D. Harvey - Modern Analytical Chemistry (794078), страница 36
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Notethat all volumes must be in the same units, thus Vs is converted from 1.00 µL to1.00 × 10–3 mL.0.1930.419= 1.00 mL 1.00 × 10 –3 mL 1.00 mL CA CA 1560ppb+ 5.00 mL 5.00 mL 5.00 mL 0.1930.419=0.200C A0.200C A + 0.312 ppb0.0386C A + 0.0602 ppb = 0.0838C A0.0452C A = 0.0602 ppbC A = 1.33 ppbThus, the concentration of Pb2+ in the original sample of blood is 1.33 ppb.It also is possible to make a standard addition directly to the sample after measuring Ssamp (Figure 5.6).
In this case, the final volume after the standard addition isVo + Vs and equations 5.5–5.7 becomeSsamp = kCAVoVs Sspike = k C A+ CSVo + VsVo + Vs Add VS of CSVoVoTotal concentrationof analyteTotal concentrationof analyteFigure 5.6Illustration showing an alternative form ofthe method of standard additions. In thiscase a sample containing the analyte isspiked with a known volume of a standardsolution of analyte without further dilutingeither the sample or the spiked sample.CACAVoVS+ CSVo + VSVo + VS5.81400-CH05 9/8/99 3:59 PM Page 113Chapter 5 Calibrations, Standardizations, and Blank CorrectionsSsampSspike=CAC A[Vo /(Vo + Vs )] + C S[Vs /(Vo + Vs )]1135.9EXAMPLE 5.5A fourth spectrophotometric method for the quantitative determination of theconcentration of Pb2+ in blood yields an Ssamp of 0.712 for a 5.00-mL sample ofblood.
After spiking the blood sample with 5.00 µL of a 1560-ppb Pb 2+standard, an Sspike of 1.546 is measured. Determine the concentration of Pb2+ inthe original sample of blood.SOLUTIONThe concentration of Pb2+ in the original sample of blood can be determinedby making appropriate substitutions into equation 5.9 and solving for CA.0.712CA1.546=5.00 × 10 –3 mL+ 1560 ppb–3–3 (5.00 mL + 5.00 × 10 mL) (5.00 mL + 5.00 × 10 mL) CA 5.00 mL0.712CA1.546=0.9990C A + 1.558 ppb0.7113C A + 1.109 ppb = 1.546C AC A = 1.33 ppbThus, the concentration of Pb2+ in the original sample of blood is 1.33 ppb.The single-point standard additions outlined in Examples 5.4 and 5.5 are easilyadapted to a multiple-point standard addition by preparing a series of spiked samples containing increasing amounts of the standard. A calibration curve is preparedby plotting Sspike versus an appropriate measure of the amount of added standard.Figure 5.7 shows two examples of a standard addition calibration curve based onequation 5.6.
In Figure 5.7(a) Sspike is plotted versus the volume of the standard solution spikes, Vs. When k is constant, the calibration curve is linear, and it is easy toshow that the x-intercept’s absolute value is CAVo/CS.EXAMPLE 5.6Starting with equation 5.6, show that the equations for the slope, y-intercept,and x-intercept in Figure 5.7(a) are correct.SOLUTIONWe begin by rewriting equation 5.6 asSspike =kCAVokCS+× VsVfVfwhich is in the form of the linear equationY = y-intercept + slope × XColorplate 2 shows an example of a set ofstandard additions and their correspondingstandard additions calibration curve.1400-CH05 9/8/99 3:59 PM Page 114Modern Analytical ChemistrySspiked114Slope =x-intercept =–CAVoCSy-intercept =kCSVfkCAVoVfVSSspiked(a)Slope = kFigure 5.7Examples of calibration curves for themethod of standard additions.
In (a) thesignal is plotted versus the volume of theadded standard, and in (b) the signal isplotted versus the concentration of theadded standard after dilution.x-intercept =–CAVoVfy-intercept =CS(b)kCAVoVf(VV )Sfwhere Y is Sspike and X is Vs. The slope of the line, therefore, is kCS/Vf, and they-intercept is kCAVo/Vf. The x-intercept is the value of X when Y is 0, or0 =kCAVokCS+× (x -intercept)VfVfx -intercept = –(kCAVo /Vf )C V= – A o(kCS /Vf )CSThus, the absolute value of the x-intercept is CAVo/CS.Since both Vo and CS are known, the x-intercept can be used to calculate the analyte’s concentration.EXAMPLE5.7A fifth spectrophotometric method for the quantitative determination of theconcentration of Pb2+ in blood uses a multiple-point standard addition basedon equation 5.6.
The original blood sample has a volume of 1.00 mL, and thestandard used for spiking the sample has a concentration of 1560 ppb Pb2+. Allsamples were diluted to 5.00 mL before measuring the signal. A calibrationcurve of Sspike versus Vs is described by1400-CH05 9/8/99 3:59 PM Page 115Chapter 5 Calibrations, Standardizations, and Blank CorrectionsSspike = 0.266 + 312 mL–1 × VsDetermine the concentration of Pb2+ in the original sample of blood.SOLUTIONTo find the x-intercept we let Sspike equal 00 = 0.266 + 312 mL–1 × (x-intercept)and solve for the x-intercept’s absolute value, giving a value of 8.526 × 10–4 mL.Thusx -intercept = 8.526 × 10 –4 mL =C AVoC × (1.00 mL)= ACS1560 ppband the concentration of Pb2+ in the blood sample, CA, is 1.33 ppb.Figure 5.7(b) shows the relevant relationships when Sspike is plotted versus the concentrations of the spiked standards after dilution.
Standard addition calibrationcurves based on equation 5.8 are also possible.Since a standard additions calibration curve is constructed in the sample, itcannot be extended to the analysis of another sample. Each sample, therefore, requires its own standard additions calibration curve. This is a serious drawback tothe routine application of the method of standard additions, particularly in laboratories that must handle many samples or that require a quick turnaround time. Forexample, suppose you need to analyze ten samples using a three-point calibrationcurve. For a normal calibration curve using external standards, only 13 solutionsneed to be analyzed (3 standards and 10 samples).
Using the method of standardadditions, however, requires the analysis of 30 solutions, since each of the 10 samples must be analyzed three times (once before spiking and two times after addingsuccessive spikes).The method of standard additions can be used to check the validity of an external standardization when matrix matching is not feasible. To do this, a normal calibration curve of Sstand versus CS is constructed, and the value of k is determinedfrom its slope. A standard additions calibration curve is then constructed usingequation 5.6, plotting the data as shown in Figure 5.7(b).
The slope of this standardadditions calibration curve gives an independent determination of k. If the two values of k are identical, then any difference between the sample’s matrix and that ofthe external standards can be ignored. When the values of k are different, a proportional determinate error is introduced if the normal calibration curve is used.5B.5 Internal StandardsThe successful application of an external standardization or the method of standardadditions, depends on the analyst’s ability to handle samples and standards reproducibly.
When a procedure cannot be controlled to the extent that all samples andstandards are treated equally, the accuracy and precision of the standardization maysuffer. For example, if an analyte is present in a volatile solvent, its concentrationwill increase if some solvent is lost to evaporation. Suppose that you have a sampleand a standard with identical concentrations of analyte and identical signals. If bothexperience the same loss of solvent their concentrations of analyte and signals willcontinue to be identical.
In effect, we can ignore changes in concentration due toevaporation provided that the samples and standards experience an equivalent lossof solvent. If an identical standard and sample experience different losses of solvent,1151400-CH05 9/8/99 3:59 PM Page 116116Modern Analytical Chemistryinternal standardA standard, whose identity is differentfrom the analyte’s, that is added to allsamples and standards containing theanalyte.however, their concentrations and signals will no longer be equal. In this case, anexternal standardization or standard addition results in a determinate error.A standardization is still possible if the analyte’s signal is referenced to a signalgenerated by another species that has been added at a fixed concentration to allsamples and standards.
The added species, which must be different from the analyte, is called an internal standard.Since the analyte and internal standard in any sample or standard receive thesame treatment, the ratio of their signals will be unaffected by any lack of reproducibility in the procedure. If a solution contains an analyte of concentration CA,and an internal standard of concentration, CIS, then the signals due to the analyte,SA, and the internal standard, SIS, areSA = kACASIS = kISCISwhere kA and kIS are the sensitivities for the analyte and internal standard, respectively.
Taking the ratio of the two signals givesSAkCC= A × A = K× ASISkIS CISCIS5.10Because equation 5.10 is defined in terms of a ratio, K, of the analyte’s sensitivityand the internal standard’s sensitivity, it is not necessary to independently determine values for either kA or kIS.In a single-point internal standardization, a single standard is prepared, and Kis determined by solving equation 5.10 C S K = IS A C A SIS stand5.11Once the method is standardized, the analyte’s concentration is given by C S C A = IS A K SIS sampEXAMPLE 5.8A sixth spectrophotometric method for the quantitative determination of Pb2+levels in blood uses Cu2+ as an internal standard.
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