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CEE 772 |
Fall 2005 |
Use the following TOX raw data to answer the following questions:
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Integrated Peak (ng) |
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Sample |
Volume (mL) |
1st column |
2nd column |
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distilled water |
50 |
605 |
555 |
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200 ug/L MCAA |
50 |
2405 |
1205 |
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100 ug/L MCAA |
50 |
1490 |
890 |
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Tap Water |
50 |
3492 |
1955 |
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Tap Water |
50 |
4059 |
1538 |
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distilled water |
50 |
510 |
660 |
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groundwater |
100 |
1288 |
799 |
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groundwater |
100 |
1892 |
865 |
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groundwater |
100 |
1484 |
990 |
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groundwater |
100 |
1553 |
921 |
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groundwater |
100 |
1399 |
1020 |
|
groundwater |
100 |
1686 |
952 |
|
groundwater |
100 |
1335 |
1100 |
|
distilled water |
50 |
495 |
600 |
For this you must first calculate an average blank value from the distilled water replicates.
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Blank Avg = |
570.8 |
Then you subtract this blank from the individual column values to get the net organic halide in ng (as Cl). Finally you add the two net values and divide by the sample volume to get the TOX concentration in ug/L. See table below for these calculations:
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Integrated Peak (ng) |
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Net OX (ng) |
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TOX |
MCAA |
% recovery |
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Sample |
Volume (mL) |
1st column |
2nd column |
Sum |
1st |
2nd |
sum |
% first |
conc (ug/L) |
conc (ug/L) |
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distilled water |
50 |
605 |
555 |
1160 |
34 |
-16 |
18 |
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200 ug/L MCAA |
50 |
2405 |
1205 |
3610 |
1834 |
634 |
2468 |
74% |
49.4 |
200 |
65.8% |
|
100 ug/L MCAA |
50 |
1490 |
890 |
2380 |
919 |
319 |
1238 |
74% |
24.8 |
100 |
66.0% |
|
Tap Water |
50 |
3492 |
1955 |
5447 |
2921 |
1384 |
4305 |
68% |
86.1 |
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Tap Water |
50 |
4059 |
1538 |
5597 |
3488 |
967 |
4455 |
78% |
89.1 |
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distilled water |
50 |
510 |
660 |
1170 |
-61 |
89 |
28 |
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groundwater |
100 |
1288 |
799 |
2087 |
717 |
228 |
945 |
76% |
9.5 |
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groundwater |
100 |
1892 |
865 |
2757 |
1321 |
294 |
1615 |
82% |
16.2 |
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groundwater |
100 |
1484 |
990 |
2474 |
913 |
419 |
1332 |
69% |
13.3 |
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groundwater |
100 |
1553 |
921 |
2474 |
982 |
350 |
1332 |
74% |
13.3 |
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groundwater |
100 |
1399 |
1020 |
2419 |
828 |
449 |
1277 |
65% |
12.8 |
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groundwater |
100 |
1686 |
952 |
2638 |
1115 |
381 |
1496 |
75% |
15.0 |
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groundwater |
100 |
1335 |
1100 |
2435 |
764 |
529 |
1293 |
59% |
12.9 |
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distilled water |
50 |
495 |
600 |
1095 |
-76 |
29 |
-47 |
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Tap Water Mean = |
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87.6 |
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Groundwater Standard
Deviation = |
2.1 |
ug/L |
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MDL = 3.14*SD = |
6.6 |
ug/L |
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The values for the two tap water samples are 86.1 and 89.1 ug/L. So the average would be 87.6 ug/L
Here you compare the net values from the first column with those from the second column. This may be done as a %age of the first compared to the total. Results of this calculation are shown in the table above under the column “% first”. These values average about 85%, and the lowest is for the first tap water sample at 70%. This is good recovery as only 30% of 30% (or 9%) would be escaping the second column, even in this worst sample.
Here you need to recognize that MCAA is only 37.5% chlorine by weight. Therefore you expect to get 37.5 ug/L and 75 ug/L for the two standards. The actual values fall well below, and they average 65.9 % of the expected value (see table above).
Based on standard methods you need 7 replicates of a low level sample. The groundwater sample serves our purpose here. First you calculate the standard deviation of those 7 samples. This turns out to be 2.1 ug/L. Then you multiply by the appropriate t-statistic (i.e., 3.14) which gives us an MDL of 6.6 ug/L (see above).
Assigned: 24 Oct 2005