Environmental Engineering Reference
In-Depth Information
Erroneous Data and Methods of Compensation
Sometimes erroneous results are obtained even when proper QA/QC protocols are
strictly followed. The analyst should then consider one of the sources of analytical
interference as discussed above in Table 9.2. Think about the chemistry introduced
at the beginning of this chapter, and try various compensation methods as
appropriate, such as background correction, use of higher temperature flame,
addition of releasing agents, use of an alternative wavelength, matrix spike, internal
standard, and standard addition calibration.
For the internal standard method, an internal standard undergoes similar inter-
ference as the analyte. Measurement of the ratio of the analyte to internal standard
signals (e.g., K/Li, Na/Li) cancels the interference. The internal standard element
should be chemically similar to the analyte elements (such as Li as the internal stan-
dard for K and Na), and their wavelengths should also be similar. In standard addi-
tion calibration, the standard is added to the sample, and so it experiences the same
matrix effects as the analyte. This method is used to account for the sample matrix
effect that may be caused by high viscosity or chemical reaction with the analyte.
The example below demonstrates how the standard addition method works to
eliminate the interference and how results should be properly reported.
EXAMPLE 9.2.
An acid digestate of sludge sample is analyzed for zinc by flame
emission spectrometry using the method of single standard addition. Two portions of digested
aliquots (1 mL each) are added to 25 mL portion of deionized water. To one portion is added
5 mL standard ZnCl
2
solution containing 100 mg/L (as Zn). The instrumental signals of the
flame emission spectrometry in arbitrary units are 23.5 and 95.6. (a) What is the concentration
of Zn in the sludge digestate? (b) If 0.5 g of wet sludge was used, and the total digestate
volume is 50 mL, what is the Zn concentration in sludge (The sludge moisture was determined
to be 35%?)
SOLUTION:
(a) Assume there is x mg Zn present in 1 mL of the digestate, then the total
milligrams of Zn in the
aliquot with added standard is:
x
(mg)þ100 mg/L
*5 mL¼xþ5*10
4
mg (Note 1 L¼10
6
mL). Therefore:
x
þ
5
10
4
95:6
23
:
5
5
10
4
95:623:5
¼ 1:6310
4
mg
x
23:5
¼
Solve for x ¼
This is the mass of Zn in 1 mL of the digestate, therefore Zn concentration in the acid digestate
is: 1.6310
4
mg/1 mL¼0.163 mg/L
(b) Note the total volume of the digestate from 0.5-g wet sludge is 50 mL. The concentration
of Zn in sludge is:
0:163
mg
L
1000mL
L
50 mL
Znðmg=kg; wet basisÞ¼
¼ 0:0163mg=g ¼ 16:3mg=kg
0:5g
16:3
mg
kg
ð10035Þ=100
¼ 25:08mg=kg
Znðmg=kg; dry basisÞ¼
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