Download Examination 1 - Idaho State University

Chemistry 112
Examination 2 Review
You must know the solubility rules from Chem 111. You must also know
strong acids and bases and polyatomic anion nomenclature. Also know the definitions of
acids and bases and the relationships between pH, pOH, pKw for acidic and basic
solutions and the definition of pH = -log(H+ ). You should be able to recognize the
conjugate base to a given acid and the conjugate acid to a given base.
Titrations involve the addition of an acid to base or a base to an acid. They give rise to
titration curves. Three cases must be considered: a strong acid-strong base, a strong
acid-weak base, and a weak acid-strong base. You should be able to sketch the
titration curve corresponding to these. The equivalence point is the point where a
stoichoimetric amount of base or acid has been added to react with the initial amount
of acid or base present in the solution. The endpoint is the point where the color
change occurs. Usually we try to get these to occur as close to each other as
possible. You should know the features of all three types of titrations. These types of
calculations are more difficult because the concentrations acid and base are changing
both due to reaction, and due to dilution. The key to performing the addition of a strong
acid or base to a buffer; the
addition of a strong base to a weak acid, and the addition of a strong acid to a weak base
is to see what ions, acids, and base is present in the solution after taking into account
the addition of the base, and using this information to write down an equilibrium
reaction. Once this is
accomplished the SAME OLD EQUILIBRIUM TABLE IS SET UP TO SOLVE FOR
THE
EQUILIBRIUM CONCENTRATIONS). A simplified approach to these types of
problems is to:
1. Assume stoichiometric reaction of the added acid or base with the initial amount
of acid present or base present to give the conjugate base or conjugate acid anion: HA +
OH- -> A- + H2O or B + H+ -> BH+ OR reaction of acid or base with initial
amount of conjugate base or conjugate acid: A- + H+ -> HA or BH+ + OH- -> B +
H2O
Remember the definition of the equivalence point and remember to work in moles, since
concentrations are changing by reaction and dilution. Calculate the amount of acid
or base and conjugate anion or cation present after this assumption.
2. Calculate the total volume of the reaction,(titration) mixture. Use this to calculate the
concentrations of the species present after assuming stoichiometric reaction.
3. Use the information gained from 1 and 2 to decide what reaction can now take place
in the solution based on the cations, anions, and water present, and the idea that you are
interested in the pH.
4. Use the concentrations from 2 as the initial concentrations in an equilibrium
calculation based on the equation you determined in 3. (i.e. use them to set up the
EQUILIBRIUM TABLE). Solve for the equilibrium pH and concentrations of ions.
Note that strong acid strong base titrations are handled simply by the stoichiometric
change in the moles and the volume change.
In the chapter that dealt with the formation of complex ions, some new terminology was
introduced. The complex ions generally consist of a metal ion surrounded by charged or
neutral species that are “complexed” to the metal. These species are known as ligands.
The overall charge on the complex ion is a sum of the charges on the ligands plus the
charge on the metal ion. Given a metal and ligands, You should be able to come up with
a formula, (including) charge for a complex-ion made from the metal and the ligands.
Remember that certain metals will only form one type of geometry, (octahedral,
tetrahedral, square planar, linear (you would be given the tables)) when they combine
with a ligand to form a complex ion.
The metal’s “d” orbitals are affected by the presence of the ligands, which results in a
splitting of the “d” orbitals. For a transition metal ion surrounded by six ligands, (an
octahedral complex) the five “d” orbitals are split into one set, the dx2-y2 and the dz2 that
are higher energy, and another set: the dxz, the dyz, and the dxy orbitals which have a lower
energy. This is known as crystal field splitting, and the energy difference between the
lower set and the higher energy set is called the crystal field splitting energy. The
identity of the ligands causes the splitting to change. The strong field ligands like CNcause a large crystal field splitting whereas weak field ligands like F- cause a very small
splitting. (You will be given the strong field-weak field ordering list if you need it.) If
the splitting is small, then as the electrons are put into the “d” orbitals, they will populate
the higher energy “d” orbitals before they pair up. If the splitting is large, (strong field
ligands), then the electrons will pair up first before they populate the higher energy
levels. This also has implications for the magnetic nature. In the low spin state, there are
more electrons paired up (more diamagnetic) as compared to a high spin state (more
paramagnetic) where the electrons are not as “paired up”.
Metals may form linear, tetrahedral, and square planar complexes in addition to
octahedral complexes, Some metals will only form one type of complex, and others can
form many types. The crystal field splitting may change for a metal if it is in a
tetrahedral complex versus an octahedral complex, etc.
The Kf, the formation constant, is just another equilibrium constant, but it is the
equilibrium constant associated with the formation of a complex ion. metal ion +
ligands - > <- complex ion
Usually these have very large values meaning that the reaction goes almost completely to
completion in the forward direction. That being the case, in order to calculate the
equilibrium concentrations of species involved with complex ion formation, often you
must assume the reaction goes to completion based on the limiting reagent and then let
the reaction go back in the reverse direction taking into account any leftover reagent
based again on complete conversion of the limiting reagent.
The next chapter contains information dealing with the insoluble or partially soluble
salts. The equilibrium constant pertaining to this dissolution process is called the
solubility product constant the Ksp. Remember that the equilibrium expression will
not contain the concentration of the pure solid. The solubility of the salt can be used to
determine the Ksp or the value of the Ksp gives the solubility of the salt. This calculation
is again perhaps best accomplished through setting up the the equilibrium TABLE, or the
ICE TABLE. One subtle difference in this type of equilibrium is the idea that at some
low concentrations of ions in solution, no equilibrium can be established such that
there is solid (precipitate present). One calculates if precipitation will occur by
looking at the value of the reaction quotient, called the ion product. If Qc is less
than Ksp, then no ppt. will be present. If Qc is > Ksp then precipitate will form
stoichiometrically until Qc = Ksp. You should be able to decide if, what ppt., and how
much ppt. will form if solution which is a mixture of salts. Of course you will need to
know the solubility rules to do this.
How does adding an acid affect the solubility of certain insoluble ionic compounds?
What about the addition of ammonia! If a complex ion can form, then this may affect the
solubility greatly. This is due to the fact that the ammonia is a ligand for forming a
complex ion from any of the metal ion that results when the ionic compound is even
minutely solublized. According to Le Chatelier’s Principle if the complex ion forms then
the solubility will be increased since metal cation is pulled out of the solution in forming
the complex ion. Also the Kf is large so the complex ion formation is essentially
complete. Remember that the overall reaction involving the insoluble ionic compound
reaction with the ammonia (or other ligand) to form the complex ion and the anion if the
insoluble compound has an equilibrium constant which is essentially the Ksp x Kf
depending on the stoichiometry.
You should know how to write a thermochemical equation. Remember the Hrxn is
based upon the reaction as written, including the phases. You should be able to use Hess’
Law, combining two or more thermochemical eqns. to get the Hrxn for the overall
reaction. Remember if you reverse the reaction, the sign of H must be multiplied by -1.
In an exothermic reaction the enthalpy of the products is lower than the enthalpy of the
reactants, Hrxn is negative, so heat is a product. The opposite is true for an endothermic
reaction. For physical processes like sublimation, vaporization, and melting, H is
positive. They require heat and are endothermic processes. The opposite processes of
condensation and freezing would be exothermic as heat must exit.
Experimentally, the enthalpy change of a reaction is determined by measuring the heat
given off or required. Since the constant pressure heat, qp = process(rxn) then one gets H
from a measurement of the heat. qp = CT; C = Ccalolrimeter + Ccontents = heat capacity;
Csubstance = specific heat * mass. Often the heat is measured in a constant volume bomb
calorimeter where qv = CT. qp = qv + ngRT. The last relation comes from the 1st Law
of thermodynamics which statesE = qin + Win, Win = -pV. Remember the subscript in
indicates that if the w is positive the work is into the system and if q is positive the work
is into the system. The 1st law is essentially a conservation of energy for the system and
surroundings. You should be able to make some simple calculations with the First Law.
To do this you have to define your system and surroundings before attempting to use the
1st Law Eqn.
The ultimate application of Hess’ Law is to use the tabulated values of the heats of
formation, Hfo, to determine the heat of the reaction under standard state conditions.
Remember the heats of formation are based on forming one mole of the molecule of
interest from the most stable form of the elemental substances at the standard state (1atm
or 1M). Remember that the Hof s of the most stable elemental substances are
themselves equal to 0. Horxn is then calculated from:
Horxn = [v Hfo (products)] - [w Hf o (reactants)]
where the v and the w are the coefficients in the balanced chemical eqn of each of the
individual reactants and products respectively. Additionally remember that for gaseous
reactions, Hrxn can be estimated from Bond Energies (BE) by:
Horxn ≈ Σ BE (bonds
broken) - Σ BE (bonds made).
The 2nd and 3rd Laws of Thermodynamics focus on the entropy which is a measure of the
randomness or disorder. The entropy can tell us about what processes are spontaneous or
naturally possible. A statement of the 2nd Law of thermodynamics is that for any
irreversible process the entropy of the universe (system plus surroundings) increases.
This statement is difficult to apply in some situations, since it is often difficult to monitor
the system of interest let alone the universe. A better indication of the spontaneity of a
process comes from a combination of the 1st and 2nd Laws, where minimization of energy
and maximization of entropy both contribute. This allows one to determine if a process is
spontaneous just by looking at the Gibbs Free Energy change of the system for processes
done at constant temperature and pressure. Gsys = sys - TSsys. (more on this later).
Remember that S is related to the heat by the expression S = q/T for a process only
involving energy transfer by heating or more. For a constant pressure process this
becomes S = H/T, and this expression works well at a phase change. For instance in a
vaporization process at constant pressure and temp.: Svap = Hvap/Tvap. A form of this
expression for the entropy change of a constant pressure process which goes through
several changes of state includes not only the phase changes, but also the heating of each
phase to the temperature of the phase change. Here we approximate H by H in particular
phase = Cphase  Because entropy is a state variable, the entropy change of the system is
independent of the path.
The third law helps us define the entropy on an absolute scale. It says the entropy of a
perfect crystalline pure substance at K is equal to 0. This defines S on an absolute scale
and when calculating So from tabulated information, we use tables of So :
Sorxn = [v So (products)] - [w So (reactants)]
As we said the change in the Gibbs free energy G=H-TS can serve as a criterion
for spontaneity of a reaction in JUST the system at constant temp. and pressure. If it is
positive, the reaction is non-spontaneous, if it is
negative the reaction is spontaneous, and if it is zero, the reaction is at
equilibrium. The standard free energy change Gof is the free-energy change that occurs
when reactants in their standard states (1 atm and/or 1M). It can be determined from the
tabulated standard free energy of formation, defined similarly to the standard
enthalpies of formation (the free-energy change that occurs when 1 mol of substance is
formed from its elements in their standard states at 1 atm and specified
temp). G =ΣnGof(prod) -ΣmGof(react). It can also be determined from the Hof and
So using G =H -TS . Assuming that H and S are indep of temp., the temp
dependendence of G is determined from the latter eqn. Also since Gorxn =-RTlnK, K
and the temperature dependence of K can also be determined.
If you want the free energy change when reactants in nonstandard states are changed to
products in nonstandard states (G), you can obtain it from the standard free energy
change G using: G=Go + RTlnQ (Q is the rxn quotient). Thus we can determine the
spontaneity of a rxn and equilibrium constant before we even carry it out from tabulated
values of the standard enthalpy
of formation, standard entropy (absolute entropy) and/or its standard Gibbs free energy of
formation. Finally, it can be shown that the maximum useful work for a spontaneous
reaction, wmax=G. A non-spontaneous reaction can occur but work must go into the
system. Reactions may be coupled one that is spontaneous, one that is not to give the
desired change spontaneously as well. This is essentially Hess’ Law. An example of this
is seen in an oxidation reduction rxn. where if the either the reduction or oxidation half is
very spontaneous yet the other half is nonspontaneous, then the overall rxn is
spontaneous.
Oxidation-reduction processes are one area of chemistry where there is a direct
application. You must know how to assign the oxidation numbers in order to understand
oxidation and reduction. You can use the half rxn method or the oxidation number
method to help you balance them. What is the reducing agent, what is the oxidizing
agent? An electrochemical cell is a system consisting of electrodes that dip into an
electrolyte and in which a chemical reaction
either generates or uses an electric current.
The two types of eletrochemical cells are: 1. Voltaic or Galvanic cell - cell in which a
spontaneous reaction generates an electric current and
2. an electrolytic cell in which an electric current drives an otherwise nonspontaneous
reaction. Examples of spontaneous voltaic cells include batteries (alkaline, Ni-Cad, Pb
storage) and fuel cells. In a voltaic cell two half-cells are connected in such a way the
electrons have the potential
to flow from one electrode to another through an external circuit while the ions flow from
one cell to another internally often through a salt bridge. Examples of electrolytic nonspontaneous cells include Downs cell and other metal deposition cells involving an
externally supplied current.
You should know general terminology like oxidation, reduction, oxidizing agent,
reducing agent, cathode, anode, and terminal. You should understand and be able to use
the notation for voltaic cells. Remember that in both the electrolytic and voltaic cell,
oxidation occurs at the anode. However, in the voltaic cell the anode is the negative
terminal, but in the electrolytic cell the positive part of the current source is hooked up
to the anode and the negative terminal of the current source is hooked up to the
cathode. (This appears to cause some confusion, but just remember that oxidation goes
on at the anode.)
You should be able to determine what type of cell you are dealing with based on the
calculated Ecell. Remember a negative Ecell means that the cell is non-spontaneous and
a positive value means that the cell is spontaneous. This is readily evident since G
= -nFEcell or Go = -nFEo cell and a negative G means that the process is spontaneous.
If the electrochemical cell is operating under thermodynamic standard state
conditions, (solute concentrations are 1M each, gas pressures are 1 atm and the
temperature has a specific value), then the Eocell can be obtained directly from the
TABLE of STANDARD ELECTRODE Reduction POTENTIALS by: Eocell =
Eored(reduction half rxn) + Eoox (oxidation half rxn) or Eocell = Eored(reduction half rxn) - Eored(oxidation half rxn) or
Eocell = Eocathode - Eoanode (Ecathode and Eanode are the std red. potent.)
Remember the elements high up in the table (very negative values of the reduction
potential (very positive values for oxidation)) are good reducing agents (easily oxidized)
and the ones at the bottom are easily reduced. Also since G =-nFE and Go =-RTlnK
then both Go and K can be determined from the STANDARD ELECTRODE Reduction
POTENTIALS. So besides getting G and K from Gof and H-TS, it can be
determined from Eocell.
For nonstandard cells the cell EMF can be determined from the Nernst Eqn:
Ecell = Eocell - (0.0592/n) log Q
(Q is the rxn quotient)
This works for half cells as well. Remember n is the number of electrons transferred in
the rxn determined from the balanced redox rxn. How can this eqn can be used to
determine the pH from a measurement of the potential?