Principles-of-Instrumental-Analysis-Solutions

Skoog/Holler/CrouchChapter1PrinciplesofInstrumentalAnalysis,6thed.Instructor’sManual1CHAPTER11-1.Atransducerisadevicethatconvertschemicalorphysicalinformationintoanelectricalsignalorthereverse.Themostcommoninputtransducersconvertchemicalorphysicalinformationintoacurrent,voltage,orcharge,andthemostcommonoutputtransducersconvertelectricalsignalsintosomenumericalform.1-2.Theinformationprocessorinavisualcolormeasuringsystemisthehumanbrain.1-3.Thedetectorinaspectrographisaphotographicfilmorplate.1-4.Smokedetectorsareoftwotypes:photodetectorsandionizationdetectors.Thephotodetectorsconsistofalightsource,suchasalight-emittingdiode(LED)andaphotodiodetoproduceacurrentproportionaltotheintensityoflightfromtheLED.WhensmokeentersthespacebetweentheLEDandthephotodiode,thephotocurrentdecreases,whichsetsoffanalarm.Inthiscasethephotodiodeisthetransducer.Inionizationdetectors,whicharethetypicalbattery-powereddetectorsfoundinhomes,asmallradioactivesource(usuallyAmericium)ionizestheairbetweenapairofelectrodes.Whensmokeentersthespacebetweentheelectrodes,theconductivityoftheionizedairchanges,whichcausesthealarmtosound.Thetransducerinthistypeofsmokedetectoristhepairofelectrodesandtheairbetweenthem.1-5.Adatadomainisoneofthemodesinwhichdatamaybeencoded.Examplesofdatadomainclassesaretheanalog,digitalandtimedomains.Examplesofdatadomainsarevoltage,current,charge,frequency,period,number.PrinciplesofInstrumentalAnalysis,6thed.Chapter121-6.Analogsignalsincludevoltage,current,charge,andpower.Theinformationisencodedintheamplitudeofthesignal.1-7.OutputTransducerUseLCDdisplayAlphanumericinformationComputermonitorAlphanumericinformation,text,graphicsLaserprinterAlphanumericandgraphicalinformationMotorRotatestochangepositionofattachedelements1-8.Afigureofmeritisanumberthatprovidesquantitativeinformationaboutsomeperformancecriterionforaninstrumentormethod.1-9.Letcs=molarconcentrationofCu2+instandard=0.0287Mcx=unknownCu2+concentrationVs=volumeofstandard=0.500mLVx=volumeofunknown=25.0mLS1=signalforunknown=23.6S2=signalforunknownplusstandard=37.9Assumingthesignalisproportionaltocxandcs,wecanwriteS1=KcxorK=S1/cxAfteraddingthestandard2xxssxsVcVcSKVV⎛⎞+=⎜⎟+⎝⎠SubstitutingforKandrearranginggives,121()ssxxsxSVccSVVSV=+−PrinciplesofInstrumentalAnalysis,6thed.Chapter13cx423.60.500mL0.0287M=9.0010M37.9(0.500mL+25.0mL)(23.625.0mL)−××=×−×1-10.Theresultsareshowninthespreadsheetbelow.(a)Slope,m=0.0701,intercept,b=0.0083(b)FromLINESTresults,SDslope,sm=0.0007,SDintercept,sb=0.0040(c)95%CIforslopemism±tsmwheretistheStudenttvaluefor95%probabilityandN–2=4degreesoffreedom=2.7895%CIform=0.0701±2.78×0.0007=0.0701±0.0019or0.070±0.002Forintercept,95%CI=b±tsb=0.0083±2.78×0.004=0.0083±0.011or0.08±0.01(d)cu=4.87±0.086mMor4.87±0.09mMPrinciplesofInstrumentalAnalysis,6thed.Chapter141-11.Thespreadsheetbelowgivestheresults(a)Seeplotinspreadsheet.(b)cu=0.410μg/mL(c)S=3.16Vs+3.25(d)13.2462.000μg/mL0.410μg/mL3.164mL5.00mLsuubccmV−×===×(e)Fromthespreadsheetsc=0.002496or0.002μg/mLPrinciplesofInstrumentalAnalysis,6thed.Chapter222-3.V2.4=12.0×[(2.5+4.0)×103]/[(1.0+2.5+4.0)×103]=10.4VWithmeterinparallelacrosscontacts2and4,2,4+6.5k111=+=(2.5+4.0)k6.5kMMMRRRRΩΩ×ΩR2,4=(RM×6.5kΩ)/(RM+6.5kΩ)(a)R2,4=(5.0kΩ×6.5kΩ)/(5.0kΩ+6.5kΩ)=2.83kΩVM=(12.0V×2.83kΩ)/(1.00kΩ+2.83kΩ)=8.87Vrelerror=8.87V10.4V100%=15%10.4V−×−Proceedinginthesameway,weobtain(b)–1.7%and(c)–0.17%2-4.ApplyingEquation2-19,wecanwrite(a)7501.0%=100%(750)MRΩ−×−ΩRM=(750×100–750)Ω=74250Ωor74kΩ(b)7500.1%=100%(750)MRΩ−×−ΩRM=740kΩ2-5.ResistorsR2andR3areinparallel,theparallelcombinationRpisgivenbyEquation2-17Rp=(500×200)/(500+200)=143Ω(a)This143ΩRpisinserieswithR1andR4.Thus,thevoltageacrossR1isV1=(15.0×100)/(100+143+1000)=1.21VV2=V3=15.0V×143/1243=1.73VV4=15.0V×1000/1243=12.1VPrinciplesofInstrumentalAnalysis,6thed.Chapter23(b)I1=I5=15.0/(100+143+1000)=1.21×10–2AI2=1.73V/500Ω=3.5×10–3AI3=I4=1.73V/200Ω=8.6×10–3A(c)P=IV=1.73V×8.6×10–3A=1.5×10–2W(d)Sincepoint3isatthesamepotentialaspoint2,thevoltagebetweenpoints3and4V′isthesumofthedropsacrossthe143Wandthe1000Wresistors.Or,V′=1.73V+12.1V=13.8V.ItisalsothesourcevoltageminustheV1V′=15.0–1.21=13.8V2-6.Theresistancebetweenpoints1and2istheparallelcombinationorRBandRCR1,2=2.0kΩ×4.0kΩ/(2.0kΩ+4.0kΩ)=1.33kΩSimilarlytheresistancebetweenpoints2and3isR2,3=2.0kΩ×1.0kΩ/(2.0kΩ+1.0kΩ)=0.667kΩThesetworesistorsareinserieswithRAforatotalseriesresistanceRTofRT=1.33kΩ+0.667kΩ+1.0kΩ=3.0kΩI=24/(3000Ω)=8.0×10–3A(a)P1,2=I2R1,2=(8.0×10–3)2×1.33×103=0.085W(b)AsaboveI=8.0×10–3A(c)VA=IRA=8.0×10–3A×1.0×103Ω=8.0V(d)VD=14×R2,3/RT=24×0.667/3.0=5.3V(e)V5,2=24–VA=24-8.0=16V2-7.Withthestandardcellinthecircuit,Vstd=Vb×AC/ABwhereVbisthebatteryvoltagePrinciplesofInstrumentalAnalysis,6thed.Chapter241.018=Vb×84.3/ABWiththeunknownvoltageVxinthecircuit,Vx=Vb×44.2/ABDividingthethirdequationbythesecondgives,1.018V84.3cm=44.3cmxVVx=1.018×44.3cm/84.3cm=0.535V2-8.=100%SrMSRERR−×+ForRS=20ΩandRM=10Ω,20=100%=67%1020rE−×−+Similarly,forRM=50Ω,20=100%=29%5020rE−×−+Theothervaluesareshowninasimilarmanner.2-9.Equation2-20isstdstd=100%rLRERR−×+ForRstd=1Ωand