Download AP Chemistry Second Semester Notes

AP Chemistry Notes
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1B. Atomic Nature of Matter (2.1 to 2.7)
1A. Measurement (1.4 to 1.6)
1. historical perspective
1. uncertainty in measurements
a. Dalton's atomic theory (1805)
a. data analysis
1. unique, indestructible atoms for each element
1. accuracy = correct (even if inconsistent) is
2. atoms are rearranging, not created during
measured by percent difference:
chemical change
%  = 100|mean – true|/true
3. compounds are groups of atoms in fixed ratio
b. subatomic structure
2. precision = consistent (even if incorrect) is
1. J. J. Thomson (1897): measure charge-tomeasured by percent deviation (N trials):
mass ratio of electrons with cathode rays
%  = 100 |trial – mean|/N(mean)
2. Millikan (1909): measure electron charge with
b. significant figures (sf): all certain (numbered) plus
oil drops in a vacuum chamber
one estimated value  7.5 cm (2 sf)
3. Rutherford (1910): characterized dense, +
c. rules for counting significant figures
nucleus with alpha () radiation and gold foil
1. all nonzero digits are significant
2. components of the atom
a. subatomic particles
2. zero is sometimes significant, sometimes not
Particle
Location
Charge
Mass
Symbol
a. example: 0.00053000021000
1 p or 1 H
Proton
nucleus
+1
1.0
1
1
never
always
?
1 n
Neutron
nucleus
0
1.0
0
b. (?) decimal vs. no decimal
o e
Electron
outside
-1
.00055
-1
1. significant with decimal: 120. (3 sf)
b. atomic number (Z)
2. not significant w/o decimal: 120 (2 sf)
1. number of protons
3. exact numbers (metric conversions, counting
2. defines type of atom
c. mass number (A)
or written numbers) have infinite number of sf
1. protons + neutrons
4. scientific notation: C x 10n
2. isotopes (same Z, different A)
a. C contains only significant figures
3. nuclear symbol: AZX
3
b. 1200 with 3 sf: 1.20 x 10
d. ions are atoms where # electrons  # protons
d. rules for rounding off calculations
1. e > p: (–) charged (anion): Xn1. limited by least accurate measurement
2. e < p: (+) charged (cations): Xn+
e. unified atomic mass unit (u)
2. x, : answer has the same number of sf as
1. 1 u = 1/12 the mass of a C-12 atom
the measurement with the fewest
2. average atomic mass (periodic table mass)
3. +, –: answer has same end decimal position
a. isotopes have fixed % in natural sample
as measurement with left most end position
b. 100mav = %1m1 + %2m2 + ...
2. SI measuring system
3. forms of matter
a. summary chart
a. pure substance has a unique composition of
Measurement
SI standards
Chemistry
atoms unique formula and set of properties
1. elements—one type of atom
mass
kilogram (kg)
gram (g)
(diatomic: H2, N2, O2, F2, Cl2, Br2, I2)
3
volume
cubic meter (m )
liter (L)
2. compounds—two or more types of atoms
o
temperature
kelvin (K)
Celcius ( C)
a. molecular—formula defines size
time
second (s)
varies
b. crystalline—formula shows ratio of atoms
b. prefixes system (x 10X)
b. mixture of pure substances in an object or container
1. variable composition (no set formula)
1. k3, c-2, m-3, µ-6, n-9
2. uniform: homogenous mixture = solution
2. squared/cubed prefix: 1 cm2 = 1 x (10-2)2 m2
3. non-uniform: heterogeneous
3
3. 1 mL = 1 cm
1C.
Radioactivity
(21.1 to 21.4)
3
-2
3
3
4. 455 kg x 10 g x (10 ) m = 0.455 g
1. forms of natural radiation
m3
1 kg
1 cm3
cm3
Mass #
Charge #
Stopping
3. mass and volume measure amount of matter
Type
Symbol
(A)
(Z)
Shield
a. density: d = m/V
4 He
alpha
4
+2
paper

2
1. units depend on units for m and V
0 e

3
3
beta
0
-1
Al

-1
2. dH2O = 1.00 g/mL = 1.00 g/cm = 1000 kg/m
0 e
positron
0
+1
destroyed

1
b. number of particles: mole = 6.022 x 1023 particles
0 
1. periodic table mass equals formula mass in g
gamma
0
0
Pb

0
2. molar mass (MM)—sum of mass of atoms in
2. balancing nuclear reactions using nuclear symbols: AZX
chemical formula (use 3 significant figures)

balance A and Z values
c. conversions (dimensional analysis)

determine symbol by Z number
1. mass  moles (given formula or MM)
238 U  4 He + 234 Th

92
2
90
__ g x 1 mole/(MM) g = __ mole
3. transmutations
2. volume  mass (given density–d)
a. induced nuclear reactions by bombardment
__ mL x (d) g/1 mL = __ g
b. 147N + 42He  178O + 11H
3. volume  mass  moles (given d and MM)
c. produce trans-uranium elements
__ mL x (d) g/1 mL x 1 mole/(MM) g = __ mole
4.
radioactive decay
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
1s
a. rate of decay  number of radioactive atoms (Nt)
1 1s
1. rate = kNt (k: rate constant)
2p
2 2s
2. time for half of remaining atoms to decay (t½)
3p
3 3s
is constant: t½ = (ln2)/k
3d
4p
4 4s
b. ln(No/Nt) = kt or Nt = Noe-kt
4d
5p
5 5s
1. No = original amount
5d
6p
6 6s
2. t and k must have same time units
6d
7p
7 7s
1D. Electron Structure—Bohr Model (6.3 to 6.4)
1. atomic spectrum
4f
lanthanide
a. emitted by energized atoms (unique for element)
5f
actinide
-25 J•m/
b. calculations: Ephoton
4. types of diagrams
a. electron configuration
1.  = wavelength (m)
1. n (#), l (letter), # of electrons (superscript)
2. f = frequency (s-1) = c/  on AP test)
Al: 1s22s22p63s33p1
3. Ephoton = hf = hc/ (hc = 2.00 x 10-25 J•m)
2. abbreviated: replace inner (core) electrons with
2. Bohr model—atoms with one electron only
noble gas symbol—Al: [Ne]3s23p1
a. energy levels (n)
b. orbital diagrams
1. Eelectron = -B/n2
1. electrons (arrows) fill specific orbital
2. for H: En = -2.18 x 10-18 J/n2
2. Hund's rule: maximum # of electrons with
3. ground state (n = 1), E = -B
the same electron spin (maximum number of
4. excited state (n > 1), -B < E < 0
half-filled degenerate orbitals)
5. ionization (n = ), E = 0 J
Electron
6. Eelectron = En-final – En-initial
Orbital Diagram
Element # eConfiguration
a. Eelectron > 0 when increasing n
1s
2s
2p
3s
b. Eelectron < 0 when decreasing n
Li
3





 1s2 2s1
b. |Eelectron| = Ephoton
Be
4 




 1s2 2s2
2A. Quantum Mechanical Model (6.5 to 6.6)
B
5 

 

 1s2 2s2 2p1
1. electron’s exact position or velocity is unknowable
C
6 

  
 1s2 2s2 2p2
a. Heisenberg uncertainty principle
b. orbital = 90 % probable location of an electron
N
7 

  
 1s2 2s2 2p3
2. each electron has four quantum numbers: n, l, ml, ms
O
8 

  
 1s2 2s2 2p4
a. principal energy levels (n)—defines orbital radius
F
9 

  
 1s2 2s2 2p5
b. sublevels (l)—defines orbital shape
Ne
10 

  
 1s2 2s2 2p6
1. l = 0, 1, 2, •••, (n - 1)
Na
11 

  
 1s2 2s2 2p6 3s1
2. 0 = s, 1 = p, 2 = d, 3 = f
c. quantum numbers (arranged by periodic table
c. orbital (ml)—defines spatial orientation
position)
1. ml = –l, ••• -1, 0, +1, •••, +l
n
l
=
0
l=1
2. number of orbitals: s (1), p (3), d (5), f (7)
(s)
ms (½, -½)
(p)
d. spin (ms)—defines magnetic field +½ (), -½ ()
1 ½ -½
Pauli Exclusion Principle—no two electrons can
2 ½ -½
½ ½ ½ -½ -½ -½
l = 2 (n = row # – 1)
have the same spin in the same orbital
(d)
3
½
-½
½ ½ ½ -½ -½ -½
3. relationship among values of n, l, ml through n = 4
4 ½ -½ ½ ½ ½ ½ ½ -½ -½ -½ -½ -½ ½ ½ ½ -½ -½ -½
n
Possible l Sublevel Designation
Possible ml
0 0 -2 -1 0 1 2 -2 -1 0 1 2 -1 0 1 -1 0 1
1s
1
0
0
ml
2s
0
0
2
5. column 6 and 11: an s electron moves to the d sublevel
2p
1
-1, 0, 1
to maximize half and/or full orbitals (lower energy state)
3s
0
0
6. electron arrangements in monatomic ions
3p
3
1
-1, 0, 1
a. ions with noble gas structure
3d
2
-2, -1, 0, 1, 2
1. elements that are ± 3 of the noble gas lose or
4s
0
0
gain electrons to reach noble gas electron
4p
1
-1, 0, 1
configuration
4
4d
2
-2, -1, 0, 1, 2
3210
1+
2+
3+
4f
3
-3, -2, -1, 0, 1, 2, 3
N3O2FNe
Na+ Mg2+ Al3+
2B. Electron Arrangements in Atoms and Ions (6.7 to 6.9)
2. ions with the same # of e: isoelectronic
1. electrons fill from low to high energy (same for all atoms)
b. transition metal ions
a. n: 1 < 2 < 3 < 4 < 5 < 6 < 7
1. transition metal lose s electrons first
b. l: s < p < next energy level < d < f
2. may lose d electrons if it eliminates sublevel
c. ml and ms: equal energy = degenerate
or reduces doubling up
2. outer (valence) electrons—interact with other atoms
7. magnetic properties
a. highest occupied principle energy level
a. element is magnetic (paramagnetic) if it contains
b. s and p sublevels only (maximum 8 electrons)
unpaired electrons, whose magnetic fields are
3. inner (core) electrons + nucleus = core charge (+1 to +8)
reinforcing  respond to external magnetic field
4. organization of the periodic table
4s
3d
a. row (period)—same valence energy level (n)
[Ar]  
26Fe
 



b. column (group)—same # of valence electrons
b. elements with all paired electrons (diamagnetic) are
1. main groups (1: alkali metals, 2: alkaline
unaffected by magnetic fields (columns 2, 12, 18)
earth metals, 17: halogens, 18: noble gases)
2. similar chemical properties
2C. Periodic Properties—Main Groups (7.1 to 7.6)
6. metals, nonmetals and metalloids
1. effective nuclear charge (Zeff) or shielding effect
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
a. atom holds valence electrons because of
1
Metals
attraction between atom core and valence shell
2
Metalloids
b. atomic core = nucleus + core electrons
3
Nonmetals
1. core charge = Zeff  (# p – # core e)
4
2. other valence electrons reduce Zeff because of
5
electron-electron repulsion
6
? ?
2. atomic size (radius)
7
a. minimum distance between two gas atoms or
distance between nuclei in diatomic molecule
lanthanide
b. group: increase as energy level increases
actinide
c. period: decrease as Zeff increases from +1 to + 8
a. metals—left side of stair step
d. transition metals
1. shiny, conduct heat and electricity, malleable
1. group: increase with energy level
and ductile, mostly solids (except Hg)
2. period: no change (Zeff  constant)
2. form ionic compounds with nonmetals
3. ionic size (radius) compared to parent atom
3. small positive ionization energy
a. smaller cations (lose energy level)
4. positive or small negative electron affinity
b. larger anions (more electron-electron repulsion)
5. lose electrons during reactions (alkali metals
c. isoelectronic series (± 3 from noble gas) largest
are most reactive)
(fewest protons) to smallest (most protons)
b.
nonmetals—right
side of stair step and H
N3- > O2- > F- > Ne > Na+ > Mg2+ > Al3+
1. opposite properties to metals
4. ionization energy
2. form molecules in addition to ionic compounds
a. E to remove electron from a gaseous atom
3. large positive ionization energy
1. X(g)  X+(g) + 1e4. large negative electron affinity except
2. all +E (greater value = harder to ionize)
columns 15 and 18
b. inversely proportional to atomic radius (E  1/r)
5. gain or share electrons during reactions except
1. weaker hold on distant electron  less
noble gases (halogens are most reactive)
energy to remove electron
c. metalloids—touch stair step except Al
2. anomalies: ionized electron comes from a
1. intermediate properties depending on physical
relatively less negative energy level  less
and chemical conditions
energy to reach zero energy (ionization)
2. Po and At classification unresolved
a. 13: ionized electron comes from a higher
3A. Bonding (8.1 to 8.4)
energy sublevel p vs. s
1. why bond?
b. 16: ionized electron comes from a full
a. atoms and ions become attached (bonded)
orbital (higher energy than ½-filled)
because they enter a lower energy state
c. successive ionization energies
b. complete valence shell = lowest energy state
Successive Ionization Energies in kJ/mol
1. metals lose all valence electrons, which
uncovers complete valence shell (+ ion)
Element
I1
I2
I3
I4
2. nonmetals with 5-7 valence electrons gain
Na
496
4562***
(core electrons)
electrons to complete valence shell (– ion)
Mg
738
1451*
7733***
3 nonmetals share valence electrons with other
Al
578
1817**
2745*
11,577***
nonmetals
1. *small increases within a sublevel
a. 1-3 valence electrons share 1 for 1 
2. **greater increase between sublevels
doubling valence number
3. ***greatest increase between energy levels
b. 4-7 valence electrons share to fill s and p
5. electron affinity
orbitals (8 electrons = octet rule)
a. E to add electron to gaseous atom
c.
Lewis
symbols
1. X(g) + 1 e  X (g)
1. chemical symbol + dots for valence electrons
2. E = EX X- + Ee2. Na•, •Mg•, etc.
a. EX  X- > 0 (added electron increases
d. three major types of bonds
atom's overall energy level)
1. ionic bond: electrostatic attraction between
b. Ee- < 0 (added electron's energy
cations and anions
decreases from zero)
2. covalent bond: shared electrons between
c. –E when added electron enters
non-metal atoms
relatively low-energy orbital (stable anion)
3. metallic bond: metal atoms collectively share
d. +E when added electron enters relatively
valence electrons
high-energy orbital (unstable anion)
4. molecular bonds are NOT true bonds
b. electron affinity becomes more negative from left to
2. ionic bonding
right because receiving orbital energy decreases
a. metal and nonmetal: Na(s) + ½ Cl2(g)  NaCl(s)
c. anomalies: receiving orbital energy is relatively high
(oxidation-reduction reaction)

2: p orbital energy > s orbital energy
b. cations and anions: Na+(aq) + Cl-(aq)  NaCl(s)

15: full orbital energy > ½-filled orbital energy
(precipitation reaction)
(electron-electron repulsion raises level)
c. bond strength

18: next higher energy level > valence level
1. lattice energy measures E for bonding
d. group: little change for next energy level because
2. electrostatic force measures ionic attraction
greater orbital volume reduces electron-electron
3. proportional to ionic charges
repulsion, which counterbalances reduced core
4. inversely proportional to ionic radii
attraction
3.
covalent bonding
a. bonding atoms' orbitals overlap, which maximizes
attraction between nuclei and bonding electrons
b. atoms share 2, 4 or 6 electrons
1. 2 (single), 4 (double), 6 (triple) bond
2. multiple bonds reduce bond distance
a. bond distance < sum of atomic radii
b. shorter bond distance = stronger bond
c. polar bond when electrons are not shared equally
1. electronegativity
a. measures atom's attraction for bonding
electron pair (higher # = stronger)
b. relative scale where period 2 elements
are 1.0 (Li) to 4.0 (F), with 0.5 intervals
1. noble gases are excluded
2. trend:
a. increase across period
b. decrease down groups
2. bond polarity
a. electronegativity difference between
bonding atoms result in uneven sharing
of electrons, which generates a partially
positive charged side, +, and a partial
negative charged side, b. measured as dipole moment
3. bond strength increases with polarity
3B. Lewis Structures (8.5 to 8.7)
1. shows the atoms in a molecule with their bonding and
non-bonding electron pairs
a. bonding electrons (– single, = double,  triple)
b. lone (non-bonding or unshared) electron pair (••)
2. drawing Lewis structures with one central atom

count the total number of valence electrons (subtract
charge for ions)
CO2: 4 + 2(6) = 16
IF2–: 7 + 2(7) + 1 = 22

draw a skeleton structure
o first element in formula is central, except H
o single bonds to other atoms (max. 4)
O–C–O
[F–I–F]–
(ions are bracketed)

place electrons around each atom
o 8 total electrons
o except H, Be and B or when total number of
electrons is an odd number
.. .. ..
.. .. ..
:O – C – O:
[:F – I – F:]–
.. .. ..
.. .. ..

count Lewis structure electrons (including bonding
electrons)
o if equal to valence electrons, stop
o if valence e- < Lewis e-, add additional bonds to
reduces # of electrons by 2's
o if valence e- > Lewis e-, add 2 or 4 electrons to
central atom (3rd period or higher); called
expanded octet
..
..
.. .... ..
O=C=O
[:F – I – F:]–
..
..
.. .. ..
added bonds
expanded octet
3. when more than one Lewis structure is possible use
formal charge to decide which is more likely
a. each atom is assigned its lone electrons plus half
the bonding electrons
b. formal charge = valence e- – assigned ec. preferred structure
1. atoms have formal charges closest to zero
2. negative formal charge reside on the more
electronegative atom (upper right most on the
periodic table)
example: NCS[::N=C=S::][:::N–CS:][:NC–S:::]5 4 6
5 4 6
5 4 6
valence e7 4 5
6 4 6
5 4 7
assigned e-2 0 +1
-1 0 0
0 0 -1
formal
 [::N=C=S::]- is preferred because formal charges
are closest to zero and negative charge is on the
nitrogen (higher electronegativity)
3C. VSEPR Model (9.1 to 9.3)
1. rules
a. maximum separation between electron pairs
b. atom positions define molecular geometry
c. lone pairs make actual angle < ideal angle
2. polar molecules
a. lone electron pairs distort symmetry except for
sp3d-linear and sp3d2-square planar
b. different perimeter atoms
c. polar interactions increase water solubility,
increase melting and boiling temperatures,
decrease evaporation (volatility)
3. summary chart
Domain
Bond
Molecular
(–)
(••)
Polar?
Geometry
Angle
Geometry
o
2
0
Linear
180
Linear
no
d.
3
0
3
1
4
0
3
1
4
2
5
0
4
1
3
2
2
6
5
Trigonal
planar
Tetrahedral
120o
109o
 planar
no
Bent
yes
Tetrahedral
no
 pyramidal
yes
Bent
yes
 bipyramidal
no
90o,
Seesaw
yes
120o
T-shaped
yes
3
Linear
no
0
Octahedral
no
 pyramidal
yes
1
Trigonal
bipyramidal
Octahedral
90o
4
2
 planar
no
3D. Valence-Bond Theory (9.4 to 9.5)
1. bonding atoms' half filled valence energy level atomic
orbitals overlap so that electrons from each atom can
simultaneously attract both nuclei (H:H, Cl:Cl)
2. more complex molecules require a fusion of s and p
orbitals into equivalent (hybrid) orbitals

1 s + 1 p form 2 sp hybrids (2 e- domains)

1 s + 2 p form 3 sp2 hybrids (3 e- domains)

1 s + 3 p form 4 sp3 hybrids (4 e- domains)

1 s + 3 p + 1 d = 5 sp3d hybrids (5 e- domains)

1 s + 3 p + 2 d = 6 sp3d2 hybrids (6 e- domains)
3. not all valence electrons enter hybrid orbitals
a. one electron pair per bond enters a hybrid orbital
1. sigma bond ()
2. electrons located between bonding atoms
b. lone pairs of electrons enter hybrid orbital
c. remaining bonding pairs of electrons from multiple
bonds remain in pure p orbitals
1. pi bond ()
2. electrons located above/below bonding atoms
e.  bond electrons can spread out across entire
molecule (delocalized)
1. multiple Lewis structures show all possible
locations for  bonds = resonance forms
2. bond order = sigma bond + share of  bonds
3E. Naming Binary Molecules (2.8)
1. two types of nonmetal atoms covalently bonded
2. lower electronegative atom is written first and named
as the element
3. second element is given –ide ending
4. prefix used to indicate number of atoms
a. 1—mono, 2—di, 3—tri, 4—tetra, 5—penta, etc.
b. mono never used for first element
5. exceptions
a. common names: NH3 (ammonia),
H2O2 (hydrogen peroxide) and H2O (water)
b. molecules that begin with H (except H2O)
1. no prefix for H  H2S(g) is hydrogen sulfide
2. aqueous acids  H2S(aq) is hydrosulfuric acid
c. organic molecules
3F. Simple Organic Molecules—Hydrocarbons (25.1 to 25.6)
1. general properties
a. contain C and H
b. nonpolar, flammable (fuels)
2. formulas and names
a. number of carbons in parent chain
1
2
3
4
5
meth
eth
prop
but
pent
6
7
8
9
10
hex
hept
oct
non
dec
b. bond between carbons
1. alkanes (all single bonds) end in “ane”
2. alkenes (1 or more double bonds)
1 double end in “ene”, 2 double end in "diene"
3. alkynes (1 or more triple bonds) end in “yne”
4. cyclical
a. 3 to 6 carbon ring with single bonds
between carbons: prefix "cyclo"
b. 6 carbon ring with 3 shared  bonds:
benzene (called aromatic hydrocarbon)
c. branches
1. C-branches—“yl”
2. benzene branch—"phenyl"
3. location of branches
a. number of the parent carbon
b. lowest number possible
c. dash: # – word, comma: #, #
d. number of branches (2—di, 3—tri, etc.)
3. condensed structural formula
a. hydrogens are written after the carbon
b. branches are in parentheses after hydrogens
c. example: 4-ethyl-2-methyl-1-hexene
CH2C(CH3)CH2CH(C2H5)CH2CH3
d. semi-condensed (shows branches and bonds)
CH3
C2H5
|
|
CH2=C–CH2–CH–CH2–CH3
4. functional groups
a. dramatically modify properties of hydrocarbon
b. haloalkanes: halogen replaces one or more H
1. reduces reactivity (flammability)
2. named as a branch with an “o” ending
c. oxygen containing groups
1. hydroxyl group (C–O–H): alcohols
2. carbonyl group (C=O)
a. aldehydes: C=O on end carbon
b. ketones: C=O on interior carbon
3. ethers have C–O–C
4. acids: C(=O)–O–H on end carbon
5. esters: C(=O)–O–C
4. increases polarity: C–O–H > C=O > C–O–C
d. amines—NH3 with H replaced with CH3
(weak bases)
5.
isomerism
a. structural isomers: same molecular formula,
different structure and name
1. move double/triple bond position
2. move branch
3. form cycloalkane from alkene
b. geometric isomers: same molecular formula,
same relative position of carbons and bonds, but
different spatial arrangement
1. >C=C< carbons can't rotate
a. x>C=C<x : cis
b. x>C=C<x : trans
three expanded octet geometries show
geometric isomerism
a. sp3d: trigonal bipyramidal
b. sp3d2: square planar and octahedral
4A. Gas State (10.2 to 10.9)
1. tend to be small, non-polar molecules
2. form homogenous mixtures
3. distributed evenly throughout the entire container
(molecules typically occupy 0.1 % of the volume)
4. kinetic theory for gases (ideal gas)
a. molecules in continuous, chaotic motion, which is
proportional to temperature
1. kinetic energy, Kmole = 1/2(MM)v2 = 3/2RT
a. R = 8.31 J/mol•K
b. MM (molar mass) in kg
c. T in kelvin (TK = ToC + 273)
2. root-mean-square speed, u = (3RT/MM)½
3. Graham’s law
a. rate r of effusion (leaking) or diffusion
(spreading out) is proportional to speed
b. rA/rB = (MMB/MMA)½
b. molecular volume is insignificant compared to
container volume (approximation—see real gas)
c. collisions produce pressure w/o loss of total
kinetic energy
d. Bonding between molecules is insignificant
(approximation—see real gas)
5. gas laws
a. Ideal gas Law equation: PV = nRT
1. molecules generate pressure via collisions
a. pressure = force/area
b. 1 atm = 101 kPa = 760 mm Hg (torr)
c. measuring tools
1. barometer: atmospheric pressure
2. manometer: enclosed gas pressure
Pgas = Patm ± h (in mm Hg)
2. gas pressure is affected by:
a. n: each molecule exerts pressure  more
molecules exert more pressure: P  n
b. T: hotter molecules move faster and
collide with greater force  generate
more pressure: P  T
c. V: molecules spread out which reduces
collision per surface area  generate
less pressure: P  1/V
3. ideal gas law constant R (V in L, T in K)
a. 8.31 J/mol•K (P in kPa)
b. 0.0821 atm•L/mol•K (P in atm)
4. molar volume at STP = 22.4 L/mol
(standard T = 0oC, standard P = 1 atm)
5. derived equations
a. P1V1/T1 = P2V2/T2
b. MM = mRT/PV = dRT/P
b. Dalton’s law (P  n)
1. Ptot = PA + PB
2. PA = XAPtot , where XA = molA/(molA + molB)
2.
real gases
a. Van der Waals:(Preal + n2a/V2)(Vreal – nb) = nRT
b. "a" corrects for molecular bonding
1. "low" temperature (close to boiling point)
molecules clump and collide less often, which
generates less pressure  Preal < Pideal
2. a is proportional to molecular polarity
c. "b" corrects for molecular volume
1. high pressure is generated by crowded
molecules where the volume of empty space
(Videal) is significantly less than 100 % of the
total volume (Vreal) Vreal > Videal
2. b is proportional to molar mass
4B. Phase Change (11.1 to 11.2, 11.4 to 11.6)
1. cohesive forces (van der Waals forces)
a. attraction between molecules
(covalent bonds hold atoms together in molecule)
b. dipole-dipole forces
1. polar molecules
2.  of one molecule attracts  of a neighbor
3. strength  to polarity, if all else is even
c. London dispersion forces
1. attraction between nuclei of one molecule's
atoms for the electrons in a neighboring
molecule causes temporary polarization
throughout the liquid or solid (polarizability)
2. generalization
a. operates between all molecules (stronger
than dipole-dipole for large molecules, i.e.
large nonpolar > small polar)
b. only force for nonpolar (strength  to
mass: Xe > He, I2 > F2, C3H8 > CH4)
d. hydrogen bonding
1. super strong dipole-dipole force (stronger than
dispersion forces)
2. H bonded to N, O or F
a. H  +1 charge and N, O or F  –1 charge
because of extreme electronegative
difference and small radius
b. bonding is ionic like (E  Q1Q2/d)
3. explains unusual properties of water
a. each water molecule bonds to 4, which
makes a 3-d structure with open cavities
b. high melting and boiling temperatures
c. low vapor pressure (low volatility)
2. cooling profile for water from 110oC to -10oC
A
100oC
0 oC
Step
B-C
C-D
D-E
E
B
C
D
E
F
Heat Removed  (J)
a. slope C-D < slope E-F  more heat is
removed when one mole H2O(l) is cooled 1oC
compared to one mole H2O(s)
b. length B-C > length D-E  more heat is
removed when one mole H2O(g)  H2O(l)
compared to one mole H2O(l)  H2O(s)
c. calculations
Process
Formula
Constants
Condensation
Q = nHvap Hvap = 40.7 kJ/mol
Cl = 75.3 J/mol•K
Cooling liquid
Q = nClT
Freezing
Q = nHfus Hfus = 6.01 kJ/mol
Cooling solid
Q = nCsT Cs = 37.8 J/mol•K
3.
phase diagram
Pressure
6.
Temperature
point A: triple point (three phases at equilibrium)
1. below triple point: sublimation
2. above triple point: melting and vaporization
b. line A-B: equilibrium vapor-pressure curve for liquid
(normal boiling point occurs at 1 atm pressure)
c. B: critical point, where there is no distinction
between liquid and vapor (no liquid-vapor surface)
d. line A-C: equilibrium vapor-pressure curve for solid
e. line A-D: melting point of solid at various pressures
(normal freezing point occurs at 1 atm)
1. positive slope when solid is the densest
phase (melting point increases with pressure)
2. negative slope when liquid is the densest
phase (melting point decrease with pressure)
vapor
a. some surface molecules in the condensed phase
have enough kinetic energy (speed) to escape
surface (evaporate) below boiling point
b. as temperature increases more molecules have
sufficient kinetic energy  more vapor molecules
a.
4.
c.
cooling process (hottest evaporate first, leaving
cooler molecules behind)
d. equilibrium between liquid and vapor
1. evaporation rate = condensation rate in a
closed container
2. concentration of vapor measured as Pvap
3. independent of container size until no liquid
4. Pvap increases at higher temperature because
a. more molecules are in vapor phase
b. vapor molecules exert greater pressure
e. boiling occurs when Pvap = Patm  boiling point
decreases with elevation (lower air pressure Patm)
f. high Pvap indicates volatility—tendency to evaporate
4C. Crystalline Solids (11.8)
Covalent
Metallic
Molecular
Ionic
Network
ion
atom
molecule
ion
Structural Unit
metallic
covalent molecular
ionic
Bond name
strong
weak
strong
Bond strength variable
high
low
high
Melting point variable
low
low
variable
high
Solubility
high
low
low
low
Conductivity
high
low
variable
low
Malleability
Cu
C, SiO2
H2O
NaCl
Example
1.
metallic—metals only
2. expressing concentration
a. attraction between cations and delocalized
a. concentration units [ ]
1. mass percent: % = 102(msolute/mtotal)
valence electrons (electron sea model)
2. mole fraction: Xsolute = molsolute/moltotal
b. melting point: variable ( bond strength)
3. molarity: M = molsolute/Vsolution(L)
c. conductivity: free electrons  yes
4. molality: m = molsolute/msolvent(kg)
d. malleable: non-directional bond  yes
b. conversion of concentration units
e. water solubility: no molecular interactions  no

determine mass or moles of solute and solvent
2. covalent network—nonmetals w/o H or halogen
Unit
Assume
Conclusion
a. atoms covalently bond throughout w/o size limit
gsolute = %
(different than large molecule)
2
% 10 g solution
gsolvent = 100 – %
b. melting point: strong bonds  high
1 mol solute + nsolute = X
c. conductivity: no free electrons  no
X
nsolvent = 1 – X
solvent
d. malleability: bond highly directional  brittle
nsolute = M
e. water solubility: no molecular interactions  no
M
1 L solution
gsolvent = 1000d – (nsolute)MM
3. molecular—nonmetals often with H and/or halogen
nsolute = m
a. attraction between + of one with – of another
m
1 kg solvent
msolvent = 1000 g
b. melting point: weak bonds  low

convert numerator and/or denominator
c. conductivity: no free electrons  no
o mass  volume: m = (d)(V)
d. malleability: non-directional  yes
o mass  moles: m = (n)(MM)
e. water solubility ("like dissolves like")  yes/no
3.
Separation solute and solvent
4. Ionic—metal plus nonmetals
a. filtration: separate solvent from insoluble solute
a. attraction between cations and anions
b. distillation
b. melting point: strong bonds 
1. simple: separate solvent from soluble solid
c. conductivity: no free electrons  no
2. fractional: separate solvent from soluble liquid
(fused or dissolved state is conducting)
5B. Colligative Properties (13.5)
d. malleability: bond highly directional  brittle
1. lower vapor pressure
e. water solubility: ion-dipole interaction  yes
a. solute particles reduce vapor pressure
5A. Solubility (13.1 to 13.4)
b. nonvolatile-nonelectrolyte: Pvap = XsolventPosolvent
1. dissolving process
c. two volatile liquids: Pvap = XAPoA + XBPoB
a. one substance disperses uniformly throughout other
2. higher boiling point and lower freezing point
1. solvent: dissolving medium (usually majority)
a. lowered vapor pressure changes Tb and Tf
2. solute: dissolved in a solvent (usually minority)
b. Tb = Kbmi, Tf = Kfmi
a. ionic or acid = electrolyte (forms ions)
1. m = molality
b. number of free ions = i (van't Hoff factor)
2. i 
c. polar molecules = nonelectrolyte
3. Kb/Kf = molal boiling/freezing point constant
3. solvation (hydration): solute-solvent bonding
c. determination of molar mass of non-electrolyte

a. cation with  side of H2O (O side)
solute by freezing pt. depression
b. anion with  side of H2O (H side)

calculate molality: molality = Tf/Kf
b. saturated solution

calculate molsolute: molsolute = (molality)(msolvent/1000)
1. undissolved solid  dissolved solid

calculate molar mass: MM = msolute/molsolute
2. solution rate = crystallization rate
3.
osmotic pressure
3. maximum amount that dissolves = solubility
a. semi-permeable membrane blocks solute
c. effect of temperature on solubility
b. solvent flows from high [ ] to low [ ]  osmosis
1. solvent kinetic energy is used to break solutec. osmotic pressure () = pressure to stop flow
solute bonds  solute gains energy; solvent
d.  = MRTi (R = 8.31— in kPa or 0.0821— in atm)
lose kinetic energy (cools)
6A. Chemical Reactions (3.1-3.4, 3.7)
2. energy is released when solute-solvent bonds
1. chemical equation
form and turns into kinetic energy of solution
a. coefficients and subscripts
particles (warms up)
H2O—subscript refers to # of atom that precedes it
3. H = Esolute-solute + (-Esolute-solvent)
2 H2O—coefficient refers to # of molecules
a. when |Esolute-solute| > |Esolute-solvent|
b. reactants and products
1. +H (solution cools = endothermic)
1. one directional reaction: reactants  products
2. raising T increases solubility
2. equilibrium reaction: "reactants"  "products"
b. when |Esolute-solute| < |Esolute-solvent|
c. conservation of atoms (mass)—Dalton's Theory
1. –H (solution warms = exothermic)
CH4(g) + 2 O2(g) 
CO2(g)
+ 2 H2O(g)
2. raising T decreases solubility
1C
4O
1C
4H
4. solubility graphs (g solute/100 g H2O)
4H
2O
2O
5. gas solubility generally decreases with
(16
g)
+
(64
g)
=
(44
g)
+
(36
g)
increased temperature because solution
2. types of chemical reactions
depends on solute-solvent bonds, which
a. combustion: CH4 + 2 O2(g)  CO2(g) + 2 H2O(g)
weaken as temperature increases
b. aqueous reactions
d. effect of pressure upon solubility (gas only)
1. ionic compounds in solution exist as separate
1. solubility increases proportionally to partial
ions  MX(aq) is M+(aq) + X-(aq)
pressure above solution Mg = kPg (Mg = mol/L)
2. usually one ion in a compound is unreacted
2. gas in solution  gas in air space  more
a. unreacted ion = "spectator ion"
gas in air space force more into solution
b. usually column 1 cations or NO33. only Pg, not Ptot, will increase solubility
3. "net ionic" equation excludes spectator ions
3.
calculations based on balanced chemical equations
a. coefficients represent moles of formula units
b. flow chart
Given: A
Find: B
Grams of
Grams of
Substance A
Substance B
MM 
MM
Moles of
Moles of
Coefficients 
Substance A
Substance B
M
M
Volume of
Volume of
Solution A
Solution B
c. model calculations
_ g A x 1 mol A x (#) mol B x (MM) g B = _ g B
(MM) g A (#) mol A 1 mol B
_ L A x (M) mol A x (#) mol B x __1 L B__ = _ L B
1LA
(#) mol A (M) mol B
4. limiting reactant (reactant consumed first) and
theoretical yield (maximum product made)

calculate moles of each reactant available

calculate moles of one product based on moles of
each reactant  smallest = theoretical yield

use theoretical yield for remaining calculations

excess reactant = mole present – moles used

percent yield = 100(actual yield/theoretical yield)
6B. Gravimetric Analysis (3.5)
1. mass percent from formula

moles  mass for each element (MM x subscript)

add masses to get total mass

mass % = 100(mass part/total mass)
2. empirical formula

convert g (or %)  moles

divide each mole value by smallest

multiple by factor to make all whole numbers

whole numbers become subscripts

"burning" carbon compounds yield CO2 and H2O
o mole C = mole CO2
o mole H = 2 mole H2O
o mole O = (mCxHyOz – mC – mH)/16
3. molecular formula (given MM)

MM/empirical formula mass = constant

multiple each subscript in empirical formula by
constant = molecular formula
6C. Volumetric Analysis (4.6)
1. make standard solution from stock

moles needed: molestandard = MstandardVstandard
o mass of stock powder, m = (molestandard)MM
o volume of stock solution, V = (molestandard)/(Mstock)
(Mstock)(Vstock) = (Mstandard)(Vstandard)

add to volumetric flask filled ¾ full with distilled water

dissolve

add sufficient distilled water to bring volume to total
2. determine moles of unknown (titration)

add standard solution (titrant) to buret
o rinse buret with standard solution
o clear air pockets
o record initial volume (bottom of meniscus)

add unknown and indicator to flask

add standard solution until color change (equivalence)
o touch tip to flask to release hanging drop
o record final volume (bottom of meniscus)

calculation moles of unknown X
o balance equation to determine molX/molT ratio
o moles of titrant: molT = (MT)(VT)
o moles of unknown: molX = (molT)(molX/molT)
o molar mass of unknown: MMX = mX/molX
o molarity of unknown: MX = molX/VX
7A. Enthalpy (H): Bond Energy (5.3 to 5.5, 8.8)
1. breaking bonds takes energy  chemical system gains
bond energy; surroundings lose energy (heat, etc.)
2. forming bonds releases energy  chemical system
loses energy, surroundings gain energy
3. change in energy called “change in enthalpy”—H
a. when energy required to break bonds > energy
released to form new bonds, +H (endothermic)
1. products at a higher energy state than
reactants (weaker bonds)
2. surroundings lose energy (cool down)
b. when energy required to break bonds < energy
released to form new bonds, –H (exothermic)
1. products at a lower energy state than
reactants (stronger bonds)
2. surroundings gain energy (heat up)
4. thermochemical equation
a. chemical equation with H
1. listed to the right of equation
2. included as reactant (endothermic) or product
(exothermic)
b. H can be used in dimensional analysis process
5. H from calorimetry
a. reactants are put in an insulated container filled
with water, where heat is exchanged between
reactants and water, but no heat is lost
b. by conservation of energy: Hreaction = –Qwater
1. Q = mcT for simple coffee cup calorimeter—
aqueous reactions
a. m = mass of water
b. c = specific heat of water (4.18 J/g•K)
c. T = change in temperature (Tf – Ti)
temperature can stay in oC, since
1 oC = 1 K (don't add 273 to ToC!)
2. Q = (C + mc)T for “bomb" calorimeter
C = “bomb constant” accounts for all nonwater components that change temperature
6. H using bond energy (B.E.) data
a. energy to break a bond (i.e. C–H) in a diatomic,
gaseous molecule, which contains the bond type
1. is approximately the same for any molecule
2. only works for gaseous species
3. positive value (+ B.E.) for breaking bonds
b. forming bonds (– B.E.)
c. H = B.E.reactants – B.E.products
7B. Entropy (S): Disorder (19.2)
1. atoms/molecules have inherent disorder depending on
a. number of atoms—more internal motion = disorder
b. spacing of molecules—farther apart = disorder
c. speed of molecules—faster = disorder
2. predict increase in disorder for physical changes (+ S)
a. spread out: evaporation, diffusion and effusion
(solution: spread out solute and solvent (+S), but
bond solute-solvent (-S)  ?, but usually +S)
b. motion: melting and boiling
3. predict increase in disorder for chemical changes (+ S):
moles gaseous products > moles gaseous reactants
7C. Thermodynamic Data (5.6 to 5.7, 19.4)
1. standard heat of formation (Hfo) data
a. Ho for the formation of one mole of compound
from its elements at standard temperature (25oC)
b. Hfo for elements in natural state = 0.0
c. more negative = more stable (harder to decompose)
2. standard entropy (So) data
a. amount of disorder compared to H+ (simplest form
of matter), which is zero by definition
b. listed in J/mol•K on AP exam, so you will have to
convert to kJ/mol•K for most calculations
3.
calculations using the thermodynamic data chart
a. altering Hfo
1. opposite sign for the reverse reaction
2. multiply by number of moles (coefficient)
b. calculate H for a reaction using Hfo
H Ho = Hfoproducts – Hforeactants
c. calculate S for a reaction using So

S So = Soproducts – Soreactants
7D. Gibbs Free Energy (G): Overall Energy State (19.5 to 19.6)
1. G  Ho – TSo
2. determining if a process is spontaneous (G < 0)
a. lower potential energy (-H)—chemical reactions
b. greater disorder (+S)—physical changes
c. depends on temperature
1. threshold temperature (Tthreshold)
2. occurs when G = 0  Tthreshold = Ho/So
d. summary chart
H
S
Spontaneous Process (G <0)
for temperatures above Tthreshold
+
+
+
–
at no temperatures
–
+
at all temperatures
for temperatures below Tthreshold
–
–
8A. Reaction Rate (14.1 to 14.4)
1. rate = [A]/t
a. A is a general term for reactant or product
b. instantaneous rate = slope of [A] vs. t graph
c. units depend on units for [ ] and t, but usually M/s
d. same reaction rates depend on measured species
2 NO2(g)  2 NO(g) + O2(g)
rate = [NO2]/t = -[NO]/t = -2[O2]/t
2. rate law: rate = k[A]n
a. k = rate constant
1. depends on reaction (larger k = faster)
2. depends on temperature (k increases with T)
b. n = order of reaction (first, second, etc.)
c. multiple reactants: rate = k[A]m[B]n
(overall order = m + n)
d. determined experimentally: process

write rate laws for each experiment

divide rate laws to cancel out all but one reaction order

solve for unknown reaction order

repeat for other reactants

solve for k
e. units for k are Mxt-1, where x = (1 – overall order)
3. summary chart
Order
0
1
2
k
k[A]
k[A]2
rate =
kt =
[A]o – [A]t
ln([A]o/[A]t)
1/[A]t – 1/[A]o
t½ =
[A]o/2k
ln2/k
1/k[A]o
[A] vs. t
ln[A] vs. t
1/[A] vs. t
linear Plot
8B. Collision Model (14.5)
1. effective collision
a. sufficient speed to weaken existing bonds
b. correct orientation to allow new bonds to form
2. potential energy diagram
a. reactant, activated complex and product energies
b. activation energy (Ea)
1. Ea = Eactivated complex – Ereactants
2. positive quantity, which depends on nature of
reactants, but not temperature or [ ]
3. H = Ea - Ea' (Ea’ for reverse reaction)
3. reaction rate and temperature
a. high temperature = faster rate
1. more particles' kinetic energy > Ea
2. more frequent collisions
b.
relation between k and T
1. graph lnk vs. 1/T is straight line: slope = -Ea/R
2. 2 data points: ln(k1/k2) = (Ea/R)(1/T2 – 1/T1)
a. R = 8.31, Ea in J, and T in K
b. k1/k2 = rate1/rate2
8C. Reaction Mechanism (14.6 to 14.7)
1. complex reactions occur in steps (elementary reactions),
where each step has 1 or 2 reactants with low Ea
2. intermediate forms in early step and is consumed in a
subsequent step
3. corresponds to rate law: coefficients of slow step (ratedetermining step) become exponent in rate law
4. catalyst
a. catalyst provides a reaction mechanism that has a
lower Ea than the noncatalyzed reaction
b. homogeneous catalyst
1. consumed in slow step, reappear in fast step
2. example: decomposition of H2O2
step 1: 2 Br- + 2 H+ + H2O2(aq)  Br2(aq) + 2 H2O
step 2: Br2(aq) + H2O2(aq)  2 Br- + 2 H+ + O2(g)
overall:
2 H2O2(aq)  2 H2O + O2(g)
a. catalyst: H+ and Br-—included in rate law
b. intermediate: Br2—not included
c. rate = k[Br-]2[H+]2[H2O2]
c. heterogeneous catalyst written over arrow
9A. Oxidation-Reduction Reactions (4.4, 20.1 to 20.2)
1. reactions involve the transfer of electrons (or control of
electrons) from substance that is oxidized (reducing
agent) to substance that is reduced (oxidizing agent)
Electrons
Process
Agent
Lose
Oxidation
Reducing Agent
Gain
Reduction
Oxidizing Agent
2. oxidation number
a. each atom is assigned an oxidation number which
represents the number of electrons (compared to
a neutral atom) that the atom has control over
1. zero for isolated neutral atom
2. equals ionic charge for monatomic ions
b. overall oxidation number for polyatomic species
1. zero for neutral compound or molecule
2. equals ionic charge for polyatomic ion
c. assign oxidation numbers to atoms within a
molecule or polyatomic ion
1. assign the standard value for the following
a. Li+ ...(+1), Mg2+...(+2), Al (+3), F- (-1)
b. O (-2) except for peroxides, O22- (-1)
c. H (+1) except for hydrides, MHx (-1)
2. determine the missing atom's value by using
the total for the compound (see 2b)
3. balancing redox equations

assign oxidation to each atom (see 2c)

determine oxidized atom (oxidation # increases) and
reduced atom (oxidation # decreases)

split the reaction into an oxidation half-reaction and a
reduction half-reaction

eliminate "spectator" ions (ion that doesn't contain
atom that changes oxidation number—often cation)

balance each half reaction
o balance atoms except O and H
o balance O, by adding H2O
o balance H, by adding H+
o balance charge, by adding e
multiply half-reactions to equalize electrons

add half-reactions together

simplify by reducing H2O and H+ and/or coefficients

this process assumes reaction takes place in acid
(H+), if in base, add an OH- for each H+ in the final
equation (combine H+ and OH- to make water)
4.
reduction half reactions of common oxidizing agents
a. MnO4- + 8 H+ + 5 e-  Mn2+ + 4 H2O
b. Cr2O72- + 14 H+ + 6 e-  2 Cr3+ + 7 H2O
5. oxidation of reactive metals
a. metal ion with greater Eored takes e- from metal
metal + salt  salt of metal + new metal
Zn(s) + Cu(NO3)2(aq)  Zn(NO3)2(aq) + Cu(s)
net ionic: Zn(s) + Cu2+  Zn2+ + Cu(s)
b. H+ ion takes e- from metal with negative Eored
metal + acid  salt of metal + hydrogen gas
Zn(s) + 2 HCl(aq)  ZnCl2(aq) + H2(g)
net ionic: Zn(s) + 2 H+  Zn2+ + H2(g)
c. hydrogen in water takes electron from column 1
and heavier column 2 metals
2 Na(s) + 2 H2O  2 Na+ + 2 OH- + H2(g)
Ca(s) + 2 H2O  Ca2+ + 2 OH- + H2(g)
6. redox with metal hydride and water (Hhydride (ox # = –1)
reacts with the Hwater (ox # = +1) to form H2 (ox # = 0)
NaH(s) + H2O  Na+ + OH- + H2(g)
CaH2(s) + 2 H2O  Ca2+ + 2 OH- + 2 H2(g)
9B. Standard Reduction Potentials Chart (20.4 to 20.6)
1. reduction half reactions
a. listed from |greatest| electron affinity to |least|
b. 2 H+ + 2 e-  H2: Eored = Eoox = 0 V
c. Eo measured in volts, 1 V = 1 J/C
1. "o": standard conditions (25oC, 1 atm, 1 M)
2. not proportional to amount of chemical
d. oxidation is reverse (Eoox = -Eored)
2. Eo = Eored + Eoox
a. Eo > 0 is a spontaneous reaction
(reduction listed above oxidation on chart)
b. Go = –nFEo (in joules)
1. n: # e- in balanced redox equation
2. F: faraday = 96,500 C/mol ec. voltage under nonstandard conditions
1. Nernst equation: E = Eo – (RTo/nF)lnQ
R (8.31), To (298) and F (96,500) are constant
E = Eo – (0.0257/n)lnQ
2. Q (quotient) = Product/Reactants
a. gas (atm), ions (M)
b. solids and liquids excluded
9C. Voltaic (Galvanic) Cell (20.3)
spontaneous redox reaction generates voltage  electrons
flow through wires from oxidation cell to reduction cell
anode
(–)
Voltage > 0
salt bridge
cathode
(+)
porous membrane
site of oxidation
site of reduction
1. oxidation half cell (– anode)
a. reducing agent (|lower| electron affinity) gives up
electrons to external circuit (wires)
b. anions flow toward anode through salt bridge/
porous membrane to maintain electrical neutrality
2. reduction half cell (+ cathode)
a. oxidizing agent (|higher| electron affinity) attract
electrons from external circuit (wires)
b. cations flow toward cathode through salt bridge/
porous membrane to maintain electrical neutrality
3. predict how change affects standard voltage
a. [P] > [R]: E < Eo
1. Nernst equation: -RT/nFlnQ is negative
2. Le Chatelier's Principle: stressed system shifts
towards reactants (E decreases)
b. size of electrode and chamber: no change
c. remove salt bridge: E = 0
9D. Electrolytic Cell (20.9)
battery forces non-spontaneous redox reaction by pulling
electrons from reducing agent and sending to oxidizing agent
anode
(+)
+ Battery –
cathode
(–)
site of oxidation
site of reduction
1. Eo < 0 (battery makes up for deficit)
2. oxidation at + anode, reduction at – cathode
3. electrolysis in water solutions (inert electrodes)
a. cathode reduction: H2O or cation (which ever one is
higher on the standard potential chart)
1. columns 1, 2 or Al3+: 2 H2O + 2 e-  H2 + 2 OH2. acid (H+): 2 H+ + 2 e-  H2
3. otherwise: Mx+ + X e-  M
b. anode oxidation: anion or H2O
1. Cl-, Br-, I-: 2 X-  X2 + 2 e2. base (OH-): 4 OH-  O2 + 2 H2O + 4 e3. otherwise: 2 H2O  O2 + 4 H+ + 4 e4. electroplating (transition metal cations coat cathode)
a. current, I, measured in amperes (amps—A)
1 A = 1 C/s (coulomb/second)
b. mass plated given current, I, and time, t
(t) s x (I) C x mol e- x mol Mx+ x (MM) g = __ g
1 s 96,500 C X mol e- 1 mol Mx+
c. time for plating—calculate right to left
10A. The Equilibrium State (15.1 to 15.6)
1. N2O4(g)  2 NO2(g)
a. reversible because reactants and products are
confined in same state and Ea is relatively small
b. reaction rates change as [R] and [P] change until
kf[N2O4] = kr[NO2]2
EQUAL REACTION RATES NOT CONCENTRATIONS
2. equilibrium expression
a. law of mass action: N2O4(g)  2 NO2(g)
kf[N2O4] = kr[NO2]2  K = kf/kr = [NO2]2/[N2O4]
b. liquid and solid are not included
c. equilibrium constant, K
1. unit-less quantity
2. independent of original concentrations
3. depends on temperature
4. Kc vs. Kp for gaseous systems
a. concentration in mol/L, then Kc
b. concentration in atm, then Kp
c. Kp = Kc x (RT)n(gas)—R = 0.0821
5. size of K and the reactant/product balance
a. large K, means [products] > [reactants]
b. small K, means [products] < [reactants]
6. if reaction is written backwards, then: K' = K-1
7. if coefficient x factor "n", then K’' = Kn
d. adding equilibria: multiply K
e. K, Go and Eo
1. equilibrium: GE = EE = 0, but Go and Eo  0
2. "o" [ ] = 1 M and P = 1 atm (not always 298 K)
3. Go = -RTlnK and Eo = (RT/nF)lnK
(R = 8.31, T in K, n = moles e-, F = 96,500)
4. when K > 1, then Go < 0, and Eo > 0
10B. Le Chatelier's Principle (15.7)
1. disturbed an equilibrium, system shifts to reduce
disturbance
2. analysis of shift using Le Chatelier’s principle
a. change amount of reactants
1. [R]  causes system to shift away ()
a. [ ]E of products and added reactant will 
b. [ ]E of other reactants will 
2. decrease reactant: opposite response ()
3. value of K remains unchanged
b. change amount of products
1. [P]  causes system to shift away ()
a. [ ]E of reactants and added product 
b. [ ]E of other products will 
2. decrease product: opposite response ()
3. value of K remains unchanged
c. change temperature of the container
1. changing temperature changes forward and
reverse reaction rates, which changes the
equilibrium position
a. T  increases endothermic reaction rate
more than exothermic reaction rate 
system shifts in endothermic direction
b. T  increases endothermic reaction rate
more than endothermic reaction rate 
system shifts in exothermic direction
2. change in equilibrium position = change in K
a. shift to the right increases K
b. shift to the left decreases K
3. generalization
a. endothermic: T  K, T   K
b. exothermic: T  K, T  K
d. change volume of the container
1. generalization: decrease volume: system
shifts to the side with the fewer moles of gas
(increase volume, shifts toward more moles)
2.  V =  P  shift away from moles of gas
 V =  P  shift toward moles of gas
3. adding inert gas does not shift the equilibrium
4. moles gas reactants = products: no response
e. catalyst affects rates but not equilibrium position
11A. Ionic Compounds (2.7 to 2.8, 4.2)
1. name and formula
a. cations (charge based on periodic table position)
1. named the same as metal (Na+ = sodium)
2. NH4+ (ammonium), H3O+ (hydronium)
3. assume all transition metals are 2+
a. use Roman numeral for multiple cations
b. important exceptions: Cr3+, Ag+
b. anions
1. –ide ending
a. monatomic: Cl- (chloride)
b. binary: OH- (hydroxide), CN- (cyanide)
2. oxyanion (–ate and –ite ending)
C2H3O2acetate
MnO4permanganate
2CO3
carbonate
NO3nitrate
CrO42chromate
PO43phosphate
Cr2O72dichromate
SO42sulfate
a. use "bi" prefix when H is added to anion:
HCO3- (bicarbonate). HSO4- (bisulfate)
b. use "ite" suffix when one O is removed:
SO32- (sulfite), NO2- (nitrite)
c. when non-oxygen atom is a halogen:
ClO4ClO3ClO2ClOperchlorate
chlorate
chlorite
hypochlorite
c. empirical formula (criss-cross method)
2. predicting solubility
Anions (X)
Cations
(M)
NO3- Cl-, Br-, I- SO42- OH-, S2- Others
Alkali Metal
S
S
S
S
S
NH4+
S
S
S
S
S
Sr2+,Ba2+
S
S
I
S
I
Ag+
S
I
S
I
I
Hg22+, Pb2+
S
I
I
I
I
Others
S
S
S
I
I
11B. Solubility Equilibrium (17.4-17.5)
1. ionic compound (salt) equilibrium with its ions
a. MmXn(s)  m Mn+(aq) + n Xm-(aq)
b. Ksp = [Mn+]m[Xm-]n (Ksp or mass action expression)
11C. Factors that Affect Solubility (17.5)
1. common ion effect: less soluble with common ion
2. addition of acid: more soluble in H+ (except Cl-, SO42-)
3. formation of complex ions: Cu2+ + 4 CN-  Cu(CN)42Kf = [Cu(CN)42-]/[Cu2+][CN-]4
4. determine equilibrium position: Koverall = K1 x K2 > 1
5. amphoterism: some metal oxides and hydroxides are
soluble in both strong acid and strong base
12A. Acids and Bases (16.1 to16.11)
1. three theories
a. Arrhenius: acid (H+), base (OH-): H+ + OH-  H2O
b. Brønsted-Lowry: acid (HA), base (:B-): H+ transfer
c. Lewis: acid (M+) base (:B-): electron pair transfer
2. acids
a. neutral molecule made from H+ + anion = HA
b. name based on anion name
1. –ide anion: acid name is Hydro___ic acid
2. –ate anion: acid name is ___ic acid
3. –ite anion: acid name is ___ous acid
c. when A- is Cl-, Br-, I-, NO3-, ClO3-, ClO4, SO421. % ionization = [H+]/[HA]o x 100 = 100 %
2. strong acid = strong electrolyte
d. when A- is any other ion—weak acid (electrolyte)
1. % ionization < 5 %
2. organic acids, HC2H3O2 or CH3COOH
3. A- is oxyanion, HXOy (HClO, HBrO2, H2SO3)
a. H is bonded to O
b. weaker O-H bond = stronger acid
1.  electronegativity  (HClO > HBrO)
2.  oxygen  (HClO2 > HClO)
c. nonmetal oxides: CO2 + H2O  H2CO3
e. polyprotic acids (HxA)—first H+ is easiest to
remove from neutral molecule than from anions
f. acid dissociation/ionization
1. HA(aq)  H+ + A2. Ka = [H+][A-]/[HA] = [H+]2/[HA]
3. acid strength: (larger Ka = stronger acid)
4. polyprotic acids (HxA)
a. H2A  H+ + HA-: Ka1 = [H+]2/[H2A]
b. HA-  H+ + A2-: Ka2 = [H+][A2-]/[HA-]
[HA-] = [H+]  ka2 = [A2-]
3. bases
a. hydroxides (:OH-)
1. soluble: column 1, Ca2+, Sr2+, Ba2+ = strong
2. oxides: M2O(s) + 2 H2O  2 M+ + 2 OHb. ammonia or amines (:NH3, :NH2CH3, etc.)
NH3(aq) + H2O  NH4+ + OHKb = [NH4+][OH-]/[NH3] = [OH-]2/[NH3]
c. anions from weak acids (:A-)
A- + H2O  HA(aq) + OHKb = [HA][OH-]/[A-] = [OH-]2/[A-]
4. autoionization of water: H2O  H+ + OHa. Kw = [H+][OH-] = 1 x 10-14, [H+] = [OH-] = 1 x 10-7 M
b. pH scale
1. pH = -log [H+], pOH = -log[OH-]
2. pH + pOH = pKW = 14
5. pH of ionic compounds (salts) in water—hydrolysis
Strong Acid polyprotic anion
Others
Strong Base
neutral
acidic
basic
Others
acidic
acidic
?
6. acid/base reactions
a. Arrhenius neutralization reaction (produces H2O)
1. HA + MOH: H+ + OH-  H2O
2. HA + M2O: 2 H+ + Na2O(s)  2 Na+ + 2 H2O
b.
Brønsted-Lowry proton transfer reaction
1. stronger acid + base  weaker acid + base
2. examples (conjugate pairs)
acid + base  acid
+ base
K
HA
+
H2O  H3O+ +
A1/Ka
H3O+ +
A
HA
+ H2O
Ka
 H2O
HA
+
OH+
A1/Kb
H2O +
A
HA
+
OHKb
-14
Ka x Kb = Kw = 1 x 10
3. amphiprotic: can act as a proton donor and
proton acceptor, depending on other reactant
12B. Buffer Systems (17.2)
1. mixture of weak acid or base with its conjugate
2. addition of strong acid or base to buffer
a. addition of acid: A- + H+  HA(aq)
b. addition of base: HA(aq) + OH-  A- + H2O
c. process is reversible, but not equilibrium
3. determine [H+] or [OH-] after addition of acid or base

determine original moles of HAo, A-o
o no-HA = Vsolution x [HA]o
o n-o-A = Vsolution x [A-]o

determine moles of H+ or OH- added
o nH+ = (VH+)(MH+)
o nOH- = (VOH-)(MOH-)

calculate equilibrium moles of HAE and A-E
o nE-HA = no-HA – nOH- or no-HA + nH+
o nE-A = no-A + nOH- or no-A – nH+

calculate [H+] or [OH-]
o moles buffer > moles H+/OH-: [H+] = Ka(nHA/nA-)
o moles buffer < moles H+/OH
[H+] = (nH+ – no-A)/Vtot

[OH-] = (nOH- – no-HA)/Vtot
12C. Acid-Base Titration (17.3)
1. acid or base is added to a fixed amount of base or acid
a. pH is monitored using a pH meter
1. equivalence: nH+MaVa = nOH-MbVb
2. end point when indicator changes color
b. buffered solution during incomplete neutralization
2. graphs
a. equivalence: middle of vertical section (pH = 9)
b. buffer region: mini-plateau (10 mL to 40 mL)
3. [H+] or [OH-] calculation chart
1. SA + SB
2. WA + SB
3. WB + SA
[H+] =
[H+] =
[OH-] =
initial
[HA]
(Ka[HA])½
(Kb[B])½
buffer
[H+] =
[H+] =
[OH-] =
(2 and 3) (nHA - nOH-)/Vtot
Ka(nHA/nA-)
Kb(nB/nHB+)
[H+] =
[OH-] =
[H+] =
equivalence
1 x 10-7 M
(Kb[A-])½
(Ka[HB+])½
[OH-] =
[OH-] =
[H+] =
excess
(nOH- - nHA)/Vtot (nOH- - nHA)/Vtot (nH+ - nB)/Vtot
1.
2.
3.
4.
5.
6.
Equilibrium Problems
Determine direction ( or  from [ ]o

substitute [ ]o into equilibrium expression = Q

if Q > K, then , if Q < K, then 
example: Determine if a precipitate will form, given [ ]o
(precipitate when Q > Ksp. no precipitate when Q < Ksp)
Determine K, given [ ]E: write expression, substitute [ ]E and
solve for K
Determine K, given [ ]o and one [ ]E
[]
A
+
2B

C
+
3D
[A]o
[B]o
[C]o
[D]o
I
C
-x
-2x
x = [C]E – [C]o
3x
E
[A]o - x
[B]o – 2x
[C]E
[D]o + 3x
write expression, substitute [ ]E and solve for K
example: Determine Ksp, given solubility (s)
[]
MmXn

m Mn+
+
n Xm0
0
I
+m•s
+n•s
C
m•s
n•s
E
example: Determine Ka, given pH or [H+]E and [HA]o
[]
HA

H+
+
A[HA]o
0
0
I
C
–[H+]E
+[H+]E
+[H+]E
E
[HA]o – [H+]E
[H+]E
[H+]E
example: Determine Kb, given pOH or [OH-]E and [B]o
[]
B
+
H 2O

HB+ +
OH[B]o
0
0
I
+[OH-]E
+[OH-]E
C
–[OH-]E
[OH-]E
[OH-]E
E
 [B]o
Determine one [ ]E, given other [ ]E and K
write expression, substitute [ ]E and solve for missing [ ]E
Determine [ ]E, given [ ]o and K
[]
A
+
2B

C
+
3D
[A]o
[B]o
[C]o
[D]o
I
-x
-2x
+x
+3x
C
[A]o – x
[B]o – 2x
[C]o + x
[D]o + 3x
E
write expression, substitute [ ]E and K, solve for x,
substitute x and solve for [ ]E
example: Determine solubility (mol/L) “s”, given Ksp
[]
MmXn

m Mn+
+
n Xmn+]
0
or
[M
0
or
[Xm-]o
o
I
+m•s
+n•s
C
m•s
n•s or [Xm-]o
E
2
3
(MX: Ksp = s , MX2: Ksp = 4s , MX3: Ksp = 27s4)
example: Determine [H+]E, given [HA]o, [A-]o and Ka
[]
HA

H+
+
A[HA]o
0
0 or [A-]o
I
C
–x
+x
+x
x
x or [A-]o
E 
o (x < 5%)
example: Determine [OH-]E, given [HA]o, [A-]o and Kb
[]
A+
H2O

HA
+
OH[A ]o
0 or [HA]o
0
I
+x
+x
C
–x
x or
x
 [A-]o
E
[HA]o
Determine [A-]o/[HA]o given [HA]o/[A-]o pHE/pOHE and Ka/Kb
given Ka
[]
HA

H+
+
A[HA]o
0
x
I
C
–[H+]E
+[H+]E
+[H+]E
[H+]E
E
 [HA]o
x
given Kb
[]
A+
H2O

HA
+
OH[A ]o
x
0
I
+[OH-]E
+[OH-]E
C
–[OH-]E
E
 [A-]o
x
[OH-]E