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General Chemistry 2

by: Guiseppe Bednar

General Chemistry 2 CHEM 1133

Guiseppe Bednar

GPA 3.93

Robert Parson

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Robert Parson
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This 99 page Class Notes was uploaded by Guiseppe Bednar on Friday October 30, 2015. The Class Notes belongs to CHEM 1133 at University of Colorado at Boulder taught by Robert Parson in Fall. Since its upload, it has received 15 views. For similar materials see /class/232183/chem-1133-university-of-colorado-at-boulder in Chemistry at University of Colorado at Boulder.


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Date Created: 10/30/15
ChEm 1133 Fall zmz Elech39uchamstxy Fan Electrochemistry Oxidationreduction reactions involve electron transfer In 39 ml nhv imllv separate the l l 39 Anode inmanam Anme mmm maze mm Figure m A galvanic all Anode Zns in a Zn2 solution Zns gt Zn2 aq 2539 Cathode Cus in a Cu2 solution Cuzaq 2239 Cus Overall reaction Cu2 aq Zns gt anaq Cus Saltbridge closes circuit Oxidation always occurs at the anodeReduction at the cathode Chem 1133 Fall 2012 Electrochemistry Part 1 Voltaic Galvanic Cell spontaneous reaction produces electrical power Cu2 aq Zns gt Zn2aq Cus Electrolytic Cell electrical power drives a nonspon taneous reaction Zn2aq Cus gt Cuz aq Zns In both oxidation at anode reduction at cathode Electrons always ow from the anode to the cathode Chem 1133 Fall 2012 Electrochemistry Part 1 Electrical Quantities and Units Charge q Units are Coulombs C 1 electron 1602X103919 C 1 mole of electrons 620X1023 mol3911602X103919 C 96485 Cmol 1 Faraday 1 F a 96485 Cmol sometimes 96500 Current I charge time I qt Units are amperes A 1 A 1 Cs Voltage or EMF 6 Energycharge Units are Volts V 1 V 1oule 1 Coulomb Power Energytime charge EMFtime P q E t I Units are Watts 1 W 115 Power delivered by a voltaic cell depends on both E and I Chem 1133 Fall 2012 Electrochemistry Part 1 Standard Cell EMFs Cell EMF depends on Cell EMF does not depend on 39 Cell constituents type of Electrode size electrode ions etc 0 Volume of solutions Concentrations of ions in soln 0 Temperature Define Standard conditions All ion concentrations 10 M Temperature 25 C 298 K Then Coceu 5 Cell EMF in standard conditions If Cocen gt 0 Cell reaction is spontaneous as written If Cocen lt 0 Reverse of Cell reaction is spontaneous Soon we will learn how to get Ccen in nonstandard conditions Chem 1133 Fall 2012 Electrochemistry Part 1 Notation for electrochemical cells If the cell reaction is Zns qu aq gt Zn2aq Cus write the cell as Zns I Zn2aq II Cu2aq I Cus shorter ZnIan II Cu2ICu Anode on left cathode on right 39 Oxidation on left reduction on right 0 Solids on outside solutions on inside denotes a phase boundary solidliq a salt bridge Later we ll put in concentrations e g Zn Zn205 M II Cu2025 M Cu If concentrations are not given assume standard 10 M Cell EMF s are additive Zn I an qu I Cu 6 C 110 V Cu I qu II Ag I Ag 6 Cquot 046 V Zn Zn2 Ag Ag 6 C 110 v 046 v 156 v We can measure the voltage of each couple against a reference electrode then determine all other cell EMF s from this list Chem 1133 Fall 2012 Electrochemistry Part 1 Reference Standard Hydrogen Electrodequot SHE H29 1 01m gt 2 Haq 10 M 2 e39 For any halfreaction An gtA define 60 as the EMF measured when the halfreaction is used as the cathode in an electrochemical cell with SHE as the anode Standard Reduction Potential Combine SHE with CuCu2 find Cu2 is reduced to Cu Pt IH21 atm I H1M II Cu21M I Cus ceu for this cell is 034 V List this as 60 for the halfreaction Cu 2 e39 Cus e 034 v Next combine ZnZn2 with the SHE find that Zn is oxidized Zn2 Zn H H IH2g PT 6 C 076 v So 60 for the oxidation of Zn by SHE is assigned the value 076 V This means that 60 for the reduction Zn22e39 gt Zn is assigned the value 076 V In other words SHE oxidizes Zn Chan 1133 Fall 2012 Eleckochemxstry Part 1 Standard Reduction Potentials at 2530 Higher halfreacfions dr39ive lower39 ones in reverse Au3 Ag gt Au Ag 5n Ni Sn Ni Cuz Ni gt Cu Niz Chem 1133 Fall 2012 Electrochemistry Part 1 Working with standard reduction potentials Mgz2e39 Mg e 237 v Nae39 a Na e 271 v Conclude that Mg2 will oxidize Na To get cell EMF turn lower reaction aroundquot and add Mg2 2e39 gt Mg 6 237 V Na gt Nae39 e 271v Mgz 2N0 Mg 2Na e 237 271 033 v gt 0 So the reaction is spontaneous as written Important notes 1 60 reverse rxn 60 forward rxn 2 DO NOT multiply 60 values by stoichiometric coefficients Chem 1133 Fall 2012 Electrochemistry Part 1 Concentration Dependence of Cell EMF Nernst Equation Zns 2 A9 aq e anaq 2 A9s How does ce11 depend on Ag and Zn2 Know two things 1 When Zn2 10 M Ag ce11 ceu standard conditions 2 At equilibrium 6 cell 0 dead battery As the cell runs concentrations move towards equilibrium and ce11 decreases For general concentrations Zn2 T Tempera rure in Kelvin R 8314 JmoIK F 96485 Cmol n moles elec rrons Transferred in The baanced chemical equa rion 0 RT 88 CVLn 2 no Ag Chem 1133 Fall 2012 Electrochemistry Part 1 General form of the Nernst Equation For the general reaction a A b B gt C C d D C Dd erte Q a At equlllbrlum Q K A Bb Then for39 any se r of concentrations 8 80 LnQ Check two special cases 1 Ifall s 10 MQ 1 ThenLnQ0and 660 2 At equilibrium Q K so 6 0 RTHF LnK But at equilibrium 6 0 so 20 RTnF LnK Which gives us a relation between 60 and K Chem 1133 Fall 2012 Electrochemistry Part 1 Derivation of the Nernst Equation not rigorous but contains the basic physical ideas From thermodynamics the Free Energy AGis the maximum work that can be gotten from a rxn at constant temperature and pressure AG0 is the free energy change under standard conditions Thermo tells us AG AG0 RT LnQ Work power time voltage current time AG e I t 601 Set q nF then AG nF 6 Plug in AG0 RT LnQ nF Set AGO nF 0 6 60 RTnF Ln Q Chem 1133 Fall 2012 Electrochemistry Part 1 Criteria for quotSpontaneousquot Change 1Kgt1 2AG0lt0 3 0gt0 These are consistent AG0 RT Ln K 60 RTnF LnK So ifK gt1 then AGO lt 0 and 60 gt 0 quotSpontaneousquot in this context really means if you start with reactants and products in their quotstandard statesquot everybody at 10 molL equilibrium lies to the right the equilibrium product concentrations are larger than the standard state product concentrations and the equilibrium reactant concentrations are smaller Chem 1133 Fall 2012 Electrochemistry Part 1 Another form of the Nernst Equation Switch to base10 Log Ln Q 2303 LogQ IfT 25 C 298 K can combine several numbers LnQ 8314298152303 LogQ 96485 l e 60 005917n LogQ and let 005917n LogKl Warning These equations are only true at T 25 C There is absolutely nothing new here This equation is a relic ofthe pre calculator era a time when people did these sorts of calculations by consulting tables of logarithms and it was a lot easier to find tables of common base10 logs than tables of natural basee logs Nevertheless you still see this in textbooks so I include it Chem 1133 Fall 2012 Electrochemistry Part 1 Calculating Equilibrium Constants from 60 values Ag e39 gt A9 6 2 07996 V AgNH32 e39 gt A9 2 NH3 60 0373 V Want K at T298 K for the reaction Ag 2 NH3 gt AgNH32 Combine halfrxns to make desired rxn then add 6 0 values Ag e39 gt A9 6 2 07996 V Ag 2 NH3 gt AgNH32 e39 60 0373 V Ag 2 NH3 gt AgNH32 60 04266 V 1 e39 transferred in balanced rxn so n1 0 RT 8314 298 g TLnK 04266 m3 196458 LnK LnK 1661 so K e721 16X107 Chem 1133 Fall 2012 Electrochemistry Part 1 Concentration Cells convert concentration difference into voltage Zns Zn 010 M II Zn 10 M Zns Same half reaction on both sides differ only in concentration Anode Zn gt Zn2 2 e39 60 076 V Cathode Zn2 2 e39 gt Zn 60 076 V Ne r Zn Zn2 gt Zn2 Zn Eoce O 6 5 RTnF Ln an1 Zn2 0 8314298296485 LnO1010 00296 v Voltages in concentration cells tend to be small they are used primarily to measure concentrations Example pH meter measures H H30 General Chemistry II CHEM 1133 Lecture and Recitation Schedule Fall 2012 Dates Topics Week 1 Recitation 1 Chemical Equilibrium Aug 27 Aug 29 Aug 31 Week 2 Sept 3 Sept 5 Sept 7 Course Introduction Review of Chemical Equilibrium Intro to Acids and Bases Autoionization and the pH scale Recitation 2 Strong and Weak Acids Labor Day No class Strong and weak acids and bases quotICEquot tables Strong and weak acids and bases ctd Week 3 Recitation 4 Acid Base Properties Sept 10 Sept 12 Sept 14 Conjugate AcidBase pairs Common Ion Effect and Buffers Buffers continued Week 4 Recitation 5 Buffers and Titrations Sept 17 Sept 19 Sept 21 Titrations Titrations continued Polyprotic acids Week 5 Recitation 6 Thermodynamics Sept 24 Sept 25 Sept 26 Sep 28 Catchup and Review for Exam 1 EXAM I 700 pm Chem 140 Entropy and the Second Law Energy entropy and chemical reactions Schedule 7 Page 1 Textbook chapters 15 review 93161 2 163 4 16 5 6 16 6 7 167 81610 171 2 172 173 173 169 181 3 184 5 General Chemistry II Chem1133 Lecture and Recitation Schedule Fall 2012 Dates Topics Chanters Week 6 Recitation 7 Redox Reactions Oct 1 Entropy quotFree Energyquot and equilibrium 186 7 Oct 3 OXidation Reduction Reactions 94 191 Oct 5 Redox reactions and Electrochemical Cells 192 Week 7 Recitation 8 Redox of halogens Oct 8 Standard cell potentials 193 4 Oct 10 Concentration dependence Nernst Equation 195 Oct 12 Concentration dependence continued 195 Week 8 Recitation 9Electrochemistry Oct 15 Practical electrochemistry batteries 196 Oct 17 Electrolysis 197 Oct 19 221 Chemistry of Transition Elements Week 9 No recitation Oct 22 Oct23 Oct 24 Oct 26 Week 10 Recitation 10 Coordination Compounds Oct 29 Oct 31 NOV 2 Catchup and review for Exam II Exam II 700 pm Chem 140 Coordination compounds of transition elements Coordination compounds ctd Solubility Equilibria Solubility Equilibria ctd Solubility and Separations Schedule 7 Page 2 22 1 2 222 174 175 176 General Chemistry II Chem 1133 Lecture and Recitation Schedule Fall 2012 Dates Topics Chanters Week 11 Recitation 11 Solubility Nov 5 Crystal eld theory 223 Nov 7 Crystal eld theory ctd 223 Nov 9 Coordination compounds in biology 225 Week 12 Recitation 12 Crystal Field Theory Nov 12 Catchup and Review for Exam III Nov 13 EXAM III 7 00 pm Chem 140 Nov 14 Chemical kinetics reactions and rates 141 3 Nov 16 Determining rate laws from experiment 144 Week 13 FALL BREAK Nov 19 through Nov 23 Week 14 Recitation 13 Reaction Rates Nov 26 Integrated rate laws 145 Nov 28 Temperature and activation energy 146 Nov 30 Reaction mechanisms 147 Week 15 Recitation 14 Mechanisms and Catalysis Dec 3 Catalysis and enzymes 148 Dec5 Nuclear chemistry and radioactivity 201 2 Dec 7 Kinetics of radioactive decay 203 Week 16 Recitation 15 Final Exam Jeopardy Review Dec 10 Induced radioactivity ssion fusion 204 6 Dec 12 Catchup and nal topics Dec 14 Review for nal Dec 18 FINAL EXAM 730 am 1000 am location TBA Schedule 7 Page 3 Chem 1133 Fall 2012 OxidationReduction Redox Reactions Historically oxidation meant react with oxygen 2 Fe 3 Oz F8203 Reduction meant lose oxygen so as to recover a metal from its ore F8203 2 Fe 3 02 But compare these reactions 2M9 02 MgO Mg F2 Mng In each case Mg became Mg2 it lost electrons Other reactantgained electrons O gt 0239 F gt F39 Modern definition Oxidation lose electrons Reduction gain electrons quotOILRIGquot Oxidation is Loss Reduction is Gain Chem 1133 Fall 2012 Extend concept to covalent compounds H2 F2 gt 2 HF 8 6 gt HF bond is polar H F or H F The electron on H is partially transferred to F when HF bond forms Hydrogen atom loses electrons H is oxidized Fluorine atom gains electrons F is reduced Define the oxidation number of an atom in a molecule as the charge it would have if all of the polar covalent bonds were fully ionic HF bond polarized towards F quotPushquot electron all the way to F H has oxidation number 1 F has oxidation number 1 Chem 1133 Fall 2012 Rules for oxidation numbers 1 Atoms in elements oxidation zero 2 Monatomic ion ox charge on ion 3 In neutral molecules ox 39s sum to zero 4 In polyatomic ions ox 39s sum to the charge 5 F always has ox 1 except in F2 6 H usually has ox 1 except in H2 and in metallic hydrides such as NaH CaHz where it39s 1 7 0 usually has ox 2 except in 02 in compounds with F and in peroxides such as H202 and 02239 where it39s 1 Chem 1133 Fall 2012 Redox Summary If something is oxidized it behaves as a reducing agent Loses electrons and increases its oxidation number If something is reduced it behaves as an oxidizing agent Gains electrons and decreases its oxidation number The Oxidizing Agent is reduced the Reducing agent is oxidized 2 Mg 02 gt 2 MgO Mg is oxidized it is the reducing agent 0 2 is reduced it is the oxidizing agent 02 oxidizes Mg and Mg reduces 02 2 Mg C02 gt 2 MgO C What is the oxidizing agent Chem 1133 Fall 2012 Famous Oxidizing Agents 02 F2 Clz Bra I2 MnO439 permanganate CrzOf dichr39oma re U5ually seen as K sal rs HNOs H2504 HOCIOg 03 ozone H202 hydrogen peroxide 02239 Famous Reducing Agents Alkali Group I metals Li Na K Alkaline earth Group 2 metals Mg Ca Metal hydrides NaH LiAH4 Chem 1133 Fall 2012 Balancing Redox Reactions 1 Break into half reactions oxidation half electrons are lost oxidation numbers increase reduction half electrons are gained oxidation numbers decrease For each half reaction 2 Balance all atoms except H and 0 3 Balance 0 by adding H20 where needed 4 Balance H by adding H where needed 5 Balance charge by adding electrons oxidation halfrxn add e39 s to right side reduction halfrxn add e39 39s to left side 6 Recombine half reactions so that electrons cancel 7 Ifsolution is basic add OH39 to cancel all H H OH39 H20 Chem 1133 Fall 2012 Check List for a balanced redox reaction All elements must balance All charges must balance No loose electrons in the full reaction In acidic solution no free OH39 In basic solution no free H In neutral solution ok to have either H or OH39 in final reaction Chem 1133 Fall 2012 Example Oxidation of Ethanol by Dichromate 1 Oxidation half rxn C2H50H gt CO2 Balance C C2H50H gt 2 CO2 Balance 0 with H20 3 H2O C2H5OH 2 CO2 Balance H with H 3 H2O C2H5OH 2 CO2 12 H Balance charge with e39 3H2O C2H5OH gt 128 C2H5OH Cr207239 gt CO2 Cr3 Reduction half rxn Cr207239 gt Cr3 Balance Cr Cr207239 gt 2 Cr3 Balance 0 with H20 Cr207239 gt 2 CP3 7 H2O Balance H with H 14 H Cr207239 gt 2 Cr3 7 H20 Balance charge with e39 6e39 14H Cr207239 gt ZCr3 7H20 Combine the halfreactions multiply reduction half by 2 to make electrons cancel 3 H20 C2H50H 12 e39 28 H 2 Cr207239 gt 2 CO2 12 H 12e39 4 Cr3 14 H20 Clean up C2H5OH 16 H 2 Cr207239 a 2 cog 4 Cr3 11 H20 In basic solution add 16 OH39 to each side to cancel H C2H5OH 5 H2O 2 Cr207239 2 cog 4 Cr3 16 OH Chem 1133 Fall 2012 Chemical Equilibrium fast and furious review Many chemical reactions do not go to completion 7 instead they approach an equilibrium state where some of the reactants are still there On the microscopic level individual reactant molecules are continually banging into each other and making products but the products are also reacting in the reverse direction to reform the reactants When the rates of reactants gt products and products gtreactants are equal you have reached an equilibrium We ignored this for most of introductory chemistry because there are lots of reactions for which the equilibrium position is pretty much all products and no reactants eg 2 H2 O gt 2 H20 7 when you burn hydrogen it all burns We said that those reactions went to completion But there are lots of other reactions where this isn t true Example hemoglobin quotHbquot binds oxygen Hb O gt HbOz at equilibrium HbOzHbOz K02 K02 is a constant that is independent of how much Hb and O we started outwith or of whether we started with HbOz or if we started with some of each Its value does depend on temperature Hb also binds carbon monoxide Hb CO ltgt HbCO The K for this reaction is much larger than K for 02 which is what makes CO poisonous large K means mostly products at equilibrium so most ofthe Hb is tied up as HbCO and is not available to bind 02 Later we will see how to calculate K from other things for now we regard it as something that we must measure More examples NO Clz NOCi2 C012 C02 Pure solids and liquids do not appear in K 2 NOClg 2 N0g Clzg K Cs C02g 2 C0g K Solving Equilibrium Problems ICE Tables Two underlying principles 1 Conservation of matter atoms of each type cannot change in a chemical reaction 2 Equilibrium constant expression 20 mol NOCI isplaced in a 40 L ask At equilibrium NOCI 048M FindK We are given NOCl at equilibrium We are not given NO or C12 But we can gure them out using conservation of matter for every two molecules of NOCl that react two molecules of NO and one molecule of C12 are formed An ef cient way to use this information is to make an quotICE tablequot Enter the initial concentrations of all reactants and products and the equilibrium concentration of NOCl Initial concentration of NOCl 20 moles40 L 050 M Initial concentration of NO zero Initial concentration of C12 zero since we start out with reactants only Equilibrium concentration of NOCl 048 M Chem 1133 Fall 2012 Concentrations M 2 NOClg 2 N0g Clzg 0 Initial 0 50 0 Change 2X 2X X Equilibrium 048 2X X Add up rows since we know both the initial and equilibrium concentrations of NOCl we can get the change quotXquot 050 72X 048 so X 001 M Then NO 2X 002 M and C12 X 001M Put these all into K N0 2 Cl K M 00220010482 17x10395 NOCl In your rst lab you will do something similar The difference is that you will be measuring the equilibrium concentration of a product What good is a K It allows you to determine what the equilibrium concentrations in a particular problem will be given initial concentrations and K How can we nd out which way a reaction will proceed Construct a quantity called the Reaction Quotient Q Q is given by a formula that has the same appearance as K ithe difference is that the concentrations in the formula are initial 39 not 1 quotquot 39 39 Q tells us quotwhat sidequot of the equilibrium we re at Compare Q to K if Qlt K an will go forward products increase reactants decrease until QK If QgtK an will go backwards until QK Things to note about Q and K 1 Q and K are unitless 7 pure numbers Concentrations in Q and K are actually ratios of concentrations relative to a reference concentrations 7 units divided out 2 Pure solids or liquids never appear in Q or K Their concentrations don t change even though their amounts do 3 Q and K depend on how the reaction is written Reaction must be balanced What happens to Q and K when you change the way the reaction is written a Reverse the reaction switch products for reactants new K 1old K b Multiply reaction through by a number 11 new K old Kn Add two reactions together multiply their K s Subtract divide the K s T itrations ACIDBASE TITRATIONS Purpose 1 Find the concentration ofan unknown acid or base 2 Find the Ka of the acid or Kb of the base Procedure Slowly add measured amounts drop by drop from a buret of a known base or acid to an unknown acid or base pH changes rapidly when the moles of base added is equal to the moles of acid that were originally present This is the equivalence point EP Four cases Discuss 1 and II in lecture others analogous 1 Strong acid titrated with a strong base 11 Weak acid titrated with a strong base 111 Strong base titrated with strong acid IV Weak base titrated with strong acid T itrations Case 1 Strong acid titrated with strong base Review Gen Chem 1 Burdgeamp0verby Chapter 9 Suppose we titrated 100 ml 010 L of an HClaq solution concentration unknown with 020 M NaOHaq Suppose the ER was reached when 50 ml ofthe NaOH was added What was the initial HCl Since both HCl and NaOH are completely ionized HCl H20 gt H30 Cl39 Na0H H20 gt OH Na The net reaction is H30 OH gt 2 H20 At EP moles base added 020 molL0050 L 0010 mol original moles acid 0010 mol HClo 010 mol010 L 010 M 7mm Strong acid Strong lJase titration pH along the titration curve Start pH Log acid In example pH rLog l 1 At equivalence point Original acid all neutralized neither H30 nor OH39 is in excess ideally Between start and EP acid in excess pH increases slowly then jumps near E Beyond EP base in excess pH rises and attens out 15 mL n1 F N20H 11mm T itrations Strong acid titrated with strong base Sample calculation Suppose 100 ml of 010 M HCl is titrated with 020 M NaOH aq What is the pH after the rst 10 ml of base is added Initial moles HCl 0010 mol All present as H30 Number ofmoles of NaOH added 020 molL0010 L 0002 mol All go in as OH39 H30 OH39 gt 2 H20 goes to completion So subtract moles NaOH added from initial moles HCl to get the moles H30 remaining at this point in the titration 0010 mol 0002 mol 0008 mol Don39t forget to keep track ofthe volume ofthe solution which changes during the titration V 100 ml 10 ml 110 ml So H30 0008 mol0110 L 007 M pH LogH30 Log007 114 After equivalence point base is in excess To get pH calculate moles base added after the ER was reached divide by the total volume to get OH39 T itrations CASE II Weak Acid titrated with Strong base At the ER the moles base added original moles acid So calculate the concentration of the unknown acid as in Case I BUT pH at the Equivalence Point is not equal to 7 Why Consider acetic acid titrated with NaOH HOAC OH39 gt H20 OAc Start out with HOAC almost no OAC39 Titrate to EP all of the original HOAC is converted to OAC39 But OAC39 is a weak base OAc39 H20 HOAC OHquot So solution at EP has excess OH39 therefore pH gt 7 T itrations SASB titration eg HCI NaOH gt NaCI H20 At EP have a solution of NaCl which is neutral WASB titration eg HOAC NaOH gt NaOAc H20 At EP have a solution of NaOAc which is basic OAc39 H20 HOAc OH39 Na H20 gtNo Reaction WBSA titration eg HCl NH3 gt NH4Cl At EP have a solution of NH4Cl which is acidic NH4 H20 NH3 H30 Cl39 H20 gt No Reaction T itrations Weak acid titrated with strong base pH along the curve Divide into four regions need to use dl erent formulas in each Region 1 Before the titration starts Region 2 Between the start and the equivalence point Region 3 At the equivalence point Region 4 Past the equivalence point T itrations Region 1 Before the titration starts Have a weak acid in water HA H20 H30 A39 Calculate pH using methods of Chapter 18 Region 2 Before the equivalence point Acid in excess OH39 limiting HA OH39 gt A39 H20 Titration has converted some ofthe HA into A39 but not all so A39 and HA are both present as major species the buffer region Can therefore use the HH equation pH pKQ Log A39 HA Or since HA and A are in the same volume pH pKa Log moles A39 moles HA T itrations Region 2 ctd Let 1n0 original moles of HA mb moles of OH39 added HA OH39 gt A39 H20 Then in Region 2 moles A39 1m and moles HA mo 1m 50 pH pKa Logmmbm anywhere in Region 2 o b Special Point mb 12 1n0 halfway to the Equivalence point 1m H 2 o 1 p pKa Logmo lm J pl 1 Log pK 1 2 0 So at the halfEP pH pKa allows determination ofKa ofacid Titrations 10 Region 2 cfd Example Titrate 100 ml of 010 M HOAC with 020 M NaOH Calculate pH at three points in region 2 between start and ER Choose 10 ml 25 ml 40 ml Note EP at 50 ml halfEP at 25 ml m3 moles of OAc39 of equilibrium H K L p p a 0g mA moles of HOAC o r eqUIIIbr39Ium A m3 moles 0H39 added OH Volume added 020 M VB mA mo m3 where mo original moles 0010 mol VB L quot13 mA pH 0010 0002 0008 414 0025 0005 0005 474 0040 0008 0002 534 Titrations l 1 Region 3 of Titration curve At the Equivalence Point At the EP nearly all 0fthe original acid has been neutralized Solution is no longer a buffer so H H equation is not valid Instead imagine a twostep process iAII HA reacts to make A39 HA OH39 gt A39 H20 ii Some 0fthe A39 reacts back A39 H20 HA OHquot To find pH at E P Find original moles acid mo Set this equal to the moles A39 in solution at the EP Divide by total volume at EP to get A390 Set up an ICE using this A39o 9969 Titrations l 2 Region 3 cfd Example Given 100 ml 010 M HOAC titrated with 020 M NaOH Original moles HOAC mo 010 M 010 L 0010 mol Find volume of NaOH added to reach the EP NaOHVNaOH moles base added mo so VNaOH moNaOH VNaOH 0010 mol020 M 0050 L 50 ml Total volume ofsolution at EP Original Volume Added Volume V 100 ml 50 ml 150 ml 0150 L So 0AC390 at EP mov 0010 mol0150 L 0067 M Now Let the OAC39 react find equilibrium Region 3 cfd Titrations 1 3 OAc39 H20 HOAc OH39 I 0067 M z 0 z 0 C x x x E 0067x x x Need Kb 0f OAC39 get from Ka of HAC Kb OAC39 10X10391418X10395 56X103910 Kb H0Ac0H390AC39 XZ0067XzX20067 x z J006756x1010 61x106 x OH39 pOH L0gx 521 pH 14 pOH 879 Titrations l 4 Region 4 Beyond the Equivalence Point Here OH39 is in excess All essentially ofthe original acid is gone Calculate how much OH39 was added after the EP was reached divide by total volume to get final OH39 get the pH from this Continue example of 100 ml 010 M HAc titrated with 020 M NaOH What is the pH when 90 ml of the NaOH have been added 50 ml of NaOH were used to get to the EP That leaves 40 ml added after the EP How many moles is this 020 M 0040 L 0008 moles of excess OH The final volume of the solution is original volume 100 ml added volume 90 ml 190 ml So OH39 is 0008 mol0190 L 42x10392 M pOH Log42x10392 138 pH 1262 Note that in region 4 it doesn39t matter if the original acid was strong or weak The pH calculation is the same Titrations l 5 Example summary 100 ml 010 M HOAC titrated with 020 M NaOH Equivalence Point reached when 50 ml of the NaOH was added pH along the curve use dl erent procedures in each region Region 1 Just ionization of a weak acid ICE table for HOAC Region 2 Can use ICE or HendersonHasselbalch equation Region 3 ICE table for OAC39 Region 4 calculate excess OH39 after the E P Vb ml mb momb pH Notes 0 0 0010 287 Before Titration 10 0002 0008 414 Region 2 25 0005 0005 474 Region 2 Half E P 40 0008 0002 534 Still Region 2 50 0010 0000 879 Region 3 EP 90 0018 0008 1262 Region 4 excess OH39 Tlh atlons Acidbase Indicators Indicator changes color over a narrow pH range suitable for different pH ranges Indicator is itselfa weak acid 2 HIn H20 1n H30 m HIn and n are different colors I pH pKaHIn Log In39HIn When pH gt pKaHIn In39 gt HIn When pH lt pKaHIn In39 lt HIn Color changes over the range pKai 1 Thvmol blue Different indicators are Ahzarm yellow Phenolphthalem mm m Titration Curve Summary Strong acid titrated with Strong Base pH at EP 7 Weak acid titrated with Strong Base pH at E P gt 7 Weak base titrated with Strong Acid pH at E P lt 7 Strong base titrated with Strong Acid pH at E P 7 Chem 1133 Fall 2012 Lecture notes Buffers BUFFER a solution containing A weak acid and a quotsaltquot of that acid example HOAc and NaOAc A weak base and a salt of that basggxample NH3 and NH4Cl How does a buffer work HA H20 H30 A LARGE SMALL LARGE Add a little strong acid H30 the A39 consumes it mostly Add a little strong base OH39 the HA consumes it mostly Either way pH stays almost constant Chem 1133 Fall 2012 Lecture notes Buffers Examples of buffers BlOOd pH 25 74 H20 H2C03 H2603 H20 HC03 H30 H2C03 and HC0339 together form a buffer that helps regulate the pH of blood Add 1 ml 10M HCl to 10 L pure water pH 20 Add 1 ml 10M HCl to 10L blood plasma pH 72 pH lt 735 acidosis can be caused by emphysema pneumonia etc pH gt 745 alkalosis can be caused by hyperventilation vomiting Chem 1133 Fall 2012 Lecture notes Buffers What you need to know about buffers Qualitative What is a buffer How does a buffer work What components do I need to make a buffer What happens when I add acid or base to a buffer Quantitative What is the pH of a given buffer solution How can I a buffer that will give a desired pH By how much does the pH ofa given buffer change when acid or base is added Chem 1133 Fall 2012 Lecture notes Buffers pH ofa buffer solution HOAC acetic acid OAC39 acetate ion Dissolve 010 moles HOAC and 020 moles NaOAc in 100 ml water HOAC H20 H30 CAC I 10 z o 20 C X x x E 10X x 20x 5130 0A6 x20 x 5 a HOAC 1 x 1398x10 Approximate X ltlt 10 M X20 Then Kg 4 10 X 18X1039520 90X10396 pH 0fthe buffer L0gx 505 Compare to pH 0f10 M HOAC 237 quotCommon ion effectquot Chem 1133 Fall 2012 Lecture notes Buffers Buffer demonstration have 3 solutions each 100 ml 0 Distilled water 1 100 M HOAC and 100 M NaOAc 2 0100 M HOAC and 0100 M NaOAc Measure the pH of each solution Add 2 ml of 10 M NaOH measure pH Add another 8 ml Le 10 ml overall measure again O 1 2 Ini rial pH 2 ml added 10 ml added Chem 1133 Fall 2012 Lecture notes Buffers Buffer demonstration calculations Initial pH of each solution Solution 1 HOAc0 100 M NaOAc0 OAC390 lHOAc H20 H30 OAc39 I 100 z o 100 C X x X E 100X x 100x H301 OAc x100 x 1 8 105 a HOAC 1 x 39 x Approximate X ltlt 100 M K N x100 5 Then 0 100 x X 18X1039 pH Logx 474 Solution 2 same ratio OAC39 HOAC so pH also 474 Chem 1133 Fall 2012 Lecture notes Buffers Effect of adding 10 ml NaOH to Solution 1 100M HOAc100 M NaOAc Do this in two steps First assume added base reacts to completion HOAc OH39 9 OAc39 H20 Initial moles of OH39 10 molL0010 L 0010 moles Therefore initial OH39 0010 m010100 L 010 M Set up a reaction table ICF not ICE lHOAc OH39 gt OAc39 H20 I 100 010 100 c 010 010 010 F 090 000 110 Chem 1133 Fall 2012 Lecture notes Buffers Second step calculate the pH of the solution using an ICE table with the new initial concentrations of acetic acid and acetate HOAc0 090 M OAC390 110 M lHOAc H20 H30 OAc39 I 090 z 0 110 C x x x E 090x x 110gtlt H300Ac xl10x 18 105 a HOAC 090 x 39 x Approximate xltlt10M xl10 Then K e 090 1ZZX x 18x10395122 pH Logx 483 Chem 1133 Fall 2012 Lecture notes Buffers Effect of adding 10 ml NaOH to Solution 2 Initial concentrations 100M HOAc0100 M NaOAC same ratio OAC39 HOAC so same initial pH 474 Again assume the added base reacts to completion Initial moles of OH39 10 molL0010 L 0010 moles Therefore initial OH39 0010 mol0100 L 010 M Set up the reaction table lHOAc OH39 gt OAc39 H20 I 0100 010 0100 c 010 010 010 F 000 000 020 Oops There39s no HOAC left It39s all been converted to OAC39 The buffer has been destroyed it could take 2 ml but not 10 ml Chem 1133 Fall 2012 Lecture notes Buffers An easier way Start with Ka H30 A39 HA Take Logs L0gKa Log H30 Log A39HA Rearrange L0g H30 L0g Ka Log A39HA pH pKa Log A39JHAD Approx A39 z A39oHA z HA pH pKa Log A39JoHAm Since A39 and HA are in the same Volume pH pKa Log moles A39 moles HA Example 010 moles HOAC and 020 moles NaOAC in 100 ml water KaU IOAC 18X10395 so pKa Log 18X10395 474 pH 474 L0g020010 504 pH pKa Log baseacid pKa Log moles basemoles acid Acidsaltbuffers HA H20 H30 A39 pH pKQ Log A39HA pKQ Log moles A39moes HA Basesaltbuffers B H20 OH39 BH pH pKQ LogBBH pKQ Logmoles Bmoles BH How do you find pKa for a basesalt buffer For any conjugate pair pKaacid prbase pKW Near room temperature pKW 14 Chem 1133 Fall 2012 Lecture notes Buffers Summarize pH of a buffer solution Every buffer contains a weak acid and the conjugate base of that same acid 0r equivalently Every buffer contains a weak base and the conjugate acid of that same base As long as both components acid and base are present in significant amounts we can use the HendersonHasselbalch Equation pH pKa Log baseacid But since the acid and the base are in the same volume pH pKa Log moles basemoles acid pH depends on the ratio baseacid What happens when base acid pH pKa quot11 bufferquot Chem 1133 Fall 2012 Lecture notes Buffers BUFFER CAPACITY Ability of a buffer to maintain pH constant Buffer 1 10M HOAc10 M NaOAc High capacity Buffer 2 010M HOAc010 M NaOAc Lower Capacity water Buffer 1 Buffer 2 Initial pH 55 474 474 2 ml added 123 476 492 10 ml added 130 483 90 Buffer Range pH pKal Log A39HA Can adjust the A39 and HA to tune the pH to a desired number but if the concentrations differ by too much the buffer will have a low capacity Useful range is about one pH unit on either side 0pra umber of moles ofweak base or acid Chem 1133 Fall 2012 Lecture notes Buffers EXAMPLE PROBLEM 1 What is the pH ofa solution made up by adding 010 mol NH4N03 to 100 ml ofa 20 M solution of NH3 and diluting to 10 L Solve two ways ICE chart and HendersonHasselbalch First let s collect some numbers we ll need for both KbNH3 18x10395 pr LogKb 474 Initial number ofmoIes of NH3 20 M010 L 020 mol Initial concentration of NH3 020 mol10 L 020 M Initial number ofmoIes of NH4 given as 010 mol Initial concentration of NH4 010 mol100 L 010 M Method 1 ICE table INH3 H20 2 OHquot NH4 I 020 R 0 010 C x x x E 020x x 010x Kb NH4OH39NH3 010xx020x Assume X small compared to 010 020 Kb 010x020 x2 X 2 Kb 218X10395 36X10395 OH39 pOH L0gX 444 pH 14 pOH 956 Method 2 HendersonHasselbalch equation pH pKa LogNH3NH4 pKa Logmoles NH3 moles NH4 pKaNH4 pKw pKaNH3 1400 474 926 Approximate the moles by the initial moles pH pKa Log020010 926 Log2 926 030 956 Chem 1133 Fall 2012 Lecture notes Buffers EXAMPLE PROBLEM 2 What is the pH 0fa solution made up by adding 50 ml of 010 M HCl to 100 ml 0fa 020 M solution 0fNH3 Solve in two steps 1 HCl strong acid reacts with NHg converts some to NH4 2 Solve the resulting buffer problem Step 1 H30 NH3 a NH4 H20 We assume that this reaction goes to completion Initial number ofmoles 0fNIIg 020 M010 L 0020 mol Initial number ofmoles 0f H30 010 M0050 L 00050 mol Since 005 gt 02 H30 is the limiting reagent So final number ofmoles 0f NH4 00050 mol final number ofmoles of NH3 0020 00050 00150 mol final volume 010 L 0050 L 0150 L final NH4 00050 m010150 L 0033 M final NH3 00150 m010150 L 010 M Step 2 NH3 H20 NH4 0H Final numbers from step one become initial numbers of step 2 Can use either or mole numbers in the HH equation since NH3 and NH4 are in the same volume pH pKCI LogNH3NH4 pKCI Logmoes NH3 moles NH4 pH pK Log0100033 pK Log01500050 pK Log30 pKNH4 pKw pKaNH3 1400 474 926 pH pK Log30 926 0477 974 Chapter 18 Entropy Free Energy and Equilibrium A very brief overview ofa complicated subject Focus on a few key results Spontaneity of Chemical reactions and physical processes in general a process is spontaneous if it will change on its own without requiring external work Examples Evaporation ofwater above boiling point burning wood rusting iron Spontaneous doesn t necessarily mean fast some quotspontaneousquot processes are very slow Later on when we do Kinetics we ll study the rates of reactions What determines whether a reaction is spontaneous AH doesn t an exothermic reaction AH lt 0 isn t necessarily spontaneous although many of them are Example water boils spontaneously at 100C but requires heat the boiling process is endothermic but spontaneous above 100 C Cut to the chase there is a statefunction called Gibbs Free Energy G which gives the answer A reaction is spontaneous at constant Temperature and Pressure if AG lt0 G HTS where H is the quotenthalpyquot review Chem I Thermochemistry and S is a new state function called quotEntropyquot What s entropy Technical definition is very abstract In this course we ll be sort of vague about it We39ll try to get a quotfeelquot for how entropy changes in a reaction Can often predict the sign of AS by looking at physical state number of particles complexity ofmolecules and other things To calculate the numbers use tables just like with AH Entropy is a measure of the number ofways that a fixed amount of energy can be arranged in a system quotLog of the number of microstates that make up a macrostatequot A crystal at absolute zero low entropy actually zero A hot gas high entropy The entropy ofa system generally increases with an increase in temperature Also depends on phase for a given compound Sgas gt Sliquid gt SSolid Entropy increases in a reaction if the number of molecules increases for exampleN204gt 2 N02 In a family of similar molecules eg hydrocarbons entropy increases as the molecule gets more complex Standard molar entropies S0 of elements and compounds under quotstandardquot conditions for gases pressure 1 at for solutes concentration 1 molar They depend on temperature usually tabulated at T298 K Unlike enthalpies H39s we don t set the entropies of elements equal to zero Instead we set all entropies to be equal to zero at T0K As a result S0 is always a positive number Example calculate the standard entropy of reaction ASOrxn for the rusting of iron 4 Fes 3 02 g gt2 Fe203s Fes S0 273 ImolK 02g S0 2050 ImolK Fe203s S0 8741molK ASOrxn 2874 4273 32050 5494 ImolK As predicted the standard entropy decreases and by a lot Mostly because the oxygen has gone from being a gas to being part of a solid Recap entropy S is a measure of how widely the energy ofa system can be dispersed around the system quotLog of the number ofmicrostates that make up a macrostatequot Historically though entropy was first introduced in the conteXt ofheat ow If the state of a system is changed quotreversiblyquot very slowly and gently then the change in entropy is AS qrevT where qrev is the heat that ows across the boundary ofa system For any real change AS gtqT Keep temperature constant to make calculation simple At constant pressure qrev AH Second Law of Thermodynamics any spontaneous change results in an increase in the total quotentropy of the universequot which is the sum of the entropy change ofyour system AS and the entropy change ofits surroundings as heat ows across the boundary Ssurr ASuniV AS ASsurr gt 0 To putin convenient form use ASsurr AHT then ASSyS gt AHT TAS AH gt 0 AH TAS lt 0 Or defining Gibbs Free Energy G H TS AG lt 0 for any spontaneous change at constant pressure and temperature This is the criterion chemists most use in practice What is the significance of G At constant pressure and temperature if AG lt0 in a change of state the change is spontaneous and AG is the maXimium amount of work that you can theoretically get from the change in practice get less If AG gt0 the change is not spontaneous and AG is the minimum amount ofwork required to make the change happen in practice need more Use This will allow us to determine how much power we can get from a battery a spontaneous chemical reaction energy released in the form of an electrical current Standard molar free energies AG0 of elements and compounds under quotstandardquot conditions for gases pressure 1 at for solutes concentration 1 molar They depend on temperature usually tabulated at T298 K Or can get from tables of AH0 and S0 As with AH0 but unlike S0 we set them equal to zero for pure elements in their stable state Recap Gibbs Free Energy G H TS At constant temperature and pressure a change will be spontaneous if A G lt0 The magnitude of A G AG tells us the maximum amount of work that we can get out of this spontaneous change For example how much electrical power we can get from a battery As long as temperature is constant AG AH T AS Now we see why it is that many exothermic reactions are also spontaneous For an endothermic reaction to be spontaneous AS has to be positive since AH is negative For an exothermic reaction AS can be negative as long as TAS is less than AH At low temperatures AH is more important than AS at high temperatures AS is more important quotentropy drivenquot IfAH lt0 and AS gt 0 exothermic entropy of system increases process is spontaneous at all temperatures If AH gt0 and AS lt 0 endothermic entropy of system decreases process is not spontaneous at any temperature IfAH lt0 and AS lt0 exothermic entropy of system decreases process is spontaneous at low temperatures and not spontaneous at high temperatures IfAH gt0 and AS gt0 endothermic entropy of system increases process is not spontaneous at low temperatures but becomes spontaneous at high temperatures Standard molarfree energy afreactl39on AGquotrxn This is AG for going from reactants to products under quotstandardquot conditions for gases pressure 1 at for solutes all concentrations 1 molar These numbers depend on temperature Just like with AHquotrxn you can get AGquotrxn for any reaction by using tables of quotfree energies of formationquot AGO which are defined to be the AG0 for the reaction that makes a compound out of elements It39s Hess39s Law all over again At constant temperature AG0 AH0 T AS OK how do we get AG at something other than quotstandard conditionsquot Without proo here is the answer AG AG0 RT LnQ where Q is the quotreaction quotientquot ie quotproducts over reactantsquot with the appropriate powers just like in the equilibrium constant except that the concentrations are not necessarily the equilibrium concentrations We are not yet ready to prove this result in this class we will do so in more advanced courses Chem 4511 or 4411 Compare Q to K review from Chem 1 If Qlt K products increase reactants decrease until QK If QgtK products decrease reactants increase until QK If QK we39re at equilibrium Reactions continue on the microscopic scale but products and reactants don39t change Compare to AG AG AG RTLnQ When all concentrations are 10 M Q1 and AG A6 If QK we39re at equilibrium SinceAG gives you the work that you can get from a chemical reaction must have AG0 at equilibrium So since QK at equilibrium At equilibrium 0 AG AG0 RT LnQ gt AG RTLnIQ This is an amazing result From tables of G39s which we can get from H39s and S39s we can calculate equilibrium constants Things to note about Q and K 1 Q and K are unitless pure numbers Concentrations in Q and K are actually ratios of concentrations relative to a reference concentrations quotunits divided out 2 Pure solids or liquids never appear in Q or K Their concentrations don t change even though their amounts do 3 Q and K depend on how the reaction is written Reaction must be balanced What happens to Q and K when you change the way the reaction is written a Reverse the reaction switch products for reactants new K 1old K b Multiply reaction through by a number 11 new K old Kn ACIDS and BASES Historically these were defined by macroscopic properties ACIDS Sour taste dissolve some rocks eg limestone burn skin BASES Bitter taste dissolve oils and fats Acids quotneutralizequot bases HCIaq NaOHaq gt NaClaq Arrhenius39 theory of acids and bases 1884 Svante Arrhenius 1884 observed that solutions of acids in water conduct electricity He proposed that Acids ionize in water to make H ions HCI 1 H 1 CI39 1 And that Bases ionize to give OH39 ions NaOH aq gt Naaq OH39aq Arrhenius39s theory explains ne HCuq NuOHuq H utralization Cl39 Nu Net ionic equation H39 OH39 gt H20 The Na and C139 are quotspectatorsquot Also the limestone rxn 2H uq 60603 gt Cuz aq H20 6029 But what is the hydrogen ionquot H aq really 7 The H immediatelyhonds to water to make the hydroniumion H30 ma aquot Inn no So what s really going on when HCl dissolves in water is more like Hydronium HCI H20 a Hao cr mewquot 3 re Alter People often still use H as shorthand for H30 A broader definition Br nstedLowry 1920 Acid Proton Donor Base Proton Acceptor H20 H30 Cl H20 NH3 NH4 OH H25 NH3 NH4 5H In this View a given compound can act as either an acid or a base depending upon what it reacts with More on this later For now we mostly talk about acids or bases in water aqueous solution Strong Acids quotStrongquot does not mean quotconcentrated A strong acid is completely ionized in aqueous solution Only a very few strong acids are important to us HCI hydrochloric HBr hydrobromic HI hydriodic HN03 nitric H2504 sulfuric HOCIOz chloric also written as H6703 HOC03 perchloric also written H6704 In water every HCl molecule becomes H30 and Cl39 WeakAcids Incomplete ionization HA H20 Hao Aquot Get an Equilibrium between the acid molecule and the ions Simple examples HF HCN H2503 H3P04 Especially important Carbnxylic acids based on COOH u u u u c c UH EHz UH inrmvcacm aceivcacm henznvc acm Acid dissociation constant Ka Quantitative measure of acid strength HA H20 H30 A Hgoqmq Equilibrium constant K E Iquot HA Since H20 hardly changes when acid dissociates in solution leave it out of the expression for K Larger K2l 2 equilibrium to right gt stronger acid Often see pKaE LogKa Acetic acid Ka 18x10394 pKa 474 What Determines Acid Strength First look at Hydrogen Halides HF weak acid Ka 68X10394 but HCl HBr HI are strong acids Look at Bond Energies HF 565 kImol HCl 427 k mo HBr 363 kImol H1 295 kImol A stronger bond to H means a weaker acid In water HCl HBr and HI are all completely ionized Levelling effect so can t see which is strongest when in aqueous solution Now look at some halogen 0X0acids Name Formula Ka Name Formula Ka Hypochlor ous HOCI 29x10398 Hypochlor ous HOCI 29x10 8 Chlor39ous HOCIO 11x10392 Hypobr omous HOBr39 2x10399 Chlor39ic HOCIOZ zl Hypoiodous HOI 1X10 Perchloric HOCIOs Strong 0H bond is what ionizes in all of these Bond strengths about the same Look at the polarity of the bond Cl is more electronegative than Br s0 OH bond in HOCI is more polar than in HOBr In water more polar bond ionizes more easily More polar bond to H gt stronger acid Summary Relative acid strengths Rule 1 Compare HA bond energies A stronger bond to Hydrogen means a weaker acid Rule 2 If the bond energies are about the same eg if H bonded to same atom in all examples a more polar the bond to H means a stronger acid Get a more polar bond if atom bonded to H is more electronegative Cl vs Br vs I more electronwithdrawing groups 0C103 VS 0C102 VS OCIO Also quotresonance stabilizationquot of the conjugate base 2 stronger acid Acetic acid CH3COOH pKa 474 H o Ethanol CszOH pK1 16 acetate Similarly have strong and weak Bases Strong bases ionize completely NaOHaq gt Naaq OH39aq Group I alkali hydroxides NaOH KOH e l39c Group II hydroxides MgOHz CaOH2 Group I oxides N020 K20 ionic compounds 0239 H20 gt 2 OHquot Weak bases NH3 H20 NH4 OH39 NH3aq solution contains both NH3 and NH4 molecules Autoionization of water H20 H20 H30 OH39 Kw H30 OH39 103914 at room Temperature In a neutral solution H30 OH39 So H30 OH39 H30 r2 103914 therefore H30 10397 In an acidic Solution H30 gt 107 CH lt 10397 In a basic solution H30 lt 107 OH39 gt 10397 pH Scale H30 concentrations cover a wide range awkward It is convenient to use a Logarithmic scale Define pH by 101 H H30 For a neutral solution H30 10397 so pH 7 For an acidic solution H30 gt 10397 so pH lt 7 For a basic solution H30 lt 10397 so pH gt 7 Invert the exponent pH Log H30 More pstuff Define pAnything E L0gAnything Examples pOH 5 Log OH39D pKa E L0gKa Since H30 OH39 KW Log H30 OH39 Log KW Expand L0gH30 Log OH39 Log KW 50 L0gH30 Log OH39 L0g Kw So At room temp Kw 1039 so pKw 14 and pOH 14 pH Acids and their Conjugate Bases o A39 HA H20 H30 A39 KHA 014 HA A39 H20 OH39 HA ampltA39 Hao1A1 OH1 HA1 KaHA 39 KbA HA A H30 OH Kw KbA39 Kw KaHA Weaker Acid lt gt Stronger Conjugate Base Ionization Of A Weak Acid HA H20 H30 A Generally are given initial concentration of HA call it co Two Standard Problems 1 Given H30 and HA find Ka easy 2 Given Ka find H30 and OH39 harder Underlying principles a Mass Balance quotICE Table b Equilibrium constant formula ICE Table simple procedure for solving equilibrium problems HA H20 H30 A Initial co z 0 0 Change x x x Equilibrium cox x x Initial concentrations HA0 co A39o 0 H30 z 0 C line determined by reaction stoichiometry E line algebraic signed sum OH and C lines Plug the E line concentrations into formula for Ka Haw A xxx HA c0 x Leads in general to ugly quadratic equation for X Kama Approximation usually HA0 gtgt H30 or A39 HA 0 weak acid 2 2 x x Then Coxzco so K Solve x2 z Ka co so x as cuKa and pH LogX Check Approximation we say it s 0k if xco lt 005 5 rule If approximation not valid solve quadratic x2 Ka x x2KnxKacoO 0 Solve x Ka i V192 41900 then use pH LogX Example 1 010 moles acetic acid HOAC dissolved in 100 ml water What is the pH of the solution Co 010 moles010 Liters 10 M Ka of HAG 18 X 10395 lHOAc H20 H30 OAc39 I 10 z 0 O x x x E 10x x x x2 x2 Kn a x z 11018 x10 s 42 gtlt103 M 10 x 10 Check XCo 42X1039310 42X10393 lt 005 so approximation is ok pH Log4 2X10393 238 Example 2 initial HAC 10394 M x2 x2 K g 10x10394 x 10x10394 x z 110 x10418 x10 s 42 X10 5 Check XCo 42X1039310X10394 042 Since XCo gt 005 the approximation is not valid and we must solve the quadratic x Ka i 1192 419 Find X 34X10395 pH LogX 446 Percent Ionization I E A39HA0 X 100 For our examples HAc Ac39 oI 10 M 42 x10393M 042 10 x10394M 35 x10395M 35 See that Percent Ionization is Larger when HAo is smaller HA H20 H3O A Adding water decreases all concentrations but increases the ratio A39JHAJo Ionization Of A Weak Base Find pH 0f010 M ammonia NH3 Kb 18X10395 NH3 H20 NH4 OH 010 M R O R O C x x x E 010x x x 2 Kb NH4 OH 1 x Approximate Kb 2quot X2010 NI la 010 x x zW01018 x 10 5 13 x 103114 Check XCo 13X10393010 0013 lt 005 so approx is ok X 0H39 Logx pOH Log13x10393 29 pH 111 AcidBase Properties 0f Salts Dissolve NaCls in Water NEUTRAL soln Dissolve NH4Cl in water ACIDIC soln Dissolve NaOAc in water BASIC soln Look at what s in the solution NH4Cl s gt NH4 Cl39 But NH4 is itselfa weak acid NH4 H20 NH3 H30 So NH4 reacts with water hydrolysis What about C139 H20 HCl OH39 No because HCl is a strong acid so Cl39 is a negligibly weak base in water So overall the solution is acidic pH lt7 Similarly for sodium acetate NaOAc s gt Na OAC39 OAC39 is the conjugate base of a weak acid HOAc and Na is the conjugate acid of a strong base NaOH So AC39 will react with water making OH39 AC39 H20 HAc OH39 but N a will not spectator ion So solution is basic What about NH4Ac NH4 and AC39 will both react with water compare K s to find out who wins Aside Highly charged metal ions make acidic solutions in water Fe3 6 H20 gt Fe H2063 FeH2063 H20 FeH205OH 2 1130 Quantify What is the pH ofa 0010 M soln of NH4Cl NH4 H20 NH3 H3O I 0010 M a o a o C x x x E 0010x x x NHs H301 x2 a NH4 0010 X KaNH4 KwKbNH3 10391418X10395 56 x1010 So XZ0010 X z X20010 56X103910 x 24x106 H30 pH Log24x106 56 Chemistry 1133 General Chemistry 2 Fall 2012 Lecture MWF 900 CHEM 140 Instructor Robert Parson RobertParsonColoradoedu Online Course materials Desire2Learn Go to httpsearncoloradoedu Log in with your Identikey Login Name and password Select CHEM1133 from the Course List Home Page various features including lecture notes Recitation Recitation 1 hour per week MWF 800 100 T 800100 Th 800 100 330 Recitation is part of this course Chem 1133 not part of lab Chem 1134 You have a recitation TA and a Lab TA they may not be the same This week if you are waitlisted pick any recitation period to attend Same for Monday sections next week Chem 11331134 Help Room Ekeley M2846 Ekeley Subbasement Prof Parson s helproom hours Mon 1100 1200 Tue 10001 1 00 Wed 11001200 TA helproom hours will be posted soon Laboratory Course Chem 1134 Lab is now a separate course strict corequisite Instructor Dr David Jonas DavidJonasColoradoedu Office Hours MW 10301 1 30 T 11301230 Checkin begins next week first experiment two weeks Monday lab sections check in in two weeks Separate grades for Lecture and Lab Chem 2 labs are longer and harderthan Chem 1 labs but there aren t as many Textbook Chemistry Atoms First vol II by Julia Burdge and Jason Overby Also needed Recitation manual Lab manual Lab Notebook goggles gloves Nonprogrammable calculator igtchker Clickers igtcickers Only 1 clicker needed for all CU classes can reuse from a previous term Register online through MyCuinfo Only need to do this once covers all your classes We will start recording clicker points on Wednesday Usually 2 points for correct answer 1 for incorrect People you need to know Robert Parson Chem 1133 Lecture instructor David Jonas DavidJonas Chem 1134 Lab instructor Your TAs assigned to both 1133 and 1134 You may have different TAs for 1133 recitation and 1134 lab Lora Ruffin Ekeley M149 Lab coordinator and lecture demonstration advisor Anne McWiIIiams Ekeley M199 Waitlist Dropadd and other administrative issueS Online homework Sapling can also get there from D2L Create an account detailed instructions are on D2L Use your CU email address when creating your account Edit your profile to include your student ID number Enrollment key is your 3digit recitation section number Access is free for the first two weeks after which you purchase an access code online not at the bookstore Start by doing Practice Assignment noncredit and Math Review Keep best 12 assignments Grading 3 Exams 300 pts total 100 each Final Exam 200 pts Homework 120 pts total 10 each Recitation 60 pts total 5 each Survey 10 pts total Clicker 60 pts total Total 750 pts Science is not a collection of facts it is a way ofthinking 1 Understanding a science is a learned skill like playing a sport N People understand concepts better by thinking actively about them and discussing them with others than by passively listening 3 Why are we doing thisquot is always a fair question I 9 1 ii 1 Mr Osborne my l be excused My brain muquot Mr Osbourne ma ylbe excused My brain is full


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