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Chapter 4: Q31P (page 93)

Suppose that you carry out an analytical procedure to generate a linear calibration curve like that shown in Figure 4-13. Then you analyse an unknown and find an absorbance that gives a negative concentration for the analyte. What might this mean?

Short Answer

Expert verified

When a negative concentration value is beyond an experimental error, there must be some errors in the analysis procedure. The reason is true for a value above 100% and it may be due to the extrapolation of the calibration curve.

Step by step solution

01

Linear Calibration Curve

The equation of a linear calibration line isy(∓sy)=[m±um]x+[b±ub], where y is the corrected absorbance.

Corrected absorbance =observed absorbance˗˗blank absorbance.

02

The reason for obtaining a negative concentration while analyzing an unknown analyte.

The reasons for obtaining a negative concentration are:

  • A negative value implies zero experimental error. If a negative value is obtained, it says that no detectable analyte is present.

  • When a negative concentration value is beyond an experimental error, there must be some errors in the analysis procedure.

  • The above reason is true for a value above 100% and it may be due to the extrapolation of the calibration curve.

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Most popular questions from this chapter

Logarithmic calibration curve. Calibration data spanning five orders of magnitude for an electrochemical determination of p-nitrophenol are given in the table. (The blank has already been subtracted from the measured current.) If you try to plot these dataon a linear graph extending from 0 to 310μg/mLand from 0 to 5260nA, most of the points will be bunched up near the origin. To handle data with such a large range, a logarithmic plot is helpful.

Overwhatrangeisthelog-logcalibrationlinear?

(a) Make a graph of log (current) versus log( concentration). Over what range is the log-log calibration linear?

(b)FindtheequationoftheLine

InTheform log(current)=m×log(concentration)+b

(c) Find the concentration of p-nitrophenol corresponding to a signal of 99.9nA.

(d) Propagation of uncertainty with logarithm. For a signal of 99.9nA, log (concentration) and its standard uncertainty turn out to be 0.68315±0.04522. With rules for propagation of uncertainty from Chapter 3, find the uncertainty in concentration.

If the difference between the two mean values were half as great as Rayleigh found, but the standard deviation were unchanged, would the difference still be significant?

If each of the four numbers 821,783,834,and855in the example is divided by 2, how will the mean, standard deviation, and coefficient of variation be affected?

What fraction of vertical bars in Figure 4-5a is expected to include the population mean (10000) if many experiments are carried out? Why are the 90 % confidence interval bars longer than the 50 % bars in Figure 4-5?

Which statement about the F test is true? Explain your answer.

(i) IfFcalculated<Ftable, there is more than a 5%chance that the two sets of data are drawn from populations with the same population standard deviation.

(ii) IfFcalculated<Ftable, there is at least 95%probability that the two sets of data are drawn from populations with the same population standard deviation.
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