4.6. Practical example

This is an example of calculating the volume and its uncertainty of liquid delivered from a self-calibrated volumetric pipette

The uncertainty of the pipetted volume u(V) has three main uncertainty components: uncertainty due to repeatability, u(V,rep); uncertainty due to pipette calibration, u(V,cal) and uncertainty due to the temperature difference from 20 °C, u(V,temp).

The estimate of the probable maximum difference of the pipette volume from the nominal volume, expressed as ±x is often used as the estimate of calibration uncertainty of the pipette (as was done in section 4.1). It is usually given by the manufacturer without any additional information about its coverage probability or distribution function. In such a case it is the safest to assume that rectangular distribution holds and to convert the uncertainty estimate to standard uncertainty by dividing it with square root of three.

Usually in the case of high-accuracy work the pipette is calibrated in the laboratory in order to obtain lower calibration uncertainty. As was seen in sections 4.2 and 4.3, if the uncertainty due to factory calibration is used, then this calibration uncertainty component is the most influential one. So, reducing it would also reduce the overall uncertainty. It is very important, that calibration and pipetting are performed under the same conditions and preferably by the same person. From the calibration data we can obtain two important pieces of information: (1) the correction term for the pipette volume Vcorrection with uncertainty (V,cal) and (2) repeatability of pipetting (V,rep). The example presented here explains this.

For calibration of a pipette water is repeatedly pipetted (at controlled temperature in order to know its density), the masses of the pipetted amounts of water are measured and the pipetted volumes of water are calculated (using density of water at the calibration temperature). Here are the calibration data of a 10 ml pipette:

 




The standard deviation is calculated according to the equation:

(4.15)

Uncertainty due to repeatability of pipetting (V, REP) is equal to this standard deviation 0.0057 ml. Pipetting is often used in titration analysis. If the solution that is titrated is pipetted then this repeatability contribution is already accounted for in the repeatability of the titration results and is not separately taken into account in the uncertainty of pipette volume.

Uncertainty of the calibration (in fact the uncertainty of the correction) has to be always taken into account. In many cases, the uncertainty due to repeatability of obtaining the correction is the only important uncertainty source and other sources can be left out of the consideration. This uncertainty is expressed (when calibration is done [1][1] When the calibration is not performed in the laboratory, the uncertainty of the calibration can be taken into account according to the manual information, that is usually given on the pipette as the tolerance range. For example, if the manufacturer has provided the tolerance range ±0.03 mL (in case of 10 mL pipette), then the standard uncertainty of the pipette correction is calculated as . Correction value on this case is 0.00 mL. ) as the standard deviation of the mean:

(4.16)

In this example the correction is -0.0080 ml and its standard uncertainty is 0.0018 ml. 

When there is a possibility that pipetting is performed at a different temperature from the calibration (and this possibility exists almost always), then an additional uncertainty source due to temperature change is introduced and it has to be taken into account. 

In this case we assume that pipette’s using temperature does not differ from the calibration temperature by more than 4 oC (Δt  = ±4˚C, assuming rectangular distribution). Water’s density depends on temperature, therefore we have to consider also the thermal expansion coefficient of water, which is γw=2.1·10-4 1/°C. So,

(4.17)

Now, when we will perform a single pipetting, the volume is 9.992 ml and its combined standard uncertainty is

(4.18)

The k = 2 expanded uncertainty [2]Later in this course (section 9.8) we will see, how to rigorously find, whether we can say that the k = 2 expanded uncertainty in a particular case corresponds to 95% (this depends on the so-called effective number of degrees of freedom). And if not then what k should be used to achieve approximately 95% coverage probability. In the case of this example the effective number of degrees of freedom is 26 and the respective coverage factor (actually the Student coefficient) with the probability of 95% is 2.06, which is only very slightly different from 2 (the expanded uncertainty would increase from 0.015 ml to 0.016 ml).  of the pipetted volume can be found as follows:

U() = uc() · k = 0.0077 · 2 = 0.0154 ml             (4.19)

As explained in section 4.5 if the first significant digit of the uncertainty is 1... 4 then uncertainty should be presented with two significant digits. Thus we can write the result:

The volume of the pipetted liquid is:

V = (9.992 ± 0.015) ml, k = 2, norm.             (4.20)

It is interesting to compare now this expanded uncertainty with the expanded uncertainty obtained in section 4.5 (eq 4.14). We see that when the pipette is calibrated in our laboratory then the uncertainty of the volume is more than two times lower. We also see that the uncertainty component due to pipette calibration, which back then was the largest uncertainty component, is now the smallest.

***

[1] When the calibration is not performed in the laboratory, the uncertainty of the calibration can be taken into account according to the manual information, that is usually given on the pipette as the tolerance range. For example, if the manufacturer has provided the tolerance range ±0.03 mL (in case of 10 mL pipette), then the standard uncertainty of the pipette correction is calculated as . Correction value on this case is 0.00 mL.

[2] Later in this course (section 9.8) we will see, how to rigorously find, whether we can say that the k = 2 expanded uncertainty in a particular case corresponds to 95% (this depends on the so-called effective number of degrees of freedom). And if not then what k should be used to achieve approximately 95% coverage probability.
In the case of this example the effective number of degrees of freedom is 26 and the respective coverage factor (actually the Student coefficient) with the probability of 95% is 2.06, which is only very slightly different from 2 (the expanded uncertainty would increase from 0.015 ml to 0.016 ml).

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