TM 5-6630-215-12
Table 2-2. Sample Sizes for Low Concentrations
Measuring - Water sample
Multiply indicator
Divide Result
Tube
size
(and buffer, if any) used by:
By
1
15 ml
2
30 ml
45 ml
4
60 ml
1
2
3
4
1
2
3
4
3
(1) Standard additions is performed by adding a small amount of
a standard solution to a previously analyzed sample and then repeating
the analysis using the same reagent,
instrument and technique.
The
amount of increase in the test result should exactly equal the amount
of standard added.
(2) For example, if a 25-ml water sample is analyzed for iron
and is found to contain 1.0 mg/l, the result can be checked by adding
0.10 ml of a 50.0 mg/l iron standard solution to another 25-ml portion
of the water sample and repeating the analysis.
The result of the
analysis on the second sample should be 1.2 mg/l iron since the
standard added has a value of 0.2 mg/l; i.e.,
0.1 ml
X 50 mg/l = 0.2 mg/l
25 ml
If 0.2 mg/l is recovered from the 0.2 mg/l addition, the analyst can
conclude that the first answer was correct and that the reagents,
instruments and method used are all working properly.
Table 2-3. Dilutions for High Concentrations
Water sample
(use pipet)
50.0 ml
25.0 ml
10.0 ml
5.0 ml
Demineralized water used
to bring volume to approximately
50 ml (use graduated cylinder)
0 ml
25 ml
40 ml
45 ml
Multiplication
factor
1
2
5
10
(3) Should the second analysis not give the correct amount of
increase in the iron content, it must be concluded that the first
answer may also be incorrect.
The analyst must then determine why the
technique did not work.
By using a logical troubleshooting approach,
the source of the problems can be determined whether the fault lies in
the reagent, the instrument and apparatus, the test procedure or an
interfering substance present in the test sample.
2-9
