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CHEM 14BL Measurements and Errors 0 Want to know how reliable a specific measurement is besides its value What can affect reliability 0 Erroruncertainty 0 There is always some error in all experimental measurement Accuracy and Precision 0 Accuracy define closeness of result to true value 0 Precision closeness of set of results to each other results should be reproducible o Deviation between data should be small Precision vs Accuracy 3 possible situations 0 1 result and data are both precise and accurate o Best result 0 2 result and data are precise but not accurate 0 3 an quotaccuratey known result may have little or no precision scientifically unacceptable Which situation does this example belong in 0 Suppose the true concentration of unknown acid is 02013 M 0 3 different sets of analysis 01915 M 02362 M 01726 M 0 Avg concentration 02013 M 0 Suppose you were also told that precision for this type of analysis is i05 0 Conclusion neither precise nor accurate because data are outside precision limit 3 o Relative standard deviation gt 9 o Unreliable results Precision and Significant Figures 0 Numbers of sig figs reported in measurement is sometimes interpreted as degree of precision 0 In lab numbers of sig figs reported by measurement should be consistent with precision limit of equipment and glassware set by manufacturer How to improve reliability 0 Instrumental errors usually minimized by calibration 0 Examples pipet and buret pH meter spectrometer Use theory on propagation of errors to estimate errors in experiment Absolute Error and Relative Error 0 Example 0 Absolute error i01 cm o Same unit as measurement o Same number of decimal places as measurement 0 Relative error in no units o iO1 cm50 cm 12 General Rules for Error Propagation 0 Multiplication or Division o CAB or CA B I A and B are experimental measurements o Relative error in C I ACCAAAABB where AA and AB are absolute errors in quantity of A and B o Absolute error in C I AC AAAABBC where AA and AB are absolute errors in quantity of A and B 0 Addition or Subtraction o CAB or CAB o Relative error in C I ACCAACABC where AA and AB are absolute errors in quantity of A and B o Absolute error in C I AC AACABCC where AA and AB are absolute errors in quantity of A and B CHEM 14BL 411 For any volumetric containers report to the hundredth decimal place For post ab average should fall in between the range of 996 mL to 1004 mL aways round to hundredth decimal place Molarity M moles of soluteLiter of solution wv weight by volume g of solute100 mL of solution For very diluted aqueous solutions 0 Parts per million ppm mg of soluteLiter of solution 0 Parts per billion ppb pg of soluteLiter of solution Gravimetric analysis weight relations 0 Very precise quantitative technique 0 Precision quantitative analysis o Numerical results must be precise 0 Disadvantages o Difficult need to selectively isolate component to be measured as solid compound o Takes a long time o Costly Volumetric Analysis 0 Quantitative analysis 05 0 Disadvantages o Takesalongtime o Costly Spectrophotometric Analysis Beer39s Law 0 Semi quantitative quot395 to 10 0 Advantages o Fast o Can easily set up quality control measures o Can be monitored online 0 Disadvantages o Absorption of light by chemical species needs to be vastly different from other constituents in solution Theory on Beer39s Law o IIo1o39 T o No units on either side 0 Io intensity of incident light beam 0 intensity of transmitted light beam 0 E molar absorptivity constant o Depends on identity of chemical species and selected wavelength 0 Cconcentration M of chemical species 0 Ldepth of solution traversed by light AbsorbanceA og TELC Beer39s Law 0 E and C are related to analyzed chemical species Beer39s Law A determined in lab during experiment C unknown quantity E physical characteristic of measured chemical species o Depends on A o Is constant if and only if I 1 must be fixed during analysis I 2sampe has low concentration ambiguous 0 L is constant Acceptable range for absorbance 01 to 10 SopeEL 2 major problems in Beer39s Law 0 Doesn39t work if concentration of solution is too high 0 Background interference o Eliminated by blank solution I Should contain everything except substance tested I Correct blank solution y intercept should equal 0 I Incorrect blank solution has y intercept that does not equal 0 Lecture 3 Midterm covers material up to next Fri Aspirin Synthesis 0 Acid catalyzed esterification rxn 0 Purification using crystallization Theory on Crystallization Recrystallization crystalizationrecrystaization 0 Goal purify impure or crude experimental product 0 Basic principle solubility of compound increases w increasing temp of solvent 0 Solubility depends on concentration polarity and temp 0 Difficulties most soluble impurities and rxn by products share solubility properties with desired product 0 2 diff solvents 9 2 diff polarities General Procedures 0 Single solvent o Dissolve crude product in minimum amount of suitable hot solvent I HEAT UP SOLVENT FIRST Filter solution to remove insoluble impurities if present Allow solution to cool Filter purified product Need nucleation for crystallization to grow OOOO I Need imperfect surface for crystal to grow Protein crystallization 0 Crystallized protein can be used to determine 3D structure by using Xray crystallography 0 Can understand protein better 0 Not all proteins can be crystallized major difficulty Example 1 Rock Candy 0 Solubility of sugar in water minimum is subjective o Room temp 180 gmL o 100 C boiling pt 250 gmL o 2 diff temp gradient for solubility of solvent ALWAYS 0 Begin w 230 g of sugar and dissolve in 100 mL of boiling water After one week volume of water is reduced to 90 mL at room temp Any sugar crystals in solution after one week 0 100 mL of boiling water can dissolve all 230 g of sugar o 250 gmL 100 mL250 g 0 Water volume drops to 90 mL after one week o 180 gmL 90 mL162 g 0 W 230 g of sugar in water 68 g will not dissolve o Sugar crystal will form o 230 g 9 162 g I Sugar doesn t evaporate water does Example 2 0 Solubility of sugar in water o Room temp 180 gmL o 100 C boiling pt 250 gmL 0 Begin w 230 g of sugar and dissolve in 150 mL of boiling water After one week volume of water is reduced to 140 mL at room temp Any sugar crystals in solution after one week 0 150 mL of boiling water can dissolve all 230 g of water o 250 gmL 150 mL375 g 0 Water volume has decreased to 140 mL o 180 gmL 140 mL 252 g 0 W 230 g 0f sugar in water there is still enough water after 1 week to dissolve all sugar o No sugar crystal will form o Concentration will increase before crystallization occurs Crystallization evaporation 0 Evaporation still has impurities 0 In crystallization filter out all impurities Choosing good crystallization solvent 0 Good solvent should Dissolve large amounts of product at high temp max solubility o Not dissolve at low temp can39t crystallize if solvent has high solubility at low temp o Dissolve impurities at all temp o No reaction with product The higher the solubility the less you get back Each solution has 2 temp high and low You want high quality and max quantity want high solubility at high temp and low solubility at low temp E E F 4 El 39339 J 1 3 f39 E E 2 3 E92 TL5 TL TLC THE TH THA 39iquoter39r39yca H high temp L low temp C has lowest solubility at low temp best 0 Maximized amount of yield 0 Get more product Solvent A low solubility at high temp Solvent B okay not best How does crystallization remove impurity 0 Example System 0 Suppose we have 9 g of A desired product and 2 g of B impurity 0 Low temp solubility o Solubility of A at 20 C1 g of A100 mL o Solubility of B at 20 C1 g of B100 mL 0 High temp solubility o Solubility of A at 100 C 10g100mL o Solubility of B at 100 C1O g of B100 mL Purification of Solution by using crystallization I B lvent I Solvent I Bolvent Crude Product all dissolve at 100 C in 100mL of solvent cool solution to 20 C first crystallization Crystals form 8g ofA 1 g of A and 1 of B and 13 of B m remains in solution Heat to 100 C and cool solution back to 20 C 2nd crystallization OR recrystallization Pure crysalform 7g g 1 g of A and 1 g 013 remains in solution Problem lose when A you try to purify 9 less product 0 Won39t do more than twice o If you do more not using right technique Mixed Solvent System 0 More flexibility 9 more ways to change composition of system 0 Typically work better than single solvent system 0 made up of 2 miscible liquids 0 One solvent will dissolve product fast than other o One should have lower solubility o Good and bad solvent 0 Experiment 0 Ethanolwater mixture for aspirin experiment 0 Aspirin is more soluble in ethanol than in water 0 Solubility 0 1 gof aspirin5mL ethanol 0 lg aspirin300mL water 25 C Aspirin is slightly polar 9 not highly soluble in water Procedures for using mixed solvent system Will not crystallize Will not crystallize Dissolve product in better solvent one with higher solubility for product at high temp Add second poorer solvent drop by drop until solution becomes cloudy Heat solution to redissolve all crystals Add few drops of better solvent if necessary to force crystals back in solution Experiment Impurities and aspirin solid crude product add min amount of hot ethanol impurities and aspririn dissolve in hot ethanol add water no more than 25mL then add solvent w low solubility for desired product white precipitate begins to form in ehtanol and water mixture add small amt of hot ethanol solid redissolves in soln wait 57 days pure aspirin crystals form on flask bottom Fast crystallization 9 poor quality crystals H H H y of Aspgrin gmL H Soll ubilit Low Temp High Poor solvent 100 ethanol Good solvent 75 water and 25 ethanol iPoor solvent 100 water To monitor acidbase rxn use an indicator 0 Commonly used to determine end point of acidbase rxn 0 Don39t provide pH readings 0 End point equiibrium o Point in acidbase rxn when indicator changes color o Doesn39t apply to pH readings o Color changes because of electron structure o pH affects electron structure 9 color change How to choose correct indicator for rxn 0 Choose indicator whose end point is closest to equivalence point pH changes color closest to equivalence point pH Equivalence pointequiibrium Need to get transition color Middle of steepest region o Equivalence point All acidbase has neutralized o All acid has reacted with base 1400 1200 1000 800 600 pH of solution 400 200 O Q Q Q Q Q LO gtQ b9 00 Volume of NaOH added mL a Strong acid titrated with strong base Methyl red changes too fast Thymol blue changes color closest to EP Melting Points crystal molecules aligned in conformation that is held together by intermolecular forces 0 Need energy to break ordered structure ie increasing temp o Not breaking chemical bonds breaks forces o Depends on how strong intermolecular forces are not of atoms 0 Can be used to find identity and purity of experimental product o Impurities are always present o Identity melting pt confirms but doesn t prove identity of given compound because many compounds may have similar melting pts o Purity sharpness of melting point indicates compound purity I Pure compounds melt 1 temp melting pt range S 1 C I Impure compounds melt over temp range Heating curve il Steam T A Heat of i altriatir1 V A Water r ateam 3353 H Llqull Water lee water Heat of FIJEIE39l Temperature ltje Heat Added time Ket Record melting pt as 2 temp range 0 Ti first sign of melting 0 Tf complete melting 0 AT melting pt range Tf39Ti S 1 C for pure substance 4 types of intermolecular forces 0 London forces very weak not permanent attractive force between electron density o Can be very effective if small intermolecular distance 0 Dipole forces EN difference 0 H bonding depends on what H is attached to 0 Ionic forces NaC Octane London forces only Melting point 568 C Cyclohexanez London forces and dipoe dipoe forces Melting point 45 C H bonding Melting pt 0 C Ionic forces Melting pt 800 C Melting pt range around 135 1355 C London forces between phenyl rings H bonding 9 why aspirin has high mp Dipoledipole forces Can interact w intermolecular Lots of intermolecular forces Effects of soluble impurities on melting pt Melting pt range depression lower end of melting pt range will occur below literature melting pt Melting pt range broadening melting pt range gt 1 C usually 2 3 C 130133 C More than 5 C range indicates compound is highly impure o o Impurities disrupt consistency and organization of crystalline structure at molecular lvl 9 non uniform melting behavior Molecules closest to impurities melt fastest Further away from impurities crystalline structure remains almost the same and melts around literature temp Titration step Volumetric Analysis 0 Can do redox rxn 0 Standard soln w precisely known concentration 0 Soln w unknown concentration 0 pH of mixture can be monitored by using pH meter Volumetric Analysis experimental technique 0 commonly used to determine concentration of unknown soln by measuring volumes require to react completely w unknown by using standard soln 0 looking for equivalence pt not end pt o end pt indicator 0 analytical technique iO5 By using technique of volumetric analysis titration graph can be created based on experimental data 0 should be able to calculate pH Example of Volumetric Analysis Titration of Vinegar w NaOH 0 objective find exact concentration of commercial vinegar soln 0 how to determine concentration of vinegar soln 0 can find concentration of soln by titrating w standard NaOH soln 0 major component of vinegar is acetic acid o weak monoprotic acid o contains only 1 acidic proton use graph to find equivalence pt Strong acids vs weak acids 0 strong acid 100 dissociated in water 0 HCI aq H20 9 H3Oaq Clquot 0 1 mole 1 mole 1 mole 0 Km H3 0iEcl I Order of magnitude of around 105 I K governs direction of rxn I Hydronium 9 acidity smaller pH 0 Weak acid significantly less than 100 dissociated in water 0 HA aq H20 9 H3Oaq Aquot o 1 mole ltlt1 mole ltlt1 mole H30A HA I Order of magnitude around 105 0 KHcI o Definition of strength to do w K not concentration I Both 1 M and 1M are strong Stron HCI HA H Cl HA H Cl HA H Cl H A Obtaining data in volumetric analysis 0 Always use pH meter as reference to properly collect experimental data in lab when performing volumetric analysis 0 Record buret volume and pH whenever pH reading changes by 0203 pH unit 0 Significant amount of NaOH is needed to increase pH 0 When near equivalence pt very little NaOH is needed to increase pH 0 Should collect lots of data pts in equivalence pt area Normality N equivalents of H 0 For acid L of solution e uivalents 0 OH 0 For base q L of solution M vs N 0 N requires knowledge of stoichiometry of rxn between reactant and products o Experimental unit o Rxn specific 0 M don t need knowledge of stoichiometry of rxn between reactants and products 0 N 2 M 0 pH doesn t match pH Compounds M N ACIDS HCI 040 moles of HCIL 040 equivalents HL H3PO4 040 moles of H3PO4L 12 equivalents HL BASES KOH 040 moles of KOHL 040 equivalents OHquotL MgOH2 040 moles of MgOH2L 080 equivalents OHquotL Assumption assume that all H and OHquot in above compounds dissociate completely 0 comes from equivalency not M Converting N to pH 0 Suppose bottle of phosphoric acid is labeled as 035 N What is pH of phosphoric acid soln based on following balanced chemical rxn o Only 2 proteins dissociate in H20 o H3PO4 2H2O 2H3O HPO4 o O35N O 035 equivalent H x 1 mole H3P04 2 mole H L 2 equivalents H 1 mole H3P04 035 mole H O f o pHlog O35MO46 Concept of Dynamic Equilibrium 0 no weak acidweak base 0 if K changes equilibrium will change 0 as soon as OHquot is added to weak acid it will react w H generated by weak acid dissociation o causes weak acid to shift equilibrium to right until rxn reaches equivalence pt 0 dissociation equilibrium for weak acid KaHA39HA 0 Ka has to be fixed Titration of 10000 mL of 01000N HA Ka110396 w 01000N NaOH in Major chemical species 1 HA initial pH El Veq 139IHII4IllI39IquotIiE1 of hlalEH m L Monoprotic acid How to calculate pH 1 Calculate equivalence pt volume for titration Macid Vacid Mbase Vbase Moles of acid moles of base Macid 01 M Vacid 100 mL Mbase01 M don39t have to be the same concentration Solve for Vbase Vbase Veq 100 mL 9 predict what graph should look like jpiltl I I Veq X 100 ml lmlumiel of l39larllH n1LI 2 pH afte 10mL of NaOH 9 limiting reagent Amounts moles HA OHquot gt H20 Aquot INITIAL 001 01M001L CHANGE 001 001 001 EQ 0009 0 0001 I Should be 0 because you used all NaOH Sodium spectator ion How much you consumehow much you produce Equilibrium line shows us that there is nonzero amt of HA and Aquot in soln HA is weak acid gt BUFFER SOLN Aquot is conjugate base of HA Buffer Soln 0 Composition buffer soln must consist of weak acid and its conjugate base or weak base with conjugate acid 0 Both components necessary 0 Purpose resist change in pH when strong acid or strong base is added to buffer soln Chemistry 0 Examples o Suppose buffer soln consists of HA and Aquot is reacted w strong base soln o Stabilization of pH when strong base is mixed w buffer o HA A Weak acid weak base excess Won39t react w NaOH Conclusion weak acid in buffer soln neutralize all strong base Neutralized strong base and produce weak base as product Suppose buffer soln consists of HA and Aquot is reacted w strong acid soln Stabilization of pH when strong acid is mixed w buffer Aquot HA Weak base weak acid excess Strong acid dissociates Conclusion weak base in buffer soln neutralize all strong acid and produce weak acid as product pH will change at slow rate Henderson Hassebach Eqn calculating pH for buffer solns only conjugate base PHPKa og weak acid Ka lt10395 Total volume 100mL 10mL 011L weak acidHA 009 moles 11L conjugate baseA 001 moles011L pH50 quotll Major chemical species 1 HA initial pH 2 HA and Aquot buffer soln Buffer region plI lmlu me of NEIEIH m L pH after 50mL of NaOH pH at equivalence pH Amounts moles HA OHquot gt H20 Aquot INITIAL 001 O1M005L CHANGE OO5 005 OO5 EQ 0005 O 0005 Half equivalence pt added half of the titrant still within buffer region 0 conjugate base weak acid pHpKa 0 most convenient way of finding Ka 0 best general purpose buffer one that gives ratio of 1 or closest to 1 Limitation of Henderson Hassebach eqn 0 Ka for weak acid S 10395 0 Applies only to buffer solns o Make sure that everything is O in ICE table except for weak acid conjugate base l P 3 H Equivalence pt i Half equivalence pt I I 39339 in 39 39 39 39 50 100 150 Holumiewoef NEIEIH n1LI pH pKa 3 pH after 100 mL of NaOH equivalence pt Amounts moles HA OHquot gt H20 Aquot INITIAL 001 O1MO1L CHANGE O1 01 z O1 Final amt O O 001 Need to consider dissociation of Aquot to find pH since this is only species that exists at equivalence pt For conjugate base EQUIVALENCE PT pH DEFINES BY HYDROLYSIS OF SALT A Amounts A H20 HA OHquot moles INITIAL 001 CHANGE x x x EQ 001X02L x02L x02L 02L total volume of soln at equivalence pt Substitute x02L as Y Kb or Kh Kw Ka 1039 10 H Y2 3 10 005M Y YOH3921O395 M 9 pH93 Ii Major chemical species 1 HA initial pH 2 HA and Aquot buffer soln 3 Aquot salt soln pl 50 100 150 lmlu me of NalH m L Half equivalence pt moles of weak acid moles of conjugate base Equivalence pt moles of weak acid moles of strong base pH after equivalence pt after 150mL of NaOH 9 pH will be determined by excess strong base in solution because only major chemical species are weak and strong base after equivalence pt 9 strong base has larger effect on pH 0 excess volume of NaOH after equivalence pt 50mL 0 total volume of son 250 mL 0 pOH ogOH39 5 0 pOH og 17 pH14pOH123 Major chemical species 1 HA initial pH 2 HA and Aquot buffer soln 3 Aquot salt soln 4 A39and OHquot excess strong base pll 1 I I I 50 100 150 lmun1eaf NHIDH n1LI 523 Acetic Acid analysis 0 Determine M of acetic acid 0 Determine experimental Ka of acetic acid Buffer analysis Find buffer limit of acetate buffer solutions 0 Examine efficiency of buffers based on buffer pH using HendersonHasselbalch eqn 0 pHpKa0gc0njugate base weak acid 0 Where is the end of the buffer 0 Effectiveness of the buffer 0 Which is the best buffer Preparation of Acetate Buffer 0 CH3COOH in excess OHquot limiting reagent 9 H20 CH3COO39 0 Create buffer soln by adding limited amt of strong base to excess weak acid soln Three methods to prepare buffers in lab 1 Mix weak acid w conjugate base 2 Mix excess amt of weak acid with limited amt of strong base DONE IN LAB 3 Mix excess amt of weak base with limited amt of strong acid BUFFER SOLN MUST NOT CONTAIN ANY EXCES STRONG ACID OR EXCESS STRONG BASE WHEN SYSTEM REACHES EQUILIBRIUM 0 Too much strong acid or base removes buffer Analysis of acetate buffer using NaOH 0 Contains both acetic acid and acetate 0 When NaOH is added weak acid in buffer will react with NaOH 0 CH3COO39 in buffer will act like spectator ion Analysis of acetate buffer using HCI 0 Contains both acetic acid and acetate 0 When HCI is added to buffer conjugate base in buffer will react with HCI 0 CH3COOH in buffer will act like spectator ion Acetic Acid analysis 0 Titrate acetic acid soln by adding strong base NaOH 0 Use volumetric analysis to determine M of soln 0 Macetic acid unknown Vaceticacid pipet MNaoH VNaoHequivalence pt volume obtained from graph 0 Ka for acetic acid can be determined experimentally using half equivalence pt when pHpKa 0 Don39t need to include spectator ion 0 Don39t need to set up ICE table o Don39t need to estimate pH 0 Can use to compare experimental and theoretical Buffer limit when one of the species is gone in the buffer Large plateau large efficiency 0 Better buffer The more buffer you have the more buffer capacity If graph is for 10m graph for 5mL is compressed 0 2x data pts for 5mL 9 data pts for 10mL Buffer limit right before rise when pH starts to rise it means you39ve lost your buffer Highest data pt on 1 derivative graph is not necessarily equivalence pt Lower pKa more willing to dissociate sp3 4 o bonds 0 TE bonds 9 tetrahedral sp2 3 o bonds 1 TE bond 9 trigonal planar sp 2 o bonds 2 o bonds 9 linear 530 Final in Ackerman Ballroom Chemical kinetics study of rates and mechanisms of chemical rxns 0 Help us to understand how chemical rxns occur and the rate at which they occur 0 Examples o Halflife concentration of radioactive materials o Metabolism or metabolic rate By monitoring rate of change of products or reactants we can determine rate of chemical rxn aAbB 9 cCdD rate 1al1b 1c1on negative being consumed positive being created Units rxn rate is in Msec concentrations are in M Factors that affect rate of chemical rxn 0 Concentration of reactants 0 Temp Catalysts enzyme Mainly focus on effect of reactants concentrations w rxn rate Rate law 0 For given overall chemical rxn of form aAbB 9 final products ratekA B m and n determined experimentally describe order of reactants Rxn order describes how much A and B comes together OOOO krate constant constant at specific temp Rxn order and unit of rate constant 0 Suppose A and B are only reactants ratekA B o If mn1 then rxn is said to be first order in A and B The overall rxn order is second order 0 Unit of k depends on overall rxn order 0 Unit of k o Concentration is always in M o For second order overall 1Msec o Not arbitrary unit Rxn rate and rate constant 0 Difference btwn rxn rate and k o Rate law ratekAquot Bquot Rate is fxn of k and reactants concentrations 0 Affected by concentration of reactants temp and catalysts K is fxn of activation of energy Ea and temp T o kAe39EaRT 9 Arrhenius eqn 0 catalysts decrease Ea of rxn Rxn mechanism 0 chemical rxns don39t occur in single step but proceed through series of intermediate steps to give overall rxn 0 each intermediate step eementary rxn o rate eqn for overall rxn can39t be derived from stoichiometry of balanced rxn o rate eqn for elementary rxn can be written in terms of stoichiometry coefficients of balanced rxn 0 sum up all steps 9 overall rxn Elementary rxn aAbB 9 products ratekAaB39 0 can write rate law directly 0 rxn mechanism consists of series of elementary rxns combined to yield rate law for overall rxn Rate determining step in rxn mechanism 0 In many rxn mechanisms one step is significantly slower than all others 9 rate determining step slow step use this step to write rate law 0 Governs rate of overall rxn since rxn cannot proceed faster than this slow step 9 control how fast you get the product Example of rxn mechanism theory K1 ZNO N202 FAST K1 Each one is elementary rxn K2 N2O2 O2 9 ZNO2 SLOW 0 Rate law based on slow stop 0 Rate k2N2O2O2 o Cannot have intermediate in rate law 0 Fast step o Forward rate reverse rate 0 kNO2k1N2O2 0 klk1NO2 N202 o Rate law k1 2 2 k 39NO O2 RatekNO2O2 o Preequilibrium approx fast step before slow I Rate o Steady intermediate Two different methods to determine experimental rate law i Calculate instantaneous initial rates or instantaneous ratesof rxn for several values of one of the reactants concentrations by holding other initial concentrations fixed from 1 rxn run to next Experiment is repeated changing one of the other concentrations and keeping others fixed a Approx rxn rate certain time w tangent line ZNO 02 ZNO2 NOmoL O2moL Initial Rate moe Ls 1o1o394 1o1oquot 281O396 1o1o394 3o1oquot 8410396 2o1oquot 3o1oquot 341o395 Order for O2 triple concentration ofO2 0 Rxn rate also triple therefore rate law is first order in 02 Order for NO double NO 0 Rxn rate increases 4x therefore rate law is 2quot order in NO ii Use integrated rate law to find order of chemical species a Based off the shape of the graph b Zero order kinetics i Rate of disappearance of A is actually independent of A ii Ratek iii At A0kt iv For zero order rate law plot of A vs time should yield a straight line with slope k c First order kinetics i Rate kA ii LnAtktnA0 iii For first order rate law plot of nA vs time should yield straight line with slope k d Second order kinetics i RatekA2 1 1 A Ale iii For 239 order rate law of 2 identical species plot of 1A vs time should yield straight line with k Experiment determine rate for rxn of dye with bleach Ratekred dyeabeach39 0 Use Beer39s law to monitor concentration of red dye at diff time intervals At LC 9 CtAtEL 0 how to determine a 0 find 6 o use Beer39s Law and absorbance data to construct dye vs time ndye vs time 1dye vs time 0 b and k Approximate rate law as pseudo rate law Ratek red dyea K pseudo rate constantkbeach39 OOO Assume bleach is constant throughout Chemistry 14BL Study Guide for Cumulative Exam 2 Spring 2014 CUMULATIVE EXAM 2 TIME amp LOCATION SHOW UP ON TIME Location Ackerman Grand Ballroom Exam must be taken in the lecture section in which you are officially enrolled NO EXTRA TIME will be given if you are late to the exam NO MAKE UP EXAM WILL BE GIVEN Exam is CLOSED book and CLOSED notes Disclaimer Please note that the questions on this study guide do NOT necessary re ect the style of the questions you will actually see on the exam It is the concepts that you should focus on when working on the following questions Please keep in mind that a question can be written in many d erent ways that relates to the same concept s L Therefore you should go back and review the midterm study guide for the previous tested materials In addition you should go through all the lecture materials study guide for the first exam SRM reading materials Mohrig s textbook writing assignment and lab reports when studying for the exam Focus on the conceptual understanding of various topics You should also review all the reports including all the experimental procedure You will need to understand the chemistry rather than simply reading the procedure without understanding the concepts behind the various experiments I I Chemical Aqueous Equilibrium and Buffer Chemistry I 1 Which of the following molecules are acids and which ones of them are bases HNO3 CH3COOH NH4Cl H2CO3 C6H5NH2 2 Write the chemical reaction with base OH39 for those molecules that you identified as acids in question 1 Write the formula of the conjugate bases for these molecules 3 Write the chemical reaction with acid H for those molecules that you identified as bases in question 1 Write the formula of the conjugate acids for these molecules 4 a Explain why the equivalence point pH7 for a strong acid and strong base titration b Explain why the equivalence point pH gt7 for a weak acid and strong base titration c Explain why the equivalence point pH lt7 for a weak base and strong acid titration 5 What volume of 0250 M NaOH is required to reach the equivalence point with 1849 mL of 0200 M HZSO4 For simplicity you may assume the 2 proton dissociate completely in H2504 6 What volume of 0500N NaOH is required to titrate with l000ml of 0120 N of HZSO4 to the equivalence point 7 Calculate the Hquot39 pH OH39 and pOH in 0030 M HCl 8 Suppose you have a 5000ml of 00100 M chloroacetic acid solution HC2H202Cl Kal 4gtltl0393 Think about what the predicted pH should be before you even start working on this question a What is the initial pH for this solution b Write down the molecular formula for the conjugate base of chloroacetic acid c What is the equivalence point pH if the chloroacetic acid was titrated with 00l00M NaOH What acidbase equilibrium principle did you use when calculating the EQUIVALENCE POINT pH What major chemical species exists at the equivalence point d Explain how the strength of the conjugate base for chloroacetic acid affects the equivalence point pH that you calculated in part c 9 Which of the following solution pairs can be used as buffers Explain and write balanced chemical reactions i e the addition of H and OHquot to the mixture to support your answer 10 Suppose you have a solution that was prepared by mixing nitric acid HNO3 with potassium nitrate KNO3 Explain whether the resulting solution is a buffer You should approach this problem using the concept of chemical equilibrium constants 11 Suppose you have a l0000ml of 00l000M methylamine CH3NH2 solution K44gtlt10394 a Calculate the equilibrium pH of this solution b Calculate the equilibrium pH after the addition of 7000mL of 0l000M HCl to methylamine using both the chemical equilibrium approach and the HendersonHasselbalch equation approach Explain why there is a difference between the two calculated values c Does the pH value based on the equilibrium approach in part b make physical sense d What is the equivalence point pH for the titration of 00l000M HCl with methylamine Does the equivalence point pH make physical sense 12 Calculate the equilibrium pH using both the chemical equilibrium approach and the HendersonHasselbalch equation approach for a buffer solution prepared by mixing 5000 mL of 02000 N NH3 Kb l8gtltl0395 with 5000 mL of 01000 N HCl 13 Suppose you want to prepare a 1 liter acetate buffer solution at pH395 which contains both 010M sodium acetate and 015M acetic acid How much volume do you need for each of these 2 solutions in order to prepare this buffer You may assume HendersonHasselbalch equation approach is applicable in this case K for acetic acid l8gtltl0395 14 Complete question 2 refer p 44 in your lab manual under week l of the acetic acid experiment 15 Complete question 3 refer to p44 in your lab manual under week l of the acetic acid experiment 16 Consider the following solutions i A solution consists of 010 moles of a weak monoprotic acid and 020 moles of NaOH ii A solution consists of 010 moles of a strong monoprotic acid and 020 moles of NaOH iii A solution consists of 010 moles of a strong monoprotic acid and 010 moles of NaOH iv A solution consists of 020 moles of a weak monoprotic acid and 020 moles of NaOH v A solution consists of 020 moles of a weak base and 020 moles of NaOH vi A solution consists of 020 moles of a weak base and 020 moles of HCl vii A solution consists of 020 moles of a weak base and 010 moles of HCl NOTE You do not need to set up a complete ICE table in order to complete the following questions All you need is to examine ALL the solutions listed above and determine what chemical species exists when the system reaches equilibrium From then on you should be able to answer the following questions a Upon mixing which one of the above solutions has a pH of 7 Explain b Upon mixing which one of the above solutions will result in a buffer solution Explain c Upon mixing which one of the above solutions will yield a weak base solution but NOT a buffer Explain d Upon mixing which one of the above solutions will yield a weak acid solution but NOT a buffer Explain e Upon mixing which one of the above solutions will exhibit no chemical reaction among the starting chemical species Explain 17 Derive the HendersonHasselbalch equation for any buffer solution containing HA and A with an acid dissociation constant between 105 and 109 ll Error Analysis Concentration Units and Beer s Law Review from Exam 1 I 18 Calculate the percent error for the following quantities a 2054i002 x O254i0003 321 i005 b 30078i0003 20174i0001 9813i0005 19 Calculate the absolute errors for the quantities in question 18 20 Describe brie y the meanings of systematic errors and random errors 21a MgOH2 is one of the active ingredients in Maalox an antacid Suppose you prepared a stock solution by dissolving 00558g weight obtained from digital balance of MgOH2 in a 50mL volumetric ask Calculate the molarity ppm and wV of the MgOH2 stock solution Your final answers should have the correct units and significant figures Molar mass of MgOH2 58322mol 21b Calculate the relative error for the molarity in the MgOH2 solution Refer to page 27of the manual for the absolute errors in lab equipment 21c Calculate the absolute error in molarity for the MgOH2 solution 22 Magnesium hydroxide reacts with acid based on the following reaction MgOH2 2Hquot 2H2O Mg2 Calculate the normality N for the MgOH2 solution prepared in 2la Show all your reasoning including any conversion factors between molarity and normality 23 Suppose a student dissolved an unknown sample containing NiII sulfate by using 25 0mL of distilled water and l00mL of 05M HCl The solution was then transferred quantitatively to a lO0mL volumetric ask If Beer s law analysis was used to determine the concentration of NiII sulfate in the solution what should be the components for the blank solution What should be used as standard solutions 24 Carefully examine the following statement Is the statement true or false If the statement is false explain WHY the statement is false Di erent solutions with the same color and identical absorbance values have identical molar absorptivity constants Ban lat 3 m 25 Explain why a red solution displays an optimal wavelength at 525nm Hfla Dra ge e M vf Bl E1 M Hi d K fellow quotquot1 am I auunm N Ul55 nm afinlet Green it 3 X 5 Blue v i39 43 nm erAEmg X7 I39ll393 UI39quotII39TquotI Review the concepts in solutions concentrations and serial dilution Know the units of ppm ppb wv normalityN molarityM and error analysis refer to lecture guide 1 for general principle on error analysis Your Beer s law postlab report is an excellent reference for Beer s law concentration units serial dilution and error analysis At this point you should be able to derive the expression for inherent error base on the equation you use in data analysis refer to the Beer s law postlab report indicators analysis postlab report and the acetic acid analysis postlab report The online tutorials Beer s law amp solution concentrations included within the report guideline for Beer s law experiment posted on CCLE may also be useful when studying for the exam lll Chemical Kinetics 26 The data below were collected for the following reaction 2N02g F2g 2N02Fg N02 p F2 O Initial Rate M S 1 0100 0100 0026 2 0200 0100 0051 3 0200 0200 0103 Write an expression for the rate law determine the average rate constant with correct unit and the overall order of the reaction 27 The halflife of a substance is the time required for the concentration to fall to one half of the initial value i Show that for a zeroorder rate law involving species A the halflife is given by t A1 2k ii Suppose the initial concentration of A ie A0 at time0 is 020M and the rate constant for the reaction is 0O030Ms391 Sketch a plot of A vs time between 0 to 70 seconds Indicate on the graph the locations of halflife Review the lecture guide your postlab report amp CHEM 14B textbook on chemical kinetics You should know how to use kinetics rate data to find out the rate law like the example we did in lecture and question 25 Know the relationship between the various integrated rate equation and the order of the chemical species in the kinetics rate law refer to lecture guide and the postlab report Review all the concepts behind the chemical kinetics experiment you did in the lab You should know how to interpret an integrated rate law plot There are plenty of questions you can practice from your CHEM 14B general chemistry textbook on chemical kinetics Focus on the topicsconcepts related to the kinetics experiment integrated rate law and using data to determine experimental rate law lV Laboratory TechniquesSkills amp Laboratory Safety I 28 List the techniques and skills you used in the acetic acid analysis assignment and the aspirin synthesis and analysis assignment 29 You should know the general definition and classification of the NFPA symbol More specifically you should know the general meaning of each region within the symbol httpwwwemlahukorkepchemsnfpa labelhtm1 V Molecular Geometry I 30 Tyrosine is a nonessential amino acid the body makes from another amino acid called phenylalanine It is a building block for several important brain chemicals called neurotransmitters including epinephrine norepinephrine and dopamine The 2D molecular structure is shown below with the use of implicit carbon and hydrogen atoms The corresponding 3D molecular structure of tyrosine is shown below The dark gray atoms are the carbon atoms The light gray atoms correspond to hydrogen atoms The red ones are the oxygen atoms and the blue atom corresponds to the nitrogen atom i What the hybridization states for carbon atoms 1 through 9 ii What are the predicted bond angles and shapes for carbon atom 2 and carbon atoms 4 through 9 iii Draw the hybridization orbitals box diagrams for carbon atom 2 and carbon atoms 4 through 9 Carbon Atom Hybriization Chart 0 gf2 gri1a5 1f Granup5 0 Ganmatryr quot unhybridiled Carbon Carhan SP3 4 D 4 4 U Tetrahedral 2 V v p p A T ganal 5393 3 1 3 3 1 Planar 5p 2 2 2 2 2 Linear Bond Angles Tetrahedral 1095 Trigonal Planar 120 Linear 180 Vl Binary Mixture Melting Point Phase Diagram 31 You should go back and review the lecture example we did in class see below on the binary mixture melting point phase diagram The example we did in class began with the system at point TX which consists of 75 A and 25 B and underwent cooling process You should understand how to determine the amount of materials ie the Values shown on the graphs that remain in solid phase and liquid phase when the system reaches the eutectic point ie 40A and 60B The three graphs on the right help you to keep track of how much liquid stills remain in the system ie the top graph and how much solids A and B form during the process ie the middle and bottom graph For simplicity you can assume there are a total of 100g materials which consists of 75g of A and 25g of B at point TX Binary Selid Melting Peint Phase iagram m L iquiEl 395 T Reaches Eefdifeatian 1 3quot Eutectic Eenijzifllete 395 Liquid EL 5 L d A iqui Limli Remaining TE A Z Sana Selid A Selid Ei el el Ameunt ef A liilele Selid A Farm 0 Fraetien 5 A n Time 9quot L A Eutectic P initj meu nt ef Selid E5 Ferm Time Jquot Answers to Practice Problems 1 2 3 4 5 7 8 9 10 Acid HN03 CH3C00H NH4Cl H2C03 Conjugate Bases N03 CH3C00 NH3 HC03 Base C5H5NH2 Conjugate Acids C6H5NH3 H OH39 H20 Nitric acid is a strong acid CH3CO0H OH39 CH3C0039 H20 NH4 OHquot NH3 H20 H2C03 OH HC0339 H20 For question 2 only net ionic equations are listed You do NOT need to include spectator ions in the equations C5H5NH2 H393939 C5H5NH3 a Final products are H20 AND a neutral salt b Final products are H20 AND a weak base c Final products are H20 AND a weak acid 296 mL 6 240 mL H 0030 pH 152 0H39 32x10 13 p0H 1248 a H 31gtltl0393M pH25l simple Ka dissociation question except you will need to use quadratic equation to solve for pH due to the high magnitude of Ka b Conjugate base C2H202Cl c The major chemical species exists at the equivalence point is C2H202Cl Equivalence point pH is determined by the HYDROLYSIS equilibrium of C2H202Cl in water 0H l9gtlt10397M pH728 the concept for this part is IDENTICAL to what we did in class for determining the equivalence point pH d Since the Ka for chloroacetic acid has an order of magnitude of around lO393 the magnitude of Kb for its conjugate base is around 1039 Remember that KagtltKb equal to Kw which is 1039 Therefore the conjugate base of chloroacetic acid should be very weak This is the reason why the equivalence point pH is almost neutral ie close to 7 A buffer mixture must consist of BOTH a weak acid and its conjugate base NH4Cl NH3 Conjugate acid is NH4 and conjugate base is NH3 NH3 H NH4 NH4 OH NH3 H20 K2C03 NaHC03 Conjugate acid is HC03 and conjugate base is C032 C032 H HC0339 Hco OH C032 H20 No the resulting mixture is NOT a buffer since nitric acid is a strong acid Ka is in the order of 102 so its conjugate base is technically neutral Kb is around 1046 practically zero Answers to Practice Problems 11 12 13 14 15 16 17 18 19 20 21a 21b 2lc a pH 1128 or 1127 b pH 1027 HHB approach 1021 chemical equilibrium approach c The pH should be lower than the pKa because this is beyond the HALF equivalence point and the solution at this point consists of more conjugate acid RNH3 than its conjugate base RNH2 pKa 14pKb 1064 d pH 647 almost neutral pH The equivalence pH should be less than 7 due to the formation of the conjugate acid of methylamine However methylamine has a Kb value of 440x10394 this means the Ka for its conjugate acid is 227x1039 With such a small Ka value we expect the equilibrium equivalence point pH to be just below 7 or very close to neutral pH 926 Need approximately 190 ml of sodium acetate and approximately 810 ml of acetic acid pH 241 acetic acid sample with no extra water added pH 271 acetic acid sample with water added No change in equivalence point volume Equivalence point pH 904 for the titration of acetic acid solution without the extra water Equivalence point pH 892 for the titration of vinegar solution the extra water added You should notice that the equilibrium pH is Ii basic than the one without the extra water added a solution iii b solution vii c solution iv d solution vi e solution v For this question you simply need to examine the moles of the starting materials and then write out the chemical reaction like what we did in class to see what final products remains at equilibrium In other words you do NOT need to solve for the entire equilibrium using Ka and or Kb You simply need to know what chemical species exist when the system reaches equilibrium Refer to the solutions for the derivation of the HendersonHasselbalch equation a 3 b 005 Report to only 1 significant figure a i 005 b i 0009 The two topics are discussed on page 18 of your lab manual 00191 M 112x103 ppm 0112 04 you will get 05 if you round off number during each step of the calculation i 00001M actual value is i 000008M Answers to Practice Problems 22 00382 N MgOH2 23 The blank solution should contain 05M HCl and distilled Water To be more specific the ratio of the H Cl to water in the blank should be roughly about 1 9 by volume Pure and dried Ni II sulfate should be used to prepare the standard solution 24 The statement is FALSE Molar absorptivity constant Varies with chemical species Different solutions with the same color AND identical absorbance Values do not necessary imply identical molar absorptivity constant Remember absorbance Value is a function of molar absorptivity constant path length and concentration Identical absorbance Values between different solutions can occur for solutions with different molar absorptivity constants and concentrations 25 A red solution should absorb in the wavelength range that corresponds to its complementary color In this case the complementary color of red is green This is why the optimal wavelength occurred at 525nm for a red solution The concept is related to the idea of complementary colors httpfirstvearchemusvdeduaucalculatorscolour wheelshtml 26 Rate kNO2F2 k average 26M391s391 SECOND order 27 t12 0693k 28 The skills are listed at the beginning of each experiment in the lab manual refer to pages 43 and 50 30 i Carbon 1 spz Carbons 2 amp 3 sp3 Carbons 4 through 9 sp2 ii Carbon 2 1095 Tetrahedral Carbons 4 through 9 120 Trigonal Planar iii Carbon 2 4 I I Hybridizafimnr 53 I I I I 25 H 15 ll 15 H Answers to Practice Problems 30 iii Carbon 4 through 9 Nvmhyrbriized pxrbital T 2 I I Hybridizaiim 25 H Hybridized mans Is H I5 H I HNO3 Nitric acid CH3COOH acetic acid HZCO3 Carbonic acid In NH4Cl ammonium chloride the NH4 can act as an acid Lewis acid an electronpair acceptor 111 C6H5NH2 aniline the NH2 group can act as a base a Lewis base an electronpair donor due to the lone pair electron on the nitrogen I a Final product is H20 and a neutral salt Conjugate base for a strong acid K around lO4or 105 is NOT basic b Final product is a weak base B39 Hydrolysis of the weak base gives out OH B H20 HB OH c Final product is a weak acid HA Hydrolysis of the weak acid gives out H HA H20 A H3O I Use M v Mb vb M 0400M NOT 0200M For molarity you will need to take into the account of the number of protons in the acid before calculating the equivalence point volume I Use N v N1 vb N 0120N NOT 0240N Normality automatically takes into account of the 2 protons in sulfuric acid I Set up the equilibrium table for the hydrolysis of chloroacetic acid HC2H202Cl 04 H30 C2H202Cl K I 00lOOM C y y y E 00lOOM y y y This is identical in concept to what we did in lecture to determine the initial pH of the weak acid However you will need to use quadratic equation in order to solve for the pH due to the relative high K value Note If you use approximation to solve for part a you will get a pH of 243 which is incorrect The actual pH is 25lwhich is actually higher due to the dissociation equilibrium of the weak acid to form the conjugate base because of the relatively high K value 2 Set up the equilibrium table in moles for the hydrolysis of chloroacetate C2H2O2Cl OH HC2H2O2Cl Kb I 00lOOMgtlt005000L C y y y E 500gtlt10394 y y y You will now need to calculate the equivalence point volume for the reaction Ma Va Mb Vb solve for Vb which should be equal to 5000mL TOTAL VOLUME 5000mL 5000mL lO000mL i e initial amount the total amount of NaOH added Divide the moles of each quantity in the ICE box by the total volume and then put them into the equilibrium expression for Kb to calculate the equilibrium pH You can also set up the equilibrium table shown above using molarity In this case you will divide all the values listed above by the total volume This is identical in concept to what we did in lecture to determine the equivalence point pH At the equivalence point all the chloroacetic acid has been converted to chloroacetate The initial amount of chloroacetic acid should equal to the amount of chloroacetate form at the equivalence point The equilibrium pH will be defined by the hydrolysis equilibrium shown above Remember whenever you use Kb you are actually solving for OH Therefore what you get is the pOH NOT pH You will need to convert pOH to pH at the end of the calculation 2 We will use RNH2 to represent methylamine This will make it easier when setting up the ICE box Initial pH Hydrolysis of the weak base RNH2 H20 RNH3 0H 1 001000M C y y y E 001000M y y y You can calculate the initial pH by using the Kb and the initial concentration of the weak base similar to what we did in class for the weak acid Kb y2 001000M y Use quadratic equation to solve for y OH Use that to find pOH and then the pH y OH 00019M pOH 272 pH 1128 If you use approximation to solve for pH you will get a pH of 1132 which is not quite correct due to relatively high magnitude of Kb The actual pH is actually lower due to the slightly smaller amount of RNH2 remains in the system because of the hydrolysis equilibrium see ICE table shown above Remember whenever you use Kb you are actually solving for OH Therefore what you get is the pOH NOT pH You will need to convert pOH to pH at the end of the calculation 2 Methylamine RNH2 is a weak base It can react with HCl to form a conjugate acid Set up the ICE box in MOLES that corresponds to the equilibrium that exists at this point RNH2 Hquot RNH3 1 001000Mx010000L 001000Mx007000L F C 001000Mx007000L 001000Mx007000L 001000Mx007000L E 300x10394 0 7000gtlt10394 This is a BUFFER solution consists of a weak base RNH2 and its conjugate acid RNH3 TOTAL VOLUME lO000mL 7000mL l7000mL i e initial amount the total amount of H Cl added You can also set up the equilibrium table shown above using molarity In this case you will divide all the values listed above by the total volume Divide the moles of each quantity in the ICE box by the total volume and then put them into the Henderson Hasselbalch equation to find the pH for the buffer pH pKa log RNH2 RNH3quot 14pKb log RNH2 RNH3quot 1027 or 1028 You will get the same pH if you the pOH form of the HendersonHasselbalch equation pOH pKb log RNH2 RNH3 372 pH 1028 Set up the ICE box in MOLES that corresponds to the hydrolysis equilibrium between the conjugate acid and conjugate base RNH2 H20 RNH3 OHquot 1 300x10394 Z 7000gtlt10394 Y C x x x E 300x10394 x 7000x10394x x You can also set up the equilibrium table shown above using molarity In this case you will divide all the values listed above by the total volume Equilibrium Concentrations RNH2 3OOgtltlO394 x Ol7OOL 000l76M y where y is in molarity x01 700L RNH3 7O0OgtltlO394x 01700L 000412M y OH x Ol7OOL y Kb OH39RNH3quot RNH2 y 000412M y 000176MM y 44gtlt10394 y OH39 000016M pOH 380 pH 1020 The pH value using chemical equilibrium approach is lower than the one calculated by using Henderson Hasselbalch equation approach This is because the chemical equilibrium takes into account the hydrolysis equilibrium between RNH2 and RNH3 In other words there are actually less RNH2 and more RNH3quot in the solution when the system reaches equilibrium due to the relatively high Kb value 2 The equilibrium pH should be lower than the pKa because this is beyond the HALF equivalence point and the solution at this point consists of more conjugate acid RNH3 than its conjugate base RNH2 pKa 14pKb 1064 2 First you will need to calculate the equivalence point volume for the reaction Ma Va Mb Vb solve for Va which should be equal to l0000mL At the equivalence point ALL the RNH2 becomes RNH3 Since the only major species at the equivalence point is the RNH3 you can set up an ICE box for the hydrolysis equilibrium of RNH3 RNH3 H20 RNH2 H3O 1 001000Mx010000L C y y y E 0001000 y y y TOTAL VOLUME l0000mL l0000mL 20000mL i e initial amount the total amount of H Cl added Divide the moles of each quantity in the ICE box by the total volume and then put them into the equilibrium expression You can set up the equilibrium table shown above using molarity In this case you will divide all the values listed above by the total volume K KwKg 100gtlt1014 44gtlt10394 y 020000L y 020000L 0001000 y 020000L Substitute x y 02000OL where x is now in the unit of M K x2 0005000 M x solve for x H3O to find pH use Ka because H3O is part of the product 2 Set up the ICE box in moles NH3 H NH4 I 02000Mx005000L 01000Mgtlt005000L C 0005000 0005000 0005000 E 0005000 0 0005000 The ICE box indicates that the final mixture is a BUFFER solution consists of a weak base NH3 and its conjugate acid NH4 This is at the HALF equivalence point same concentration for both the conjugate acid and the conjugate base Therefore pH pKa 14pKb 925 or 926 Suppose V1 amount in Liters of acetic acid and V2 amount in Liters of sodium acetate are required to mixed the buffer Moles of acetic acid in the final buffer HOAc O15MxV1 Moles of sodium acetate in the final buffer NaOAc OAc 01OMgtltV2 We also know that the final volume of the buffer is V1 V2 1L Equation 1 Concentrations of the weak acid and conjugate base in the buffer solution Divide the moles of each quantity in the ICE box by the total volume which is 1L and then put them into the HendersonHasselbalch equation pH pKa log OAc HOAC pKa log 010V21L 015V11L Equation 2 With two equations and two unknowns you can now solve for V1 amp V2 Questions 14 amp 15 are almost identical to the various examples that we did in lecture for a reaction between a weak acid and a strong base For question 14 hydrolysis of the weak acid determines the initial pH refer to the lecture material for the calculation of INITIAL pH For question 15 refer to the lecture material on how to determine the pH at the equivalence point on the titration graph Note Water added to the acetic acid solution had no effect on the moles of acidic proton in the solution This is why the equivalence point volumes between the two different samples of acetic acid remain unchanged The added water will affect the overall pH since pH depends on the concentration of acidic protons in the solution The solution with the water added should have a higher starting pH in question 14 less acidic and a lower equivalence pH in question 15 less basic for the equivalence point pH The ONLY MODIFICATION you will need is to include the volume of water in the ICE table when finding the total volume of the solution This is the only difference between the solution with no water added and the one with water added For question 14 Initial pH Total volume of the initial solution Volume of acid Volume of water for one of the solutions only For question 15 Equivalence point pH Total volume at the equivalence point Volume of acid Volume of NaOH to reach the equivalence point Volume of water for one of the solutions only 2 An equilibrium pH equal to 7 occurs when mixing an equal amount of STRONG ACID with an equal amount of STRONG BASE I A buffer solution can be prepared in three different ways I mix a weak acid with its conjugate base I mix an EXCESS amount of weak acid with a LIMITED amount of strong base I mix an EXCESS amount of weak base with a LIMITED amount of strong acid You should always remember that a buffer solution MUST consist of both a weak acid and its conjugate base or a weak base with its conjugate acid when the system reaches equilibrium A buffer solution MUST not contain any excess strong acid or excess strong base when the system reaches equilibrium I Since the solution contains ONLY a weak base this can only happen when you mix an equal amount of weak acid with a strong base The strong base neutralizes the weak acid in the solution to form a weak base or the conjugate base of a weak acid This is NOT a buffer solution since it contains only a weak base I Since the solution contains ONLY a weak acid this can only happen when you mix an equal amount of weak base with a strong acid The strong acid neutralizes the weak base in the solution to form a weak acid or the conjugate acid of a weak base This is NOT a buffer solution since it contains only a weak acid I Any solution that has an acid mix with another acid or a base with another base exhibits no chemical reaction Note Refer to your CHEM 14B textbook Q SRM 08 SRM09 and SRMlO for common STRONG ACIDS and STRONG BASES Suppose that the amount of dissociation is z X Y are the INITIAL concentration of the HA and Aquot respectively Set up the ICE box in terms of X Y and 2 HA HOH H3O A I X Y C Z z z E Xz z Yz X Y amp 2 have units in molarityM K A39 H3O HA Yzz Xz If K2 is less than or equal to 1039s we can then assume that zltltY amp zltltX therefore K z Yz X Take the log on BOTH SIDE OF THE EQUATION log K log Yz X log Y log 2 log X This equation can now be simply to using the definitions of pKa and pH pKa log Y pH log X pH pKa log Y log X pKa log YX pKa log A HA Note If the Ka becomes too high the dissociation cannot be neglected During the process of deriving the buffer equation refer to the derivation shown above our assumption is that the dissociation of the weak acid can be ignored If the dissociation cannot be neglected the assumption that we made during the derivation will fail and so does the buffer equation If the Ka of the weak acid in the bu er is too high one will need to determine the bu er pH using the chemical equilibrium approach The chemical equation shown below indicated that the reaction involved a neutralization of two hydroxide ions in magnesium hydroxide MgOH2 2H 2H2O Mg Therefore we can convert the molarity to normality using the following relationship 00191 M x 2 equivalents of OH39 1 mole MgOH2 0382N Use trials l amp 2 to find the order of N02 in the reaction Use trials 2 and 3 to find the order of F2 in the reaction 2 You can determine the order of N02 and F2 based on the data obtained for the three trials However it is not necessary to always fix one chemical species and vary the other in order to solve for the order In this case as long as you can determine the order for one of the species you can determine the order for the second species using simple ratio Halflife refers to the time that is required to reduce the amounts of a specific material by one half We can derive the halflife expression starting with the integrated form of the zero order rate law At Ao 1ltfr to If to 0 and t tug at t t12 A A0 2 this is the halflife relation with concentration We can now rewrite the equation as Al0 A102 ktl2 Halflife t12 0 p Use the integrated rate law equation At A0 kt to to sketch the plot of A vs time Refer to next page for the sketch of A vs time plot II 012 Hiallif life Idnenrea ers UJ5 with dlecreas img cuncelnctratian A1 m fl39 l U 10 ill t half lift 3rid I39maJIf life D an 4b 60 39 an Tllmles 5312 First plot the graph using the integrated rate law equation At ktt ta A0 to 0 A0 020M k 00030Msquot You should notice that in this case the slope of the line is the rate which is equal to the rate constant for a zero order rate law ie Rate kA0 You can use any time points between 0 to 70 seconds to plot the graph The points on the graph are just for illustration You do not need that many data points since the graph is linear Second determine the locations of the halflife points using the halflife expression starting with the initial concentration FIRST HALF LIFE 020M 2gtltOOO30Ms391 33sec SECOND HALF LIFE 010M 2gtlt00O30Ms391 17sec Time 17sec 33sec 50sec Remember total time is additive during a reaction THIRD HALF LIFE 005M 2gtltOOO30Ms391 8 sec Time 50sec 8sec 58sec Remember total time is additive during a reaction 0 Temp lowest melting ptfreezing pt o Always lower than melting point of pure A and pure B 0 Composition 40 A and 60 B Important example of eutectic mixture 0 Menthol and phenol 0 Menthol low Tm and camphor high Tm 0 Can be applied as lotion at room temp 9 use in topical cream for itch relief