D. Harvey - Modern Analytical Chemistry (794078), страница 23
Текст из файла (страница 23)
The absolute uncertainty in the charge issR = R × 0.0672 = (18) × (±0.0672) = ±1.2Thus, we report the total charge as 18 C ± 1 C.4C.4 Uncertainty for Mixed OperationsMany chemical calculations involve a combination of adding and subtracting, andmultiply and dividing. As shown in the following example, the propagation of uncertainty is easily calculated by treating each operation separately using equations4.6 and 4.7 as needed.1400-CH04 9/8/99 3:54 PM Page 67Chapter 4 Evaluating Analytical DataEXAMPLE 4.7For a concentration technique the relationship between the measured signaland an analyte’s concentration is given by equation 4.5Smeas = kCA + SreagCalculate the absolute and relative uncertainties for the analyte’s concentrationif Smeas is 24.37 ± 0.02, Sreag is 0.96 ± 0.02, and k is 0.186 ± 0.003 ppm–1.SOLUTIONRearranging equation 4.5 and solving for CACA =Smeas − Sreag24.37 − 0.96== 125.9 ppmk0.186 ppm-1gives the analyte’s concentration as 126 ppm.
To estimate the uncertainty inCA, we first determine the uncertainty for the numerator, Smeas – Sreag, usingequation 4.6sR = (0.02)2 + (0.02)2 = 0.028The numerator, therefore, is 23.41 ± 0.028 (note that we retain an extrasignificant figure since we will use this uncertainty in further calculations). Tocomplete the calculation, we estimate the relative uncertainty in CA usingequation 4.7, givingsR=R2 0.028 0.003 + 23.41 0.186 2= 0.0162or a percent relative uncertainty of 1.6%.
The absolute uncertainty in theanalyte’s concentration issR = (125.9 ppm) × (0.0162) = ±2.0 ppmgiving the analyte’s concentration as 126 ± 2 ppm.4C.5 Uncertainty for Other Mathematical FunctionsMany other mathematical operations are commonly used in analytical chemistry,including powers, roots, and logarithms. Equations for the propagation of uncertainty for some of these functions are shown in Table 4.9.EXAMPLE 4.8The pH of a solution is defined aspH = –log[H+]where [H+] is the molar concentration of H+.
If the pH of a solution is 3.72with an absolute uncertainty of ±0.03, what is the [H + ] and its absoluteuncertainty?671400-CH04 9/8/99 3:54 PM Page 6868Modern Analytical ChemistrySOLUTIONThe molar concentration of H+ for this pH is[H+] = 10–pH = 10–3.72 = 1.91 × 10–4 Mor 1.9 × 10 –4 M to two significant figures. From Table 4.9 the relativeuncertainty in [H+] issR= 2.303 × s A = 2.303 × 0.03 = 0.069Rand the absolute uncertainty is(1.91 × 10–4 M) × (0.069) = 1.3 × 10–5 MWe report the [H+] and its absolute uncertainty as 1.9 (±0.1) × 10–4 M.Table 4.9Propagation of Uncertaintyfor Selected FunctionsaFunctionsRR = kAs R = ks AR= A+BsR =s 2A + s 2BR= A–BsR =s 2A + s 2BR= A×BsR=R sA sB + A BABsR=R sA sB + A BR=2222sAAR = ln( A)sR =R = log( A)s R = 0.4343 ×R = eAsR= sARsR= 2.303s AR s sR= k A R AR = 10 AR = AksAAaTheseequations assume that the measurements A and B areuncorrelated; that is, sA is independent of sB.4C.6 Is Calculating Uncertainty Actually Useful?Given the complexity of determining a result’s uncertainty when several measurements are involved, it is worth examining some of the reasons why such calculations are useful.
A propagation of uncertainty allows us to estimate an ex-1400-CH04 9/8/99 3:54 PM Page 69Chapter 4 Evaluating Analytical Datapected uncertainty for an analysis. Comparing the expected uncertainty to thatwhich is actually obtained can provide useful information. For example, in determining the mass of a penny, we estimated the uncertainty in measuring massas ±0.002 g based on the balance’s tolerance. If we measure a single penny’s massseveral times and obtain a standard deviation of ±0.020 g, we would have reasonto believe that our measurement process is out of control. We would then try toidentify and correct the problem.A propagation of uncertainty also helps in deciding how to improve the uncertainty in an analysis.
In Example 4.7, for instance, we calculated the concentration of an analyte, obtaining a value of 126 ppm with an absolute uncertaintyof ±2 ppm and a relative uncertainty of 1.6%. How might we improve the analysis so that the absolute uncertainty is only ±1 ppm (a relative uncertainty of0.8%)? Looking back on the calculation, we find that the relative uncertainty isdetermined by the relative uncertainty in the measured signal (corrected for thereagent blank)0.028= ±0.0012, or ± 0.12%23.41and the relative uncertainty in the method’s sensitivity, k,0.003= ±0.016, or ± 1.6%0.186Of these two terms, the sensitivity’s uncertainty dominates the total uncertainty.Measuring the signal more carefully will not improve the overall uncertaintyof the analysis. On the other hand, the desired improvement in uncertaintycan be achieved if the sensitivity’s absolute uncertainty can be decreased to±0.0015 ppm–1.As a final example, a propagation of uncertainty can be used to decide whichof several procedures provides the smallest overall uncertainty.
Preparing a solution by diluting a stock solution can be done using several different combinations of volumetric glassware. For instance, we can dilute a solution by a factorof 10 using a 10-mL pipet and a 100-mL volumetric flask, or by using a 25-mLpipet and a 250-mL volumetric flask. The same dilution also can be accomplished in two steps using a 50-mL pipet and a 100-mL volumetric flask for thefirst dilution, and a 10-mL pipet and a 50-mL volumetric flask for the second dilution. The overall uncertainty, of course, depends on the uncertainty of theglassware used in the dilutions. As shown in the following example, we can usethe tolerance values for volumetric glassware to determine the optimum dilutionstrategy.5EXAMPLE 4.9Which of the following methods for preparing a 0.0010 M solution from a1.0 M stock solution provides the smallest overall uncertainty?(a) A one-step dilution using a 1-mL pipet and a 1000-mL volumetricflask.(b) A two-step dilution using a 20-mL pipet and a 1000-mL volumetric flaskfor the first dilution and a 25-mL pipet and a 500-mL volumetric flask forthe second dilution.691400-CH04 9/8/99 3:54 PM Page 7070Modern Analytical ChemistrySOLUTIONLetting Ma and Mb represent the molarity of the final solutions from method(a) and method (b), we can write the following equationsM a = 0.0010 M =M b = 0.0010 M =(1.0 M)(1.000 mL)1000.0 mL(1.0 M)(20.00 mL)(25.00 mL)(1000.0 mL)(500.0 mL)Using the tolerance values for pipets and volumetric flasks given in Table 4.2,the overall uncertainties in Ma and Mb are sR = R Ma sR = R Mb22 0.006 0.3 + 1.000 1000.0 222= 0.006 0.03 0.2 0.3 0.03 + + + 25.00 500.0 1000.0 20.00 2= 0.002Since the relative uncertainty for Mb is less than that for Ma, we find that thetwo-step dilution provides the smaller overall uncertainty.4D The Distribution of Measurements and ResultsAn analysis, particularly a quantitative analysis, is usually performed on severalreplicate samples.
How do we report the result for such an experiment when resultsfor the replicates are scattered around a central value? To complicate matters further, the analysis of each replicate usually requires multiple measurements that,themselves, are scattered around a central value.Consider, for example, the data in Table 4.1 for the mass of a penny. Reportingonly the mean is insufficient because it fails to indicate the uncertainty in measuringa penny’s mass. Including the standard deviation, or other measure of spread, provides the necessary information about the uncertainty in measuring mass.
Nevertheless, the central tendency and spread together do not provide a definitive statement about a penny’s true mass. If you are not convinced that this is true, askyourself how obtaining the mass of an additional penny will change the mean andstandard deviation.How we report the result of an experiment is further complicated by the needto compare the results of different experiments. For example, Table 4.10 shows results for a second, independent experiment to determine the mass of a U.S.
pennyin circulation. Although the results shown in Tables 4.1 and 4.10 are similar, theyare not identical; thus, we are justified in asking whether the results are in agreement. Unfortunately, a definitive comparison between these two sets of data is notpossible based solely on their respective means and standard deviations.Developing a meaningful method for reporting an experiment’s result requiresthe ability to predict the true central value and true spread of the population underinvestigation from a limited sampling of that population.
In this section we will takea quantitative look at how individual measurements and results are distributedaround a central value.1400-CH04 9/8/99 3:54 PM Page 71Chapter 4 Evaluating Analytical DataTable 4.10Results for a SecondDetermination of the Mass of aUnited States Penny in CirculationPennyMass(g)123453.0523.1413.0833.0833.048–Xs3.0810.0374D.1 Populations and SamplesIn the previous section we introduced the terms “population” and “sample” in thecontext of reporting the result of an experiment.
Before continuing, we need to understand the difference between a population and a sample. A population is the setof all objects in the system being investigated. These objects, which also are members of the population, possess qualitative or quantitative characteristics, or values,that can be measured.
If we analyze every member of a population, we can determine the population’s true central value, µ, and spread, σ.The probability of occurrence for a particular value, P(V), is given asP(V ) =populationAll members of a system.MNwhere V is the value of interest, M is the value’s frequency of occurrence in the population, and N is the size of the population. In determining the mass of a circulatingUnited States penny, for instance, the members of the population are all UnitedStates pennies currently in circulation, while the values are the possible masses thata penny may have.In most circumstances, populations are so large that it is not feasible to analyzeevery member of the population.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
