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Solids - TSS Duplicates
CJ Felice
Posted: Thursday, October 11, 2012 11:15 AM
Joined: 10/11/2012
Posts: 2


I'm in the process of reviewing our analytical procedures and came across something in the Solids section that I'd like to get some clarification on. 

 

Standard Methods, 21st Ed., Method 2540D TSS, end of Section 3.c.: "Duplicate determinations should agree within 5% of their average weight."  Does this mean that the two final weights should be averaged, and each weight compared to that average must be within 5% of that average?  How can this critera best be handled?

 

If I recall correctly, this statement appears in all of the SM Solids analyses. I'm working with TSS first and will make adjustments to other analyses if needed. 

 

Any assistance to clarify this criteria would be appreciated.

 

Thanks!

 


17526110
Posted: Tuesday, October 30, 2012 12:21 PM
Joined: 1/18/2012
Posts: 6


I agree that it is written ambiguously. I take it to mean the RPD or relative percent difference should be 5%.

Analysis =R1

Duplicate =R2

 

(R1-R2)/(average)  * 100

 

You can't really compare the weights and need to compare the TSS value. Unless of course, you filter the same volume for both analysis, then comparing the weights should be fine.


01636251
Posted: Thursday, May 16, 2013 9:30 AM
Joined: 12/3/2009
Posts: 1


If the two sample volumes are the same, the Weight of the Dry Residue of both the sample and duplicate should fall within 5% of their average.

 

If the sample and duplicate volumes are not exactly the same, convert the Weight of the Dry Residues from mg/L to g/L:

 

Weight of Dry Residue mg/L/(Sample volume mLs/1000).

 

Then compare the two Weight of Dry Residues in g/L to 5% of the average of the two.

 

This should be indicated somewhere on the benchsheet.