Chemistry: Titrations, Complex Ions, and Electrochemical Cells | Exams Chemistry

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

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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:

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-^ + H 2 O 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 + H 2 O 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.
Calculate the total volume of the reaction,(titration) mixture. Use this to calculate the concentrations of the species present after assuming stoichiometric reaction.
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.
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 CN-^ cause 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.

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

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 non-spontaneous 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) + E o ox (oxidation half rxn) or E o cell = E o red(reduction half rxn) - E o red(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 Nerst 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?

Chemistry: Titrations, Complex Ions, and Electrochemical Cells, Exams of Chemistry

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