water due

to LDFs. Water induces a dipole in electron

cloud. Solubility increases with mass of gas

due to greater distortion.

When induced forces between molecules are

very weak, the solid will sublime (solid to gas)

Liquids

Molecules are in constant motion, molecules

close together

Liquids are almost incompressible

Evaporation

-To evaporate, molecules must have sufficient

energy to break IMFs

-Condensation is reverse (remove energy and

make IM bonds)

Vapor Pressure

Heat of Vaporization heat required (at

constant P) to vaporize the liquid

Equilibrium vapor pressure & the

Clausius-Clapeyron Equation

o Used to find ∆vapH˚

o Logarithm of vapor pressure P is

proportional to ∆vapH˚ and to 1/T

o lnP = -(∆vapH˚/RT) + C

o Surface Tension leads to spherical

liquid droplets

Properties resulting from IMFs

Viscosity: resistance of a liquid to flow

Surface Tension: energy required to increase

the surface area of a liquid

Intermolecular forces lead to capillary action

and concave meniscus for a water column

-Capillary Action: movement of water up a

piece of paper depends on the H-bonds

between H2O and the OH groups of the

cellulose in the paper

Summary of the page:


Cohesion: attraction of molecules for other

molecules of the same compound

Adhesion: attraction of molecules for a

surface

Meniscus: curved upper surface of a liquid in

a container; a relative measure of adhesive and

cohesive forces

London Dispersion Forces

-Increase with increasing molar weight,

increasing # of e-,

increasing # of atoms

-“Longer” shapes (more likely to interact with

other molecules)

Phase Changes

-Endothermic: melting, vaporization,

sublimation

-Exothermic: condensation, freezing,

deposition

Structures of solids

Amorphous: without orderly structure (ex.

Rubber, glass)

Crystalline: repeating structure; have many

different stacking patterns based on chemical

formula, atomic or ionic sizes, and bonding

Types of Crystalline Solids:

-Atomic: Properties: poor conductors, low

melting point

-Molecular: Properties: poor conductors, low

to moderate melting point

-Ionic: Properties:hard and brittle, high

melting point, poor conductors, some

solubility in H2O

-Covalent (a.k.a. covalent network):

Properties: very hard, very high melting point,

generally insoluble, variable conductivity

-Metallic: Properties: excellent conductors,

malleable, ductile, high but wide range of

melting points

Molecular

Ionic Covalent Network Metallic

Credits: Google Images

Summary of the page



Vocabulary

Molarity: measure of concentration in

solutions, mol/L

Solvation: dissolving; the interactions

between solute and solvent

Crystallization: process by which

solute particles leave solvent

Saturation

Saturation: solution that is in

equilibrium with undissolved solute

Unsaturated solution contains less

solute than saturated solution

Supersaturated solution contains

more solute than saturated solution

but appears unsaturated

Factors Affecting Solubility

Miscible liquids mix; both are polar or

both are nonpolar

Covalent network solids do not

dissolve in polar or nonpolar

solvents

Increasing temperature increases

solubility for most solids, but decreases

solubility for gases

Describe the steps with the lab equipment needed to make 0.1 M

100 ml NaOH? (Hint: volumetric flask)

How are supersaturated solutions created?

Dissolve solute with heat, then cool solution slowly. Solution is

“tricked” into appearing unsaturated

At 40oC, the solubility of KNO3

in 100g of water is 64 g.

What is the solubility of KCl at

10oC?

30 g

At which points would an

unsaturated solution appear?

Supersaturated solution?

How much KClO3 needs to be

added to 10g of KClO3 at 60oC

to make a saturated solution?

20 g

What is normal

boiling point of

ethanoic acid?

117 oC

solution Liters

solute molM




Reaction rate: A measure of the (average)

speed of a reaction

Reaction rate is affected by:

1. Concentration of reactants

2. Temperature of the reaction

3. Presence/absence of a catalyst

4. Surface area of solid or liquid

reactants and/or catalysts

-Average Rate: Rate of a reaction over a

given period of time

-Instantaneous Rate: Rate of reaction at

ONE given point of time.

-Initial Rate: Rate of reaction at t=0. (its

instantaneous rate at t=0)

Reaction order: the exponents in a rate

law (can be fractions)

Rate law: shows how the rate of reaction

depends on the concentration of

reactant(s); determined experimentally.

Cannot be determined by the coefficients

of a balanced reaction (unless in an

elementary step)

Rate = k[A]m

[B]n

*Units of k change w/order of the rxn

To Find Rate Laws:

1. Using Initial Rates

2. Integration Method: Determining Rate

Law by Determining the Change in

Concentration of reactants over time

gives rate law either graphically or

by calculations.

Rate with regards to other

reactants and products

2N2O5 (g) → 4 NO2 (g)

+ O2(g)

If D[O2]/Dt = 5.0 M/s, what is D[N2O5]/Dt?-10.0 M/s

Orders of Reactions & Related Equations:

Using Initial Rates Method: data given

Expe

rimen

t

[A] [B] Initial Rate of Formation of

C in M

1 0.60 0.15 6.3´10-3

2 0.20 0.60 2.8´10-3

3 0.20 0.15 7.0´10-4

Integration Method (Using graphs or equations) Zero Order

[A]t = -kt + [A]0 (k) = M/s

First Order

ln[A]t = -kt + ln[A]0 or log[A]t = -kt / 2.303 + log[A]0

(k) = s-1

t

C

][

c

1 Rate

t

B

b

][1

t

A

][

a

1

t

Rate

A0][Rate Ak



The Collision Model

-Reactants must collide, and with the right

orientation and energy for an effective

collision

-Elementary steps: a single event or

step (reaction) in a multi-step reaction

-Molecularity: the # of molecules

participating as reactants in an

elementary step

-Catalyst: Substance that changes the

rate of a reaction without undergoing a

permanent chemical change itself

Check for Permissible Rxn. Mechanism

1. Balanced eq.

2. Rate Determining Step (RDS) is

the slow one.

-Radioactive decay follows a first order

kinetics. Half life (t1/2) can be calculated,

if rate constant is known or vice versa.

-Activation energy ( Ea ): minimum

energy required to initiate a chemical

reaction

Second Order

Half Life

Activation Energy

For a graph between ln k

and 1/T, the slope can be

used to calculate Ea—

activation energy.

Remember to use 8.314

j/mol. K for R.

1/21/2 t

0.693

t

2ln k

R

ESlope a

1 = kt + 1

[A] [A]0



Chapter 15 Summary Notes


Chemical Equilibrium

Occurs when rate of forward reaction = rate of reverse

reaction. Ex. Vapor pressure: rate of vaporization = rate

of condensation, Saturated solution: rate of dissociation =

rate of crystallization. Conc. of reactants and products

does not have to be equal at the equilibrium, only the rates of forward and reverse rxn become equal.

Express concentration in Partial Pressure for gases and

molarity for solutes in liquids

Rate = kforward [A]

Rate = kreverse [B]

at equilibrium

-If Kc > 1, then more products at equilibrium

-If Kc < 1, then more reactants at equilibrium

-If Kc = 1, then almost equal concentrations of products and reactants

There is a spontaneous tendency towards equilibrium.

(spontaneous ≠ quickly, spontaneous = always moving

towards equilibrium)

It is possible to force equilibrium one way or the other

temporarily by altering the reaction conditions, but once

this “stress” is removed, the system will return to its

original equilibrium.

Law of Mass Action :

a A + b B ↔ c C + d D

Concentrations of pure solids and pure liquids are not

included in Keq

The equilibrium expression is:

For a heterogeneous equilibrium:

CaCO3 (s) ↔ CaO (s) + CO2 (g)

RT

PA A][

RT

PB B][

RT

PkRate A

f

RT

PkRate B

r

r

f

k

k

[A]

[B] eq

r

f

A

B Kconstantk

k

P

Por

ba

dc

c[B] [A]

[D] [C] K or b

B

a

A

d

D

c

Cp

)(P)(P

)(P)(PK


][CaCO

)(P [CaO] K

3

CO

eq2

2COp P K


14100.1 wab KKK

00.14 wab pKpKpK


Acids and Bases Definitions

Arrhenius definition

o Acids: produce protons; increase [H+]

o Bases: produce hydroxides; inc. [OH-]

Brønsted-Lowry definition

o Acids: H+ donor

o Bases: H+ acceptor

Lewis definition

o Acids: electron pair acceptor

o Bases: electron pair donor

pH Scale

pH = -log[H+] = -log[H3O

+] or [H

+]= 10

-pH

pOH = -log [OH-] or [OH

-] = 10

-pOH

pH + pOH = 14.00

7 is neutral, <7 is acidic, >7 is basic

Key Terms

Amphoteric: capable of acting as either

an acid or base. Common amphoteric

substances are H2O, transition metal

cations (Al3+

, Fe3+

etc.), transition metal

hydroxides Fe(OH)3, Cu(OH)2 etc. and

conjugate bases of polyprotic acids such

as HSO3- etc.

Polyprotic acids have at least one H+

Hydrolysis: acid/base reaction of ion with

water to produce H+

or OH-

Anion (A-) = a conjugate base

A- (aq) + H2O (l) ↔ HA (aq) + OH

- (aq)

Cation (B+) = a conjugate acid

B+

(aq) + H2O (l) ↔ BOH (aq) + H+

(aq)

Acid-Base Behavior

Strong acids and bases fully ionize in

water (equilibrium is shifted “entirely”

toward ions).

Stronger acids, HA, have:

H with a higher d+

Weaker H-A bonds (smaller bond

energies)

More stable conjugate bases A-

Stronger oxyacids, HxOz-Y, have:

Central nonmetal “Y” with higher

electronegativity

pH + pOH = 14.00

Zinc Oxide acting amphoteric:

In acid: ZnO + 2H+ → Zn

2+ + H2O

In base: ZnO + H2O + 2OH- → [Zn(OH)4]

2-

Weak Acid Ionization

HA (aq) ↔ A- (aq) + H

+ (aq)

Use ICE chart to calculate pH of a weak acid or base.

Weak Base Ionization

B (aq) + H2O (l) ↔ BH+ (aq) + OH- (aq)

eq

aHA

AHK

][

]][[

100%[HA]

][HIonization %

init

eq

eq

bB

OHBHK

][

]][[


More O atoms




Common Ion Effect: solubility of solids decrease because of

Le Chatlelier’s Principle

Ksp: solubility-product constant

Molar Solubility: moles of salt per liter of solution in a

saturated solution

To predict whether precipitate will form:

o Compare ion-product, Q, with Ksp

o If Q > Ksp precipitation occurs

o If Q = Ksp solution is at equilibrium

o If Q < Ksp solid dissolves (until Q= Ksp)

Buffers: solutions that resist changes in pH1. Solutions that resist

drastic changes in pH upon additions of small amounts of acid or

base.

-Buffers are two component systems: each buffer consists of a weak

acid and its conjugate base or weak base and its conjugate acid.

-There are acidic buffers and basic buffers. The buffer with weak

acid/conjugate buffer is called an acidic buffer and a weak

base/conjugate acid is called basic buffer. Ex. Acidic buffer- HA/A- ,

Acetic acid/Sodium Acetate buffer HC2H3O2 + NaC2H3O2

(HC2H3O2/C2H3O2-), basic buffer – B/BH+ NH3 + NH4Cl

(NH3/NH4+)

Henderson-Hasselbalch equation: used for calculating the pH

of buffer.

Titration Curve: graph of pH vs volume of titrant added

Every titration has 4 points:

1. The initial pH

2. Between the initial pH and the equivalence point: pH is

determined by amount of solution not yet neutralized

3. The equivalence point: moles of aid = moles of base,

leaving a solution of the salt produced in the acid-base

neutralization

4. After the equivalence point: pH is determined by amount of

excess titrant

pH Indicators: chemicals with acid and base forms

with significant color differences

o Phenolphthalein: colorless to pink at ph of 8.5 to 9.5

o Methyl Red: red to yellow at pH from 4.2 to 6.3

Ex: What is the molar solubility of CaF2 in a

1.0 L solution of 0.10 M Ca(NO3)2?

Ksp(CaF3) = 3.9 x 10-11

Ans: 9.9 x 10-6

mols CaF2 are soluble per L

Ex: Will a precipitate form when 0.10 L of

3.0 x 10 -3

M Pb(NO3)2 is added to 0.400 L

of 5.0 x 10-3

M Na2SO4?

Ksp(PbSO4) = 1.6 x 10-8

Ksp(NaNO3) = 112.12

Ans: Since Q > Ksp, PbSO4 will precipitate

Ex. What is the pH of a buffer containing

0.12mol of NH3 and 0.095mol of NH4Cl in

250mL of water?

Titration Curve

Ex: A volume of 15 mL of 0.40 M NaOH is

added to 25 mL of 0.40 M HF solution.

Assume volumes are additive. Calculate the

pH of the solution.

][

]][[

HA

AHKa

][

][][

A

HAKH a

][

][log

acid

basepKpH a


Important Equations Acid Hydrolysis: HA (aq) + H2O (l) H3O

+ (aq) + A

- (aq)

, Ka = [H3O

+][ A

-]

[HA]

Base Hydrolysis:

B(aq) + H2O (l) BH+

(aq) + OH

-(aq), Kb = [BH

+][ OH

-]

[B]

Auto Ionization of Water: Kw = [H+][OH

-]= 1.0 x 10

-14

Cation Hydrolysis: A

+ + H2O (H

+OH

-) AOH + H

+ Acidic, Ka =[AOH][H

+]

[A+]

Anion Hydrolysis: B

- + HOH HB + OH

- Basic, Kb =[HB][OH

-]

[B-]

SA-SB Titration

WA-SB Titration

Ans: 3.32

Practice Problem:

0.100 M NaOH is added to 50.0 mL of 0.100

M CH3COOH; calculate pH at regular intervals

of titration (Ka = 1.8 x 10-5):

After 30.0 mL of 0.100 M NaOH has been

added: 4.92

After 50.0 mL of 0.100 M NaOH has been

added: 8.72

After 70.0 mL of 0.100 M NaOH is added: 12.22

Four main characteristics:

- Low starting pH

- Sharp change in pH near

the equivalence point

- Equivalence point at pH 7

- Sharp rise in pH after

equivalence point

Four main characteristics:

higher starting pH

Slow change in pH near

the equivalence point due

to buffering action

Equivalence point at >pH

7

Sharp rise in pH after

equivalence point



1st Law of Thermodynamics: In any

process, energy is neither created nor

destroyed.

When a system changes from one

state to another (DE = q + w), it

-Gains heat (+ q) or loses heat (- q)

and/or

-Does work (- w) or has work done on

it (+ w)

Reversible reaction: can proceed

forward and backward along same

path

Irreversible reaction: cannot proceed

forward and backward along same

path

Gaussian Distribution: Bell-curved

shape of distributions that shows

probably states of molecules. Tells us

that the most likely arrangement of

molecules is an even distribution.

Thus, we call a process spontaneous

if it produces a more probable

outcome, and non-spontaneous if it

produces a less likely one.

Entropy (S): a measure of molecular

randomness or disorder

-S is a state function: DS = Sfinal -

Sinitial

+ DS = more randomness

- DS = less randomness

And for reversible processes at

constant T:

Gaussian Distribution:

Sgas > Sliquid > Ssolid

T

q S rev

system



Redox: reactions with a change in oxidation

number

Oxidation: Loss of electrons; oxidation

number increases

Reduction: Gain of electrons; oxidation

number decreases

Oxidizing Agent (Oxidant): causes

another substance to be oxidized; gains

electrons

Reducing Agent (Reductant): causes

another substance to be reduced; loses

electrons

Anode: half-cell where oxidation occurs

Cathode: half-cell where reduction

occurs

Voltaic (Galvanic) Cells: A device that

spontaneously produces electricity by

spontaneous redox reactions

Remember the pneumonic

AnOx RedCat

In a voltaic cell: anode is (-), cathode is

(+)

In an electrolytic cell: anode is (+),

cathode is (-)

Electrochemical cells refer to both

types- voltaic/galvanic or electrolytic

cells

-Standard Cell Potential, E0 cell: the

electric potential difference of the cell

(voltage)

E0 cell = E

0r cathode – E

or anode

-If the E0

cell is positive, the reaction

occurring is spontaneous.

-If the E0 cell is negative, the reaction

occurring is non-spontaneous

Spontaneity of Redox Reactions o Gibbs free energy and emf:

DGo = - n F E

o

-Balancing Redox Reactions:

Assign oxidation numbers to all species; based on this, split

the half-reactions of reduction and oxidation.

For each, balance all elements except H and O.

Balance oxygen by adding H2O to the opposite side.

Balance hydrogen by adding H+

to the opposite side.

Balance charges by adding e- to side with overall positive

charge.

Multiply each half-reaction by an integer so there are an

equal number of e- in each.

Add the half reactions; cancel any species; check final

balance.

Voltaic or Galvanic Cell Diagram:

Ex. What is the standard potential of the cell represented below:

Ans: + 0.34 V


Electrolysis: the process of supplying

electrical energy to force a nonspontaneous

redox reaction to occur

Current (I): a measure of the charge (Q)

per unit time (t)

I = Q/t

-Electroplating:

Electrolytic Cell:

35

Ex: Na+ (l) + Cl- (l) → Na (l) + Cl2 (g)

Oxidation half-cell: 2 Cl- (l) → Cl2 (g) + 2 e-

Reduction half-cell: Na+ (l) + e- → Na (l)

anode

Pt Pt

(-)(+)

cathode

e- flow

NaCl (l)

Cl2 (g) Na (l)

• In a voltaic cell: anode is (-), cathode is (+)

• In an electrolytic cell: anode is (+), cathode is (-)

Ex. A current of 0.452 A is passed through an electrolytic cell

containing molten NaCl for 1.5 hours. Write the electrode

reactions. Calculate the mass products formed at each

electrodes. Ans:

Oxidation: 2 Cl- (l) → Cl2 (g) + 2 e-

Reduction: Na+ (l) + e- → Na (l) 0.58 g Na and 0.90 g Cl2

Ex.If the voltage on an aqueous mixture of CuCl2 and ZnCl2 in


an electrolytic cell is slowly increased, what products will form

at each electrode? Ans: Cu (s) > Zn (s) > H2 (g)

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