AP Chemistry: Total Notes Review
Table of Contents
1. Stoichiometry
2. Electronic Structure
3. Bonding and Molecular Structure
4. Intermolecular Forces
5. Properties of Solutions
6. Kinetics
7. Equilibrium
8. Thermodynamics
9. Electrochemistry
10.
Nuclear Chemistry
11.
Organic Chemistry
12.
Reaction Writing
#1: STOICHIOMETRY
o Found in Chapters 3 and 4 of the textbook
o Limiting Reactants: divide # moles by whole number coefficients – the
smallest is the limiter (use that to find the resulting moles of product)
o 4 basic steps:
1: write the balanced chemical equation
2: convert to moles
3: use the mole: mole ratio
4: convert to asked for units
But that’s for easy stuff like, “determine the number of [moles/grams/liters/etc.]
in [substance] for [number and units] of [element] according to the equation
below”
If you want to get to the really hard-core stuff, look at”
Empirical Formulas and Analyses
This stuff tells you what the empirical formula of a reactant is based on its
products. For example:
“Ascorbic Acid contains 40.92% H and 54.50% O by mass. What is the empirical
formula of ascorbic acid?”
How to Solve:
1. Assume that there’s a 100g sample of the substance (that means that the
percents equal their mass)
2. convert samples to moles
3. divide each by the smallest number of moles
4. make a ratio
5. use the ratio to write the equation
Take a Look:
Sometimes with analysis, you have to find the molecular formula. That’s like the
empirical formula, but it’s not in its lowest ratio. For example:
“Mesitylene, a hydrocarbon that occurs in small amounts in crude oil, has an empirical
formula of C3H4. The experimentally determined molecular weight of this substance in
121 amu. What is the molecular formula of mesitylene?”
How to Solve:
1. find the empirical formula weight (use the formula given in the question)
2. put numbers into the equation
whole # multiple = molecular mass / empirical mass
3. solve
4. multiply subscripts by whole numbers
Take a Look:
It’s important to remember that chemistry doesn’t always do the problems the way
examples show them. Most of the time you have to manipulate equations and whatnot
using the techniques you learn from the examples. Look out for questions that combine
these two concepts or otherwise flip them around.
Combustion Analysis
This is a lot like the two things above, but it’s different at the same time (that’s chem. For
you). Okay, here’s what it is: the question will tell you the products of the combustion of
usually a hydrocarbon (there are about a million different kinds, so be prepared. Oh, and
look at the organic chem. Chapter if you don’t believe me). For example:
“Isopropyl alcohol, a substance sold as rubbing alcohol, is composed of C, H, and O.
Combustion of 0.255g of isopropyl alcohol produces 0.561g of CO2 and 0.306g of H2O.
Determine the empirical formula of isopropyl alcohol.”
How to Solve:
1. Find the mass of C and H in the substance:
i.
convert CO2 and H2O to moles
ii.
use mol: mol ratios to isolate C and H
iii.
change moles to grams
2. subtract the mass of C and H from the total mass to find the mass of O
3. convert grams to moles
4. divide by the smallest
Take a Look:
Limiting Reactants
Might as well do one of these so you don’t screw it up on the test. These have a reactant
that limits the amount of product produced because one of the reactants runs out. For
example:
“A 55.8g sample of pure iron is submerged in a solution containing 339.0g of silver
nitrate. Fe(NO3)3 and solid silver are the products. a) write the balanced chemical
equation b) determine the limiting reactant c) determine the moles of each product
produced d) determine the mass of excess reactant remaining”
How to Solve:
1. write the balanced chemical equation
2. convert reactants to moles
3. find the limiter (divide # of moles by the coefficient and the smallest is the
limiter)
4. use the limiter in an I.C.E. table to find the moles of products and excess reactant
remaining
5. convert to units
Take a Look:
Theoretical Yeilds
The quantity of product that is calculated to form when all of the limiting reactant reacts
is the theoretical yield. The amount that actually reacts is the actual yield.
Percent yield = actual amount / theoretical yield x 100
It’s not too hard, so we won’t do an example.
Solution Stoichiometry (AKA: titrations)
Use this to determine how much (volume or concentration) of a particular ion there is in a
solution.
Equivalence point: pretty much the most important part of a titration, it’s when the
moles of “A” equal the moles of “B.” It’s also called the “neutralizing point.”
Use the same steps as every other stoich. problem to solve these (they just look scarier).
For example:
“A sample of iron ore is dissolved in acid, and the iron is converted to Fe+2 . The sample
is titrated with 47.20 mL of 0.02240 M MnO4- solution. The oxidation-reduction reaction
that occurs during the titration is as follows:
a)
b)
c)
d)
how many moles of MnO4- were added to the solution?
How many moles of Fe+2 were in the solution?
How many grams of iron were in the sample?
If the sample had a mass of 0.8890g, what is the percentage of iron in the sample?
How to Solve:
Use the basics:
1. use the balanced equation
2. convert to moles (MnO4-)
3. use mol to mol ratio (Fe+2)
4. convert to units
Take a Look:
#2: ELECTRONIC STRUCTURE
o chapters 6: Electronic Structure of Atoms
Terms to Know:
o Electromagnetic Radiation (radiant energy): moves through a vacuum at the
speed of light (c = 3 x 108) Includes: Gamma rays, X rays, Ultraviolet, visible
light, infrared, microwaves, and radio frequency
o Electronic structure: describes the energies and arrangements of electrons
around an atom
Electrons behave as both particles and waves (wavicles)
o Wave characteristics described with wavelength (λ) and frequency (ν)
λxν=c
o Photon: wavicle
o h = Planck’s constant (6.62 E -34 J/s)
o quantum theory: energy is quantized; it can only have certain values
o Photoelectric Effect: emitting electrons from a metal’s surface by light
(proposing that light behaves as if it consists of quantized packets – photons)
o minimum amount of energy that an object can gain or lose is related to the
frequency of radiation
E=hxν
For example:
“A yellow Na vapor lamp (street light) has a beam with λ = 589 nm. Calculate a)
the frequency of the light b) the energy in a mole of photons of the light c) how
many photons are in a 10 mJ burst of this light.”
How to Solve:
1: use λ x ν = c to solve for frequency
2: use E = h x ν to solve for J/mol
3: convert 10 mJ to J and use ratios to find # of photons
Take a Look:
More terms:
o Spectrum: the result of dispersion of radiation into its component
wavelengths
o Continuous spectrum: a spectrum containing all of the wavelengths
o Line spectrum: a spectrum that only contains certain wavelengths
o Bohr proposed a model explaining H’s line spectrum:
~ energy of H depends on n (the quantum number)
~ n must be a positive integer, each value corresponding to a different
specific energy
~ energy of an atom increases as n increases
~ ground state: n = 1
~ if an electron drops from a lower energy state to a higher one, it emits
light; light must be absorbed to do the opposite
~ frequency of light emitted must be such that hv equals the difference of
energy between the two allowed states of the atom
Equations that represent the movement of an electron from one “allowed
energy state” to another
E = -2.18E-18
n2
1 = (RH)( 1 - 1 )
λ
n2i n2f
where n is a whole # (principle quantum #)
Rydgerg eq.: calc. wavelength of the spectral line of H)
RH = 1.096776 E 7 m-1 (Rydberg constant)
Some questions that might come up:
“Calculate the energy of an electron in the H atom when n = 3”
How to Solve:
Plug and chug in Bohr’s equation.
Take a Look:
“Find the wavelength of radiation emitted when an electron moves from n = 3 to n
= 1 in the H atom. What region of the electromagnetic spectrum is the emitted
radiation in?”
How to Solve:
Plug and chug in Rydberg equation
Take a Look:
More and More Terms:
o De Broglie: matter (like electrons) should have wave-like properties
o The characteristic wavelength of an object depends on momentum:
λ = h / m x v (m = mass in kg; v = velocity)
(more mass, less visible wavelength)
o Uncertainty principle: limit to accuracy with which the position and
momentum of a particle can be measured at the same time
For example:
“What is the wavelength of an electron moving with a speed of 5.97 E 6 m/s?
(electron mass = 9.11 E -28g)
How to Solve:
Plug and chug into equation.
Take a Look:
Orbitals and all their Wonder
o Wave Functions: ψ; behavior of the electron in the quantum mechanical
model of the H atom
~ each has a precisely known energy
~ location can’t be exactly determined
o
Probability density: ψ2; the probability that the electron will be at a
particular point in space
~ Electron density: a map of the probability of finding the electron at all
points in space
o Orbitals: the allowed wave functions of an H atom
~ combo. of an integer and a letter
~ n: principle quantum number (positive integer)
~ l: azmuthal quantum number (s: 0, sphere; p: 1, dumbbell; d: 2, flower;
f: 3, rare), defines the shape of the orbital
~m1: magnetic quantum number; (-1 to 1); describes orientation of the
orbital
~m2: spin magnetic quantum number; (+1/2 or -1/2), two directions of an
electron spinning about an orbital axis
o Electron shell: the set of all orbitals with the same value n ex) 3s, 3p, 3d
o Subshell: set of one or more orbitals with the same n and l values ex) 3s, 3p,
and 3d are each subshells of the n = 3 shell
~ there is one orbital in an s subshell
~ 3 orbitals in a p subshell
~ 5 orbitals in a d subshell
~ 7 orbitals in an f subshell
o Radial probability function: the probability that an electron will be a certain
distance away from the nucleus
o Node: an area where there is 0 probability that the electron will be found
Electron Configuration and the Periodic Table
o Different subshells of the same shell have different energies
~ For a given value of n, the energy of the subshells increases as the value of l
increases (ns < np < nd < nf)
~ degenerate: orbitals within the same subshell, hence the same energy
o Pauli exclusion principle: no two electrons in an atom can have the same values
for n, l, m1, and ms
~limits the amount of electrons that can occupy an orbital to 2
o Electron configuration: describes how the electrons are distributed among the
orbitals fo an atom
o Hund’s Rule: the lowest energy is attained by maximizing the number of electrons
with the same electron spin (so don’t pair up until an electron of the same spin
occupies each orbital)
o Elements in the same group on the Periodic Table have the same type of electron
arrangement in their outermost shells ex) F, [He]2s22p5; and Cl[Ne]3s23p5
~ Outer-shell electrons: those that lie outside the orbitals occupied in the nextlowest noble gas element ex)[He]2s22p5
~ Valence electrons: outer-shell electrons involved in chemical bonding (elements
with atomic number 30 or less, all outer-shell electrons are valence electrons)
~ Core elements: electrons that aren’t valence electrons
o Representative (main group) elements: elements in which the outermost subshell
is an s or a p subshell (alkali metals and noble gases)
o Transition elements: elements in which the d subshell is being filled
o Lanthanide (rare earth) elements: elements in which the 4f subshell is being filled
o Actinide elements: elements in which the 5f subshell is being filled
~ Lanthanide and Actinide elements: f-block metals
You can use these facts to utilize the Periodic Table to write electron configurations
of elements.
Example:
“Write the electron configuration of Al.”
How to Solve:
1: write the n value that’s closest to the nucleus
2: write the l value
3: Go to the next n value and write the l value
4: repeat steps until you reach Al on the Periodic Table
Take a Look:
This tells you that there are: 2 electrons in the 1s sublevel, 2 electrons in the 2s sublevel,
6 electrons in the 2p sublevel, 3 electrons in the 3s sublevel, and 1 electron in the 3p
sublevel.
Effective Nuclear Charge
o Valence Orbitals: elements in the same group have the same number of electrons
in their valence orbitals
~ there are differences between elements of different groups because their valence
orbitals are in different shells
o Effective Nuclear Charge: a representation of the averaged electrical field
experienced by the electron; the average environment created by the nucleus and
other electrons in the atom; expressed as Zeff
~ As you get closer to the nucleus, Zeff increases; as you get farther away, there
are more electrons “shielding” the one on the outside, so it experiences a smaller
Zeff
~ P.T. trend: left to right -> decreased Zeff (more electrons shielding)
So, F has a lower effective nuclear charge than C
Sizes of Atoms and Ions
o Bonding atomic radius: used to determine the size of an atom
~ P.T. trend: radii increase as you go down a column (increase n value) and
decrease as you go right across a period (fewer protons in elements on the left, so
smaller Zeff)
~ the charge of the nucleus on an atom or monatomic ion is the same as the
atomic number
o Cations: smaller than parent atoms (lose an electron)
o Anions: bigger than parent atoms (gain an electron)
o Ions with the same charge: size increases as you go down a column
o Isoelectronic Series: a series of ions with the same number of electrons
~ size decreases with increasing nuclear charge (electrons are more strongly
attracted to the nucleus)
~ Larger ions have lower atomic numbers
Ex) Na+1, Mg+2, Al+3, F-, and O-2
For example:
“Arrange these atoms and ions in order of decreasing size: Mg+2, Ca+2, and Ca”
How to Solve:
1: know that cations are smaller than parent atoms, so Ca+2 is smaller than Ca
2: Ca is under Mg on the periodic table, so Mg+2 is bigger than Ca+2
3: write it out
Ca > Mg > Ca+2
Ionization Energy
o Ionization energy: the minimum amount of energy needed to remove an electron
from an atom in its gaseous phase (forms a cation)
~ first ionization energy: taking the outermost electron (or first electron)
~ second ionization energy: removing a second electron
~ always positive (need to put energy into the system to take the electron out)
~ sharp increase in ionization energy after removing all of the valence
electrons (higher effective nuclear charge)
~ P.T. Trend: increase as you go left to right and decrease as you go down
(because smaller atoms’ electrons are closer to the nucleus) (opposite of atomic radii)
o Writing electron configurations for ions: write the configuration of the neutral
atom (see page 11), then remove or add the right number of electrons
~ add electrons to orbitals with the lowest value n
~ remove electrons from the highest value n
For example:
“Arrange the following in order of decreasing first ionization energy: Ne, Na, P, Ar, and
K”
How to Solve:
1: locate each element on the periodic table
2: Na, P, and Ar are in the same period, so they must vary in the order Na < P < Ar
3: Ne is above Ar, so Ar < Ne
4: K is below Na, so K< Na
5: write it out
Ne > Ar > P > Na > K
“Write the electron configuration of Ca+2”
How to Solve:
1: Write out parent atom’s electron configuration: [Ar]4s2
2: Remove 2 electrons from orbitals with the highest value n
[Ar] or 1s22s2p63s2p6
Electron Affinities
Electron affinity: the energy change upon adding an electron to an atom in the gas phase
(opposite of ionization energy)
~ more negative electron affinity means the anion’s more stable
~ P.T. Trend: more negative as you go left to right (until you get to noble gases)
and more negative as you go up
~ Noble gases are always positive because they would have to gain a new subshell
to gain an electron
~ Halogens have the most negative electron affinities (Cl is the most negative)
~ Enthalpy measurement: ΔH = kJ/mol
#3: BONDING AND MOLECULAR
STRUCTURE
o Chapters 8 and 9
Chemical Bonds
3 types:
1: Ionic
2: Covalent
3: Metallic
Ionic Bonding
o Electrostatic attraction between ions (one takes the other’s electron)
o Metal + metal, metal + non-metal, metal + polyatomic ion, or 2 polyatomic ions
o Lattice Energy: the energy required to completely separate a mole of a solid ionic
compound into its gaseous ions
Coulomb’s Law:
Eel = K Q1 x Q2
d
where Q is charge and d is distance between ions
~ increases with the charge of the ions and decreasing size of ions (charge plays
bigger role, though)
For example:
“Arrange the compounds from least to greatest lattice energy: NaF, CsI, CaO”
How to Solve:
1: look at charges (Na+, F-; Cs+, I-; Ca+2, O-2)(CaO has more than all, because stronger
charges)
2: Compare distances (Cs is bigger than Na, and I is bigger than F, so NaF > CsI)
CsI < NaF < CaO
Covalent Bonding
o Atoms share electrons (usually 2 non-metals)
o Electronegativity: how much an atom “wants” an electron
~ P.T. Trend: increases from bottom left to top right
~ differences in electronegativity between bonded things causes polarity
o Polarity: unequal sharing of electrons
~ greater the electronegativity difference, the more polar
Lewis Electron Dot Structures
o Show molecules with all electrons, bonded and non-bonded
o Steps to Making one:
1: Sum the valence electrons (use Periodic Table) (anion – add electron; cation – subtract
electron)
2: make the central atom the least electronegative one
3: Fill the octets
4: use double and triple bonds as necessary
o Formal charges: subtract the amount of electrons on the periodic table (for that
element) from the electrons you drew in
~ 0 means right on
~ the negative charge should be on the most electronegative atom
o Resonance: when one Lewis structure can’t accurately describe a molecule (due
to something like a double bond that “resonates” between atoms as in ozone or
benzene); the electrons in the shifting bond are delocalized
o Exceptions to the octet rule:
1: Odd number of electrons: rare, but you’ll just have to cope with it
2: Fewer than 8 electrons: if filling the octet of the central atom results in a negative
charge on the central atom and a positive charge on the more electronegative outer
atom, don’t fill the octet of the central atom (as in BF3)
3: More than 8 electrons: atoms in the 3rd row or below on the Periodic Table can
have expanded octets due to empty d orbitals; always check formal charges to make
sure it all works
Bond Enthalpy
o Strength measured by how much energy is required to break the bond
o Called: Bond Enthalpy
o Always positive because bond breaking is endothermic
Enthalpies of Reaction:
Estimate ΔH: ΔH = Σ(bond enthalpies of reactants (or broken)) – Σ(bond enthalpies
of products (or formed))
Molecular Shapes
o Shape plays a key role in reactivity
o Can predict the shape of a molecule by noting the number of bonded and nonbonded pairs of electrons
~ electron domains: pairs of electrons
o VSEPR (Valence Shell Electron Pair Repulsion Theory): the best arrangement
of a given number of electron domains is the one that minimizes the
repulsions among them
Electron
Domains
Non-bonded
domains
Bonded Domains
Bond Angle
2
3
3
4
4
4
2
3
2
4
3
2
0
0
1
0
1
2
5
5
5
5
6
6
6
5
4
3
2
6
5
4
0
1
2
3
0
1
2
180
120
<120
109
107
104.5
90, 120
<120 equit. 90
90
180
90
<90
90
Name
linear
trigonal planar
bent
tetrahedral
trigonal pyramidal
bent
trigonal
bipyramidal
seesaw
t-shape
linear
octahedral
square pyramidal
square planar
o Polarity: just because a molecule has polar bonds, doesn’t mean the whole
molecule is polar; you can tell whether a molecule is polar or not by its symmetry
Hybridization
o Overlap: increased overlap brings electrons closer together while simultaneously
decreasing electron-electron repulsion
~ BUT: if atoms get too close, the internuclear repulsion greatly raises the
energy
o Once you know the electron-domain geometry, you know the hybridization state
of the atom
(count on your fingers to find out the hybridization: sp, sp2, sp3, sp3d, sp3d2)
o Two ways orbitals can overlap:
Sigma bonds (σ): single bonds; head-to-head overlap; stronger
Pi bonds (π): form with double or triple bonds; side-to-side overlap
#4: INTERMOLECULAR FORCES
o Chapter 11
o Intermolecular Forces: forces between molecules in non-metals
~ often confused with ionic bonds, but those are intramolecular forces
How things “Stick Together”
LDFs (London Dispersion Forces): momentary polarization due to electrons’
revolution about the nucleus (at one instant, electrons can all be on one side creating
an instantaneous dipole so that the molecule can attract to other molecules
undergoing the same thing); everything has LDFs; they’re not super strong; bigger
molecule leads to greater LDFs because there are more electrons; shape of molecules
affect LDFs’ strength (surface area)
Dipole-Dipole Forces: negative and positive ends of two polar molecules attract to
each other; some LDFs involved; only between polar molecules; not instantaneous;
has greater effect than LDFs
Hydrogen Bonding: a kind of dipole-dipole, but stronger; only with “FON” (Fluorine,
Oxygen, and Nitrogen); electronegativity of F, O, and N causes a super strong dipoledipole moment
Ion-Dipole: ion attracted to molecule with a dipole moment
How strongly things stick together
o Depends on LDFs and D-D Forces
For example:
“List the substances BaCl2, H2, CO, HF, and Ne in order of increasing boiling
points.”
How to Solve:
1: identify what types of forces hold the molecules together
BaCl2: ionic bonds; H2: LDFs (non-polar); CO: D-D (polar); HF: H-bonds
(FON); Ne: LDFs
2: recognize differences between atoms, ions, and molecules as well as the
strength of the forces
LDFs are the weakest, then D-D, then H-bonds
H2 < Ne < Co < Hf < BaCl2
Phase Changes
o Phase changes strain intermolecular forces (the strength of intermolecular
forces determines when the substance will change phases)
o Condensed states: solids and liquids (that’s why it’s harder to go from liquid
to gas than solid to liquid)
o Every phase change involves a change in enthalpy
Solving for enthalpy of phase changes:
ΔH = q= mcΔT
phase change: ΔH = q = ΔHvaporization x #mole
= ΔHfusion x #mole
For example:
“Calculate the enthalph upon converting 1.00 mol of ice at -25 C to water vapor at 125 C
under a constant pressure of 1 atm. The specific heat of ice, water, and steam are 2.09
J/g-K, 4.18 J/g-K, and 1.84 J/g-K, respectively. For H2O, ΔHfus = 6.01 J/g-K and ΔHvap =
40.67 kJ/mol.”
How to Solve:
1: find the enthalpy change for going from 25 C to 0 C
2: find the enthalpy of the phase change (use ΔHfus)
3: find enthalpy change for going from 0 C to 100 C
4: find enthalpy of the phase change (use ΔHvap)
5: find enthalpy change for going from 100 C to 125 C
6: add all of those enthalpies up to find the total enthalpy
Take a Look:
Some Key Words:
o Solid: most condensed phase; doesn’t flow; closely packed
o Liquid: another condensed phase; less closely packed; flows into a container
o Gas: not condensed; free moving molecules (see chapter
o Melt: going from solid to liquid (use heat of fusion)
o Freeze: going from liquid to solid (also use heat of fusion)
o Vaporize: liquid to gas (use heat of vaporization)
o Condense: gas to liquid (also use heat of vaporization)
o Sublime: solid to gas (use heat of sublimation)
o Depose: gas to solid (also use heat of sublimation)
o Evaporate: molecules escaping from the surface of a liquid into the gas phase
o Volatile: evaporates quickly/readily
o Normal boiling point: the boiling point of a liquid at 1 atm pressure (occurs when
vapor pressure equals the pressure of the atmosphere around it)
o Vapor pressure: the pressure exerted by the vapor of a liquid when the liquid and
vapor states are in dynamic equilibrium
o Equilibrium: when evaporation and condensation occur at equal rates
Phase Diagrams:
o AB line: vapor interface (each point is the boiling point at that pressure)
o A: triple point (all three states in equilibrium; liquids can’t exist below it)
o B: critical point (above this, liquid and vapor are indistinguishable from
eachother)
o AD line: liquid/solid interface (each point is the melting point at that pressure)
o AC line: solid/gas interface (each point is the sublimation point at that
pressure)
Structures of Solids:
o Crystalline (structured) or amorphous (no order)
Types of Questions You Can Expect:
o Questions about comparing intermolecular forces
~ which is stronger?
~ what type of force does the substance have?
o Calculating Enthalpies of Phase Changes
o Phase diagrams
~ what phase is the substance in?
~ what do certain points mean?
~ interpretation
o Questions about the different properties of Crystalline Solids
#5: PROPERTIES OF SOLUTIONS
o
o
o
o
Chapter 13
Solution: homogenous mixture of two or more pure substances
A solute is dispersed throughout a solvent
Formation of Solutions: solvent pulls solute particles apart and surrounds
(solvates) them
Types of Solutions
State of Solution
Gas
Liquid
Liquid
Liquid
Solid
Solid
Solid
State of Solvent
Gas
Liquid
Liquid
Liquid
Solid
Solid
Solid
State of Solute
Gas
Gas
Liquid
Solid
Gas
Liquid
Solid
Example
Air
oxygen in water
alcohol in water
salt in water
hydrogen in palladium
mercury in silver
silver in gold
Energy Changes in Solution
3 Processes Affect Energetics of Solutions
1: separation of solute particles
2: separation of solvent particles
3: new interactions between solute and solvent
ΔHsoln = ΔH1 + ΔH2 + ΔH3
Entropy: the amount of randomness in a solution; the more random a solution, the lower
the energy
So, although enthalpy increases, if the system becomes more disordered (entropy
increases), the overall energy can decrease.
To What Extent is the Solution a Solution?
o Saturated: solvent holds as much solute as possible at that temperature
o Unsaturated: less than the maximum amount of solute for that temperature is
dissolved in the solvent
o Supersaturated: solvent holds more solute than is possible for the given
temperature (usually unstable)
Factors Affecting Solubility
o Temperature: solubility of solids in liquids usually increases with increasing
temperature; solubility of gases usually increases with decreasing temperature
Remember! “Like dissolves like”
~ polar substances dissolve in polar solutions
~ non-polar substances dissolve in non-polar solutions
(think oil and water)
Why?
More similar intermolecular forces
Gases
o The solubility of a gas increases in direct proportion to its partial pressure above
the solution
So, pretty much, the more pressure a gas exerts over a solution, the more
soluble it is (I guess that’s why you keep soda in highly pressurized cans or bottles,
eh?)
Henry’s Law:
Sg = kPg
Sg : solubility of the gas
of the gas
k: Henry’s constant for that gas solute Pg: partial pressure
The Wonderful World of Concentrations
Ways to express it:
o Mass percentage
Mass of A in solution x 100
Total mass of solution
o Mole Fraction:
Moles of A
Total moles in solution
o Molarity (M):
Mol solute
L of solution
o Molality (m):
Mol of solute
Kg of solution
Note: you can change from M to m if you know the density (g/L) of the solution
Colligative Properties
Some properties change when you make solutions. (The changes depend on the
number of solutes, not their identity)
Such as:
o Vapor Pressure Lowering: higher concentration on non-volatile solutes makes it
harder for the solvent to escape into the vapor phase
Raoult’s Law: PA = XAP°A (find the new vapor pressure)
o Boiling Point Elevation: solutions want to stay in the liquid phase longer; the
elevation is proportional to the molality of the solution
ΔTb = Kb x m
(Kb is the boiling point elevation constant)
o Freezing Point Depression: solutions want to stay in the liquid phase longer; the
depression is proportional to the molality of the solution
ΔTf = Kf x m
o Osmotic Pressure: the net movement of solvent is always toward the solution with
the higher solute concentration; obeys a law similar to the ideal-gas law
πV = nRT
(π = osmotic pressure)
so, π = (n/V)RT = MRT
~ isotonic: two solutions of identical π
~ hypotonic: a more concentrated solution
~ hypertonic: a more dilute solution
The van’t Hoff Factor
Reassociation is more likely with high concentrations, so you can’t really expect
something like NaCl, for example, to totally split up into 2 moles of ions.
That’s why we have this equation:
ΔTf = Kf x m x i
(i = how much the substance dissociates)
NOTE: you can determine the molar mass of a substance using colligative properties
by manipulating the equations
Colloids
Suspensions of particles larger than individual ions or molecules, but to small to be
settled out by gravity
o Tyndall Effect: the effect of colloidal suspensions scattering rays of light
o Hydrophyllic: water-loving colloids
o Hydrophobic: water-hating colloids (need to be stabilized in water)
Types of Questions you can Expect
o Different concentration calculations
o Colligative Property calculations
~ vapor pressure, boiling point, freezing point, and osmotic pressure
~ molar mass
o Predicting whether or not a substance will dissolve
o Calculate molality and Molarity from each other
#6: KINETICS
Chapter 14
4 Factors that Affect the Rate of a Reaction
1: nature of the reactants
2: Concentration/pressure of the reactants
3: Temperature
4: presence of a catalyst
Reaction Rates
Rate = Δ concentration
Δ time
(Δ = final – initial)
Use mole ratios to determine the different rates
Rate and Concentration
Rate = k[A]x[B]y (x and y = exponents of rate-determining step; k = rate constant)
o Orders:
Exponent to which the specie is raised
o Overall order:
Add the exponents (orders) of each specie
Now let’s do a problem….
Find the Rate of a Reaction
“The initial rate of a reaction A + B  C was measured for several different
concentrations of A and B. The results are below. Use the data to determine:
a) the rate law for the reaction b) the magnitude of the rate constant c) the rate of the
reaction when [A] = 0.050M and [B] = 0.100M”
Experiment
Number
[A] (M)
1
2
3
[B] (M)
0.1
0.1
0.2
0.1
0.2
0.1
Initial rate (M/s)
4.0 x 10^-5
4.0 x 10^-5
16.0 x 10^-5
How to Solve
1: write out the basic formula for a rate law (rate = k[A]x[B]y)
2: use given values to determine “x”
3: use given values to determine “y”
Now you know the rate law
Be careful, because sometimes they like to make it hard for you. Keep your mind open
to different ways to manipulate the equations (sorry, I can’t find any super awesome
examples so you’ll have to cope with that great piece of advice).
4: Use an “initial rate” value from the table and the corresponding [A] and [B] values
to solve for “k”
Now you know the magnitude and units of k
Make sure to be careful with the units of k. Check the table below to see what they
usually are.
5: plug and chug all of the values you just found to get the value when [A] and [B]
are what they ask for
Now you’re done!
Take a Look
Change of Concentration Over Time (integrated rate laws)
Rate Law
Equation
Integrated Rate
Law
Zero Order
First Order
Second Order
Rate = k[A]0
A  products
[A] = -k x t + ln[A]0
Rate = k[A]1
A  products
ln[A] = -k x t +
ln[A]0
Rate = k[A]2
A  products
[A]-1 = k x t +
[A]0-1
t1/2 = [A]0 x k
2
y = mx +b
t1/2 = ln(1/2) = 0.693
k
k
y = mx + b
t1/2 = 1
k[A]0
k = slope
[A] vs. time
k = slope
ln[A] vs. time
k = slope
1/[A] vs. time
M/s
1/s
1/Ms
y = mx + b
Half-Life
Equation
Graphically,
what makes it
linear
k and its units
The first and second order integrated rate laws are on the bottom left-hand corner of
the crappy periodic table.
Most of the problems involving these will be simple plug-and-chuggers, so we won’t
do any samples. Just memorize the equations.
Energy Profile Diagrams
o vertical axis: Energy
o horizontal axis: reaction progress
o Activation energy: the minimum amount of energy needed to start a reaction
o lower activation energy (Ea) means faster reaction
o Catalysts: provide alternate pathways for a reaction, thus lower its activation
energy
The Arrhenius Equation
The increase in rate with increasing temperature is nonlinear. The Arrhenius Equation
relates 3 factors that affect reaction rates:
1) fraction of molecules possessing Ea
2) collisions that occur per second
3) fraction of collisions that have appropriate orientation factor
k = A x e ^ -Ea/RT
k = rate constant; A = free factor; -Ea = activation energy;
R = gas constant; T = temperature in Kelvins
OR
lnk = -Ea x 1/RT + lnA (This one’s on the crappy on the very bottom left)
Note: the slope = activation energy
Reaction Mechanisms
o coefficients need to match between element steps and exponents of the rate
law
o the slow step is the rate determiner
o intermediates may be used as long as they’re used up by the end
o rate laws are only determined experimentally
o elementary reactions: single steps
o molecularity: indicates the number of molecules that act as reactants in a
reaction
o unimolecular: A 
o bimolecular: A + B 
o termolecular: A + B + C 
o multi-step mechanisms: always have to add up to give the equation of the
overall process (like Hess Law in Thermodynamics)
For example,
“The conversion of ozone into O2 proceeds by a two-step mechanism.
O3  O2 + O
O3 + O  2O2
a) describe the molecularity of each elementary reaction in this mechanism
b) write the equation for the overall reaction
c) identify the intermediates.”
Take a Look
# 7 EQULIBRIUM
Chapters 15, 16, and 17
What is equilibrium?
o All reactions are reversible, but we want to know to what extent they are
reversible.
o Equilibrium occurs when a reaction and its reverse proceed at the same rate
o Depicted by a double arrow
The oh-so-magnificent Rate Constant
o Symbolized by K (the ratio of the rate constants at that temperature)
Finding K:
Rate of forward reaction: rate= kf[reactants]
Rate of reverse reaction: rate = kr[reactants]
So, at equilibrium: Ratef = Rater => kf/kr = products over reactants = K
Generally, aA + bB  cC +dD
Kc = [C]c[D]d
[A]a[B]b
o Only gases are used in the expression
o The reverse and forward equilibrium constant expressions are reciprocals
o If you multiply a reaction, the K is raised to that power (always do one
operation above)
o Pressure and concentration are proportional, so K can also be written
according to pressures (Kp)
Kp = Kc(RT)Δn (equation to change from concentration to pressure or vice
versa)
The meaning of K:
K tells you how much something dissociates. Basically, it tells you whether the
reaction is product-favored (k >> 1) or reactant-favored (k << 1).
Different kinds of K:
K: equilibrium constant kf: forward rate kr: reverse rate Kc: equilibrium with
concentrations Kp: equilibrium with pressures Kw: equil. of water Ka:
dissociation of an acid Kb: dissociation of a base Ksp: solubility product constant
We’ll discuss all of these in this chapter.
For example:
“Oxalic acid, H2C2O4, is a diprotic acid with K1 = 5.36 x 10-2 and K2 = 5.3 x 10-5.
For the reaction below, what is the equilibrium constant?
H2C2O4 + 2 H2O  2H3O+ + C2O4-2”
How to Solve:
1: write out the two equations involved in the dissociation of oxalic acid
2: manipulate those equations until they look like the one in the problem (like
Hess’s Law)
3: if you multiplied or divided any equations, you have to either raise K to that
power or take the root of that power; if you reversed them, take the inverse of K
(K-1)
4: once it’s all situated, multiply the K values together
Take a Look:
OR: “A closed system containing 1.000E-3 M H2 and 2.000E-3 M I2 at 448 C is
allowed to reach equilibrium. Analysis of the mixture shows that the concentration of
HI is 1.87E-3 M. Calculate Kc at 448 C for the reaction
H2 + I2   2 HI”
How to Solve:
1: make an I.C.E table with the initial concentrations
2: add the change in concentration (1.87E-3) to the product
3: use stoichiometry to subtract that amount from the reactants
4: now you know the concentrations, so put them in to the equilibrium constant
expression
Take a Look
I.C.E. Tables
o Initial Change End Tables demonstrate the concentrations of substances
undergoing equilibrium
o They can be used for about some really important things:
~ calculating K
~ using with a known K value to find concentrations
o Remember: in titrations, the values always have to be in moles (we’ll get to
that later, though)
(we do about a million of these, so I don’t think I’ll do an example here)
The Reaction Quotient
o Q is the reaction quotient. It gives the same ratio as K, but the system isn’t at
equilibrium.
o You can compare K to Q to figure out how far along the reaction is
Q > K: there’s too much product, so the reaction shifts to the left (toward the
reactants)
Q < K: there’s too much reactant, so the reaction shifts to the right (toward the
products)
Q = K: it’s at equilibrium
If a question asks about the concentrations of a system when it’s not at equilibrium, plug
it into a Q expression. Usually they have you compare it to K and ask what this indicates
about the reaction.
Le Chatlier’s Principle
o This is used so much. I’m not even lying. It comes up everywhere.
o Here’s what Le Chat had to say:
“If a system at equilibrium is disturbed by a change in temperature, pressure, or the
concentration of one of the components, the system will shift its equilibrium position
so as to counteract the effect of the disturbance.”
Let’s break it down:
Change in Concentration:
~ increase: shift to opposite side
~ decrease: shift to that side
~ I kind of think of it like a Star Wars spaceship going into hyperspace. It charges
up the concentration, then splurts to the other side. Well, that’s a bad analogy, but
think of it as splurting to one side or the other based on how much stuff there is.
Change in Pressure:
~ increase: shift to the side with the fewest moles of gas
~ decrease: shift to the side with the most moles of gas
Change in Temperature:
~ if it’s exothermic
Hotter  shift to reactants
~ if it’s endothermic
Hotter  shift to products
(write heat as either a product or reactant, depending on ΔH, and think about it like
concentration)
Note: this is the only one that affects K
Catalysts:
~ DO NOT AFFECT EQUILIBRIUM!
~ they only provide an alternate pathway, which lowers the activation energy, and
speeds up the reaction
Acids and Bases
o Arrhenius acids and bases:
Acid: substance that increases the concentration of H+ in solution when
dissolved
Base: substance that increases the concentration of OH- in solution when
dissolved
o Bronsted-Lowry acids and bases:
Acid: proton donor
Base: proton acceptor
o Lewis acids and bases:
Acid: electron acceptor
Base: electron donor
o Amphoteric: can be either an acid or a base (HCO3-, HSO4-, H2O)
Note: in any acid-base reaction, the equilibrium will favor the reaction that moves
the proton to the stronger base.
pH and pOH
o
o
o
o
o
“p” just means to take the negative logarithm of something
pH = -log[H+] and pOH = -log[OH-]
pH + pOH = 14
[H+][OH-] = 1.0 x 10-14
All of these equations are on the crappy and they’re used a lot
Dissociation
The equilibrium expression for the general dissociation of an acid or base is Ka or
Kb. The higher the value, the stronger the acid or base (because it dissociates more).
Ka x Kb = 1 x 10-14
pKa + pKb = 14
You can use these values to calculate pH and percent ionization or vice versa. But
keep your eyes peeled, because chemistry likes to throw new stuff at you (it’s either to
make sure you know what you’re doing or make you feel like an idiot, I’m not sure
which yet).
Percent Ionization
[H3O+]eq x 100
[HA]initial
[OH-]eq x 100
[B]initial
Use these equations to find out how much of a substance is ionized. If the percent
ionization > 5%, you can’t ignore x when solving for concentrations with K.
Let’s do Some Problems
Finding Ka
“A solution of 0.10M formic acid (HCHO2) is found to have a pH = 2.38. Calculate
the Ka and the percent ionization in 0.10M solution.”
How to Solve:
1: write the balanced chemical equation
2: make an I.C.E. table
3: fill in the values (you know the pH, so you know the [H+], so you can figure out
the rest of the values from that)
4: use the end values from the I.C.E. table and plug them into the Ka expression
5: find the percent ionization by plugging and chugging the values you found into the
equation
Take a Look
Use Ka to calculate pH
“The Ka of Niacin is 1.5E-5. What is the pH of a 0.01M solution of niacin?”
How to Solve:
1: write out the chemical equation (you don’t need to know all of the chemical formulas)
2: write out the Ka expression and make an I.C.E. table
3: use “x”s to express the concentrations gained or lost
4: plug the end values (xs) into the equilibrium constant expression and solve for x
5: x = [H+] (be careful with stoichiometry here, though, because it’s not always 1:1) so
take the –log of that to get pH
Take a Look
Use Ka to calculate percent ionization
“Calculate the percent ionization of HF molecules in a 0.010 M solution of HF.”
How to Solve
1: write out the equation and set up an I.C.E. table to find the [H+] at equilibrium
2: solve the I.C.E. table using xs and the Ka value as in the previous example
3: plug in the value of [H+] that you found as well as the initial concentration of the
solution to the equation and solve
Take a Look
You can do all of this with bases, too. Just make sure you use Kb and solve for [OH-]
instead of Ka and [H+]. Also, if you need to find pH or Ka, you can use the equations
above to go back and forth between them.
Polyprotic Acids
o These acids have more than one proton on them (H2SO4, H3PO4, etc.)
o These acids dissociate more than once, thanks to these extra protons, so they have
multiple dissociation constants
o It’s always easier to remove the first proton from a polyprotic acid than to remove
the second (so you’ll have a lower concentration of the final conjugate base, and
almost all of the [H+] comes from the first dissociation)
Calculating the pH of a polyprotic acid solution
“The solubility of CO2 in pure water at 25 C and 1 atm is 0.0037M. The common
practice is to assume that all of the dissolved CO2 is in the form of carbonic acid
(H2CO3), which is produced by a reaction between the CO2 and H2O:
CO2 + H2O   H2CO3
What is the pH of a 0.0037M solution of H2CO3? And what is the [CO3-2] in that
solution?”
How to Solve
1: write out the first dissociation and make an I.C.E. table
2: put in values and use x-s to indicate the change in concentration (as before)
3: use the Ka1 value to find the [H+], and thus the pH (because the first dissociation
creates more H+)
4: write out the second dissociation and make another I.C.E. table. This time, put in
the [H+] you found in the first dissociation for its initial value instead of leaving it as
0.
5: solve through for x again using the ka2 constant to find the [CO3-2]
Take a Look
Acid-Base Properties of Salt Solutions
Acid + Base  salt + water
o Salts can act as acids or bases when dissolved in water. Here’s a little
summary of how that all works
1) if an anion is the conjugate base of a strong acid, it won’t affect the pH
2) if an anion is the conjugate base of a weak acid, it will cause an increase in the
pH (more basic)
3) if a cation is the conjugate acid of a strong base, it won’t affect the pH
4) if a cation is the conjugate acid of a weak base, it will cause a decrease in pH
(more acidic)
5) other metal ions (that don’t form strong bases) will cause a decrease in pH
6) when a solution has both the conjugate base of a weak acid and the conjugate
acid of a weak base, the ion with the larger Ka or Kb will have the greater
influence on pH
o pretty much, when you see a salt, ask “what are its parents?” and if its parent
is strong, it won’t affect the pH. If it’s weak, it will either make the solution
more acidic (if it’s a weak base) or more basic (if it’s a weak acid).
For Example:
“List the following inorder if increasing pH: 0.1M Ba(C2H3O2), 0.1M NH4Cl,
0.1M NH3Br, 0.1M KNO3”
How to Solve
1: split up each salt into its ions
2: ask what acid or base each ion came from, and use that information to surmise
whether or not the salt acts basic or acidic
Take a Look
Factors that Affect Acid Strength
o Polarity (difference in electronegativity)
o Bond Strength (strong, but not too strong)
o Stability of conjugate base
Four kinds of Acids
1) Binary: H-X
~HBr, HCl, HI
~ bond strength affects whether or not these acids are strong
2) Oxyacids: non-metal – O - H
~ HClO, HClO2, HClO3, HClO4, etc,
~ the strength of these acids is affected by polarity (adding more O makes the bond
stronger because it’s more polar and can donate an H+ better)
2) Carboxilic acids: hydrocarbon – carboxyl group (COOH)
~ HC2H3O2 (acetic acid)
~ these are affected by the strength of their conjugate bases and the number of
electronegative ions in the acid
3) Amino Acids: “base acids” (can react with themselves)
Common Ion Effect
This is related to Le Chatlier’s Principle. If you add a strong electrolyte to a weak
electrolyte and they have an ion in common, the extent of ionization of the weak
electrolyte is decreased. This is because you increase the concentration of an ion on
one side of the equation, which will move the equilibrium to the left, away from the
products.
NaF in an HF solution.
NaF  Na+ + F- and HF  H+ + FF- is a common ion, so there will already be some in the HF dissociation from the
NaF. That will shift the reaction to the left.
For example,
“Calculate the pH of a solution containing 0.085M nitrous acid (HNO2; Ka = 4.5x14) and 0.10M potassium nitrite (KNO2).”
How to Solve
1: identify the major species in the solution and decide whether they’re strong or
weak
2: identify the equilibrium of interest that affects [H+]
3: use an I.C.E. table to find the concentrations of the ions involved
4: use K to find [H+] and pH
Take a Look
Buffers
A buffer is a solution that resists a change in pH. It contains either a weak acid and its
conjugate base or a weak base and its conjugate acid.
Some equations:
Dissociation constant
Ka = [H3O+][X-]
[HX]
Henderson-Hasselbalch Equation
pH = pKa – log ([HX] / [X-]) or pOH = pKb + log ([X-] / [HX])
generally,
pH = pKa + log ([base] / [acid])
Use this to calculate the pH of a buffer.
For example:
“How many moles of NH4Cl must be added to 2.0L of 0.10M NH3 to form a buffer
whose pH is 9.00? (assume no volume change)”
How to Solve
1: set up the Hendersen-Hasselbalch equation equaling 9 (that’s the pH you want)
2: you want to know the [acid], so set it up using the values you know and solve for x (or
[A])
Take a Look:
“A buffer is made by adding 0.300mol HC2H3O2 and 0.300mol NaC2H3O2 to enough
water to make 1.0L of solution. The pH of the buffer is 4.74. Calculate the pH of this
solution after: a) 0.020 mol NaOH is added b) 0.020mol HBr is added.”
How to Solve:
a) 1: write out the balanced chemical equation and I.C.E. table it
2: NaOH is the limiter, so use 0.020M as the change value
3: notice how there’s a weak acid and its conjugate base left. Use the HendersenHasselbalch equation to find the pH
Careful! NaOH is a strong base, so it will be added to HC2H3O2. Be sure you plug
in the [A] and [B] into the appropriate spots in the equation.
b) 1: write out the balanced chemical equation and I.C.E. table it
2: HBr is the limiter, so use 0.020M as the change value
3: this is a buffer too, so use the H-H equation to find the pH
Careful! HBr is a strong acid, so it will be added to C2H3O2-. Be sure you plug
in the [A] and [B] values into the right spots!
Take a Look
Titrations
A + B  +ion + -ion + HOH
Some trends to remember:
o If both are strong: ions don’t affect pH and the pH is 7
o If it’s a weak acid + strong base: slightly basic (due to the weak acid’s conjugate
base) and the pH > 7
o If it’s a strong acid + weak base: slightly acid (due to the weak base’s conjugate
acid) and the pH < 7
o If it’s a weak acid + weak base: depends on the strength of the weak base vs. the
weak acid (write out an I.C.E. table and equilibrium of interest)
Equivalence point: moles acid = moles base
Half-way to the equivalence point is the area of most buffering
Some graphs:
Note: Always ask yourself, “What drives the pH?” It could be a buffer, a strong
base/acid, or a weak base/acid. Each has a different way of solving for the pH.
~Strong acid/base: I.C.E. table it to find the [H+] or [OH-] and then the pH
~ Buffer: Henderson-Hasselbalch equation
~ weak conjugate base/acid: I.C.E. table it and use Kb/Ka values to find the [H+] or
[OH-]
For example:
“Calculate the pH when the following quantities of 0.10M NaOH solution are added to
50.0mL of 0.100M HCl: a) 49.0mL b) 50.0mL c) 51.0mL.”
How to Solve:
1: find the moles of A and B (NaOH and HCl)
2: write the chemical equation of A + B (NaOH + HCl  HOH + NaCl)
3: I.C.E. table it for each of the different quantities of NaOH
a) There’s fewer moles of NaOH, so it’s the limiter. Use this as the change
value. There’s some HCl left, so find the [H+] to find the pH.
b) The moles of NaOH and the moles of HCl are equal, so the pH is 7
c) HCl is the limiter, and there is some NaOH left. So, find the [OH-] to find the
pOH, then the pH.
Take a Look
“Calculate the pH when the following quantities of 0.10M NaOH solution are added to
50.0mL of 0.100M HC2H3O2. Ka = 1.8x10-5. a) 45.0mL b) 50.0mL c) 51.0mL”
How to Solve:
a) 1: write out the chemical equation and make an I.C.E. table
2: find the number of moles of NaOH and acetic acid
3: there are fewer moles of NaOH, so use 0.0045 as the change value
4: a weak acid (acetic acid) and its conjugate base are left … it’s a buffer!
5: use the H-H equation to find the pH
b) 1: write out the chemical equation and make an I.C.E. table
2: find the number of moles of each
3: the #moles are equal, so use 0.005 as the change value
4: you’re left with a weak conjugate base (C2H3O2-), so react it with water and
I.C.E. table it to find the [OH-] and the pH (you need to find Kb here, which is
just 1E-14 / 1.8E-5)
c) 1: write the chemical equation and I.C.E. table
2: find the #moles of each
3: there are fewer moles of acetic acid, so it’s the limiter. Use 0.005 as the change
value
4: you’re left with a strong base (NaOH) in solution. NaOH and OH- are in a 1:1
ratio, so find the [NaOH] by dividing moles by total L and that equals the [OH-].
You can find the pH using that.
Take a Look
“Calculate the pH at the equivalence point when: a) 40.0mL of 0.025M benzoic acid is
titrated with 0.050M NaOH (Ka = 6.3 E -5)”
How to Solve:
1: set up the chemical equation (you don’t need to know the formula for benzoic acid)
2: find the #moles of benzoic acid. This is equivalent to the #moles NaOH, so you can
find the volume of NaOH to use later. Also, use this number as the change value.
3: look at what’s remaining in solution. The only thing that drives the pH is the conjugate
base of benzoic acid
4: find the [X-] by adding up the volumes of NaOH and benzoic acid (you found V
NaOH is step 2) and dividing the #moles by that quantity
5: change the Ka into Kb (because it’s a conjugate base) and solve for x ([OH-]). Use this
to find the pH
Take a Look:
Solubility Equilibria
This is pretty much the dissociation of non-soluble salts. You can tell just how much nonsoluble salts dissociate by looking at the dissociation constant as with acids and bases.
… but this time it’s different. Simple K is now Ksp!
Non-soluble salt  + ions + - ions
Ksp = [+ ions][- ions] (not divided by reactants because they’re solid)
What affects solubility?
3 factors:
1) the presence of common ions
2) the pH of the solution
3) the presence of complexing agents
4) amphoterism (haha, and you thought there were only 3)
Common Ions
Look on page 41 for more info. about this effect. Pretty much, when you have slightly
soluble salt and you add another solute that brings along a common ion when it
dissociates, the solubility of the slightly soluble salt will decrease. (Think like Le
Chatlier)
pH of the Solution
The solubility of slightly soluble salts containing basic anions increases as [H+] increases
(as pH lowers). The more basic the anion, the more the solubility is influenced by the pH
of the solution.
Complex ion formation
A complex ion is a metal ion and Lewis bases bonded to it. The solubility of metal salts
increases in the presence of suitable Lewis bases (NH3-, CN-, or OH-). The stability of a
complex ion in solution is determined by its equilibrium constant of formation (Kf).
Amphoterism
Substances that can act as either an acid or base are amphoteric. These substances
dissolve better in strong acid and strong base solutions than in water (yeah, they dissolve
well in both acidic and basic solutions). It’s because of the formation of complex anions
with a bunch of OH- ions bound to a metal.
For example:
“What is the solubility of LaF3 in moles per liter if Ksp = 2 E -19?”
How to Solve:
1: write out the equation and make an I.C.E. table
2: ignore the reactant (it’s solid) and use “x”s to show the change and end values
3: put the “x” values into the Ksp expression and solve for x to get the [LaF3]
Take a Look:
“Calculate the molar solubility of CaF2 in a solution that is: a) 0.0100M in Ca(NO3)2 b)
0.010M in NaF (Ksp = 3.9 E -11”
How to Solve:
a) 1: write equation and I.C.E. table
2: use “x”s and Ksp expression to find [CaF2]
b) 1: write equation and I.C.E. table (Notice the common ion, F-, from the NaF.
Make sure that’s accounted for as an initial value)
2: same thing as before
Take a Look
Note: the [CaF2]in part b was less than part a due to Le Chatlier’s Principle.
Will a Precipitate Form?
These are the last types of questions you need to know about for equilibria. Woot! These
ask you, “Will a precipitate form when…” or “What concentration is necessary to form a
precipitate when…”
For example:
“ Will a precipitate form when 0.10L of 8.0 E -3 M Pb(NO3)2 is added to 0.40L of 5.0 E 3 M Na2SO4? (Ksp = 6.3 E -7)”
How to Solve:
1: write out the chemical equation to determine whether an insoluble salt forms or not
2: write out the equilibrium of interest
3: find Q (page 33) Be careful with additive volumes! The concentrations of ions don’t
stay the same because it’s a new volume!
4: compare Q with Ksp
Take a Look:
“A solution contains 1.0 x 10-2M Ag+ and 2.0x10-2M Pb+2. When Cl- is added to the
solution, both AgCl and PbCl2 precipitate form. What concentration of Cl- is necessary
to begin the precipitation? Which salt will precipitate first?”
How to Solve:
1: find the Ksp values of the two insoluble salts
2: use the Ksp values to find the [Cl-] necessary
3: the smaller [Cl-] means that it precipitates first
Take a Look:
Wow, that’s a lot of stuff on equilibrium, so here’s a little summary of how to solve basic
pH problems.
Another helpful trick is knowing how to estimate logarithms.
Logs to Remember
log 1 = 0
log 3 = 0.5
log 5 = 0.7
log 7 = 0.85
log 10 = 1
Inverse Logs to Remember
10^-0.1 = 0.8
10^-0.3 = 0.5
10^-0.5 = 0.3
10^-0.7 = 0.2
10^-0.9 = 0.12
When you have to do a logarithm or inverse of a logarithm without a calculator, break up
the logarithm, find each log by estimating, then add them together.
For example:
Find pH: [H+] = 3.2 x 10-5
Do that all of the time and you’ll be set.
#8: THERMODYNAMICS
Chapters 5 and 19
Are you ready to switch gears completely? Too bad, because we are.
Thermochemistry Basics
o 1st Law of Thermodynamics: energy is neither gained or lost, it is conserved
o Energy: the ability to do work or transfer heat
o Joule: the unit used to express energy (calories are used too, but let’s face it, we
have enough of those already) J = 1 kg x m2/ s2
o Work: energy used to move something (w = F x d)
o Potential energy: energy an object possesses by virtue of its chemical composition
o Kinetic energy: energy an object possesses by virtue of its motion (KE = 0.5mv2)
o State Function: something that only depends on the present state of something—not
the path it took to get to that state
o System: the thing we study
o Electrostatic energy: energy that holds atoms together (k x Q1Q2 / d) (page 15)
o ΔH: enthalpy; heat at constant pressure
o Exothermic: a system that releases energy to surroundings
o Endothermic: a system that takes in energy from surroundings
o Thermochemical Equation: shows the chemical equation of a reaction and gives the
ΔH for it in J or kJ (it shows how much energy is given off or taken in by a mol of
each of the reactants or products)
AB2  A + 2B ΔH = +100kJ
kJ per mol.
So, when AB2 decomposes, it takes in 100
5 Ways to Find the Enthalpy of a Process
1) Information from the Equation
2) Calorimetry
3) Hess Law
4) Enthalpy of Formation
5) Bond Enthalpies
Information from an Equation
“How much heat is released when 4.50g of methane gas is burned in a constant
pressure system?
CH4 + 2O2  CO2 + 2H2O ΔH = -890 kJ”
How to Solve:
1: convert to moles
2: solve for kJ
This isn’t unlike stoichiometry.
Take a Look:
Note: it’s not always a 1:1 ratio. Also, they might ask questions about the reverse of
the reaction, in which case you just switch the sign of ΔH.
“What is the ΔH when 5.0g of H2O2 decomposes?
2H2O2  2H2O + O2 ΔH = -196 kJ”
Solve this one on your own…
Answer: -14.4 kJ
Calorimetry
Equations:
q = m c ΔT
q = -Ccal x ΔT
Cp = ΔH / ΔT
coffee-cup
bomb calorimetry
Heat capacity: the temperature change experienced by an object when it absorbs a
certain amount of heat
Specific Heat: the heat capacity of one gram of a substance (c) (water’s is 4.184 J /
gK)
For example:
“When a student mixes 50mL of 1.0M HCl and 50mL of 1.0M NaOH in a coffee-cup
calorimeter, the temperature of the resultant solution increases from 21.0 C to 27.5 C.
Calculate the enthalpy change for the reaction, assuming the calorimeter loses only a
negligible quantity of heat, that the total volume of the solution is 100mL, that its density
is 1 g/mL, and that its specific heat si 4.184 J/gK”
How to Solve:
Plug and chug, my friend. Plug and chug. (q = m c ΔT)
Take a Look:
“A 0.5865g sample of lactic acid (HC3H5O3) is burned in a bomb calorimeter whose
heat capacity is 4.182 kJ/C. The temperature increased from 23.10 C to 24.95 C.
Calculate the heat of combustion of the lactic acid a) per gram b) per mole”
How to Solve:
1: plug and chug values into bomb calorimetry equation
2: divide by mass to get kJ/g (a)
3: multiply by the molar mass to get kJ/mol (b)
Take a Look:
Hess Law
If a reaction is carried out in a series of steps or reactions, the enthalpy change for the
overall reaction will equal the sum of the enthalpy changes of each individual step.
Basically, you have a bunch of different equations, but you only want to know the ΔH for
one. That one is the sum of all of the others, but you may need to do some manipulation
of those individual ones to end up with the equation you want. If you do manipulate an
equation, make sure you change its ΔH the same way.
For example:
“Calculate the ΔH for the reaction: NO + O  NO2 given the following information:
NO + O3  NO2 + O2 ΔH = -198.9 kJ
O3  3/2 O2 ΔH = -142.3 kJ
O2  2O ΔH = 495.0 kJ”
How to Solve:
1: flip the O2  2O and multiply it by ½ (do the same with ΔH)
2: flip the O3  3/2 O2 equation (do the same with the ΔH)
3: add and cancel the chemical equations to get the equation they’re asking for
4: if it all cancels out correctly, add up the ΔHs for all of them
Take a Look:
Note: this is a lot like the reaction mechanisms we did on page 30.
Standard Enthalpies of Formation
ΔHrxn = Σ ΔHf products – Σ ΔHf reactants
ΔHf = enthalpy of formation (usually the problem gives this to you)
“Using the given values, determine the enthalpy of reaction for the combustion propane
gas to form carbon dioxide gas and liquid water.
C3H8
-103.85 kJ
CO2
-393.5 kJ
H2O
-285.8 kJ
How to Solve:
1: write out the balanced chemical equation for the reaction
2: plug in the ΔHf values into the ΔHrxn (be sure to include coefficients!)
Take a Look:
Note: these problems don’t always simply ask for the ΔHrxn, sometimes they ask for the
ΔHf values of an individual species. In that case, just use algebra. Also, the ΔHf value for
elements in their standard state is 0.
Bond Enthalpies
ΔHrxn = bonds broken – bonds formed
Note: This is reactants – products, which is kind of weird. To solve questions about this,
write out the Lewis electron dot configuration of each of the reactants and products (page
16), look up the bond enthalpies of each bond that exists, sum them up, and plug them
into the equation.
For example:
“Use the bond enthalpies given to estimate the enthalpy of the reaction for the
combustion of gaseous ethane, producing carbon dioxide gas and water.”
C–H
413 kJ
C–C
348 kJ
O=O
495 kJ
C=O
799 kJ
O-H
463
How to Solve:
1: write out the balanced chemical equation for the reaction
2: re-write the equation with Lewis electron dot configurations instead of chemical
formulas
3: plug in the bond enthalpies of each bond in each element (careful for coefficients!) into
the equation
Take a Look:
Now let’s get a little more complicated…
Spontaneity
Spontaneity tells you whether or not a reaction will occur naturally. If a reaction is nonspontaneous, it needs some energy put into it to occur. Just because a reaction is
spontaneous, doesn’t mean it’s rapidly occurring
There are 3 variables that describe this:
ΔH: enthalpy (heat at constant pressure)
ΔS: entropy (disorganization)
ΔG: free energy (tells whether or not a reaction is spontaneous)
We already talked a lot about ΔH, so we’re going to move on.
Entropy ΔS
o 2nd Law of Thermodynamics: nature wants entropy to increase (kind of like your
room: it will keep getting unorganized unless you put some energy into cleaning
it)
o 3rd Law of Thermodynamics: the entropy of a pure crystalline substance at
absolute zero is 0.
S = k lnW (but you don’t really need to know that) The only thing you really need to take
away from this is that W = possible # of microstates.
o Microstates: the possible position and energy of the individual gas molecules.
More of these means higher entropy.
How to get more microstates:
~ increase temperature
~ increase volume
~ increase the number of independently moving particles
The system gets MORE DISORDERED!!!
Calculating Entropy
ΔS = ΔSfinal – ΔSinitial
ΔS increases when:
~ gases are formed from solids or liquids
~ liquids are formed from solids
~ more gas molecules are created during a chemical reaction
That means that the ΔS value is positive, and the system is more disorganized. A negative
ΔS value means the system is getting more organized.
For example:
“Calculate the ΔS for the synthesis of ammonia from N2 and H2 at 298K:
N2 + 3H2  2NH3”
NH3
192.5 J/mol K
N2
191.5 J/mol K
H2
130.6 J/mol K
How to Solve:
1: plug in the values into the ΔS equation above (watch out for coefficients!)
Kablamo, you’re done.
Take a Look:
Note: This is like the ΔH questions, they don’t always ask for the whole reaction;
sometimes they ask for an individual species.
Entropy change in surroundings:
ΔSsurr = -qsys / T (-qsys = ΔHrxn)
but that’s also not really important
ΔSuniverse = ΔSsys + ΔSsurr (when this is positive, the reaction is spontaneous)
Free Energy
This was proposed by Josiah Gibbs to show how to use ΔH and ΔS to predict the
spontaneity of a reaction. It has a really useful equation that you can find on the left side
of the crappy periodic table equation sheet.
ΔG = ΔH - T ΔS
If ΔG is negative, the reaction is spontaneous in the forward direction; if it’s 0, the
reaction is at equilibrium; if it’s positive, the reaction is spontaneous in the reverse
direction.
Note: in any spontaneous process at constant temperature and pressure, the free energy
always decreases.
Calculate the Free Energy of a Reaction
ΔGrxn = ΔGproducts - ΔGreactants
The ΔGf for an element at standard conditions is 0.
You don’t always have to calculate ΔG, though. Most of the time, they ask you to predict
the ΔG based on ΔH or ΔS values that you find. In that case, use the G = H – TS equation
to estimate:
~ ΔH - and ΔS +: reaction is spontaneous (ΔG is negative)
~ ΔH + and ΔS -: reaction is non-spontaneous (ΔG is positive)
~ ΔH + and ΔS +: spontaneity depends on T (high T makes it spontaneous)
~ ΔH – and ΔS -: spontaneity depends on T (low T makes it spontaneous)
What if it’s not at standard conditions?
Then the equilibrium constant comes in.
ΔG = ΔG° + R x T x lnQ
We can manipulate this when it’s at equilibrium (ΔG = 0):
0 = ΔG° + R x T x ln K
ΔG° = -R x T x ln K
K = e ^ - ΔG° / RT
(R = gas constant; use 8.314 J / mol K)
Let’s do some problems now…
“The Haber Process: N2 + 3H2  2NH3. a) Predict the direction in which ΔG° for this
reaction changes with increasing temperature b) calculate the values of ΔG for the
reaction at 25 C and 500 C c) calculate the equilibrium constant K”
How to Solve:
a) 1: determine whether the ΔS value will be positive or negative
2: because ΔS is negative, the -T ΔS part of the equation will get bigger as T
increases, so ΔG also gets bigger
b) 1: calculate the value of ΔH for the reaction using standard enthalpies of formation
2: calculate the value of ΔS for the reaction using standard entropies of formation
3: plug these values along with the T into the equation
c) 1: establish which equation to use (the third one from the previous page)
2: plug in values to solve for K
Take a Look:
“Calculate the K for PbCO2 at equilibrium at 773 K.”
How to Solve:
1: write out the balanced chemical equation
2: notice how it’s not at standard conditions. This means that you need to calculate the
ΔG from the ΔH and ΔS values, so find them using products – reactants
3: use the G = H – TS equation to find ΔG
4: plug in the ΔG value you just found into the equation to find the K
Take a Look:
# 9: ELECTROCHEMISTRY
Chapter 20
Redox Reactions Basics
o Oxidation: loss of electrons (bigger oxidation number)
o Reduction: gain of electrons (smaller oxidation number)
o Redox Reaction: a reaction in which one or more elements gains an electron (is
reduced or oxidizing agent) and one or more elements loses an electron (is
oxidized or reducing agent)
o Assigning Oxidation Numbers:
Rules:
1: if an atom is in elemental form, its charge/oxidation number is 0
2: for any monatomic ion, the charge is whatever is on the ion
3: O is almost always -2 (except in H2O2, so look out for that in identifying redox
reactions)
4: H is always +1 with a non-metal and -1 with a metal
5: F is always -1
6: the sum of oxidation numbers in a neutral compound is 0
7: the sum of oxidations numbers in an ion equals the charge of the ion
Practice assigning Oxidation Numbers:
1) H2S 2) S8 3) SCl2 4) Na2SO3 5) SO4-2
Answers:
1) H = +1, S = -2 2) S = 0 3) S = +2, Cl = -1 4) Na = +1, S = +4, O = -2 5) S = +4, O
= -2
o Reduction Potentials: use these to tell what’s reduced and oxidized in a
spontaneous redox reaction; the more positive the reduction potential, the better
it’s reduced (this is also known as standard electrode potential, electromotive
force, emf, E°, and is measured in V)
Electrochemical Cells
Two half-cells, a salt bridge, and a wire
Two kinds: Galvanic (Voltaic) and Electrolysis
Galvanic (Voltaic)
Some Terms
o Produce energy through a spontaneous reaction
o Anode: where oxidation occurs (“an ox”)
o Cathode: where reduction occurs (“red cat”)
o Electrode: strip of metal
o ½ Cell: electrode in solution
o Salt Bridge: used to neutralize ions in solution
o Electrode potential: how much an electrode wants to gain or lose electrons (use
reduction potentials)
o Coulomb (C): unit of charge (how many electrons)
o Faraday’s Constant: 96,500 C / mol (on crappy)
o Volt: measure of cell potential ( V = J / C)
o Ampere: unit of current (rate of electron movement) (A = C / sec)
o Cell Potential = Ecell ox (cathode) – Ecell red (anode)
o Line notation: anode anode solution cathode solution cathode (double line
represents a porous barrier)
Balance Redox Reactions:
1) assign oxidation numbers
2) divide into 2 half-reactions (one for oxidation and one for reduction)
3) balance all species except H and O in each reaction
4) balance all Os by adding HOH as needed
5) balance all Hs by adding H+ as needed
6) balance the charge by adding electrons as needed
7) multiply the half-reactions by integers if necessary so that the number of electrons
are equal
8) add the half-reactions (simplify when possible)
9) check for mass and charge
10) if it’s basic, add OH- s to each side
Watch:
MnO4- + C2O4-2  Mn-2 + CO2
For example:
Electrode A is made of lead and the solution is lead (II) nitrate
Electrode B is made of manganese and the solution in manganese (II) nitrate
1) Which is the most easily reduced metal?
2) What is the balanced equation showing the spontaneous reaction that occurs?
3) What is the maximum emf the cell can produce if everything exists at standard
conditions?
4) What is the direction of the flow of electrons in the wire?
5) What is the direction of the positive ion flow in the salt bridge?
6) Which electrode is increasing in size?
7) Which electrode is decreasing in size?
8) What is happening to the concentration is Electrode A’s solution?
9) Identify the
i: anode
ii: cathode
iii: positive electrode
iv: negative electrode
10) If Electrode A were replaced with a S.H.E., what would be the
i: emf of the cell
ii: direction of electron flow
Some Quick Hints for Acing Those Types of Questions:
o Write absolutely everything you know on the picture (reduction potentials,
equations, positive/negative charges, electron flow, everything)
o In the salt bridge: cations move to the cathode and anions move to the
anode
o Red cat and an ox
o Emf = cathode – anode
o The salt bridge keeps the solutions neutral (they like to ask questions
about it)
o S.H.E.s (Standard Hydrogen Electrode) have a reduction potential of 0 V
o Pay attention to what ions they’re asking about and the charge of the ion
Free Energy and Redox
ΔG = -nFE
(n = moles electrons, F = Faraday’s constant, E = cell potential, emf)
o When E is positive, the reaction is spontaneous (G is negative)
o When E = 0, ΔG = 0 (it’s at equilibrium… hold on to your socks)
ΔG = -nFE = -R x T x lnK
So,
Ecell = R x T x lnk = 0.0592 log K
nF
n
When it’s not at standard conditions (when there isn’t a concentration of 1M):
Nernst equation: E = E° - 0.0592 / n log Q
For example:
“Find E°, ΔG, and K for 2 Ag + ½ O2 + 2H+  2Ag+ + H2O”
How to Solve:
1: Find E by looking up reduction potentials and doing cathode – anode
2: find ΔG by using the –nFE equation
3: find K using the ΔG = -RTlnK equation
Take a Look:
“Calculate the emf at 298K generated by: CrO7-2 + 14H+ + 6I-  2Cr+3 + 3I2 + 7H2O
When the concentrations are 2.0M, 1.0M, 1.0M, and 1 E -5 M, respectively.”
How to Solve:
1: write out the balanced half-reactions
2: look up reduction potentials and find the E°
3: this isn’t at standard conditions, so write the Q expression ([products] / [reactants])
4: use the given concentrations to find Q
5: plug in the values into the Nernst equation
Take a Look:
Note: They usually ask you how changing the concentration of a species affects E or K.
You can just use Le Chatlier’s to describe this, but the Nernst equation works too. Also,
they might ask to find the concentration of an individual species and give you the emf, so
just manipulate the equation.
For example:
“The voltage of a cell is 0.45V at 298K when [Zn+2] = 1.0M and PH2 = 1.0 atm, what is
the [H+]?”
How to Solve:
1: write out the Q expression and plug in the values you know (leave [H+] as a variable)
2: find the E°
3: write out the Nernst equation using the values you know, leaving log Q as a variable
4: find what Q (on its own) equals with those values
5: put in the expression you found in step 1 for Q and solve for [H+]
Take a Look:
Note: You can use this to find pH, too.
Electrolysis
o Electrolysis: using electricity to create a non-spontaneous charge
o E° = negative and ΔG = positive
o Use Faraday’s constant a lot with these problems
o Use it a lot with molten salts
5 Basic Steps to Solve:
1) balance half-reactions
2) convert to charge
3) use Faraday’s constant to find moles of electrons
4) use ratios to find moles of substance
5) convert to units
Sounds a lot like stoichiometry, eh?
For example:
“Calculate the number of grams of aluminum produced in 1.00 hours by the electrolysis
of molten AlCl3 if the electrical current is 10.0A.”
How to Solve:
1: Write the balanced half-reactions for Al and Cl
2: convert Amperes and hours into Coulombs
3: use Faraday’s constant to find the moles of electrons
4: use the ratio from the reaction in step 1 to find the moles Al
5: convert to grams
Take a Look:
# 10: NUCLEAR CHEMISTRY
Chapter 21
Just the Basics:
o As an atom gets bigger, it needs more neutrons to stabilize it
o When an atom decays, it breaks down into other elements
o 5 kinds of decay:
1: β (beta) particle emission (an electron emitted from the nucleus)
2: α (alpha) particle emission (Helium-4 particle 42He emitted from
nucleus)
3: gamma radiation (represents energy lost when remaining nucleons
become more stable; not shown in nuclear equations)
4: positron emission (a particle with the same mass as an electron, but the
opposite charge)
5: electron capture (nucleus captures an electron from the electron cloud)
o annihilation: when a positron and an electron run into each other (releases a
ton of energy)
o things usually decay into lead because it’s super stable
o N stands for the amount of radioactive particles
o Half-life: the amount of time it takes for half of the neutrons in an element to
decay (use lots of algebra to figure out those questions or just use the first
order half-life equation on page 28)
o Geiger counter: used to detect radioactivity
o Nuclear fission: large nuclei split
o Nuclear fusion: small nuclei fuse together (like in the Sun)
o Subcritical: when you have less radiation than you need
Nuclear Problems:
These don’t show up to much, but if and when they do, you just have to make sure that
the sums of the mass numbers and atomic numbers (numbers on the top and bottom)
equal each other on both sides.
#11: ORGANIC CHEMISTRY
Chapter 25
Organic Chemistry is the chemistry of all living or once-living things.
All of them have Carbon!
Structures
o Governed by bonds on C (always 4 bonds)
2 classifications: saturated and unsaturated
Alkanes: saturated (4 single bonds); -ane
Alkenes: unsaturated (at least 1 double bond); -ene
Alkynes: unsaturated (at least 1 triple bond); -yne
Aromatic: form a ring (benzene)
Alkanes
CnH2n+2
Methane (CH4), Ethane (C2H6), Propane (C3H8), etc.
How to Make a Structure:
1: write out Cs
2: add H
Isomers: same molecular formula, but different arrangement.
Naming:
1: count how many Cs there are on the longest chain (to find prefix)
2: number the longest chain
3: add the methyl (CH3), ethyl (C2H5), or propyl (C3H7) groups to the C with the
smallest number
For example:
“2-methylpentane”
That means there’s a methyl group off of the 2nd C in the pentane chain.
This is an isomer of hexane (6 C)
o Cycloalkanes: exist in a ring of single bonds
Alkenes
CnH2n
-cis
-trans
Some common Alkenes are
Ethene (ethylene)
Propene (propylene)
For isomers of these, you need to indicate where the double bond is by numbering the
longest C chain and putting it on the smallest (like naming Alkane isomers)
Alkynes
CnH2n-2
Common Alkynes:
Ethyne (acetylene)
propyne
Reactions with Alkenes/Alkynes
Something like a Halogen will break a double bond and replace it (Halogenation)
Aromatic
Cyclic (form a ring)
Contain benzene
Function Groups
Functional Group
type of compound
suffix/prefix
example
systematic name
alkene
word - ene
ethene
alkyne
word - yne
ethyne
alcohol
word - ol
methanol
ether
ether
dimethyl ether
haloalkanes
halo-
chloromethane
amine (act basic)
word - amine
ethylamine
aldehide
word - al
ethanal
ketone
word - one
propanone
carboxilic acid
word - oic acid
ethanoic acid (acetic
acid)
ester
word -oate
methylethanoate
amide (good base)
word - amide
ethanamide
THE EXAM
2 Parts
Part I: 50%
o 75 multiple choice questions in 90 minutes
o No calculators
o Periodic Table provided
o No equation sheet
o No penalty for guessing (like the ACT)
Part II: 50%
o 6 Free Response questions in 95 minutes
o 2 “Sub-parts”
1: 55 minutes; 20%
#1: equilibrium
#2: equilibrium, thermodynamics, kinetics, or electrochemistry
#3: equilibrium, thermodynamics, kinetics, or electrochemistry
You can use your calculator, periodic table, and equation sheet on this
sub-part.
2: 40 minutes; 30%
#4: write net-ionic equations and answer short questions for 3 reactions
#5: atomic theory, bonding, intermolecular forces, or colligative properties
#6: atomic teory, bonding, intermolecular forces, or colligative properties
You can use your periodic table and equation sheet on this part, but no
calculator.
Of #2, #3,#5, or #6, one will have to do with a lab