Titrations in Analytical Chemistry S Harles D

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comments by Fatemeh Masdarolomoor Shahrood University of Technology Titrations inJuly 2020 CHAPTER 13 Analyticalal Chemistry

ˬΎϫΩϳγ΍ ̵έϳ̳ ϩί΍Ωϧ΍ ̵΍έΑ ϥϭϳγ΍έΗϳΗ ̵ΎϬηϭέ ˬ̵ίϠϓ ̵ΎϬϧϭϳ ˬΎϫ ϩΩϧϫΎ̯ ˬΎϫ ϩΩϧγ̯΍ ˬΎϫίΎΑ έ̴ϳΩ Ω΍ϭϣ ί΍ ̵έΎϳγΑ ϭ Ύϫ ϥϳ΋Ηϭέ̡ Titrationstions are widelyy used in analyticalanaly Ю chemistry toź determine acids,acids, bases, oxidants,oxiidants, reductants,rreducttants, metaltal ions,ionsionns, proteins, andnd manymanm other species. TTitrationsŞ are based on a reaction between the ana- lytetee aand a standard reagent known as the titrant. The reaction is of known and reproducible stoi- chichchiometry.hihi The volume, or the mass,Ɖ of the titrant needed to react completely withw the analyte is dedetermined and used to calculate the quantity of analyte. A volume-based titrationio is shown in this ffigure in which the standard solutionķ is added from a buret, and the reaction occursooccc in the Erlen- meymeyer flask. In some titrations,ɧ known as coulometric titrations, the quantityntity ofoffch charge requiredi to compcompletelyԮ consume the analyte is obtained. In any titration, the point of chemicalchehem equivalence,eq len calledlled the end point when determined experimentally, is signaled by an indicatoricatcat color changechaa orr a changchange in an instrumental response. This chapter introduces the titration principlerin and thethh calcu- Ҏlatilations that determine the amount of the unknown.own. TitrationTit curves, which showw theth progressogr of the titration, are also introduced. Such curves will be used in several of the followingg chapters.chach pte

:αΎγ΍ Charles D. Winters Εγ΍ ϡίϻ ΕϳϟΎϧ΁ ΎΑ ϝϣΎ̯ εϧ̯΍ϭ ̵΍έΑ Ϫ̯ ϡϭϠόϣ ΕυϠϏ ΎΑ έ̴ηϧ̯΍ϭ ϥ΍ίϳϣ ϥϳϳόΗ ź Titrationn methodmethodsmethoods arearre basedb on de- itration methods, often called titrimetric methods,ethods,odsds includeinc a largerge anda powerful group termininggq the quantityqquantua ti y of a reagent g ofƈ T of quantitative procedures based on measuringsurinuringringnng theth amountam of a reagent of known known concentrationntratra that is requiredq ɧ concentration that is consumed by an analyte in a chemicalchemchchemicalem or electrochemicale reaction. to react comcompletelypletely with the analyte. ֌ The reagent may be a standard solution Volumetric titrations involve measuring the volumeolume of a ssosolutionsolutiooluto of known concentration of a chemical or an electricࣰ current of that is needed to react completely with the analyte.nalyte.yt In ГGravimetGravimetricavi titrations,Ż the mass of known magnitude. Ƽ the reagent is measured instead of its volume. Inn coulometcoulomecoulometric͇ titrations,Ƹ the “reagent” In volumetric titrations, the volume is a constant direct electrical current of known magnitude҅nitudenitud that consumes the analyte. For of a standard reagentź is the measured this titration, the time required (and thus the total charge) to complete the electrochemical quantity. reaction is measured (see Section 22D-5). In coulometric titrations, the quantity This chapter provides introductory material that applies to all the different types of of charge required to complete a reac- titrations. Chapters 14, 15, and 16 are devoted to theĹ various types of neutralization titration with the analyte is the measured tions in which the analyte and titrants undergo acid/base reactions. Chapter 17 provides quantity. information about titrations in which the analytical reactions involve complex formation or formation of a precipitate. These methods are particularly important for determining a variety of cations. Finally, Chapters 18 and 19 are devoted to volumetric methods in which the analytical reactions involve electron transfer. These methods are often called redox titrations. Some additional titration methods are explored in later chapters. These methods include amperometric titrations in Section 23B-4 and spectrophotometric titrations in Section 26A-4.

13A SOME TERMS USED IN VOLUMETRIC TITRATIONS1 A standard solution (or a standard titrant) is a reagent of known concentration that A standard solution is a reagent is used to carry out a volumetric titration. The titrationtitration is performedperformed byby slowlyslowly add-add- of known concentration. Standard ing a standard solution from a buret or other liquid-dispensing device to a solution solutions are used in titrations and in many other chemical analyses. of the analyte until the reaction between the two is judgedjudge complete. TThe volume or mass of reagent needed to complete the titrationonn is determineddetermdetedetdee from the difference between the initial and final readings. A volumetriclumetrictric titrationtitrattittratrationon processpproceoce s iss depicteddep in FigureFigure 13-113-1. It is sometimes necessary to add an excess off thet standardstandarst rdd titrantti ntt anda d thenth determine the excess amount by back-titrationbbaack-titrationon withth a seconds ndd standardstans dtd titrant. For Back-titration is a process in which example, the amount of phosphate in a sample canann beb determineddeterminddeterm byb adding a mea- the excess of a standard solution used sured excess of standard silver nitrate to a solutionn of theth sample,sam whichwh leads to the to consume an analyte is determined by titration with a second standard formation of insoluble silver phosphate:hate:te: solution. Back-titrations are often έ΍ΩϘϣ ϪΑ Ϫ̯ Ε΍έΗϳϧ ϩέϘϧ Ωέ΍ΩϧΎΗγ΍ ϝϭϠΣϣ Ю requiredq d when the rate of reaction 1 1 32 S ź .Ωϭη ̶ϣ ϪϓΎο΍ ΕϳϟΎϧ΁ ϪΑ "̵ΩΎϳί" Ύϣ΍ ϡϭϠόϣ 3Ag3Ag PO4 Ag3POO4(s) Ş between the analyteyg and reagent is slow or when the standardstandard solutionsolution lacks The excess silver nitratee is thethen backbacback-tback-titratedackckk with a standardƉ solution of potassium stability.stability. thiocyanate: Ε΍έΗϳϧ ϩέϘϧ ̵ΩΎϳί ˬαϭ̰όϣ ϥϭϳγ΍έΗϳΗ έΩ 1 1 2 S ķ .Ωϭη ̶ϣ έΗϳΗ ΕΎϧΎϳγϭϳΗ Ωέ΍ΩϧΎΗγ΍ ϝϭϠΣϣ ΎΑ Agg SCN AgSCN(Ƹs) ɧ The amountunt of silvesilver nnitratetratete is chemicallychemic͕Ԯ equivalent to the amount of phosphatete ion plus the amountammount of thiththiocyanateocyanateyanate used fofor the back-titration. The amount of phos- ththe amount of silver nitrate and thehe amountam of ڢphate is thenn the differenced ren bbetweentween thiocyanate.ate. Ҏ 13A-1 EquivalenceA-1 Equivale1 qu Points and End Points ź TheThhe equivalenceequivuivauivivalencalen point in a titrationĹ is a theoretical point reachedr when the amountmountounun Thehe equivalenceequivaequivivalivvalenclen point is the point in of addedaddeadddded titrtitranttitrtr is chemically equivalent to the amount of analyte in the sample.ple.e. ForFoF a titrationationtion whenwh the amount of added example,examplemp e the equivalence point in the titration of sodium chloride with silver nitratenitiitra standardd reagent is equivalent to the ƈ amountam of analyte. occursursr after exactly oneɧ mole of silver ion has been added for each mole off chchlchlochlo-hlohhlloo- ֌ ride ion in the sample. The equivalence point in the titration of sulfuric acid withh sodium hydroxide is reached after introducing 2 moles of base for eachach mole Ż of acid. ࣰ ͇Г Ƹ We cannotƼ determine the equivalence point of a titration experimentally. Instead,nstead,҅ we canź only estimate its position by observing some physical change associated with the condition of chemical equivalence. The position of this change is called the end point The end point is the point in a titra- for the titration. We try very hard to ensure that any volume or mass difference between tion when a physical change occurs the equivalence point and the end point is small. Such differences do exist, however, as that is associated with the condition of chemical equivalence. a result of inadequacies in the physical changes and in our ability to observe them.Ĺ The difference in volume or mass between the equivalence point and the end point is the titration error. In volumetric methods, the titration Indicators are often added to the analyte solution to produce an observable physi- error, Et, is given by cal change (signaling the end point) at or near the equivalence point. Large changes in 5 2 the relative concentration of analyte or titrant occur in the equivalence-point region. Et Vep Veq These concentration changes cause the indicator to change in appearance. Typical where Vep is the actual volume of reagent required to reach the end point and Veq is the theoretical volume necessary to reach the equivalence point. 1For a detailed discussion of volumetric methods, see J. I. Watters, in Treatise on Analytical Chemistry, I. M. Kolthoff and P. J. Elving, Eds., Part I, Vol. 11, Chap. 114. New York: Wiley, 1975.

Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 304CHAPTER 13 Titrations in Analytical Chemistry s harles D. Win h Charles D. Winters Charles D. Winters Charles D. Winters Ch Charles D. Winters Typicalpicall setupsetu forf carryingyinging DetailDetDЮ of the buret gradua- Before the titration begins. oout a titration.itratio TThe ap- tions. Normally, the buret is The solution to be titrated, paparatparatusaratusus consistsconsicon of a buret,uret,et, filled with titrant solution an acid in this example, is a buretretet standsta and clamp to within 1 or 2 mL of the placed in the flask, and the withthh a whitew porcelain base Ɖzero position at the top. indicator is added as shown to proviproprovide an appropriate The initial volume of the in the photo.ot TheT indicator backbacbackgroundackgackac for viewing buret is read to the near- in this casease is phenolphtha-phen indicinindiindicatordic changes, and a ɧķest 0.01 mL. The reference lein, whichwhihichhiichch turnstut pinpink in wide-mowide-mwwide-mouth Erlenmeyer point on the meniscus and basicasicc solution.solutisolutluution.utiuttti flasksk contacontainingconԮ a preciselyƸ the proper position of the known͕n volumvolumev of the solu- eye for reading are depictedted tion to beb ttitrated. The so- in Figure 2-21. is normally delivered ڢlution intoҎo the flask using a pipet, Ĺas shown in Figure 2-22. ź ɧƈ ֌ ࣰ Ż źƼ Г Charles D. Winters Charles D. Winters

During titration. The titrant Titration end point. The end point is achieved when the is added to the flask with barely perceptible pink color of phenolphthalein persists. swirling until the color of The flask on the left shows the titration less than half a the indicator persists. In the drop prior to the end point; the middle flask shows the initial region of the titra- end point. The final reading of the buret is made at this tion, titrant may be added point, and the volume of base delivered in the titration is rather rapidly, but as the calculated from the difference between the initial and final end point is approached, buret readings. The flask on the right shows what hap- increasingly smaller por- pens when a slight excess of base is added to the titration tions are added; at the end mixture. The solution turns a deep pink color, and the end point, less than half a drop point has been exceeded. In color plate 9, the color change of titrant should cause the at the end point is much easier to see than in this black- Figure 13-113-1 The titration process. indicator to change color. and-white version.

indicator changes include the appearance or disappearance of a color, a change in color, or the appearance or disappearance of turbidity. As an example, the indicator used in the neutralization titration of hydrochloric acid with sodium hydroxide is phenolphthalein, which causes the solution to change from colorless to a pink color once excess sodium hydroxide has been added. We often use instruments to detect end points.ointsts.tss. TheseThT instruments respond to properties of the solution that change inn a characteristiccharacchararactacterist waywaw duringdurinurin the titration. Among such instruments are colorimeters,orimeters,ters,ers,, turbidimeters,turbturbidimurbbid metersters, spectropho-spectrspect tometers, temperature monitors, refractometers,ters,rs, voltmeters,voltmetoltm ers,rsrs,s current c rrentent meters,metersm te and conductivity meters.

13A-2 Primary Standards A primary standard is a highlyg purifiedrifiedfied compoundco d thatththahatt servessserv asa a reference mate- A primary standard is an ultrapure rial in titrations and in otherherer analyticalana alyticalal methods.methome Thehe accuracyaccuacccuracy Юof a method critically compound that serves as the reference ź material for a titration or for another depends on the properties off thethhe primaryprimar standard.d.. ImportantImpoIm requirements for a Ş type of quantitative analysis. primary standard are the following:fo wing:ng: Ϫϳϟϭ΍ ̵ΎϫΩέ΍ΩϧΎΗγ΍Ɖ ΕΎϳλϭλΧ 1. High purity. Establishedblishedlished methodsmethethodsethohods forf confirming purity should be available. 2. Atmospheric stability.ty. ķ 3. Absence off hydhydrate wateraterr soo thatth the composition of theɧ solid does not change with variationsriationsions in humidity.h miditydi y. Ԯ Ƹ 4. Modestst cost. ͕ medium.m ڢReasonableableblee solubilityso y iin thethee titration .5 6. Reasonablyablyy largelargg molarola massma so thattha the relative error associated withh weighingweigh the standarddardard is minimized.minm nimizniminimmized.ed Ҏ

Veryy feww ccompoundscompou meet or even approach these criteria, and onlyly a limitedte numbermberber ofo prprimaprimary-standard Ĺsubstances are available commercially. As a conse-nse ź quenquence,ence,ce lesslelessss pureppu compounds must sometimes be used in place of a primary standard.ndard.ndarddadarddarard The puritypurpuurityity of such a secondary standard must be established by careful analysis.ysis.s. A secondaryndd standard is a compound ƈ whosew purity has been determined ɧ by chemicalc analysis.֌ The secondary standardstandstan serves as the working standard 13B STANDARD SOLUTIONS materialm for titrations and for many otherot analyses.Ż Standard solutionsࣰ play a central role in all titrations. Therefore, we mustust con-͇Г Ƹ sider theƼ desirable properties for such solutions, how they are prepared, and҅d how theirź concentrations are expressed. The ideal standard solution for a titrimetrictric method will ϝ΁ ϩΩϳ΍ Ωέ΍ΩϧΎΗγ΍ ϝϭϠΣϣ ΕΎϳλϭλΧ 1. be sufficiently stable so that it is necessary to determine its concentration only once; Ĺ 2. react rapidly with the analyte so that the time required between additions of reagent is minimized; 3. react more or less completely with the analyte so that satisfactory end points are realized; 4. undergo a selective reaction with the analyte that can be described by a balanced equation.

Few reagents completely meet these ideals. The accuracaccuracyy of a titration can be no better than the accuracyaccuracy of the concen-concentration of the standard solution used. Two basic methods are used to establish the

concentration of such solutions. The first is the direct method in which a carefully determined mass of a primary standard is dissolved in a suitable solvent and diluted to In a standardization, the concentra- a known volume in a volumetric flask. The second is by standardization in which the tion of a volumetric solution is deter- titrant to be standardized is used to titrate (1) a known mass of a primary standard, mined by titrating it against a carefully (2) a known mass of a secondary standard, or (3) a measured volume of another stan- measured quantity of a primary or secondary standard or an exactly known dard solution. A titranttranannt thatthath is standardized is sometimes referred to as a secondary-secondary- volume of another standard solution. standard solutiosolutionionn. Thee concentrationcconceco atiti of a secondary-standard solution is subject to a larger uncertaintyrtainty thanhanan isis ththehe concentrationcononcentcentrationation of a primary-standard solution. If there is a choice, then,n, solutionstio aareree bebbest preparpreppreparedaredd by the direct method. Many reagents, however, lack the propertiesproper requiredreqrequirre foror a primary standard and, therefore, require standardization.n 13C VOLUMETRICVOLV METRETRRIRICIIC CALCULATIONSCACЮ ź AsA wewe indindicatedindic in SectionSecti 4B-1, we canŞ express the concentration of solutions inn sevseveseveraleveraeveev ways. For thethh standard solutions used in most titrations, either molamolarolalara concentration, c, orƉ normal concentration, cN, is usuallyuallyal used. Molar coconcenconcconcentration is the number of moles of reagent contained in onone litliter of solu- tiontion,ioi and normal concentrationķ is the number of equivalentsalentsts of reagentreagenr in the samesamsaamea vvolume. ɧ ThrougThroThroughoutԮ this text,Ƹ we base volumetric calculationstionss exclusivelyusivelyve y on mmomolarolarar ccoconcon-on-- centration͕rati and molar masses. We have also includedded in Appendixnd 7 a discussionddiscudis ussionssion how vovolumetric calculations are performed based on normalmal concentratiocconcentrationrat andnd ڢof equivalentҎquiva masses because you may encounterncounterounter these termsms and theirheir useses ini theth in- dustrial and health science literature.

13C-1 Some Useful Relationshipsps mA ź 5 nA ĹMost volumetric calculations are based on twowoo pairspaippa of simplempleple equationsequeqquatuau that are M ❯ A derived from definitions of the mole, the millimole,llimole,mo aand the molamolmolar concentration. where nA iss the amountamamomo of A, M For the chemical species A, we can write mA is the massasss oof A, and A is ɧƈ the molar mass of A. ֌ mass A (g) amount A (mol) 5 (13-1) ࣰ molarolar massГ A (g/mol) Ż n ͇ A Ƹ 5 Ƽ5 3 c orn V c A V A A ❯ ҅mass A (g) ź amount A (mmol) 5 (13-2)(13-2) millimolar mass A (g/mmol)

Any combination of grams, The second pair of equations is derived from the definition of molar concentration, moles, and liters can be expressed ❯ that is, in milligrams, millimoles, and Ĺ milliliters. For example, a 0.1 M 5 3 amol Ab solution contains 0.1 mol of a amount A (mol) V (L) cA (13-3)(13-3) species per liter or 0.1 mmol per L milliliter. Similarly, the number of moles of a compound is equal mmol A to the mass in grams of that 5 3 a b amount A (mmol) V (mL) cA (13-4)(13-4) compound divided by its molar L mass in grams or the mass in milligrams divided by its where V is the volume of the solution. millimolar mass in milligrams. Equations 13-1 and 13-3 are used when volumes are measured in liters, and Equations 13-2 and Equations 13-4 when the units are milliliters.

13C-2 Calculating the Molar Concentration Ωέ΍ΩϧΎΗγ΍ ϝϭϠΣϣ ϪϳϬΗ of Standard Solutions The following three examples illustrate how the concentrations of volumetric reagents are calculated.

EXAMPLE 13-1

Describe the preparation of 2.000 L of 0.0500 MMA AgNOAg 3 (169.87( 87 gg/mol)mol) from the primary-standard-grade solid. Solution 5 3 amount AgNO3 Vsoln(L) cAgNO3(mol/L)

0.050.0500 mol AgNOO3 5 2.00.00 L 3 5 0.1000. mol AgNO L Ю 3 To obtain the mass of AgNO3, we rearrangeear Equation 13-23 toƉ give 169.87 g AgNO 5 3 3 mass AgNONO3 0.1000.1010010 mol AgNO3 mol AgNO3 5 16.987.987 g AgNO3

Therefore, the solution shouldsh ld be preparedprepa by dissolving 16.987 g of AgNO3 in water and diluting to the markrk in a 2.000 L volumetricv flask.

EXAMPLEEXAXAMPLEE 13-2 1 ź A standardtandnd rd 0.01000. M solution of Na is required to calibrate an ion-selectiveiveve electrodeelec de method to determine sodium. Describe how 500 mL of this solutionion cann beb preparedp fromɧƈ primary standard Na2CO3 (105.99 g/mL). Solution ֌ We wish to compute the mass of reagent required to produce a species concentrationntration of 0.0100 M. In this instance, we will use millimoles since the volume is in milliliters.lli Ż ࣰ 1 Г Because Na2CO3 dissociates to give two Na ions, we can write that the numberber͇ of Ƹ millimolesź of Na2CO3 needed is 0.0100 mmol Na1 1 mmol Na CO amount Na CO 5 500 mL 3 3 2 3 2 3 mL 2 mmol Na1 5 2.50 mmol Ĺ From the definition of millimole, we write

mg Na CO 5 3 2 3 mass Na2CO3 2.50 mmol Na2CO3 105.99 mmol Na2CO3 5 264.975 mg Na2CO3

Since there are 1000 mg/g, or 0.001 g/mg, the solution should be prepared by dissolving

0.265 g of Na2CO3 in water and diluting to 500 mL.

EXAMPLE 13-313-3 How would you prepare 50.0-mL portions of standard solutions that are 0.00500 M, 0.00200 M, and 0.00100 M in Na1 from the solution in Example 13-2? Solution The number of mimillimolesll mol of Na1 taken from the concentrated solution must equal the number in thee dilute solutions.ut Thus,T ,

amountt NNa1 fromoomm concdd ssoln 5 amount Na1 in dil soln

Recall that the number oof milmillimoles is equal to the number of millimoles per milliliter times the numberer ooff mimillmillilmilliliters,i that is,

3 5 3 Vconcd cconcd Vdil cdil

wherere Vconcd and Vdil areare the volumes in milliliters of the concentrated and diluted 1 solutions,utioututi respectively, and cconcd and cdil are their molar Na concentrations. For the 0.000.00500-M000 solution, this equation can be rearranged to

V 3 c 3 1 5 dil dil 5 50.0 mL 0.005 mmol Na /mL 5 Vconcd 1 25.02 mL cconcd 0.0100 mmol Na /mL Therefore,͕refo to produce 50.0 mL of 0.00500 M Na1, 25.0 mL of the concentratedconconc ed ssolutiontion .be diluted to exactly 50.0 mL ڢshould ReRepeat the calculation for the otherer ttwo molarities to confirmfirm thatat dilutingd Ҏ10.0 and 5.00 mL of the concentrated solutiosolution to 50.0 mL producesucesces the desired Ĺconcentrations. ź 13C-3 Working with Titration Dataa ƈ Two types of volumetric calculations are discussedcucussedussedsssed here.here In the first, we compute ɧ concentrations of solutions that have been standstandardizedandandndardddaardizedar d agagainst either֌ a primary- standard or another standard solution. In thehe second,secondecondnd,nd we calculatecca the amount of analyte in a sample from titration data. Bothothh types of calculationcalccalcual areŻ based on three ࣰ algebraic relationships. Two of these are Equationsquationsations͇Г 13-2 andan 13-4, which are based Ƽ on millimoles and milliliters. The third relationshipnship is theth stoichiometricƸ ratio of the number of millimoles of the analyte to the number҅er of millimoles of titrant. Calculating Molar Concentrations from Standardization Data Examples 13-4 and 13-5 illustrate how standardizationĹ data are treated. EXAMPLE 13-4 A 50.00-mL portion of an HCl solution required 29.71 mL of 0.01963 M

Ba(OH)2 to reach an end point with bromocresol green indicator. Calculate the molar concentration of the HCl. Solution

In the titration, 1 mmol of Ba(OH)2 reacts with 2 mmol of HCl:

1 S 1 Ba(OH)2 2HCl BaCl2 2H2O

Thus, the stoichiometric ratio is

2 mmol HCl stoichiometric ratio 5 1 mmol Ba(OH)2

The number of millimoles of the standard is calculateded by b substitutingsub into Equation 13-4:

mmolm BaBa(OH)OH) 5 3 2 amount Ba(OH)2 29.71 mL Ba(OH)2 0.019630 01 mLm Ba(OH)(OH 2

To find the number of millimoles of HCl, we multiplyp thishiisi resultresul by the stoichiometric ratio determined from the titration reaction:

5 3 3 2 mmol HCl amount HCl (29.7171 0.01963)963) mmol Ba(OH)OH)2 1 mmol Ba(OH)Şź2 To obtain the number of millimoles offH HHCl per mL, we divide by the volume of the acid. In determining the number of Ɖ ❮significantficanca figures to retain in Therefore, volumetricumetricet calculations,c the (29.71.71.71 3 0.01963 3 2) mmol HCl stoichiometricio ratio is assumed 5 to bebe knownknoknk wnw exactlyexac without cHCl ɧ 50.0 mL HCl Ƹ uuncertainuncertainty.ncertairtaainty.ain Ԯmmmol HCl 5 0.023328 5 0.02333 M mL HCl

ΎΑ ϥ΍έΗϳΗ ϝϭϠΣϣ ϥΩέ̯ έ΍ΩϧΎΗγ΍ Ĺ Ϫϳϟϭ΍ Ωέ΍ΩϧΎΗγ΍ ź EXAMPLEEXXAMPLAMPLPLEPLEE 113-5 Titrationra i of 0.2121 ƈg of pure Na2C2O4 (134.00 g/mol) required 43.31 mL of KMnOK 4. What isɧ the molar concentration of the KMnO4 solution? ThTheh ֌ chemical reaction is

2MnO 2 1 5C O 22 1 16H1 S 2Mn21 1 10CO 1 8H O Ż Ƽࣰ4 2 4 2 2 ͇Г Ƹ Solutionź From this equation we see that

2 mmol KMnO stoichiometric ratio 5 4 5 mmol Na2C2O4 Ĺ

The amount of primary-standard Na2C2O4 is given by Equation 13-2

1 mmol Na C O 5 3 2 2 4 amount Na2C2O4 0.2121 g Na2C2O4 0.13400 g Na2C2O4 (continued)

To obtain the number of millimoles of KMnO4, we multiply this result by the stoichiometric ratio:

2 mmol KMnO 5 0.2121 3 4 amount KMnO4 mmol Na2C2O4 0.1340 5 mmol Na2C2O4

The concentrationn ofo KMnOKMK 4 is then obtained by dividing by the volume consumed. Thus,

a 0.212121212 3 2b mmolm KMnO4 5 0.13400133 5 5 cKMnO4 0.01462 M 43.31 mL KMnO4

NoteNNotote thatth units are carriedcar through all calculations as a check on the correctness of thehee relationshipsrelrerelal used in Examples 13-4 and 13-5. CalculatingCalCaCalcal the Quantity ofķ Analyte from Titration Datata AsA sshownsh by the examples thatɧ follow, the systematic approachroachchh justjuuststt describeddedesc isi also usedsed to ccompute analyteƸ concentrations from titration data.a. Ԯ͕ڢ ҎEXAMPLE 13-6 εϫΎ̯-εϳΎγ̯΍ ϥϭϳγ΍έΗϳΗ ί΍ ϩΩΎϔΗγ΍ ΎΑ ϥΩόϣ ̲ϧγ έΩ ϥϫ΁ ϥϳϳόΗ A 0.8040-g sample of an iron ore is dissolvedsolv in acid. The iron iss thenhehen reduced 21 to Fe and titrated with 47.22 mL of 0.02242 M KMnO4 solution.ioon.on. CalculateC the results of this analysis in terms of (a) % Fe (5(55.847 g/mol)mol)ol) anddd(b (b)b % Fe3O4 ź Ĺ(231.54 g/mol). ƈ Solution ɧ The reaction of the analyte with the reagent is describedsc byyt theh equation֌ 2 1 21 1 1 S 21 1 31 1 MnO4 5Fe 8H MMn 5Fe5FFe 4H2O iɰȧ ࣰ 1 Г 2 5 mmol FFe Ƽ (a) stoichiometric ratio 5 ź 1 mmolmol KMnOK 4 0.02242 mmol KMnO 5 3 4 amount KMnO4 47.22 mL KMnO4 mL KMnO4 5 mmol Fe21 21 5 3 3 amount Fe (47.22 0.02242) mmol KMnO4 1 mmol KMnO4

The mass of Fe21 is then given by

g Fe21 mass Fe21 5 (47.22 3 0.02242 3 5) mmol Fe21 3 0.055847 mmol Fe21

The percent Fe21 is

(47.22 3 0.02242 3 5 3 0.055847) g Fe 21 % Fe21 5 3 100% 5 36.77% 0.8040 g sample

(b) To determine the correct stoichiometric ratio, we note that

21 ; 2 5 Fe 1 MnO4

Therefore,

; 21 ; 2 5 Fe3O4 15 Fe 3 MMnOO4

and

5 mmolol Fe O stoichiometricc ratiorat 5 3 4 3 mmol KMnO4

As in part (a),

47.227.227 mLm KMnO 3 0.02242 mmol KMnO 5 4 4 amountoun KMnO4 mL KMnO4 5 mmol Fe O 5 3 3 3 4 amountt FFe3O4 (47.22(4 2 0.02242)0.022 mmol KMnO4 3 mmol KMnO4 g Fe O 5 a 3 3 5b 3 3 4 mass Fe3O4 47.2247.24 0.02242 mmol Fe3O4 0.23154 3 mmolol Fe F 3O4 5 a47.22 3 0.02242 3 b 3 0.23154 g Fe O 3 3 4 ź 5 3 5 % Fe3O4 Ĺ 100% 50.81%% 0.8040 g sample ֌ FEATURE 13-1 Another Approach to Example 13-6(a) ͇Г ƸŻ Some people find it easier to write out the solution to a problem in such a way҅ that theź units in the denominator of each succeeding term eliminate the units in the numerator of the preceding one until the units of the answer are obtained.2 For example, the solution to part (a) of Example 13-6 can be written

0.02242 mmol KMnO 0.055847 g Fe 3 4 3 5 mmol Fe 3 47.22 mL KMnO4 Ĺ mL KMnO4 1 mmol KMnO4 mmol Fe 1 3 3 100% 5 36.77% Fe 0.8040 g sample

2This process is often referred to as the factor-label method. It is sometimes erroneously called dimensional analysis. For an explanation of dimensional analysis, perform a search at the Wikipedia website. In earlier texts, the factor-label method was sometimes called the “picket fence” method.

EXAMPLE 13-7 A 100.0-mL sample of brackish water was made ammoniacal, and the sulfide

it contained was titrated with 16.47 mL of 0.02310 M AgNO3. The analytical reaction is

1 22 S 2Ag 1 S Ag2S(s)

Calculate the concentrationcecenntrn n oof H2S ini the waterw in parts per million, cppm. Solution At the end point,

1 mmol H S stoichiometricstostoicchio ratio 5 2 2 mmol AgNO3 mmol AgNO 5 3 3 amount AgNONO3 16.47 mL AgNO3 0.02310 mL AgNO3 1 mmolmo H S 5 3 3 2 amount H2S (16.47 0.02310) mmol AgNO3 2 mmolol AgNO3 g H S 5 a 3 3 1b 3 2 mmass H2S 16.47 0.02310 mmol H2S 0.0340814081811 2 mmolm oll H2S ͕ 2 5 3 3 6.483 10 g H2S 23 ڢ 6.483 3 100 g H S 5 2 3 6 Ҏ cppm 100 ppmp 100.0 mL sample 3 1.00 g sample/mLsam sample 5 64.8 ppm ź Ĺ .Ωϳϧ̯ ϪΟϭΗ ppm ϑϳέόΗ ϪΑ ɧƈ ֌ FEATURE 13-2 ࣰ Rounding the Answer to Example 13-7 ͇Г ƸŻ Ƽ Note that the input data for Example 13-7 all contained҅ontained four or more significant fig- ź ures, but the answer was rounded to three. Why? We can make the rounding decision by doing a couple of rough calculations in our heads. Assume that the input data are uncertain to 1 part in the last significant figure. The largest relative error will then be associated with the sample size. In Example 13-7, the relative uncertaintyĹ is 0.1/100.0. Thus, the uncertainty is about 1 part in 1000 (compared with about 1 part in 1647 for the volume of

AgNO3 and 1 part in 2300 for the reagent concentration). We then assume that the calculated result is uncertain to about the same amount as the least precise measurement, or 1 part in 1000. The absolute uncertainty of the final result is then 64.8 ppm 3 1/1000 5 0.065, or about 0.1 ppm, and we round to the first figure to the right of the decimal point. Thus, we report 64.8 ppm. Practice making this rough type of rounding decision whenever you make a computation.

EXAMPLE 13-8 αϭ̰όϣ ϥϭϳγ΍έΗϳΗ ί΍ ϩΩΎϔΗγ΍ ΎΑ ̶ϫΎϳ̳ ̵΍ΫϏ ̮ϳ έΩ έϔγϓ ̵έϳ̳ ϩί΍Ωϧ΍

32 The phosphorus in a 4.258-g sample of a plant food was converted to PO4 and precipitated as Agg3PO4 byy adding 50.00 mL of 0.0820 M AgNO3. The excess AgNO3 was back-titrated with 4.06 mL of 0.0625 M KSCN. Express the results of this analysis in terms of % P2O5. Solution The chemical reactions are

1 S 32 1 1 P 2O5 9H2O 2PO4 6H3O 32 1 1 S 2PO4 6 Ag 2Agg3POPO4(s) excess A g1 1 SCNN2 S AgSCN(s)

The stoichiometric ratios arere

1 mmol P2O5 1 mmol KSCN and 6 mmolmmo AgNO3 1 mmol AgNO3

mmol AgNO3 total amount AgNO 5 50.000 mL 3 0.0820 5 4.100 mmol 3 mL mmol KSCN amount Ag AgNONO consumedmed by KSCKSCN 5 4.06 mL 3 0.0625 3 mL

1 mmol AgNO3 3 mmol KSCN 5 0.2538 mmol

1 mmol P O ź 5 Ĺ2 3 2 5 am amountmoumo P2O5 (4.100 0.254) mmol AgNO3 6 mmol AgNO3 5 0.6410 mmol P2O5 ֌ 0.1419 g P2O5 0.6410 mmol 3 5 mmol 3 5 % P2O5 100% 2.14% Ż ź 4.258 g sample ͇Г Ƹ EXAMPLE 13-9 ̵έΗϣϭΩϳ εϭέ

The CO in a 20.3-L sample of gas was converted to CO2 by passing the sample over iodine pentoxide heated to 150°C: Ĺ S I2O5(s) 1 5CO(g) 5CO2(g) 1 I2(g)

The iodine was distilled at this temperature and was collected in an absorber

containing 8.25 mL of 0.01101 M Na2S2O3.

22 S 2 22 I2(g) 1 2S2O3 (aq) 2I (aq) 1 S4O6 (aq)

The excess Na2S2O3 was back-titrated withw 2.16 mL of 0.00947 M I2 solution. Calculate the concentration of CO (28.01 g/mol) in mg per liter of sample. (continued)

Solution Based on the two reactions, the stoichiometric ratios are

5 mmol CO 2 mmol Na S O and 2 2 3 1 mmolmm I2 1 mmol I2

We divide the firstt ratio by thehe secondon too get a third useful ratio

5 mmolmm COC

2 mmolol Na2S2O3

This relationship revealsveals thatha 5 mmol of CO are responsible for the consumption of

2 mmolmol ofo Na2S2O3. Thehe totaltota amount of Na2S2O3 is

mmol Na S O 5 3 2 2 3 amount Na2S2O3 8.25 mL Na2S2O3 0.01101 mL Na2S2O3 5 0.09083 mmol Na2S2O3

TheTTh amount of Na2S2O3 consumed in the back-titration is

mmol I 2 mmolmo Na S O 5 3 2 3 2 2 3 aamountm Na2S2O3 2.16 mL I2 0.00947 ͕ mL I2 mmolm I2 5 mmol Na2S2O3 0.04091 ڢҎ The number of millimoles of CO can thenthe be calculated by usingsising tthe third stoichiometric ratio:

5 2 3 5 mmolmmolC CO ź Ĺ amount CO (0.09083 0.04091) mmol NaN 2S2O3 2 mmolmm l Na2S2O3 ɧƈ 5 0.1248 mmol CO ֌ 28.01010 mg CO mass CO 5 0.1248 mmol CO 3 5 3.4956 mg ࣰ mmГol COO Ż 3.4956 mg COO mg CO mass CO 5 5͇ Ƹ Ƽ 0.1720.17 ź vol sample 20.3 L sample L sample

13D GRAVIMETRIC TITRATIONSĹ Mass (weight) or gravimetric titrations differ from their volumetric counterparts in that the mass of titrant is measured rather than the volume. Therefore, in a mass titration, a balance and a weighable solution dispenser are substituted for a buret and its markings. Gravimetric titrations actually predate volumetric titrations by more than 50 years. With the advent of reliable burets, however, mass titrations were largely supplanted by volumetric methods because the former required rela- Remember that for historical tively elaborate equipment and were tedious and time consuming. The availability reasons we often refer to weight ❯ or weighing, but we really mean of sensitive, low-cost, top-loading digital analytical balances and convenient plastic mass, although most of us cannot solution dispensers has changed this situation completely, and mass titrations can bring ourselves to say massing. now be performed as easily and rapidly as volumetric titrations.

13D-1 Calculations Associated with Mass Titrations The most common way to express concentration for mass titrations is the weight

concentration, cw, in weight molar concentration units, Mw, which is the number of moles of a reagent in one kilogram of solution or the number of millimoles in one

gram of solution. Thus, aqueous 0.1 Mw NaCl containsntaainsns 0.10 mol of the salt in 1 kg of solution or 0.1 mmol in 1 g of the solution.

The weight molar concentration cw(A) off a solutionution ofo a solutesosolute A iss computedcompcompu using either one of two equations that are analogousogous tooE EquationEquaEqE tionionon 4-2:4-2:

no.o. molol A no.non mmolm lA A weight molar concentration 5 5 (13-5) no. kg solutionoolulu onn no.non g solutions

5 nA cw(A)A) msoln Ю ź where nA is the number of molesles off specspecispecies A and msoln iis the mass of theŞ solution. Gravimetric titration dataata ccan thenhehenennnb beb treated by using theƉ methods illustrated in Sections 13C-2 and 13C-33C-3 afteaftererr subssubstisubstitutionbsb of weight concentration for molar concentration and grams and kilogramskilogramramamsam for milliliters and liters.ɧķ 13D-2 AdvantagesAdvantagvanta s offG GravimetricGravimeGravim͕Ԯ TitrationsƸ In additionion too greater speedspesp edd and convenience,convenconv mass titrations offer certain otherr counterparts:countڢ advantages overer theirtht volumetricvolumeolu tricricҎ 1. Calibrationlibrationration off glasswareglasswggla glasswa and tedious cleaning to ensure properper drainageraina area completelytely eliminated.elimin 2. Temperaturemp atu correctionsco are unnecessary because the mass (weight)ht) molarla concenconcentrationoncencentcenentratiorat does not changeĹ with temperature, in contrast to the volumeumm ź momolarlar conconcentration.con This advantage is particularly important in nonaqueousqueousueoeoueo titratititrationstr tioio because ofƈ the high coefficients of expansion of most organic liquidsliqquidsquidqu (abo(aboutb 10 times thatɧ of water). ֌ 3. Mass measurements can be made with considerably greater precision and accuracyy than can volume measurements. For example, 50 g or 100 g of an aqueousueousou solu- 6 6 Ż tion can beࣰ readily measured to 1 mg, which corresponds to 0.001 mL. ͇ThisГ greaterƼ sensitivity makes it possible to choose sample sizes that lead to significantlyficantlyicantly Ƹ smallerź consumption of standard reagents. ҅ 4. Gravimetric titrations are more easily automated than are volumetric titrations.Ĺ 13E TITRATION CURVES As noted in Section 13A-1, an end point is signaled by an observable physical change near the equivalence point of a titration. The two most widely used signals involve (1) changes in color due to the reagent (titrant), the analyte, or an indicator and (2) a change in potential of an electrode that responds to the titrant concentration or the analyte concentration. To understand the theoretical basis of end point determinations and the sources of titration errors, we calculate the data points necessary to construct titration curves Titration curves are plots of a for the systems under consideration. A titration curve is a plot of some function of concentration-related variable versus the analyte or titrant concentration on the y axis versus titrant volume on the x axis. titrant volume.

13E-1 Types of Titration Curves The vertical axis in a sigmoidal Two general types of titration curves (and thus two general types of end points) titration curve is either the ❯ p-function of the analyte or occur in titrimetric methods. In the first type, called a sigmoidal curve, important titrant or the potential of an observations are confined to a small region (typically 60.1 to 60.5 mL) surround- analyte- or titrant-sensitive ing the equivalencee point.ppoinoin A sigmoidal curve in which the p-function of analyte electrode. (or sometimes thee titrant)titratrant)traranra isi plotted as a function of titrant volume is shown in Figure 13-2aa. The vertical axis of a linear- In the secondond typeypepee of curvccurve,urve,urveve ccalledalledled a linearline segment curve, measurements are segment titration curve ❯ is an instrument signal made on both sidesidesdes of,f, butb wellwelw l awaya y from,fromfrom,, theth equivalence point. Measurements that is proportional to the near equivalencence arere avoided.av dedd. InI this typet of curve, the vertical axis represents concentration of the analyte an instrument readingaadindinggt thatt isi directlydirecd proportional to the concentration of the or titrant. analytee oro the titrant.nt.t. A typicaltypicypiypi linearline segment curve is found in Figure 13-2b. Thee sigmsigmoidalsig typepepe offersoffoffeofffersfers thet advantages of speed and convenience. The linear segmentseegmentnt typety is advantageousvantageoantageeous for f Ю reactions thatź are complete only in the presence of a considerableconconsiderabsider excess of theth reagent or analyte.Ş Innnth tth thithis chapter and several that follow, we deal exclusively with sigmoidal titration curves.curvevesveese We explore linear segmentƉ curves in Chapters 23 and 26. p-function Equivalence ķ point 13E-2 Concentration13E-13E13 ɧ Changes During Titrationsrationsionsons Reagent volumeme Thehe equequivalenceequivԮ point Ƹin a titration is characterizedd by majorr changeschanha ges in thetht e rela-relrerela-- (a) Sigmoidal curvee tive concenconconcentrations͕ of reagent and analyte. Table 13-13-1 illustratesustratesessst thist phenphenomenon.henenomno enon.non datadat in the second column of the tablele show the changess in ththe hydroniumon ionon ڢThe concentrationҎonce as a 50.00-mL aliquot of a 0.100.1000 M solutionion of hydrochloricydrrochlor acacid is titrated with 0.1000 M sodium hydroxide.roxide.de. TheTh neutralizationn on reactiontionioion is describeddedescrib by the equation

1 1 2 S H3O OH 2HH2O (13-6) ź EEquivalence Instrument reading Ĺ point To emphasize the changes in relative concentrationntrationatatio thatth occurrin ini the equivalence ReagentReageea volume ƈ point region, the volume increments computedtedd areaare thoset requiredr to cause tenfold (b) Linear segment curve ɧ 1 decreases in the concentration of H3O (or tenfoldnfolfoldfo d increasesincrease in hydroxide֌ ion con- Figure 13-2Two types of titration centration). Thus, we see in the third columnolumnumn that aann adaaddition of 40.91 mL of 1 curves. ࣰ base is needed to decrease the concentrationtionn of HГ3O by one orderŻ of magnitude Ƽ from 0.100 M to 0.0100 M. An addition of only͇ 8.11 mmL is requiredƸ to lower the concentration by another factor of 10 to 0.00100҅00 M M; 0.89 mL causes yet another ź 2 tenfold decrease. Corresponding increases in OH concentration occur at the same time. End-point detection then depends on this large change in the relative concentration of the analyte (or titrant) that occurs at the equivalence point for every type of titration. Feature 13-3 describes how Ĺthe volumes in the first column of Table 13-1 are calculated. The large changes in relative concentration that occur in the region of chemical equivalence are shown by plotting the negative logarithm of the analyte or the titrant concentration (the p-function) against reagent volume, as seen in Figure 13-3. The data for these plots are found in the fourth and fifth columns of Table 13-1. Titration curves for reactions involving complex formation, precipitation, and oxidation/reduction all exhibit the same sharp increase or decrease in p-function in the equivalence-point region as those shown in Figure 13-3. Titration curves define the properties required of an indicator or instrument and allow us to estimate the error associated with titration methods.

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TABLE 13-1 Concentration Changes During a Titration of 50.00 mL of 0.1000 M HCl 1 Volume of 0.1000 M [H O ], Volume of 0.1000 M NaOH to Cause 3 1 NaOH, mL mol/L a Tenfold Decrease in [H3O ], mL pH pOH 0.00 0.1000 1.00 13.00 40.91 0.0100 40.9191 2.00 12.00 49.01 1.000 3 1023 8.111 3.00 11.00 49.90 1.000 3 1024 0.89899 4.00 10.00 49.99 1.000 3 1025 0.09000090 5.0000 9.00 49.999 1.000 3 1026 0.009009 6.000 8.00 50.00 1.000 3 1027 0.001001 7.00 7.00 50.001 1.000 3 1028 0.0010011 8.00 6.00 50.01 1.000 3 1029 0.0090.009 9.00 5.00 50.10 1.000 3 10210 0.09099 10.00 4.00 51.10 1.000 3 10211 0.91.9 11.00 3.00 61.11 1.000 3 10212 10.100.10 Ю 12.00ź 2.00 Ɖ Ş 14 pH 12 ɧķ 101 Ƹ H Ԯ 8 ͕

ڢ H or p 6 pH or pOH p 4 Ҏ pOH 2 Figuree 13-313-33-333 TitrationTitraTi curves of pH andd pOHOH versusversusrsrsus volumev of base for 0 ź 0 10203040506070Ĺ thee titrationitratioionioon ofo 0.1000 M HCl with ɧƈ Volume NaOH, mL 0.10000 M NaOH.N ֌ FEATURE 13-3 Calculatingࣰ the NaOH Volumes Shown in the First Column of Table 13-13 Г Ż 1 ͇ Ƹ Prior toƼ the equivalence point, [H3O ] equals the concentration of unreacted҅ HCl (cHClź). The concentration of HCl is equal to the original number of millimoles of HCl (50.00 mL 3 0.1000 M) minus the number of millimoles of NaOH added 3 (VNaOH 0.1000 M) divided by the total volume of the solution:

50.00 3 0.1000 2 V 3 0.1000 5 3 14 5 NaOH cHCl H3O Ĺ 1 50.00 VNaOH

where VNaOH is the volume of 0.1000 M NaOH added. This equation reduces to

3 14 1 3 14 5 2 50.00 H3O VNaOH H3O 5.000 0.1000VNaOH

Collecting the terms containing VNaOH gives

1 3 14 5 2 3 14 VNaOH(0.1000 H3O ) 5.000 50.00 H3O (continued)

5.000 2 50.003H O14 5 3 VNaOH 1 3 14 0.1000 H3O

1 5 Thus to obtain [HH3O ] 0.0100 M, we find

5.000.0 2 50.0000 3 0.0100 V 5 5 40.91 mL NaOHOH 0.10000 1 0.00100

Challenge: Use the samemme reasoningeas to show that beyond the equivalence point,

50.0003OH24 1 5.000 V 5 NaOHNaO 0.1000 2 3OH24

Spreadsheet SummaryChapter 7 of Applicationsns of Microsoft® Excel in Analytical Chemistry, 2nd ed., deals with plotting tititration curves. Several types of titrations are presented and ordinary titrationn curvescu are plotted aalong with derivative plots and Gran plots. The stoichiometricc approapproachpprprproa developedlo in ͕thishisԮ chapter is used andƸ a master equation approach is explored. ڢҎ WEBWE Look up titration in Wikipedia, the online encyclopedia.loped Give the definition ofo titrationtit WOWORKS found there. Is a chemical reaction necessary for a quantitative procedure to be cacalcalledlledd a titration? From what Latin word does titration derive? Who developeddev the first buretburet and in what Ĺyear? List five different methods to determine the endd pointp of a titration.itratioation. DefineD the ź ɧƈterm acid number, also called the acid value. How are titrationsatiatiotio applied too biodieselbiodid ֌ fuels? QUESTIONS AND PROBLEMS 13-1. Define ࣰ REQUIRED)ED)) to expressГ this ratio for calculationŻ of the *(a) millimole.Ƽ percentage of ͇ Ƹ (b) titration. *(a) hydrazine in҅ rocket fuel by titration with standard *(c) stoichiometricź ratio. iodine. Reaction: (d) titration error. 1 S 1 2 1 1 13-2. Write two equations that—along with the stoichiometric H2NNH2 2I2 N2( g) 4I 4H factor—form the basis for the calculations of volumetric titrations. (b) hydrogen peroxideĹ in a cosmetic preparation by 13-3. Distinguish between titration with standard permanganate. Reaction: *(a) the equivalence point and the end point of a titration. 1 2 1 1 S (b) a primary standard and a secondary standard. 5H2O2 2MnO4 6H 13-4. Briefly explain why the concentration units of milli- 1 2Mn2 1 5O (g) 1 8H O grams of solute per liter and parts per million can be 2 2 used interchangeably for a dilute aqueous solution. # *(c) boron in a sample of borax, Na B O 10H O, by 13-5. Calculations of volumetric analysis ordinarily consist 2 4 7 2 titration with standard acid. Reaction: of transforming the quantity of titrant used (in chemical units) to a chemically equivalent quantity of analyte 2 1 BO 2 1 2H 1 5H O S 4H BO (also in chemical units) through use of a stoichiometric 4 7 2 3 3 factor. Use chemical formulas (NO CALCULATIONS

(d) sulfur in an agricultural spray that was converted *13-13. Describe the preparation of

to thiocyanate with an unmeasured excess of (a) 1.00 L of 0.150 M KMnO4 from the solid reagent. cyanide. Reaction: (b) 2.50 L of 0.500 M HClO4, starting with a 9.00 M solution of the reagent. S(s) 1 CN2 S SCN2 (c) 400 mL of a solution that is 0.0500 M in I2, start-

ing with MgI2. After removal of the excess cyanide, thee thiocyanatethiocythiocyaocyacyanateyany (d) 200 mL of 1.00% (w/v) aqueous CuSO4 from a was titrated with a standard potassiumm iodateate solutionsolusolutioutionution 0.218 M CuSO4 solution. in strong HCl. Reaction: (e) 1.1.50 L of 0.215 M NaOH from the concentrated ccommercial reagent [50% NaOH (w/w), sp gr 2 1 2 1 1 1 2 S 2SCN 3IO3 2H 6Cl 1.525]. 1 22 1 2 1 2 1 (f) 1.50 L of a solution that is 12.0 ppm in K , start- 2SO4 2CN 3IClCll2 H2O ing with solid K4Fe(CN)6. 13-14.1 A solution of HClO was standardized by dissolving 13-6. How many millimoles of solutelutete are ccontainedd inn 4 3 23 0.4008 g of primary-standard-grade HgO in a solu- (a) 2.00 L of 2.766 101 M KMnOKM 4? Ю tionź of KBr: (b) 250.0 mL of 0.0423.0423042233 M KSCN?KSCN Ş 2 2 2 (c) 500.0 mL of a solutionionn contcocontainingontainintant 2.97 ppm CuSCSCuSO ? 4 HgO(s) 1 4Br 1 H O S HgBr 2 1 2OH (d) 2.50 L of 0.352.35 M KClKCKCl?Cl?l?? Ɖ 2 4 *13-7. How many millimolesmillimo ofo solutesolu are contained in 2 The liberated OH consumedonsumedum 443.75 mL of the acid. (a) 2.95 mL of 0.0789 MMK KHK PO ? 2 4 Calculate the molar concentration of the HClO (b) 0.201120 L of 0.0564644 M HHgCl ? ķ ar concentrationnccentrenentr o 4. 2 ɧ *13-15. A 0.4723-g sampleplee offfp primary-standard-gradeprimarprimappr y-stast dard-ard (c) 2.56 L of a 47.555p ppm solutisolutsolution of Mg(NO3)2? Ƹ Na2CO3 requiredquireded 34.7878 mLmL of an H2SOO4 solutionsolutioolutio to (d) 79.8 mLm oof 0.1379379 M NHԮ4VO3 (116.98 g/mol)? ͕ reach the end pointoint in thehe reactiontionionon 13-8. Whatatt mass of solutesolusol te in milligramsmilligram isi contained in

(a) 26.0 mL off 0.2500 0 M sucrosesucros (342 g/mol)? 22 1 S 1 1 ڢ2 CO 2HH H O COC (g) 3 4 3 2 2 (b)b)) 2.922..9292 L of 5.235.2 10Ҏ M H2O2? (c) 67367 mmLL of a solution that contains 5.76 ppm WhatWha is t ncentrationtiotion ofo the SO ? Pb(NOP ) (331.20( g/mol)? What is the molar concentration of the H2 4 3 2 13-16. A 0.5002-g sample thathatat assayedassassay 96.4% Na SO (d)) 6.75 mL of 0.0426 M KNO ? 2 4 3 required 48.63 mL of a barium chloride solution. *13-9.3-9.-9.9. WhatWhat massma of solute in grams is contained in equiredd 448.6 L off a bariumarariumum ź Reaction:ction:n:n (a)(a(a)) 450.0454 mL of 0.0986Ĺ M H O ? 2 2 (b)(b) 26.4 mL of 9.36 3 1024 M benzoic acid (122.1 Ba21 1 22 S g/mol)? ƈ B SO4 BaSO4(s) (c) 2.50 L ofɧ a solution that contains 23.4 ppm Calculatealculateulatlaatea theth analytianalyanalytical molar֌ concentration of BaCl SnCl ? 2 2 inn the solutionsolution.on.n. (d) 21.7 mL of 0.0214 M KBrO3? *13-17.*13-17. A 0.4126-g sasamplesampl of primary-standardprimarŻy-standard Na2CO3 was 13-10. Calculateࣰ the molar concentration of a solution that is Г atedted with 40.4 Ƽ50.0% NaOH (w/w) and has a specific gravity of 1.52. treated͇ with 40.00 mL Ƹof dilute perchloric acid. The ϪϠ΋γϣ solutionon was boiled to remove CO , following which *13-11. Calculate the molar concentration of a 20.0% solu- ҅ 2 ź 13-13 the excess HClO4 was back-titrated with 9.20 mL of tion (w/w) of KCl that has a specific gravity of 1.13. 13-12. Describe the preparation of εϳ΍έϳϭ dilute NaOH. In a separate experiment, it was established that 26.93 mL of the HClO4 neutralized the (a) 500 mL of 0.0750 M AgNO3 from the solid έΗ ̶ϣϳΩϗ reagent. NaOH in a 25.00-mL portion. Calculate the molari- (b) 2.00 L of 0.325 M HCl, starting with a 6.00 M ties ofĹ the HClO4 and NaOH. 13-18. solution of the reagent. Titration of 50.00 mL of 0.04715 M Na2C2O4 (c) 750 mL of a solution that is 0.0900 M in K1, required 39.25 mL of a potassium permanganate solution. starting with solid K4Fe(CN)6. (d) 600 mL of 2.00% (w/v) aqueous BaCl from a 2 2 1 1 1 S 2MnO4 5H2C2O4 6H 0.500 M BaCl2 solution. (e) 2.00 L of 0.120 M HClO from the commercial 21 1 1 4 2Mn 10CO2( g) 8H2O reagent [60% HClO4 (w/w), sp gr 1.60]. 1 (f ) 9.00 L of a solution that is 60.0 ppm in Na , Calculate the molar concentration of the KMnO4 starting with solid Na2SO4. solution.

*13-19. Titration of the I2 produced from 0.1142 g of C6H5COOH (122.12 g/mol). An end point was ob- primary-standard KIO3 required 27.95 mL of sodium served after addition of 43.25 mL of base. thiosulfate. (a) Calculate the molar concentration of the base. (b) Calculate the standard deviation of the molar 2 1 2 1 1 S 1 IO3 5I 6H 3I2 3H2O concentration if the standard deviation for the 2 2 2 mass measurement was 60.3 mg and that for the I1 2S O 2 S 2I 1 S O 2 2 2 3 4 6 volume measurement was 60.02 mL. (c) c) AssumingAssuminsumin an error of 20.3 mg in the mass mea- Calculate the concentration of the Na S O . 2 2 3 surement,sursuurement,ement ccalculate the absolute and relative sys- 13-20. A 4.912-g sample of a petroleum product was burnedurneded tematictemattematic errorerre in the molar concentration. in a tube furnace, and the SO produced was col-ol- 2 *13-25.* -2525.25 TheT ethylthylhy acetate concentration in an alcoholic solu- lected in 3% H O . Reaction: 2 2 tionti was determined by diluting a 10.00-mL sample to 100100.00 mL. A 20.00-mL portion of the diluted solu- SO (g) 1 H O S H SOO 2 2 2 2 4 tiotion was refluxed with 40.00 mL of 0.04672 M KOH:

A 25.00-mL portion of 0.008730873873 M NaOHOH wasw intro-o- Ю 2 S 1 źCH3COOC2H5 OH duced into the solution of H2SOSO4, followingfollowinollowi which Ş 2 1 the excess base was back-titrateded withwithth 15.1715 mL of CH3COO C2H5OH 0.01102 M HCl. Calculatelateate theth sulfurulfuulfurlfufuuru concentrationc Ɖ 2 in the sample in parts per million.millionm on.n. After cooling, the excess OH wawas back-titratedb

*13-21. A 100.0-mL sample of spring waterwatwaw was treated with 3.41 mL of 0.05042 M H2SOO4. CalculateCal the to convert any ironi presentsent too Fe21. Addition of ɧķ amount of ethyl acetate (88.1188.111 g/mol)g/mg/mmol)mom in the ooriginal 25.00-mL off 0.0025170.0 0.002 M K2Cr2O7 resultedre in the Ƹ sample in grams. reaction ͕Ԯ 13-26. A 0.1475-M solutionon of BBa(OH)a(OH)H)2 waswaw useduusedd to titrattitratetitratitrate the acetic acid (60.0505 g/mol)ol) in a diluted aqueaqueousaqueoqueueouuee solu-olu- 21 1 22 1 1 S tion. Thehe followingf resultsults werere obtained.oob ڢ6FeFe Cr2O7 14H 31 1 31 1 6Fe66FFe 2Cr2C Ҏ 7H2O SampleampleeS Sampleamp Volume, mL Ba(OH)H)2 Volume,VolumVVolume, mL 1 50.0050.0 43.173.3 Thehe excess K2Cr2O7 wwas back-titrated with 8.53 mL 21 2 49.509. 42.6842.642 of 0.00949.0094900009 M Fe solution. Calculate the concentra- 3 25.005.0 21.4721.4221.21 tionionon of ironiroon in theth sample in parts per million. ź Ĺ 4 50.00500. 43.334 13-22. Thee arsenicaarsennicic in a 1.203-g sample of a pesticide was convertedconconvev to H3AsO4 by suitable treatment. (a) Calculatee thethhhe meanmem w/vw percentage of acetic acid Thehe acidaca was then neutralized,ƈ and 40.00 mL of 13-18 ɧ in the sample.mpleple.lee 0.05871 M AgNO3 was added to precipitate the ar- ֌ 1 (b) Calculateate theheh standardstanandardandandard deviationded for the results. senic quantitatively as Ag3AsO4. The excess Ag in (c) Calculateulateate the 90%%co% confidencec interval for the the filtrate and in the washings from the precipitate Ż ࣰ mean. Г was titrated with 9.63 mL of 0.1000 M KSCN, and (d) At the 90%0%%͇ confidenceconfidenconfide level,Ƹ could any of the re- the reactionƼ was ź sults be discarded?҅cardedarded *13-27. (a) A 0.3147-g sample of primary-standard-grade Ag1 1 SCN2 S AgSCN(s) Na2C2O4 was dissolved in H2SO4 and titrated with 31.67 mL of dilute KMnO4: Find the percentage of As2O3 in the sample. *13-23. The thiourea in a 1.455-g sample of organic mate- 2 1 22 1 1 S 2MnOĹ4 5C2O4 16H rial was extracted into a dilute H SO solution and 2 4 1 21 2 1 1 titrated with 37.31 mL of 0.009372 M Hg via the 2Mn 10CO2( g) 8H2O reaction Calculate the molar concentration of the KMnO4 1 21 S 21 solution. 4(NH2)2CS Hg [(NH2)2CS]4Hg (b) The iron in a 0.6656-g ore sample was reduced quantitatively to the 12 state and then titrated with Find the percentage of (NH2)2CS (76.12 g/mol) in the sample. 26.75 mL of the KMnO4 solution from part (a). Calculate the percent Fe O in the sample. 13-24. A sosolutionlution ooff BBa(OH)a(OH))2 was standardized against 2 3 0.1215 g of primary-standard-grade benzoic acid, 13-28. (a) A 0.1527-g sample of primary-standard AgNO3 13-21 was dissolved in 502.3 g of distilled water.

Calculate the weight molar concentration of Ag1 13-30. A solution was prepared by dissolving 367 mg of in this solution. K3Fe(CN)6 (329.2 g/mol) in sufficient water to give (b) The standard solution described in part (a) was 750.0 mL. Calculate used to titrate a 25.171-g sample of a KSCN (a) the molar analytical concentration of K3Fe(CN)6. 1 solution. An end point was obtained after add- (b) the molar concentration of K . 32 ing 24.615 g of the AgNO3 solution. CCalculatealcu (c) the molar concentration of Fe(CN)6 . the weight molar concentration of the KKSCNKSCKS (d) the weight/volume percentage of K3Fe(CN)6. 1 solution. (e) the number of millimoles of K in 50.0 mL of (c) The solutions described in partsarts (a))a) andanndn (b)(b)) ththis solution. # 32 were used to determine the BaCll2 2HH2O in ((f) ) th the concentration of Fe(CN)6 in ppm. a 0.7120-g sample. A 20.102-g sampleple of thee 13-31.13 ChallengeC Problem: For each of the following acid/ 1 2 AgNO3 was added to a solution of thehee sample,samample,mmple, base titrations, calculate the H3O and OH concen- and the excess AgNO3 was back-titratedraratedteded withth trations at equivalence and at titrant volumes corre- 7.543 g of the KSCN solution.ution Calculate thehee pepper- sponding to 620.00 mL, 610.00 mL, and 61.00 mL # of equivalence. Construct a titration curve from the cent BaCl2 2HH2O in the sample.samp Ю *13-29. A solution was preparedpreparreparered by dissolvingdissoldis 7.48.48 g ofof data,ź plotting the p-function versus titrant volume. # # (a) 25.00 mL of 0.05000 M HCl with 0.02500 M KCl MgCl2 6H2O (277.85(27(277.77.85.85 g/mol)g/g in sufficientfficieffici Ş water to give 2.000000 L. CalculateCalculatlcuculaululalatelat NaOH. # Ɖ (b) 20.00 mL of 0.06000000 M HCl with 0.03000 M (a) the molarr analyticalanalyanalyticaal concentrationconceconcenncecence of KCl MgCl2 in this solution.olution.utio NaOH. 21 (b) the molar concentrationatioat of Mg . (c) 30.00 mL of 0.07500500M0 M H2SO4 with 0.1000 M 2 ķ (c) thehe molarm concentrationcentrntrationntratiattion ofo Cl . ɧ NaOH. # # (d) 40.00 mL off 0.0250025000 M NNaOHaOH withwith 0.050000.0500.0500 M (d)) the weighweiweight/volumeolumeume percenpercentagpercentageԮ of KCl MgClƸ2 HCl. 6HH2O. ͕ 2 (e)e) theth number ofo millimoles of Cl in 25.0 mL of (e) 35.00 mL of 0.200.2000.20000 M Na2COO3 withthh 0.20000.200.20 20000 M .HCl ڢ tthishis solution.sos on. (f ) theth concentratioconceconconcentrationonceoncentr ofҎ K1 in ppm. Ĺ ź ɧƈ ֌ ࣰ Ż źƼ ͇҅Г Ƹ Ĺ

FEATURE 17-2 Determination of Hydrogen Cyanide in Acrylonitrile Plant Streams

} { Acrylonitrile, CH2 CH C N, is an important chemical in the production of poly- acrylonitrile. This thermoplastic was drawn into fine threads and woven into synthetic fabrics such as Orlon, Acrilan, and Creslan. Althoughg acrylicacry fibers are no longer produced in the US, they are still made in many countries.untries.s. Hydrogend cyanidec is an im- purity in the plant streams that carry aqueous acrylonitrile.onitrilitrrir Thehe cyanidenid iss commonlycom

determined by titration with AgNO3. The titration reactionctio is 1 1 2 S 2 Ag 2CN Ag(CN)N)2 In order to determine the end point of the titration,on thethheh aqueousaqu sample is mixed with a basic solution of potassium iodide before the titration.titraratiratra oon. Before the equivalence 1 point, cyanide is in excess, and all thee AgAg is complexed.ed. AsAs soons as all the cyanide 1 has been reacted, the first excessxce of Ag causesca a permanentmanent turbiditytur to appear in the solution because of the formation of the AAgI precipitatete accordingac to 1 2 Ag 1 I S AgI(s)

8 10 .Curve C involves a unidentatentatetaate ligand,l A, that forms MA4 in four steps with suc- cessive formationtionion coconstants off 1008, 106, 104, and 102. These curves demonstrate that a much sharperrper end pointpo iss obtainedbtained with a reactionr that takes place in a single step. Foror this reason,n, multidentatemu tidentate ligandslliggandsnds are usuallyusu preferred for complexometric titrations. -titration with a unidentate ligandnd is the titra ڢThe mostt widelywidelwide usedd complexometriccompc plexometrilexometr tion of cyanideyanide wwithth silverssilve nitrate,ni a method introduced by Liebig in thee 1850s.1850s This Ҏ 2 methodod involvesnvolve theeefoef formationformforma of soluble Ag(CN)2 , as discussed in Featureature 17-2. OthOther commonmmon inorganicorg comcomplexing agents and their applications are listed in Tableablel 17-1. ź SpreadsheetSprSp SummaryĹ The complexometric titration of Cd(II) withthh CCl2 is considered in Chapter 9 of Applications of Microsoft ® Excel in Analyticalica Chemistry, 2ndɧƈ ed. A master equation approach is used. ֌ 17B-2 Precipitation Titrations Ż Precipitationࣰ titrations are based on reactions that yield ionic compoundsnds of lim-Г ited solubility.Ƽ Precipitation titrimetry is one of the oldest analytical techniques,es, ͇dat- Ƹ ing backź to the mid-1800s. The slow rate at which most precipitates form, however,҅ever, limits the number of precipitating agents that can be used in titrations to a hand- ful. We limit our discussion here to the most widely used and important precipitating reagent, silver nitrate, which is used for the determination of the halogens, the ϩΩϧϫΩ Ώϭγέέ̴ηϧ̯΍ϭ ϥϳέΗϣϬϣ TABLE 17-1 Ĺ Typical Inorganic Complex-Forming Titrations Titrant Analyte Remarks 2 2 2 2 Hg(NO3)2 Br , Cl , SCN , CN , thiourea Products are neutral Hg(II) complexes; various indicators used 2 2 2 AgNO3 CN Product is Ag(CN)2 ; indicator is I ; titrate to first turbidity of AgI 2 22 NiSO4 CN Product is Ni(CN)4 ; indicator is Agl; titrate to first turbidity of AgI 21 21 21 22 KCN Cu , Hg , Ni Products are Cu(CN)4 , Hg(CN)2, and 22 Ni(CN)4 ; various indicators used

halogenlike anions, mercaptans, fatty acids, and several divalent inorganic anions. Titrations with silver nitrate are sometimes called argentometric titrations. The Shapes of Titration Curves̵ϭϗ ίΎΑ/̵ϭϗ Ωϳγ΍ ϥϭϳγ΍έΗϳΗ ̶ϧΣϧϣ ϪΑΎηϣ ̶ϧΣϧϣ ϝ̰η Titration curves for precipitation reactions are calculated in a completely analogous way to the methodsds describedddescridesc in Section 14B for titrations involving strong acids and strong bases.es. The ononlyonll differencedif is thathat the solubility product of the precipitate is substitutedd for thee ion-production-ionion-pron productproddu t constantconstonstantnt for water.w Most indicators for argentometric titrations respondespondondd to changeschangechangesngege inin the concentrationsconcentrcco en of silver ions. Because of this response, titrationionn curvescurves forr precipitationprecipppr ationion reactionsreae usually consist of a plot of pAg

versus volumee of the silverlvverv reagenreareagentgen (usuallyy AgNO3). Example 17-1 illustrates how p-functions are obtainedbtabtainedbtaineainedd for thetht preepreequivalence-point region, the postequivalence- point regioregiregion, and thee equivalenceeqequiveququivalencequivalenqu Ю pointpo for aź typical precipitation titration. Ε΍έΗϳϧ ϩέϘϧ ΎΑ ΩϳέϠ̯Ş ϡϳΩγ ϥϭϳγ΍έΗϳΗ ̶ρ >Ag+@ϪΑγΎΣϣ EXAMPLEEXXAXXAMA 17-1 N1V1Ɖ=N2V2 .Ωϭη ϪΑγΎΣϣ ̵ίέ΍ ϡϫ ϪρϘϧ ϡΟΣ ̶ΗγϳΎΑ ΍ΩΗΑ΍ CalculateCaC the silver ion concentration in terms of pAg duringng the titration ofof 50.00 mL of 0.05000 Mķ NaCl with 0.1000 M AgNO3 after theth addition of thet ffollowingg volumes ofɧ reagent: (a) in the preequivalencence pointppoipo regiono at 10.00 mL, (b) at the equivalenceƸ point (25.00 mL), (c)c) afterter theth equivalenceui enc Ԯ 5 3 210 point͕nt aat 26.00 mL. For AgCl, Ksp 1.82 10 . ίέ΍ ϡϫ εϳ̡ ρΎϘϧ̵: ڢSolutiSolution - (a) Preequivalence-Point Data ̶ϓΎο΍ Cl ϭ AgCl Ώϭγέ Ҏ 1 At 10.00 mL, [Ag ] is very small and cannotnot be computed from stoichiometrichihiomhiome con-

siderations, but the molar concentration of chloride,h cNaCl can be obtainedned readily.re The equilibrium concentration of chloride is essentially eqequal to cNaClCl. 2 2 ź original no. mmol Cl no. mol AgNOgNOO addedad Ĺ 2 3 [Cl ] < c 5 NaCl total volumem of solution 3 2 3 ƈ 5 (50.00 0.05000 10.00 0.1000)0.0.10. 5 ɧ 1 0.025000 M 50.00 10.00 ֌ 210 Ksp 1.82 3 100 [Ag1] 5 5 5 7.28.28 3 1029 M ࣰ [Cl2] 0.025000 52 3 29 5 Ƽ pAg log (7.28 1010 ) 8.14 ź Additional points in the preequivalence-point region can be obtained in the same way. Results of calculations of this kind are shown in the second column of Table 17-2. TABLE 17-2 2 Changes in pAg in Titration of Cl with Standard AgNO3 pAg 50.00 mL of 0.0500 M NaCl 50.00 mL of 0.005 M NaCl

Volume of AgNO3 with 0.1000 M AgNO3 with 0.0100 M AgNO3 10.00 8.14 7.14 20.00 7.59 6.59 24.00 6.87 5.87 25.00 4.87 4.87 26.00 2.88 3.88 30.00 2.20 3.20 40.00 1.78 2.78

Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 17B Titrations with Inorganic Complexing Agents 409 :̵ίέ΍ ϡϫ ϪρϘϧ .Εγϳϧ ̶ϓΎο΍ ̭έΗηϣ ϥϭϳ ̵ϭΎΣ ϭ Εγ΍ ϩΩηέϳγ Ώϭγέ ί΍ ϝϭϠΣϣ (b) Equivalence Point pAg 1 5 2 1 2 5 5 3 210 5 1 2 At the equivalence point,, [Ag ] [Cl ], and [Ag ][Cl ] Ksp 1.82 10 [Ag ] 1 5 " 5 " 3 210 5 3 25 [Ag ] Ksp 1.82 10 1.35 10 25 pAg 5 2log (1.35 3 10 ) 5 4.87 (c) Postequivalence-Point Region :̵ίέ΍ ϡϫ α̡ ρΎϘϧ 1 + At 26.00 mL of AgNO3, Ag is in excess so Εγ΍ ̵ΩΎϳί Ag ϭ Ώϭγέ ̵ϭΎΣ ϝϭϠΣϣ (26.00 3 0.1000 2 50.00 3 0.05000)00..05 ) [Ag1] 5 c 5 5 1.3232 3 1023 M AgNO3 76.00 23 ΏϳέϘΗ ί΍ Ύϳ΁ Ύϣη έυϧ ϪΑ pAg 52log (1.32 3 10 ) 5 2.882 ˮΕγ΍ ϩΩη ϩΩΎϔΗγ΍ Additional results in the postequivalence-point regionregionon arear obtained in the same way and are shown in Table 17-2. Thehe titrationt curveve canca alsoals be derived from the charge-balance equation as shown for an acid/baseac titrationtratioonn ini FeatureF 14-1.

The Effect of Concentrationntratiotrat onnTitn TitrationTTi Curves The effect of reagentt and analytelyteytete concconcentration on titration curves can be seen in the data in Table 17-2 and the twotwt curves shown in Figure 17-2. With 0.1000 M

AgNO3 (Curveve A), the changeangengege inn pAg in the equivalence-point region is large, about 2 pAg units.ts. With theth 0.0100001000000 M reagent,reagreagenԮ the changeƸ is about 1 pAg unit, but still pronounced.ced. Ann indicatoindicator thathat producesproduce͕ a signal in the 4.0 to 6.0 pAg region shouldd solution.solso For the more dilute chlorideide solution ڢgive a minimalmalal errorere forr thet strongertronger (Curve B), thee changechachan e in pAg inn theh equivalence-pointq region wouldd be drawndraw out Ҏ , over a fairlyirly largelar e volumevolvolumvo of reagent ( 3 mL as shown by the dashedshed lines in the 10 Ĺ ź ƈ A 8 ɧ ֌ ࣰ B Г Ż Ƽ ϭ ϥ΍έΗϳΗ B έΩ ͇ Ƹ ź 6 έΑ΍έΑ 10 ϩΩϧϭηέΗϳΗ ҅ .Ωϧ΍ ϩΩη ϕϳϗέ Potential end-point Equivalence pAg region point 4 Ĺ έ̳ΎγΎϧη ΏΎΧΗϧ΍έΩ Ωϳϔϣ

2 Figure 17-2Titration curve for (A), 50.00 mL of 0.05000 M NaCl titrated with 0.1000 M AgNO3, and (B), 50.00 mL of 0.00500 M NaCl titrated with 0.01000 M AgNO3. Note the 0 increased sharpness of the break at the 0.00 10.00 20.00 30.00 40.00 end point with the more concentrated

Volume AgNO3, mL solution.

figure) so that to determine the end point accurately would be impossible.. The effect here is analogous to that illustrated for acid/base titrations in Figure 14-4. The Effect of Reaction Completeness on Titration Curves εϧ̯΍ϭ ϥΩϭΑ ϝϣΎ̯ έΛ΍ A useful relationship can be Figure 17-3 illustrates the effect of solubility product on the sharpness of the end derived by taking the negative ❯ logarithm of both sides of a point for titrationsns withwith 00.1 M silver nitrate. Note that the change in pAg at the solubility-product expression. equivalence pointointt becombecomesomesom grgreater as thehe ssolubility products become smaller, that Thus, for silver chloride, is, as the reactionction betweenbetwetwetwwew enn thet aanalytanalytealyte and sissilver nitrate becomes more complete. 1 2 By choosing ann indicatorndicatordicatori ato thatthatt cchchangesngesges colorco r in the pAg region of 4 to 6, titration 2log Ksp 5 2log ([Ag ][Cl ]) 5 2log [Ag1] 2 log [Cl2] of chloride ionsns shoshouldhouldld be possiblpospossibleos withth a mminimal titration error. Note that ions forming precipitatesipitatess withwiithi solubi solubsolsolubilityubi products much larger than about 10210 do 5 1 pKsp pAg pCl not yield satisfactorytoryororyry endendd points.poipopoinp This expression is similar to the :̶ϫϭέ̳ ϑϳϠ̰Η Titrationationtion CurvesC foror MixturesMMixMiixturesxtur of Anions acid-base expression for pKw Ю Ύϫ ϥϭϳϧ΁ ρϭϠΧϣ ϥϭϳγ΍έΑϳϟΎ̯ ̶ϧΣϧϣ 5 1 TheT methodsmetho developedelopedoped innE Example 17-1 for constructing precipitation titration pKw pH pOH ź cucurvecurvesurvess can be extendeddeded to mixtures thatŞ form precipitates of different solubilities. To iillusillustrate,illullllulus consider 50.00Ɖ mL of a solution that is 0.0500 M in iodide ion and 0.08000.08080000 M in chloride ion titrated with 0.1000 M silver nitrate. The curveur for the initial stagstagestagtagesag of this titration is identical to the curve shown for iodide in Figuregu 17-317 because silvesilsilvsilverve chloride, with its muchɧķ larger solubility product, doeses notot beginbbeegegin to precipitateprec untilun wwell into the titration.Ƹ It͕ isԮs interestingintei to determine how much iodide iss precipitatedcipitatededd beforebbeforere appreciableapprapppreciablciabble amountsounts of silver chloride form. With the appearancee of thee smallestllestlel amountuntnt ofo solidolid chchloride, the solubility-product expressionsxpre for both precipitatesrecipitatescip apply,ap andndd ڢsilver divisionҎvisi of one by the other provides the useful relationship 1 2 217 Ksp(AgI) [Ag ][I ] 8.3 3 10 5 5 5 4.56 3 101027 1 2 3 210 Ksp(AgCl) [Ag ][Cl ] 1.82 10 Ĺ [I–] 5 (4.56 3 101 –7)[Cl)[ –] ź From this relationship, we see that the iodide concentrationconcentraonncen decreasescrea to a tiny frac- ɧƈ tion of the chloride ion concentration beforee silvsilverilvilve chlorichloride begins to֌ precipitate. 16.0 Ż ࣰ K = 8.3 × 10 –17 sp Г I– ͇ Ƹ źƼ 14.0 ҅ 12.0 :3-17 ϝ̰η K = 5.2 × 10 –13 Br– sp ̵ίϳΗ ̵ϭέ Ksp έΛ΍ 10.0 ϥϭϳγ΍έΑϳϟΎ̯ ̶ϧΣϧϣ K = 1.8 × 10 –10 Cl– sp Ĺ 8.0 pAg ΏΎΧΗϧ΍ ϡϭίϟ K × –8 IO– sp = 3.0 10 6.0 3 ΏγΎϧϣ έ̳ΎγΎϧη ϪρϘϧ ϥϳϳόΗ ̵΍έΑ 4.0 ϥϻ΍ϭ ̶̯΍ FigureFigure 17-317-3 Effect of reaction com- K = 5.7 × 10 –5 – sp pleteness on precipitation titration BrO 3 curves. For each curve, 50.00 mL of 2.0 a 0.0500 M solution of the anion was titrated with 0.1000 M AgNO3. Note 0.0 that smaller values of Ksp give much 0.00 10.00 20.00 30.00 sharper breaks at the end point. Volume 0.1000 M AgNO3, mL

So, for all practical purposes, silver chloride forms only after 25.00 mL of titrant have been added in this titration. At this point, the chloride ion concentration is approximately 2 50.00 3 0.0800 c 2 < [Cl ] 5 5 0.0533 M Cl 50.00 1 25.00 Substituting into the previous equation yields

[I–] 5 4.56 3 10–7[Cl–] 5 4.56 3 10–77 3 0.05330.0530.05533 5 2.433 3 100–8 M The percentage of iodide unprecipitated at thisis pointointnt canca bebe calculatedcacalculatedated asa follows:followoll amount I2 unprecipitated 5(75.00 mL)(2.43 31028 mmolmm I2/mL/m ) 51.82822 31026mmol original amount I2 5 (50.00 mL)(0.0500000 mmol/mLmmool/mLl/mL) 5 2.502.5 mmol 1.821.8 3 1026 percentage I2 unprecipitatedipit 5 3 100%10100000% 5 7.3 3 1025% 2.50 Ю ź Thus, to within about 7.3 3 1010–55 percentpercpe of the equivalencequivqui point forŞ iodide, no silver chloride forms. Upp tto tthiss point,pointpoipooino the titration curve isƉ indistinguishable from that for iodide alone,e, as shownsh n in FigureFigFi 17-4. The data points for the first part of the titration curve, shownown by theesoes solidsossol line, were computed onķ this basis. As chloride ionio beginsns to precipitate,precprecireecip however, the rapidɧ decrease in pAg ends abruptly at a level thatth cann beb calculated ccalculat from the solubilityƸ product for silver chloride and the computedomput chloridechlorideride concentrationconcentratconcen͕Ԯ (0.0533 M): 210 Ksps (AgCl)(AgCl) 1.821.1 3 10 29 3 5 5ڢ 5 1 [Ag ] 3.41 10 M [Cl2] Ҏ 0.0533 pAg 5 2log(3.41 3 10–9) 5 8.47 Thee susuddensudd n eend to the sharp decrease in [Ag1] can be clearly seen in Figuregure 17-47-4 ź at pAgpAg 5 8.47.8.478.48 Further additionsĹ of silver nitrate decrease the chloride ionn con-conon centration,centratcentraration,on anda the curve then becomes that for the titration of chloride byy itself.tse 16.0 ƈ ɧ I – + Cl– ֌ A ࣰ14.0 Г Ż 12.0 ͇ Ƹ Ƽ – – Br + Cl ҅ ź B 10.0

8.0 pAg Ĺ 6.0 I – + Cl– and Br– + Cl– 4.0

2.0

0.0 Figure 17-4Titration curves for 0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 50.00 mL of a solution 0.0800 M in 2 2 2 Volume 0.1000 M AgNO3, mL Cl and 0.0500 M in I or Br .

For example, after 30.00 mL of titrant have been added,

2 50.00 3 0.0800 1 50.00 3 0.0500 2 30.00 3 0.100 c 2 5 [Cl ] 5 5 0.0438 M Cl 50.00 1 30.00 In this expression, the first two terms in the numerator give the number of millimoles of chlorideoriddee anand iodide, respectively, and the third term is the number of millimoles off titrant.trant.t.. Therefore,TThere 1.821.81 82 3 1010210 [AgA 1] 5 5 4.16 3 1029 M 0.0404438 pAg 5 8.38 The remainder of thehe dadatadataa popoints poip for this curve can be computed in the same way as for a curveurve of chloridee byy iitself.itselfit Curverve A in Figureuree 17-4,17-17-444, whichwhiwh Ю is the titration curve for the chloride/iodide mix- turetuturure justust considered,cocon d, is a compositeccom of the individualź curves for the two anionic species.ciesies.. TwoTw o equivalence poipoints are evident. ŞCurve B is the titration curve for a mixture offbrom brobromiderorom and chloride ions.Ɖ Note that the change associated withh thet first equiva- lenceleence pointp becomes less distinct as the solubilities of the two precipitatprecipitatesip approach onenne another. In the bromide/chlorideķ titration, the initial pAgAg valuesueues are lowerlow than theythethheyh area in the iodide/chlorideɧ titration because the solubilitybilitytyy off silversisil ilveil bromidebromidb ex- ceedseeds thatththa of silver iodide.Ƹ Beyond the first equivalencece point,oint, however,howeverhowevwever,ver wherewhw eree chlo-chlchhlo-o- ride ͕iononԮ is beingb titrated, the two titration curves aree identical.identicatical. TitrationTitratioTitra curves similar to those in Figure 17-4 canan be obtainedn experimentallyexperimerimim allyllyڢ by measuringmea the potential of a silver electrodeectrodectro immersed in the analytenaly solutionolu (see SectionҎ 21C). These curves can thenn be use to determined thehe concentrationcentrationentrntrationtr oof eacheac of the ions in mixtures of two halide ions.ionons. ΎΑ Ύϫ ϥϭϳϧ΁ ρϭϠΧϣ ϥϳϳόΗ έΩ ̶Ϡϣϋ ϩΩΎϔΗγ΍ End Points for Argentometric Titrationsns ̶ϧΣϧϣ έΩ ϥϻ΍ϭ ̶̯΍ ρΎϘϧ ϥΩέϭ΁ ΕγΩ ϪΑ ̵έΗϣϭϳγϧΎΗ̡ ϥϭϳγ΍έΗϳΗ εϭέ ΎΑź ĹChemical, potentiometric, and amperometric endndd pointsppoin arere usedd inin titrationsttit with silver nitrate. In this section, we describe onee offtf the chemicalall indicatorinind methods. ƈ In potentiometric titrations, the potential differencefferencefereerencerenc betweenbetw a silver electrode and ɧ a reference electrode is measured as a functiononn ooffft titrantt t vovolume. Titration֌ curves similar to those shown in Figures 17-2, 17-3,-3,3, and 17-417-7-47 4 are obtained.o Potentiomet- ric titrations are discussed in Section 21C.C. In amperometricamperomamperometmetme titrations,Ż the current ࣰ generated between a pair of silver electrodes iss measured͇Г andan plottedƸ as a function of Ƽ titrant volume. Amperometric methods are considerednsidered҅sidered ini Section 23B-4. ź Ϫϧϣ΍ΩέΩ ̶ΗγϳΎΑ έ̳ΎγΎϧη Chemical indicators producep a color change or occasionallycca the appearance or disap- ϪϳΣΎϧέΩ pAg ί΍ ̵ΩϭΩΣpearanceϣ of turbidity in the solution being titrated. The requirements for an indicator for .ΩϫΩ ̲ϧέ έϳϳϐΗ ̶ϧΣϧϣ ίϳa Ηprecipitation titration are that (1) the color change should occur over a limited range in p-function of the titrant or the analyte and (2) the color change should take place within the steep portion of the titration curve for the analyte.Ĺ For example, in Figure 17-3, we see that the titration of iodide with any indicator providing a signal in the pAg range of about 4.0 to 12.0 should give a satisfactory end point. Note that, in contrast, the end-point signal for the titration of chloride would be limited to a pAg of about 4.0 to 6.0. The Volhard Method.The Volhard method is one of the most common argentometric methods. In this method, silver ions are titrated with a standard solution of thiocyanate ion: ϥ΍έΗϳΗ Ag1 1 SCN2 8 AgSCN(s) Iron(III) serves as the indicator. TheT solution turns red with the first slight excess of thiocyanate ion due to the formation of Fe(SCN)21.

The most important application of the Volhard method is the indirect determination of halide ions. A measured excess of standard silver nitrate solution is added to the sample, and the excess silver is determined by back-titration with a standard thiocyanate solution. The strongly acidic environment of the Volhard titration is a distinct advantage over other titrations of halide ions because such ions as carbon- ate, oxalate, and arsenate do not interfere. The silververr saltssalts of these ions are soluble in acidic media but only slightly soluble in neutralral media.a Silver chloride is more soluble than silverr thiocyanate.yanatanateanaat Ass a result,reesult,ult, ini chloridechlochlori determinations using the Volhard method, the reactioneactiontion AgCl(s) 1 SCN2 8 AgSCN(N(s) 1 Cl– occurs to a significant extent near the end of the back-titration.baback-ack-titratiitrationitratitratio This reaction causes the end point to fade and results in overconsumptionvercoverc n ofo thiocyanatethiocyanatthiocth ion. The result- ing low results for chloridedee cacan be overcomovercomeverco by filterinfilteringlteriningngg the Юssilver chloride before undertaking the back-titration.ation.tionn. FiltrationF ration is not requiredquire forfof other halidesź because they form silver salts that are lessss solublessoluolubleuble thant silver thiocyanate.hiocyhioc Ş + Εγ΍ Ag ϥ΍έΗϳΗ Other Argentometricicc MethMMethods.ods.ss. InI the MohrMohr methodmethod,Ɖ sodium chromate serves as the indicator for thehe argentometricargentomarg tometritomomet titration of chloride, bromide, and cyanide ions. Silver ions react with chromatematmatt to form the brick-red silverķ chromate (Ag2CrO4) precipitate in the eequivalence-pointnce-pe-poinointoio reregion. The Mohr methodɧ is now rarely used because Cr(VI)VI) is a carcicarcarcinogen.gen.n. Ƹ Ԯ Adsorptionsorptorp n indicatorsinindicandicatorsatorstors werewewere firstfirsfi The Fajans methodmeth useusessess aan adsorptioadsorption͕orp ion indicator, an organic compound thatat ❮ describedcri by K. FaFajans,ans,ns, a PolisPPoPolisholi surfacesurfac of the solid in a precipitation titration. Ide- chemisthemistmi in 1926.926 TitrationsTi ionson ڢadsorbs ontotoo or desorbsd fromfr m the ally, the adsorptiondsorppptiontion or dedesorpdesorptionon occoccurs near the equivalence point andnd resuresults not involvinglvingvingg adsorptionadsorptioadso indindicators only in a color chchangehangeanangenge butb also inҎ the transfer of color from the solutionutionn to the ssolid are rapid,id,d, accaaccurate,urate,urau andan reliable, or vicece versa.sa. Εγ΍ Ag+ ϥ΍έΗϳΗ butt theireirir applicatioapplappapplicationppl is limited to thehe feweww precipitationprepprecip titrations thathatatt formfof mmcol colloidalcololl precipitatesź Ĺ rapidlyrapidly.apidlydly.dlyy SpreadsheetSp Summary In Chapter 9 of Applications of Microsoftoftft® Excel in Analytical Chemistry, 2nd ed., we plot a curve for the titrationn of NaCl with AgNOɧƈ3. A stoichiometric approach is first used and then a masterstetetere equation approach is explored. Finally, the problem is inverted, and the volume neededdedde ֌ to achieveƼ aࣰ given pAg value is computed. ͇Г ƸŻ 17Cź ORGANIC COMPLEXING AGENTS ҅ Several different organic complexing agents have become important in analytical chemistry because of their inherent sensitivity and potential selectivity in reacting with metal ions. Organic reagents are particularly useful in precipitating metals, in binding metals so as to prevent interferences, in extracting metals from one solventĹ to another, and in forming complexes that absorb light for spectrophotometric determinations. The most useful organic reagents form chelate complexes with metal ions. Many organic reagents are useful in converting metal ions into forms that can be readily extracted from water into an immiscible organic phase. Extractions are widely used to separate metals of interest from potential interfering ions and for achieving a concentrat- ing effect by transfer of the metal into a phase of smaller volume. Extractions are appli- cable to much smaller amounts of metals than precipitations, and they avoid problems associated with coprecipitation. Separations by extraction are considered in Section 31C. Several of the most widely used organic complexing agents for extractions are listed in Table 17-3. Some of these same reagents normally form insoluble species