QUANTITATIVE ANLYS CHEM 321
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This 27 page Class Notes was uploaded by Carmela Kilback on Wednesday September 9, 2015. The Class Notes belongs to CHEM 321 at University of Washington taught by Jaromir Ruzicka in Fall. Since its upload, it has received 17 views. For similar materials see /class/192548/chem-321-university-of-washington in Chemistry at University of Washington.
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Date Created: 09/09/15
C321 Learning Objectives July 24th 2002 These objectives will be used to prepare the exam The problems are as indicated and will be similar to those on the exam This set of homework will not be turned in for extra credit The extra credit homework for this week has been assigned from Chapter 14 and includes problems and written quotanswersquot to the learning objectives Chapter 16 problem 3 Section 161 5 3391 U 6 7 Draw a schematic of the titration of iron with cerium Explain the function of the indicator and reference electrodes in the iron titration For the overall reaction identify Fe and Ce in terms of whether they are oxidized reduce and whether they are oxidizing or reducing agents Understand the indirect measurement that is being made in this reaction really involving two galvanic cells Use the blue box on page 417 to aid in your explanation Calculate the voltage of the redox titration at various points along a titration curve as in problem 3 Draw a representative titration curve for a redox titration Calculate the concentration of iron in an unknown given titration data Chapter 15 problems 2 27 32 34 39 40 wPH 90899 0 Delineate the difference between a metal electrode and an ion selective electrode Describe the basic components of a potentiometric measurement Draw a picture of how an indicator and a reference electrode can be contained in a single housing Write the Nemst equation for an ISE for chloride as well as one for calcium State the factors that in uence electrode response De ne the term junction potential Calculate the value of S at temperatures other than 25 degrees De ne the term selectivity coef cient and be able to predict potentials of various solutions given certain k values such as in problem 39 Be able to use the method of standard addition in anion selective electrode measurement you will be expected to combine concepts from the upcoming Logarithmic Diagrams An overview by Ilkka La39hdesm ikj CHEM 321 1 Introduction Traditionally chemical equilibria acid base complexometric redox and precipitation are treated algebraically That is to say you set up an equation with equilibrium constants and initial concentrations Solving the equation then gives you the unknown quantity that you wanted to find out Logarithmic diagrams are a graphical way of presenting chemical equilibria and offer an alternative way for nding out your unknown quantities at equilibriumi Logarithmic diagrams provide no additional information compared to the algebraic approach so why bother using them The answer is that they can considerably simplify the treatment of problems in the following ways Most people find it easier to understand a graphical explanation rather than formulas and text This is probably because a purely algebraic approach tends to draw attention toward mathematical manipulation rather than toward the chemistry Pictures and graphs are a powerful way of transmitting information a single picture can contain information that would take up pages if explained in terms of equations and text 39 0 As soon as you start tackling problems involving several components a mixture of NazHPO394 and NH 4Cl for example the calculations get very hairy very quickly Log diagrams provide an excellent way of reducing these problems into a simple form and solving them Log diagrams have found most of their use in acid base and complexometric problems Examples of their specific use are estimating pH of weak acids and bases as well as solutions of their salts sketching out crude pH and complexometric titration curves 0 selection of the appropriate titration indicator for acid base complexornetric and precipitation titrations evaluation of titration errors for acid base complexometric and precipitation titrations Log diagrams are by no means an absolutely better method than the arithmetic way Which one is better is determined by the problem at hand Generally it could be said that log diagrams are fast and easy but not necessarily very accurate and as such best suited for backof theenvelope calculations However a great deal maybe even most of the pH and complexation calculations you will do will be back Ofthe envelope type of problems where an accuracy of 01 log units is suf cient and consequently can be solved using logarithmic diagrams The greatest advantage oflog diagrams once you have mastered them is that they will give you a true ngertip feeling of any equilibrium problem 2 Theory This chapter presents the theoretical basis for construction and use of log diagrams Knowledge of the theory is not necessary for basic use of the log diagram so you can skip it if you wish However it is needed in order for you to make the most of the diagrams Also knowing the theory you will avoid erraneous use of the diagram Since log diagrams are most often encountered in connection with acid base equilibria we will use acid base concepts for deriving the theory You should note though that the obtained results are equally applicable to complexometric equilibria the only change you will need is switching the names of the variables 21 Mathematical description of an acidbase system Consider a 001 M solution of a weak acid HA with K5106 The acid will dissociate to some extent according to HAZHA In order to describe the exact equilibrium composition of the solution we need to nd out the following concentrations HA A39 PF OH39 From freshman chemistry we know that OH KWH so we really only have three unknowns HA A and Hquot We need three simultaneous equations to determine their values let s set out to find these First of all we have the expression for the dissociation constant K Hutr 1 HA Secondly we know that the sum of the acid and the basic forms must always amount to the total acid concentration CHA in this case 001 M HAA Ca 2 The third equation is obtained from the so called proton balance which needs some futher explanation The species that you start out with in our case HA and H20 are said to form the zero level of your acid base system As soon as the zero level species are mixed together they will undergo protonation and deprotonation reactions to greater or lesser extent and the system is shifted from zero level over to chemical equilibrium The proton balance states that the number of protons released must equal the number of protons taken up or remaining free in solution in other words the number of basic species formed from zero level must equal the number of acidic species formed from zero level As already stated in our case the zero level species are HA and H20 In order to be able to write down the proton balance we need to nd all acidic and basic species that can be formed from the zero level species possible formed acidic species H possible formed basic species A39 OH The proton balance is then obtained by stating that concentrations of formed acids are equal to concentrations of formed bases HlA390H 3 22 Construction of the log diagram Now that we have all the necessary equations we could solve H HA A and OH of our 001 M HA solution purely mathematically However we wanted to show how it is done in a graphical way To do that we choose to draw the logarithm of concentrations of all acidbase species H OH39 HA A as a function of pH in the same diagram This is the logarithmic diagram its yaxis is logY where Y H OH39 HA or A39 and its x axis is pH ie logH An empty diagram is shown in Fig 1 pH 109 Y 439 Fig 1 An empty logarithmic diagram In order to draw the traces for logH logOH logHA and logA we first need to express them in terms of pH LogH and pH have the relationship logtH pH In our log diagram this equation is a straight line with a slope of 1 and intersection at 00 Fig 2 shows what it looks like The trace for OH is obtained just as easily We have the relationship OH39 KwII taking logarithms of both sides and rearranging gives logOH pHlogKW pH 14 This too is a straight line but with a slope of 1 and intersection at 140 Also this line is shown in Fig 2 Since H and OH39 are always present in aqueous acid base systems you always need these lines in your diagram That is the reason why you always start out with a diagram like the one in Fig 2 pH 3909 Y Fig 2 A logarithmic diagram showing logHI j and IogOH as a function of pH Deriving expressions for logHA and logA is more complicated and is omitted here The only thing we need to know about the expressions is their graphical appearance which is shown in Fig 3 pH pKa 1 log CHA gt2 3 E 4 D ltgt 5 v 6 7 Fig 3 A full logarithmic diagram describing the different acid base species in an aqueous solution of the acid HA The dot shows the location of the system point of the HAA acidbase pair The lines for HA and A39 both consist of a horizontal part that bends into an sloping part Note how they cross at the point pH p12 10gYJ logCHA This is a very important point in the diagram called the system point of the HA A39 acid base pair You will later see how the system point will serve as a guide for drawing up lines in the 39 diagram Now the diagram is complete What it shows is the distribution of the di erent acid base species as afunction opr For example it is easy to see that at pH8 the concentration ofHA is about 10 1 M and the concentration of A39 ile392 M 23 Determining equilibrium composition Now that we have constructed the diagram the only remaining task is to use the proton balance to find out which point in the diagram represents equilibrium As shown above the proton balance for our model system of 001 M HA is 11 14 0H39l Let s draw a log diagram for these species only log CHA gt 109 Y Fig 4 Log diagram for the proton balance species Since we have a solution of an acid pH must be much less than 7 and consequently OH39 is negligibly small compared to Hquot and A39 The correctness of this assumption can be easily verified by looking at the pHlt7 region of the log diagram the lines for H and A39 are clearly above the OH 39 line The proton balance reduces to Hl A39 I The equation is satisfied at the intersection of the curves for IF and A39 The equilibrium pH of 001 M solution of HA is the pH corresponding to this point ie 40 As a general rule the equilibrium point is located at the topmost intersection of an acid curve and a base curve This is the graphical equivalent of stating that you will only consider the largest concentrations on each side of the proton balance equation Usually the error introduced by neglecting the rest of the concentrations is negligible 24 Improvements in the use of the diagram 24 Accurate determination of equilibrium point In our example we determined the equilibrium pH by reading it off the pH axis This is a completely legitimate way of doing it but requires you to draw the lines very accurately Any inaccuracies in slopes or offsets will be re ected in the value for equilibrium pH A more reliable way is obtained by using some geometry consider the triangle abS in Fig 5 Since it is made up by lines with slopes 1 1 and 0 it has angles 45 90 and 45 Consequently the pH at point 2 lies exactly halfway between the pH values of points a and S Thus pH pCHA pKa2 2060 2 40 log CHA gt log Yl Fig 5 Geometric arrangement in a log diagramr The traces outline a triangle abS The beauty of all this is that even very crudely sketched lines will do for the diagram The lines are only used to determine the approximate location of the equilibrium point after which the exact equilibrium pH is calculated with the aid of geometry A general formula for calculating the equilibrium pH would be PHaPH5 b 2 4 pHeq pH where pH1 pH corresponding to point a in the triangle pHs pH corresponding to point S in the triangle pI L and pHS always coincide with known numeric values in the above case pHa coincides with pCHA and pHS coicides with pKa Thus the final determination of equilibrium pH involves no visual readings it is done based on the known numeric values of pCHA and pKa That way the accuracy of pHeq is just as good as quot r the accuracy of pCHA and pl 3 Summary The theory section was rather long and full of details so we will summarize the most important points discovered there Log diagram is a diagram where x axis pH y axis 10gY YH OH HA Aquot System point for an acidabase pair is a point in the log diagram where PH PKa 10gY log C C total concentration of acidbase Logarithmic expressions for H OH HA A39 as functions of pH Appearance in log diagram Expression long PH logOH39 pH 14 pH lt pKa l logHA log C logA 2 pH constant pH gt pKa l logHA pH constant logA39 log C straight line w slope 1 straight line w slope 1 horizontal line straight line w slope 1 straight line w slope 41 horizontal line Zero level original species in solution Proton balance acidic species formed 2 from zero level basic species formed from zero level Equilibrium point point in the diagram that satisfies proton balance 10 4 Instructions for Using Log Diagrams We will new list the steps you need to take when using a log diagram to solve an acid base problem Since this is best done with an example let s take a look at a solution of 001 M acetic acid pKa 48 Step 1 Get a log diagram containing the curves for H1 and OH39 Such a diagram is shown in Appendix Fig 21 Step 2 Locate system points for each acid base pair see Appendix Fig b For HAc Ac pK48 and log C 2 gt system point at 48 2 Step 3 Use the system point for each acid base pair to draw39the traces for the acid form and the conjugate base form in the diagram Showu in Appendix Fig c Step 4 Specify the problem you want to solve IF pH is known and you are out to nd the concentrations in other words you want to answer a question like What are the concentrations of HAc and Ac39 at pH 8 just read the concentrations off the diagram as shown in Appendix Fig d At pH 8 we get HAc z 10quot M A0 10 2 M IF pH is what you want to find out go to Step 5 Step 5 Write down the zero level of your system That is list the acidic and basic species originally present in the system HAc I le Step Q Write down all the acids and bases that can be formed from the zero level species HAc gt H acid Ac39 base H20 gt H acid OH39 base Step 7 Write down the proton balance formed acids formed bases H AC39 OH39 Step 8 Modify the log diagram so that it only contains traces for the species appearing in the proton balance In other words this almost always means removing the traces of the zero level species In our example the curve for HAO will be removed see Appendix Fig e 11 Step 9 Locate the topmost intersection between and acid curve and a base curve This is the equilibrium point The pH of the solution is the pH corresponding to this point The pH can either be read off the x axis visually or more accurately calculated using eq 4 pH pHS 20 48 H P 2 2 34 Location of equilibrium point is shown in Appendix Fig 1 12 5 Special Applications 51 Sketching titration curves A titration curve can be sketched reasonably well based on the following four points F stands for fraction titrated 1 pH before start ie at F0 2 pH at first in ection point ie at F05i 3 pH at end point equivalence point ie at F10 4 pH beyond end point age at F20 100 overtitrated Using a logarithmic diagram you can very easily find these key points We will take a look at an example where 001 M acetic acid pKa476 is titrated With a strong base A graphical representation of the example is shown in Fig g and h in Appendix 19 We have a 00 M solution of the weak acid HAG zero level HAc H20 formed species HAC gt H acidAc39base H20 gt H acid OH39 base proton balance HAc OH HquotAc Proton balance shows that the equilibrium pH will be found at the intersection of H and Ac lines at pH 34 F415 The added strong base has turned half of HAc into Ac HAc NaOH gt NaAc H20 This means that HAO 2 Ac39 05 CMAC and the equilibrium pH will be found at the intersection of HAc and Ac39 lines at pH 13K 48 F1 The added strong base has turned all of HAc into Ac meaning that we have a 001 M solution of the weak base Ac39 zero level Ac39 H20 formed species Ac39 gt HAc acid HZO gt H acid OH39 base proton balance HA0 H OI139 gt HAc z OH Proton balance shows that the equilibrium pH will be found at the intersection of HA0 and OH lines at pH 84 F20 Strong base has been added 001 M in excess and will determine the pH of the solution Graphically the pH at F20 is located at the intersection of the vertical line at logCHAc and the line for ngOH39 We now have all four points and can sketch the titration curve as shown in Fig h Determining each point took several steps de ning zero levels proton balances etc so how is this any easier or quicker than calculating the points algebraically The answer is now that you have gone through the steps once you never have to do it again Any titration of a weak acid with a strong base will give you a similar pattern so you just draw the log diagram for the acid nd the key points at the intersections and sketch the titration curve 6 References General theory and use of logarithmic diagrams de Levie R Critical Reviews in Analytical Chemistry 1997 27 5176 Sill n L in Kolthoff I and Elving P eds Treatise on Analytical Chemisny part I v01 1 Ch 8 Interscience New York 1959 Hansen E Kemisk Reaktionslaere Polyteknisk Forlag Lyngby 1987 W inninen E Analytisk Kemi Turku Abo 1987 Logarithmic diagrams in complexometric titrations Johansson A Winninen E in Kolthoff I and Elving P eds Treatise on Analytical Chemistty part 1 vol 11 ch 117 Interscience New York 1975 Estimation of titration errors with logarithmic diagrams Wannjnen E Talanta 1980 27 2932 Appendix Ema V 23 in Examples of log diagram use I is quot Example 1 Find the pH of a 01 M solution of NH4C1 t pl Get a log diagram containing the curves for H and OH Step 2 Locate system points for each acid base conjugate pair tor 11 N113 131 913 analog C 1 gt system pomt at 913 Step 3 Use the system point for each acid basa species to draw the traces for the acid form and the conjugate base form in the diagram Step 4 Specify the problem you want to solve IS pH is known and you are out to find the concentrations No IS pH is what you want to find out Yes gt Go to step 5 Step 5 Write down the zero level of your system NH H10 39Step 6 Write down all the acids and bases that can be formed from the zero level species NH4 gt H acid NH3 base H20 gt H acid OH base Step 7 Write down the proton balance formed acids 2 formed bases H NH3 OH Step 8 Modify the log diagram so that it only contains lines for the species appearing in the proton balance Step 9 Locate the equilibrium point at the topmost intersection between and acid curve and a base curve Calculate exact equilibrium pH using geometry and known numerical values pH pHS 10 925 q z 5 1 H P e 2 2 jmvm QM Egamplc 1 Em Eco Emquot Example 2 Find the pH of a 01 M solution of NaH2P04 Step 1 Get a log diagram containing the curves for FF and OH Step 2 Locate system points for each acid base conjugate pairs For39HgP043939 HPOfquot pK 216am log C l 39 dmiwstempomtart Zilofl For H2190 HPof39 pig 721 and log C 1 2 system point at 721 1 For HPO 1 PO43 pKaz 1232 and log C 1 2 system point at 1232 1 Step 3 Use the system point for each acid base species to draw the traces for the acid form and the conjugate base form in the diagram Snem Specify the problem you want to solve IS pH is known and you are out to nd the concentrations No IS pH is what you want to nd out Yes gt Go to step 5 Egg Write down the zero level of your system 152130 H20 m Write down all the acids and bases that can be formed from the zero level species HZPO gt H acid H3P04 acid HPof base POf base H20 gt H acidOH base Note Whether an amphoteric species is de ned as an acid or a base depends on the zero level species In this case the zero level species is IlZPO439 Because HPOf is formed from it by release of a proton HPOf is defined as a base In some sense you could say that HPOf is a base relative to HZPO4 Step 7 Write down the proton balance formed acids formed bases HEPO4 H1 HPOf39 POf OH Step 8 Modify the log diagram so that it only contains lines for the species appearing in the proton balance Step 9 Locate the equilibrium point at the topmost intersection between and acid curve and a base curve Calculate exact equilibrium pH using geometry and known numerical values I 39 szq pH pHs z 216721 47 77 Ogm 39 109M lugIY gMas r EKQWLIO C Z oalYl Iogm 1232 14 72 721 1232 14 l logarithmic diagram u n n u u Step 1 locate system points Step 2 set cdncentration level pH Log C I SYSTEM POINT 390 H OH39 Fig2 Experiment 7 Ni Notes Preparation I Submit a weighing bottle for your unknown I Dry the NiO standard and unknown for 3 hours in a 125 degree oven Experiment I Carry a blank through the entire procedure This means to do the entire digestion process on a blank eg some Dl HZO DON T FORGET 7 roughly 20 of students have to start over I Evaporation to dryness takes a long long time Do it in the hood You must monitor the boiling acid solutions at all times The TA will turn off any unattended hot plates I Do not leave digested and dried samples in the 125 degree oven for much more than 1 hour The Ni may oxidize I Use cork stoppers to cap test tubes NOT parafilm which is soluble in chloroform Review Question Reminder This review question is not required 7 it is designed to extend your knowledge about the experiment at hand Find information on aqua regia a mixture HCl and HNO3 How does it work How does it convert NiO into soluble nickel cation PreLab Calculation The answer is include units
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