Spring ’06 Exam Study Guide for LBS 172

Professor: Dr. Robert LaDuca

This guide covers lectures from 1/9/06 to 4/28/06

The information presented here is a composition of Dr. LaDuca’s lectures and the study guides for the previous guides for exams one through three. The author would like to thank Dr. LaDuca and the authors of the other guides for allowing him to use their information.

Practice problems are provided that cover concepts from Chemical Equilibrium to Nuclear Chemistry

Table of Contents Page Gases……………………………………………… 2 Liquids……………………………………………. 5 Solids……………………………………………... 6 Molecular Orbital Theory………………………… 7 Solubility………………………………………….. 10 Chemical Kinetics………………………………... 11 Chemical Equilibrium……………………………. 16 Gaseous Equilibrium……………………………… 17 Predicting Reactions……………………………… 17 Finding Concentrations at Equilibrium…………… 17 Acid-Base Chemistry……………………………… 18 Titration Curves…………………………………… 24 Solubility Equilibrium…………………………….. 26 Reduction-Oxidation Reactions…………………… 28 Electrochemistry…………………………………… 29 Thermodynamics Part II…………………………… 32 Nuclear Reactivity…………………………………. 35 Final Exam Point Breakdown……………………… 39 Gases  The hardest to see but easiest to understand  Close connection between microscopic characteristics and physical properties  At the molecular level: o Gas molecules are not in constant contact o Molecules move randomly o Molecule/molecule and molecule/container wall collisions are elastic  At the macroscopic level: o Very low density, measured in grams per liter o Gases mix completely and evenly o Exert pressure which can be used for work o Are compressible o At STP (25°C and 1atm) the volume of a gas is 24.5L  Torricelli o Invented the barometer o 1atm of pressure raises the Hg column by 760mm Hg o 1atm = 760mm Hg = 760 torr  Relationship Laws o Boyle’s Law . Pressure (atm) and temperature (K) have an inverse relationship . Increasing temperature gives molecules more energy, which then hit the walls of the container with more force

. V1P1 = V2P2 for the same gas at two different pressures/temps o Avagadro’s Law . Amount of gas (moles) and pressure are directly related . The more stuff you have, the larger of a container you’ll need

. V1 n1  V2 n2 for the same has at two different volumes/amounts o Charle’s Law . Temperature and Volume (L) have a direct relationship . Lowering the temperature removes energy from the molecules which don’t move as fast or as far, so less space is required

. V1 T1  V2 T2 for the same gas at two different volumes/temps o The Ideal Gas Law . If one combines these three relationship, you can relate pressure, temperature, amount and temperature all at once . PV = nRT L  atm . R is the Ideal Gas Law Constant and is equal to .08206 mol  K . It is important to note that n represents the number of moles of all gases in the container. 12mol of pure oxygen and a mixture of 6mol hydrogen/6mol nitrogen will exert the same pressure when volume and temperature are constant. o Relating molar mass (M ) to the Ideal Gas Law . Since the molar mass of something is in units of mass in grams per mole, it can be rewritten as n = m/M m  R T . Substitute this for n in the Ideal Gas Law to get PV  M m  R T which is rewritten to M  PV . Since density (D) is the mass of something divided by the D  R T volume, the equation finally becomes M  P o Dalton’s Law of Partial Pressures . The total pressure of a mixed gas is the sum of the pressures of the individual gases

. Ptotal = P1 + P2 + P3 + ….. . Using this knowledge and the Ideal Gas Law, we can express the relationship between the total pressure of a container and

the pressure of an individual gas as Pgas A = XA*Ptotal

. XA is the “mole fraction” of that gas and = nA ntotal o Diffusion & Effusion of a Gas . Diffusion – A gradual mixing of gases due to random thermal molecular motion . Effusion – The escape of a gas through a narrow hole . The equation to relate the average molecular speed of a gas (u ) 3RT to the temperature and molar mass of a gas is u  M

. So, heavier gases move more slowly and vice versa o Exceptions to the Ideal Gas Law . The IGL does not apply at either very high pressures or very low temperatures

 8.0 moles of Cl2 gas in a 4.0L container at 27°C should have a pressure of 49.2 atm according to the IGL. It actually has a pressure of 29.5 atm  8.0 moles of Ar gas in a 4.0L container at 27°C should also have a pressure of 49.2atm according to the IGL but its actual pressure is 47. . Both of these examples have lower pressures than they should, but chorine’s pressure is much lower. Why?  At high pressures, collisions turn inelastic and the molecules become susceptible to Inter-Molecular Forces (IMFs). The extreme difference is due to the fact that chlorine’s IMFs exceed those of Ar.  Inter-Molecular Forces (IMFs) o These are attractive forces between neighboring molecules, they are not chemical bonds. There are five main types o They determine physical properties such as natural state, boiling point, melting point, viscosity, surface tension, and chemical properties such as solubility and reactivity o Ion/Ion Forces . Present in ionic compounds (a metal combined with a non- metal having an electronegativity greater than 1.5) . Caused by the positive/negative attraction forces . Strongest of all the IMFs o Dipole/Dipole Forces . Attraction of a polar molecule to another resulting from the asymmetric distribution of charge . Less strong than Ion/Ion forces . Larger electronegativity values create stronger DP/DP forces o London Dispersion Forces . Weak IMFs caused by temporary dipoles . The more electrons and/or surface area a molecule has, the stronger it’s LDFs will be . All molecules have LDFs o Hydrogen Bonding . An IMF, not an actual bond . A special and strong kind of dipole/dipole interaction between a hydrogen atom in an H-X molecule and an X atom in a neighboring molecule. Not as strong as ion/ion forces . X can be nitrogen, oxygen or fluorine . Water is a very good example of this type of force and the strong bond explains the high boiling point of water and the empty space caused by this force explains why ice is much less dense than water

o Ion/Dipole Forces . Causes the solubility or non-solubility of compounds . In water, the cation of an ionic compound is attracted to the oxygen while the anion is attracted to the hydrogen . If the ion/ion forces are greater than the ion/dipole forces, the compound will remain solid in solution . “Like Dissolves Like” Rule  Polar solvents dissolve polar solutes and non-polar solvents dissolve non-polar solutes  Some exceptions such as oxygen. The polar water induces temporary dipoles in the oxygen molecule allowing a small amount to dissolve. This is what allows fish to breath in water o Soap . Soap is composed of a large non-polar segment (usually a long C-H chain) with a polar segment on the end . The non-polar section attaches to grease while the polar section attaches to the water molecule . The soap encloses the grease in a “micelle” and it gets swept away by the water

Liquids  Molecules are in contact, but have enough energy to slide past each other  Higher density than gas but less dense than a solid  Are able to flow o Viscosity – the resistance to flow . Higher viscosity means higher IMFs  Have Vapor Pressure o VP is the pressure of the molecules above the surface of a liquid o Since temperature is a measurement of the average kinetic energy, some liquid molecules have enough energy to enter the gas phase o Surface molecules are partially exposed, creating lower IMFs which allows them to make the transition to gas easier o VP depends on IMF strength, surface area and temperature o The temperature at which the VP = the external pressure is called the boiling point o Plotting the natural log of VP vs. the inverse temperature gives a

straight line whose slope = H vap R where R is 8.314 J K  mol

o ∆Hvap is the heat of vaporization which is the energy input required to transform one mole of a substance from liquid form to gaseous form  Have Surface Tension o This is a result of surface molecules having lower IMFs by being partially exposed o In response, they pack together to from a “tough skin” of sorts  Have Capillary Action o This is the slight creeping up of a liquid where the surface touches the wall of its container o Due to IMFs between a liquid and its container’s walls . For example, water can H-Bond to glass

Solids  Are a covalent network held together with chemical bonds  Can be 2-D such as mica or graphite, or 3-D such as glass or diamond  Can be amorphous or crystalline o Amorphous means “without shape . Lacks a regular, long-range atomic order such as the random

coils of SiO2 in glass . Somewhat of a cross between liquids and glass o Crystalline . Has a regular, long-range atomic ordering . The smallest, symmetric, non-repeating unit in a crystal is called the “unit cell” . If the size and shape of a unit cell, number and type of atoms in the cell can be found, we can find the atomic structure . Shooting x-rays at a crystal will cause the rays to shatter into a geometric pattern which reveals the size and shape of a unit cell and the position of the atoms in it  Unit Cells o Atoms in crystals can be shared amongst atoms if in special positions o Corner atom . Cut in half by three faces so only 1/8 of the atom is in the cell . Shared by 8 atoms

o Edge atom . Cut in half by two faces so ¼ in the cell. Shared by four atoms o Body atom . Fully inside the cell. Unshared o Simple Cubic Unit Cell . 8 corner atoms equaling one atom per cell

. Only observed for Po(s) and Hg(s) . Rare due to there being so much empty space o Body Centered . Four corner and one body atom so two atoms per cell . Structure of iron and alkali metals o Face Centered . Atoms at corners and on each face, so four atoms per cell . Most common for metals (Al, Ag, Au, Cu, etc…)

Molecular Orbital Theory  An inside look at the formation of chemical bonds  To find number of molecular orbitals o Take atomic orbitals and add or subtract the wave functions to get the molecular orbitals o # of original orbitals = # of molecular orbitals

 The molecular orbitals for H2:

o H2 forms one bonding, one antibonding orbital (σ and σ*, respectively), corresponding to the original two s-orbitals. o σ* bonds have a higher energy than σ bonds because the electrons cannot exist between the atoms (due to the nodal plane).  The molecular orbitals for O2:

o In addition to the two s-orbitals, there are the three p orbitals o These six p orbitals create three bonding and three anti-bonding bonds o There is one pair of σ bonds and two pairs of π bonds o The σ and π bonds have less energy than their anti-bonding counterparts o The π2p bonds have a lower energy than the σ2p because they have fewer nodes (1 vs. 2) o Similarly, the π*2p have a lower energy than the σ*2p anti-bonds since they too have more nodes  From this data a Molecular Orbital Diagram can be constructed, seen here:

 Using the MO diagram, the Bond Order can be determined using this #bonding e  #antibonding e formula: BO = . 2

 The MO for B2 is on the left while the MO for F2 is on the right  4  2   8  6   So, the BO for B is    1, and the BO for F is    1. 2  2  2  2 

o Partial Bond Order values are possible  8  3 o Example: NO =    2.5  2   Orbitals with unpaired e– will result in an attraction to a magnetic field

o This is called “paramagnetism” (MO for B2)  Orbitals with fully paired e– result in no attraction to a magnetic field

o This is called “diamagnetism” (MO for F2)  Two orbitals are especially important

o The Highest Occupied Molecular Orbital (HOMO)

o The Lowest Unoccupied Molecular Orbital (LUMO)

o These two orbitals account for much of the way elements react  Why is CO so bad for your body? o CO has a bond order of 3 which gives it a π* orbital very similar to the d-orbitals on Fe. The CO attaches to Fe in your body and makes

the Fe unable to attach to O2  Band Theory

o Molecular Orbital Theory taken to its "extreme"

o The overlap of orbitals in a large crystal of an element results in bonding and antibonding "bands" of molecular orbitals closely spaced in energy.

o If the electrons fill one band fully and another partially, it is a metal

o If the electrons fill a band, and there is a large energy gap to the next band, it is an insulator

o If the electrons fill a band, and there is a small energy gap to the next band it is a semiconductor

 Conductivity in metals terms of temperature

o One might expect no change with an increase or decrease of temperature since it is very easy to promote electrons to higher bands in metals and semiconductors

o However, at higher temperatures, the metals atoms have increased vibration due to the increase in energy which impedes the flow of electrons  Superconductors

o These are special types of metal which conduct electricity with near- zero resistance

o All of the electrons travel in pairs with opposite spin

o They are perfectly diamagnetic and repel all magnetic fields

o YBa2Cu3O7 superconducts at 95°K

Solubility  Solution – homogenous mixture of two or more purse substances  Solute – the substance in lesser amount  Solvent – the substance in greater amount  Henry’s Law

o Relates gas solubility to the pressure it is at

o The solubility of gasses is directly proportional to the pressure

o Sm = kHP

o Sm is solubility of the moles

o kH is the Henry’s Law Constant . This is dependant on the gas and the temperature  Effect of Temperature on Solubility

o The solubility of gasses decreases with increasing temperature

o The solubility of most liquids and solids increases with increasing temperature  Colligative Properties of Solutions

o Solutions have different physical properties than pure liquids

o The difference depend only on the number of dissolved particles in the solution . Raoult’s Law  Vapor pressure lowering

 Psol’n = Xsolv Psolv

nsolvent nsolvent  Xsolv = = ntotal nsolvent  (nsolute )(i) . “ i ” is the Degree of Ionization . For molecular compounds that dissolve but don’t dissociate it is 1 . For ionic compounds, it is how many ions the solute dissolves into (e.g. i for NaCl is 2) . Boiling Point Elevation

 ΔTB = (m)(i)(kB) . kB is the boiling point constant and is different for all solvents  mol solute  . m is the molality    kg solvent  . Freezing Point Depression

 ΔTf = (m)(i)(kF) . kF is the freezing point constant and is different for all solvents . Osmotic pressure  π = (M)(R)(T)(i) . M is molarity L  atm . R = 0.08206 K  mol . T is temperature in Kelvin

Chemical Kinetics  Most chemical reactions occur in solution or gas o Easier for molecules to mix o Molecules can move  Reactions occur at molecular level due to collisions of random motion with enough energy, in correct orientation  Reaction Rates o The change in concentration of reactant or product with respect to time M  1  [A] o =    where n is stoichiometric coefficient of A s  n  t  o The rate for the reactants is negative since they are disappearing o The rate for the products is positive since they are appearing o Example  2A → B + 3C  1  [A]  [B]  1  [C]  Rate =             2  t   t   3 t 

 Rate Laws o x y Rate = k[R1] [R2] o x and y are the order of the reactants o The rate law can show how the reaction is occurring at the microscopic level (the reaction mechanism) i.e."look inside the arrow" o 0 Zero Order Reaction – Rate = k[R1] [A] .  k t .  [A]   k t .  d[A]   k dt

. [A]t [A]0  kt

. [A]t  [A]0  kt

o 1 First Order Reaction – Rate = k[R1] [A] .  k[A] t [A] .   k t  [A]  d[A] .   k dt  [A]   [A]t  . ln   kt [A]0 

. ln[A]t  kt  ln[A]0 . Nuclear decay is first order

 t1/2 is half-life, can be used to find rate of reaction ln 2  k t1/ 2 o  Half-lives of various elements 238 9 o U = 4.5 x 10 yrs 14 o C = 5715 yrs 95 o Tc = 20 hrs 2 o Second Order Reaction – Rate = k[R1] [A] .  k[A]2 t . … skipped steps, you can look at the class notes or just do the integration 1 1 .  kt  [A]2 [A]0

o In conclusion:

. The rate order is 0 if the plot of [A]t vs. time is a line . The rate order is 1 if the plot of ln[A]t vs. time is a line 1 . The rate order is 2 if the plot of vs. time is a line [A]t  All reactions need energy to start

 This is known as the Activation Energy (Ea)

o Rate and rate constant (k) depend on Ea

. Ea↑ k↓ Rate↓ . Ea↓ k↑ Rate↑

Ea  Ea RT , also ln(k)   ln(A) o k  Ae RT . A = collisional frequency factor . R = 8.314 J/(K mol)  Catalysts

o A catalyst speeds up the reaction without being consumed

o Catalysts provide alternate mechanisms of reaction with a lower Ea

o The Rate Law is determined by the slowest step in the mechanism (the Rate Determining Step – RDS.)

o If a substance's concentration appears in the rate law, it is involved in the RDS (or in a pre-equilibrium step right before). o The decomposition of H2O2 can be sped up with the addition of Iodine  1 . H O I  H  O 2 2 2 2 2

. Rate = k [H2O2] [I]    . RDS is: H 2O2  I  H 2O  IO  H 2O  I  O  2O  O2 . The reactive intermediate, IO-, must break down to return the I-

to its original state. This is the second hump in the Ea diagram.

 Proposed Rate laws can be determined for more than one mechanism. The actual mechanism is determined by the experimental determination of the rate law, thus finding out what goes on “inside the arrow.” - -  CH3I + Br → CH3Br + I o Mech. 1: - - . CH3I + Br  CH3Br + I -  Br- “attacks” carbon and knocks off the I - . Rate = k[CH3I][Br ] . Only one step, easiest mechanism o Mech. 2: slowest + - . CH3I  CH3 + I + - fastest . CH3 + Br  CH3Br . Rate = k[CH3I] . Two steps, and the slowest reaction determines rate o You determine which mechanism is used by comparing the expected rate law against the actual experimentally observed rate law o In this case, Mech. 1 is the correct method o With three CH3 group instead of the 3 H's, Mech. 2 occurs.

 NO2 → NO + O2 o Mech. 1 slowest . NO2  NO + O• fastest . 2O•  O2 . Rate = k[NO2] o Mech. 2 equilibrium . NO2  NO + O• . O• + NO2 → NO + O2 . In this case, you cannot measure the concentration of reactive intermediate, so you must substitute for it in the rate law.

. Rateforward = kforward[NO2] . Rateback = kback[NO][O•]

 Because this reaction is at equilibrium, these two rates are equal, thus: o kforward[NO2] = kback[NO][O•] k [NO ] o [O•] = forward 2 kback [NO]

. Rate2 = k2[O•][NO2]

 Substitute the equilibrium concentration of [O•]:  k [NO ]  forward 2   Rate2 = k2  [NO2 ]  kback [NO]  k[NO ]2 Rate = 2  [NO] o Mech. 3 slowest . NO2 + NO2  NO + OONO fastest . OONO  O2 + NO 2 . Rate = k[NO2] o Mech. 4 equilibrium . NO2 + NO2  NO + OONO . OONO → O2 + NO . Again, substitutions must occur 2 . Rateforward = kforward[NO2] . Rateback = kback[NO][OONO] 2  kforward[NO2] = kback[NO][OONO] 2 k forward [NO2 ]  [OONO] = kback [NO] 2 2  k forward [NO2 ]  [NO ] k   k 2 . Rate = k2[OONO] = 2   =  kback [NO]  [NO] o Notice how the Rate laws for Mech. 2 & 4 are the same, even though the proposed mechanisms are different – more complex tests would be necessary to determine the difference between mechanisms. o Mech. 3 is the correct mechanism in this instance. (actual Rate law is 2 Rate = k[NO2] …determined by experiment) Chemical Equilibrium  Many chemical reactions can proceed forward and reverse. Equilibrium is where the forward and reverse reaction rates are equal

o Example: aA + bB ↔ cC + dD . a b Rateforward = [A] [B] . c d Ratereverse = [C] [D] c d . Ratefor [C] [D] K rxn   a b Rate rev [A] [B] . K is the equilibrium expression constant and has units, but for all intents a purposes, we don’t worry about them

o Once equilibrium is reached, there is no net change in the amount of reactants or products. The reactions are still occurring however

o Pure solids and/or liquids are not included in the K expression  LeChatlier’s Principal

o If an external stress is applied to a system at equilibrium, the system will adjust to attempt to relieve the stress

o Stressors can be change in temperature, pressure, volume, concentration or addition of a catalyst +3 - +2 o Example: Fe (yellow) + SCN ↔ FeSCN (red) . Stress: Addition of SCN- Response: Equilibrium shifts right to consume SCN- . - Stress: Addition of HPO4 - +3 Reaction: The HPO4 reacts with the Fe lessening the concentration of it. Equilibrium shift to the left to try and make more Fe+3 +2 - -2 o Example: Co(H2O)6 (pink) + 4Cl ↔ CoCl4 (blue) + 6H2O . Forward rxn is endothermic and reverse is exothermic . Stress: Increase in temperature Response: Equilibrium shifts right to consume added heat . Stress: Decrease in temperature Reaction: Equilibrium shifts left to make more heat

o Example: 2NO2 (g) ↔ N2O4 (g) . Stress: Volume of container is lowered Response: This causes the pressure to increase. The equilibrium shifts right to lower the number of moles of gas in the system to attempt to lower the pressure  If a reaction takes more than one step, the K for the overall reaction is the product of the K values for the individual reaction steps

Gaseous Equilibrium

 Example: 2NO2 (g) ↔ N2O4 (g)

[N 2O4 ] o Kc deals with concentration, so K c  2 [NO2 ] P N2O4 In a similar manner, K deals with pressure so K p  o p (P ) 2 NO2

o While Kc can be used with either liquids or gases, Kp can only be used for gas equilibrium o To convert between Kc and Kp, an equation derived from a

combination of the Kp expression and the Ideal Gas Law. It is: Kp = (b-a) Kc(RT) where R is the same R as the IGL and “a” and “b” are the moles of reactant and product gas respectively

Predicting Reactions

 If K for a reaction is very large (>1000) then it is product favored

 If K is very small (<0.001), then the reaction is reactant favored

 If K is somewhere between these values, we need to use Qc to tell o Qc is the equilibrium expression for a reaction where the initial values of the products/reactants are plugged in o Example: aA ↔ bB b b [B] [B]i . K c  a while Qc  a [A] [A]i o If Qc > Kc the equilibrium shifts left due to too much product o If Qc < Kc the equilibrium shifts right due to too much reactants o If Qc = Kc the reaction is at equilibrium

Finding Concentrations at Equilibrium

 To do this, an I.C.E table is used (Initial, Change, Equilibrium) o On the top row, the reaction itself is written o The second row (I) holds the initial concentrations o The third row (C) shows how the reaction is changing from initial values to equilibrium values. The reactants change by –x, multiplied by any coefficients while the products change by +x, also multiplied by any coefficients o The bottom row (E) is the second and third rows combined o To solve for x, write a K expression where K equals the values of the bottom row. Then use the quadratic formula to find the two possible x values. Choose the one that logically works o Plug x in to the bottom row and these are the concentrations of each reactant/product at equilibrium o - Example: A 1.5M solution of HSO4 is added to a 0.75M solution of - SO4 . What are the concentrations at equilibrium? (Kc = 0.012) Acid-Base Chemistry +  This type of chemistry deals with the transfer of the H ion

 Brønsted definitions: o Acid – gives off H+ in an aqueous solution o Base – accepts H+ in an aqueous solution

 Acids o Strong acids include HCl, HBr, HNO3, HI, H2SO4 and HClO4 and always react completely (one directional reaction) o All other acids are weak and only partially react, resulting in an equilibrium reaction

 Bases o - - - Strong bases include H , OH , NH2 , and negatively charged C, also always react completely o All other bases are weak and only partially react

 Conjugates o Conjugate acids result when a base accepts an H+ ion o Conjugate bases result when an acid looses an H+ ion  Amphoteric substances can act as an acid or a base. The best example of this concept is water

 Kw o When water reacts with water, the equilibrium expression is   K c  [H 3O ][OH ] since pure liquids are not involved. This Kc has a special name, the Auto-Ionization Constant for Water and is -14 represented by the symbol Kw. It is equal to 1.0x10 o When any acid reacts with water, its K value is called KA and when

any base reacts with water, its K values is called KB o For any acid or base, Kw=KA*KB

 pH o How acidic or basic a solution is measured in the amount of + - hydronium (H3O ) or hydroxide (OH ) ion present respectively o Since these values are usually very tiny, the negative log of these concentrations are used instead. A lower case p represents that the negative log was taken of whatever follows the p o + + - So pH3O (or pH) = –log[H3O ] and pOH= –log[OH ] o + –pH - –pOH Also, [H3O ] = 10 and [OH ] = 10 o A pH of 0 is extremely acidic and a pH of 14 is extremely basic o For all liquids, pKw = pH + pOH = 14

 Finding the pH of a solution o For strong acids or bases, the initial concentration is equal to the concentration of hydronium or hydroxide respectively at equilibrium o For weak acids or bases, use an equilibrium expression and set it

equal to KA or KB depending on whether the species reacting with water is an acid or a base o There is a special “cheat” that can sometimes be used. If 100 times the K value is less than the initial concentration of the species reacting with water, then the “–x” in the denominator can be dropped o pH of a 0.25M HCl

o -5 pH of a 1.35M weak base (KB =1.8x10 )  Relating KA and KB to Acid/Base Strength o The stronger an acid is, the larger its KA is and the smaller the KB of its conjugate is o The stronger a base is, the larger its KB is and the smaller the KA or its conjugate is o Extreme cases of these situations occur with strong acids/bases

 Here, the KA or KB value is nearly infinite, so the conjugate KB or KA is nearly zero

 This means that the conjugate of a strong acid or base has no base or acid strength

 Predicting the pH of a salt solution (ionic compound) o Alkali metal cations have no effect on pH (Na+, K+, etc…) o Conjugate base of a strong acid is non-basic so no effect o Conjugate acid of a strong base is non-acidic so no effect o Conjugate base of a weak acid causes pH to rise o Conjugate acid of a weak base causes pH to lower o Transition metal cations are slightly acidic, causes pH to lower o Examples

 NaCl

 Alkali metal and conj. of a strong acid

 No effect / No effect, so solution is neutral  NaCH3COO

 Alkali metal and conj. base of a weak acid

 No effect / basic, so the solution is basic

 FeBr2

 Transition metal and conj. of a strong acid

 Acidic / No effect, so solution is acidic o If one part of a salt is acidic and the other part is basic, one cannot tell the general pH by inspection

 Neutralization Reactions o Here, an acid is added to a basic solution or a base is added to an acidic solution until the products are in equimolar amounts o This means that all of the reactants are used up, and only products remain o Since the volume of the solution is changing by adding acid or base, a M.I.C.E table is used (the M stands for moles)

 This functions exactly like an I.C.E table, but the moles are utilized instead of concentrations

 To find the new pH, you need to find the new concentration of your conjugate then react it with water in an I.C.E table to find the new hydronium or hydroxide concentration, and finally the new pH o A M.I.C.E table is used whenever a base is added to an acidic solution or vice versa. Not just in neutralization reactions. For these situations, remember to find which of your reactants is the limiting reactant o Example

 What is the pH of a solution resulting from the reaction of 30.0mL 1.50M CH3COOH with 15.0mL of 3.00M NaOH? (KA =1.8x10-5)

 Buffers o A buffer is a solution that resists pH change o It consists of a weak acid or base and its conjugate o Strong acids or bases cannot form buffers since their conjugates are non-acidic and non-basic o To find the pH of a buffer, use the Henderson-Hasselbach equation [conj. base] . For acids: pH  pK A  log   [acid]  [conj. acid] . For bases: pOH  pK B  log   [base]  . Note that since the second part is a concentration over a concentration, the volumes can cancel out and the moles of the acid and base can be used instead o Example

. If you have solution that is 1.00M in CH3COOH and 0.50M in – -5 CH3COO , what is the pH? (KA=1.8x10 )

. If 5.00mL of a 0.100M HCl solution is added to 100.0mL of the above buffer, what is the new pH?

o How to make a buffer of a desired pH

. Pick a weak acid who’s pKA is close to the desired pH . If given the desired molarity of one component, find the needed molarity of the other component . From the molarities and the required volume of buffer, find the mass required of each component o Example . If a buffer with a 4.25 pH is required and the concentration of the acid will be 0.75M, what is the required concentration of base given the following?

Acid KA Value Acetic acid 1.8x10-5 Benzoic acid 6.4x10-5 Propinoic acid 1.3x10-5

. How many grams of the sodium salt of the base will you need if you’re making 750mL of solution?

Titration Curves  This is a graphical representation of adding a strong base to a weak-acid solution or a strong acid to a weak-base solution. The x-axis is the volume added while the y-axis is the pH of the solution  There are four distinct sections of a titration curve that require four different methods to find the pH. The good news is that all four methods have already been covered!  Example o This is the titration curve for adding a base to an acidic solution. The same principals apply for adding an acid to a basic solution, just use logic to switch acid with base and vice versa when necessary

o The key to titration curves is to always think about what is in the solution at the point in question and always start from scratch o At the blue point, no base has been added yet. The solution only contains weak acid therefore a weak acid I.C.E table is used to determine the pH. o The points along the red line are where some base has been added. The base reacts with the weak acid and is completely consumed while making some conjugate base. Since there is weak acid and its conjugate base, the solution along here is a buffer so the Henderson-Hasselbach equation is utilized to find the pH o The green point is the equivalence point. This is where the acid has been neutralized by an equimolar amount of base. Use a M.I.C.E. table to find the moles of conjugate base created, find the new molarity of the base

and use a KB I.C.E. table to get the new pH o At the points along the teal line, we have the weak conjugate base and the strong base added both in solution. Since the strong base is so much more potent, we can ignore the weak base. We assume the hydroxide concentration is equal to the concentration of the strong base, so find the pOH from that information, then find the pH Solubility Equilibrium  In high school chemistry, it was learned that compounds were either soluble or insoluble. This was a lie!  The truth is not that extreme. All compounds are soluble to some extent  Since these are only partial reactions, equilibrium expressions and I.C.E tables are utilized  Example o Some solid compound MX dissolving into M+ and X– o Since solids are not part of equilibrium expressions, the one for this + – reaction is: Ksp =[M ][X ] mol  Using Ksp we can calculate the molar solubility of a compound (Smol in /L). g From this, we can then calculate the grams solubility (Sg in /L) by multiplying by the molar mass of the solid  Example o “Gold Bond” foot powder which uses ZnO to absorb moisture o –14 Given a Ksp of 1.8x10 , what is the Sg value for Gold Bond?

 The Common Ion Effect o The solubility of a solid will decrease if one or more of the ions it will dissolve into is already in solution  Example o Looking at Gold Bond powder again, what is the grams solubility in a 0.01M solution of NaOH?

 Note that there is a “cheat” for these problems also. If 100*Ksp is less than the molarity of any ions in solution, the +s for that ion can be ignored. (See the example if confused)

 How can “insoluble” compounds be made more soluble? o If a base anion is present in the compound, add acid. This will lower the concentration of that anion in solution so the equilibrium will shift right to compensate, making more of the solid dissolve o Many ionic compounds can be made more soluble with the addition of

NH3 or organic amines (R–NH2) . The lone pair on the nitrogen combines with the metal ion. The equilibrium will shift right to try and compensate o These principals are used to treat toxic waste-water. Soluble ionic compounds are added that make the toxic elements become insoluble when they combine. Usually, the compounds are added in such a way so that only one toxic element becomes insoluble at a time. This aids in the separation of toxic elements from any elements that might want to be recovered o Example . You have waste water with Ba+2, Ag+ and Na+ ions in waste water. What anions should be added in what order so that the Ba+2 and the Ag+ can be removed separately? Hint: Use Dr. LaDuca’s purple sheet . Answer: There are many solutions to this problem, so don’t worry if your answer doesn’t match exactly. It is easiest to remove the silver than the barium so add OH–, Cl–, Br-, or I– to + –2 – –2 –3 precipitate the Ag , then added SO4 , F , CO3 or PO4 to precipitate the Ba+2

Reduction–Oxidation Reactions  Whereas Acid-Base deals with the transfer of H+ ions, Red-Ox reactions deal with electron transfer  Reduction – electron gain  Oxidation – electron loss o Remember LEO the lion says GER o Loose electrons = oxidation, Gain electrons = reduction  Oxidizing agent – Causes oxidation and is reduced  Reduction agent – Causes reduction and is oxidized  Example o Fe + O2 → Fe2O3 (rust) o Fe0 → Fe+3 is the oxidation reaction o O0 → O–2 is the reduction reaction  To balance a red-ox reaction, both the atoms and the charges must balance out. To do this follow these steps (these are covered on the yellow and pink sheets given out during class) 1. Assign oxidation numbers (see yellow sheet) 2. Separate the reaction into an oxidation half-reaction and a reduction half-reaction 3. Balance all non oxygen and hydrogen atoms first + 4. Add H2O to balance the oxygen atoms. The add H atoms to balance the hydrogen atoms 5. Add enough electrons to balance the charge in each reaction 6. Multiply both reactions by whatever is necessary to make the number of electrons added to each side equal 7. Add the two half-reactions together and fully or partially cancel out anything that does. All electrons must cancel. Check to make sure atoms and charges balance 8. If you have an acidic solution, you’re done 9. If you have a basic solution add OH– to equalize the H+. Combine

them to make H2O and cancel anything that cancels. Check to make sure atoms and charges balance.  Example

o What is the balanced red-ox, single replacement reaction of: –2 Zn + Cu(OH)2 → Zn(OH)4 + Cu ?

Electrochemistry  This is the section of chemistry that studies the reactions that take place between an electronic conductor and an ionic conductor (e.g. an electrochemical cell or a battery)  Red-Ox chemistry is the driving force behind electrochemistry  This is an electrochemical cell:

o The left side of the cell is where reduction occurs (cathode)

o The right side of the cell is where oxidation occurs (anode)

o Electrons flow along the wire from anode to the cathode

o The salt bridge sends its positive and negative ions to the cathode and anode respectively to balance the charges

o The anode will shrink because it is being made into ions

o The cathode will grow because its ions are turned to solid

o The voltage of the individual half cells cannot be measured, only the entire cell at once + o To solve this problem, scientists have defined the reduction of 2H + – 2e → H2 (g) as having a voltage of 0.0

o All other half cells are based off of this definition

o Ecell = Ereduction + Eoxidation  In the example above, 1.10V = 0.34V + 0.76V

o For any half-cell, the oxidation and reduction voltages are equal but opposite in sign

o The voltage for a half cell is constant, molar coefficients have no bearing on it

o If Ecell is greater than 0, the reaction will want to occur

o Voltages for half-cells are usually given in the reduction form. To get the oxidation voltage, reverse the sign as said above  Example + – o Ag + e → Ag(s) Ered = 0.80V +2 – o Cu + 2e → Cu(s) Ered = 0.34V

o Combine and balance the above half-reactions so that the reaction will want to occur

nd o The 2 possibility is the one with a positive Ecell so it is the reaction that will want to occur

 Batteries work under the same principal as seen below  Batteries run out because the cathode eventually runs out of solid to turn into ions

o Nickel-Cadmium batteries work by these half-reactions: – – Red: NiO(OH) + H2O + e → Ni(OH)2 + OH – – Ox: Cd + 2OH → Cd(OH)2 + 2e

o The Ecell for this reaction is 1.25V

o By applying a voltage greater than 1.25V, the above processes can be reversed resulting in a recharging of the battery

Thermodynamics Part II  Quick Review

o ΔH is the symbol for enthalpy ° ° o ΔH = [H f (products)]  [H f (reactants)]

o Reactions that give off heat are exothermic (ΔH < 0)

o Reactions that take in heat are endothermic (ΔH > 0)  New Terms

o Spontaneous – a reaction will occur

o Non-Spontaneous – a reaction will not occur  Entropy (S)

o This is the measure of inherent disorder in a system nd o The 2 Law of Thermodynamics states: The entropy of the universe is always increasing

o The entropy of a system can increase (S > 0) or decrease (S < 0)

o Systems tend toward more disorder unless work is done to it q J o S = /T meaning heat divided by temperature (units of /K)

o For a solid compound at 0°K, S = 0

o Since the T in the entropy definition is in Kelvin, S can never be negative. However, ΔS can be negative

o When comparing states of matter, Sgas is much larger than Sliquid which

is only slightly larger than Ssolid

o It is relatively easy to predict the sign of S by inspection. The formula for finding ΔS is the same as the formula for ΔH, with entropy substituted for enthalpy

o Example

. What is the sign of S for: CaO(s) + CO2 (g) → CaCO3 (s) ? . Answer: Since the S of gas is so much larger than that of liquids or solids, the entropy on the left is much larger. Therefore ΔS in this case will be negative

 Gibbs Free Energy (G) o Through some manipulation and substitution this:

ΔSUniverse = ΔSSystem + ΔSSurroundings becomes this:

–TΔSUniverse = –TΔSSystem + ΔHSurroundings o We represent –TΔSUniverse with ΔG (Gibbs Free Energy) o So, ΔG = ΔH – TΔS o For spontaneous reactions, ΔG < 0 o For non-spontaneous reactions ΔG > 0 . If ΔG > 0, the reverse reaction is spontaneous o At standard conditions (25°C, 1atm, 1.0M), ΔG= ΔG° o ΔG° has been measured or computed for many substances o If asked to find ΔG at standard conditions for a reaction use ΔGf° in

the same way you would use ΔHf° for finding enthalpy change o If asked to find ΔG at non-standard conditions, use the equation derived at the top of this page KJ . Remember to convert S values to units of /K . Remember to multiply values by their molar coefficients o Example . Given this reaction:

. Fe2O3 + 3CO → 2Fe + 3CO2 . Find ΔG at standard conditions . Now find ΔG at 227°K

Fe2O3 CO Fe CO2 kJ Hf° ( /mol) -825.5 -110.525 0 -393.509 J S° ( /K*mol) 87.40 197.674 27.78 213.74 kJ Gf° ( /mol) -742.2 -137.168 0 -394.359 o As stated before, for spontaneous reactions, ΔG < 0 and for non- spontaneous reactions ΔG > 0 o If ΔG = 0, then the reaction is at equilibrium and this is the point where the reaction switches between spontaneous and non- spontaneous

. To find the temperature at which this happens (Tc = crossover

H f temperature), use Tc = S

. Note that Tc is still in K, so if ΔHf or ΔS is negative while the other is positive, the result is negative Kelvin

. This obviously is a problem. However, if Tc is negative, it simply means the reaction in question is always spontaneous or always non-spontaneous . To find which, use the Gibbs Free Energy equation to find the sign of ΔG o Using complicated Physical Chemistry calculations (don’t worry about them) we know that ΔG = ΔG° + RT*ln(Q) . Q is the reaction quotient (see Predicting Reactions) and R is J 8.314 /K*mol . At equilibrium, ΔG=0 and Q(c or p) = K(c or p) so the equation is rewritten as ΔG° = –RT*ln(K) o ΔG can also be tied into Electrochemistry ° . If an electrochemical (red-ox) process occurs, E > 0 . Since ΔG < 0 for spontaneous reactions, we can say that ΔG is proportional to –E° . To make an equation relating these two variables, introduce constants to get ΔG = –nF E°  n is the moles of electrons being transferred (consult the red-ox equation) J  F is the Faraday constant (96500 /V*mole electrons)

Nuclear Reactivity  Deals with reactions that occur in the nucleus  Relevant to medicine, energy, warfare and the environment  1896 – Becquerel is working with uranium o Noticed that it exposed the photo paper in his dark desk. Why?  1899 – Marie Curie proposes the idea of radioactivity o Radioactivity is the energy emitted when nuclei decay  1907 – Rutherford finds radioactivity can convert one element into another element  Isotopes o A Represented by Z X . X is the elemental symbol . Z is the atomic number (# of protons) . A is the atomic mass (# of protons + # of neutrons) o Two elements with the same Z value but different A values are called isotopes 1  Protons are represented by 1 p 1  Neutrons are represented by 0 n 0  Electrons are represented by -1 e  In nuclear reactions, both A and Z must balance  Rutherford founds two types of radioactivity o When heavy, slow, positive particles are given off, this is called α (alpha) radiation 4 4 . 2 He particles are given off, often written as 2  4 . 2  are formed via ejection from a larger nucleus . Able to be stopped by a piece of paper o When light, fast, negative particles are given off, this is called β (beta) radiation . An electron is ejected from a neutron . The neutron then turns into a proton 0 . Represented by -1  . Able to be stopped by a thin sheet of heavy metal  Villard discovered a third type of radiation o When extremely high energy photons are given off, this called γ (gamma) radiation o 0 These particles are represented by 0  o When alpha or beta particles are given off, the nuclear configuration becomes extremely unstable o When the nucleus reorganizes, γ rays are given off o These rays are horribly toxic  rays . H2O  H• and OH• . OH• shreds DNA in cells  Nuclear Reactions o Two main types 238 4 234 . 92U  2   90Th  Alpha rays decrease Z by 2 14 0 14 . 6 C  1  7 N  Beta rays increase Z by 1 o Two rarer types . Positron emission  A positron is an anti-electron (anti-matter) 0  Represented by 1   A proton turns into a neutron and a positron 11 0 11  6 C  1  5 B  Z decreases by 1  Positron Emission Tomography – PET scan . Electron capture  A proton in the nucleus grabs an electron from the 1s shell and they combine to form a neutron 55 0 55  26 Fe  1e  25 Mn  Z decreases by 2  Fission o Occurs when heavier nuclei are bombarded with smaller nuclei o The heavier nuclei then fragments  Fusion o Occurs when a larger nuclei is assembled from two or more smaller nuclei 235 1 fusion 236 fission 92 141 1  92U  0 n  92 U  36 Kr  56 Ba  30 n o The uranium resulting from the fusion reaction is extremely unstable and immediately undergoes fission o The three resulting neutrons fly off and cause a chain reaction by creating three new reactions of their own  The masses of the product particles are only slightly smaller than the mass of the reactant particles (about 0.007% smaller) o The missing mass is converted to light via E=mc2 o So even though the mass is extremely small, since it is being multiplied by the speed of light squared, E become very large  A chain reaction in a controlled environment = nuclear reactor  A chain reaction in an uncontrolled environment = Chernobyl  A chain reaction in a purposely uncontrolled environment = bomb  How Nuclear Reactors work  Fusion 235 4 239 o 92U  2  94 Pu

. Pu is used in weapons 239 4 240 1 1 o 94 Pu 2  95 Am1p 20 n

. Am is used in smoke detectors o Fusion powers the sun 1 4 0 . 41 H 2 He 21   

. Results in 0.7% mass loss 8 . Temperature is at about 10 °K 9 kJ . Energy release from this is equal to about 10 /mol  Radioactive Decay o Always first order and temperature independent o Depends only on the nuclear stability of the element/compound

[A]0  o ln   kt  [A]t 

. [A]0 is the initial amount of radioactive material . [A]t is the amount left at time t . t1/2 is the half life of the substance . How long it takes for half of the substance to decay ln(2) . k  t 1/ 2

. Make sure all times are in the same units o Example 239 . The half life for Pu is 24,100 years. Find the time necessary for a sample of Pu to decay to 10% of the original amount . For this problem, the actual amount of material doesn’t need to

be known. We set [A]0 to 1 (100%) and [A]t to .10 (10%)

 Medical Imaging 99 o 43Tc is a beta-emitter with a half life of six hours

. Used in imaging techniques such as a bone, heart, liver, kidney or neurological exams . The radiation given off is easily traceable and hardly damaging to the body since it is gone so son 131 o 53 I is also a beta-emitter

. Used in thyroid tests . If thyroid removal surgery is required, high levels of this can be injected which will destroy the thyroid . No invasive surgery required! 18 o 9 F is a positron emitter 18 . There are 9 F tagged glucose molecules

. Allows the examination of metabolism with the use of glucose in the brain via a PET scan 198 o 79 Au is a strong beta-emitter

. Implanted as a radioactive “seed” into tumors to reduce them . Often used with prostate cancer 60 o 27 Co is an extremely potent gamma-emitter

. Effective against brain tumors  Device worn around the head focuses the gamma radiation on specific spots and removes the tumor . Some collateral damage, but overall beneficial . These rays can also be used for food safety  Meat is passed on a conveyor belt past a contained sample of cobalt-60  The gamma radiation shreds the DNA of any bacteria  This method of sterilization is known as “irradiation” so the meat is labeled “Irradiated meat"

Point Breakdown for the Final Exam  200 points total  20 points – Gas Law  36 points – Acid Base Titration (Multi-Step)  20 points – Thermodynamics  15 points – Nuclear Chemistry  25 points – Red-ox Reactions  15 points – Intermolecular Forces  25 points – Kinetics  20 points – Colligative properties  24 points – Short Answer Questions about Equilibrium