- #1
PedroAndrade
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Hi!
In my lab we're preparing a toxicity bioassay using nitrate (NO3-) as the test toxicant. The tests we're doing are range-finding tests, to help determine the lethal concentration (LC50) for that species with nitrate.
We are following the lead of another work (http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V74-4F02KWG-B&_user=10&_rdoc=1&_fmt=&_orig=search&_sort=d&view=c&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=172b82c36ff657803f003c89ddaca62d") which used sodium nitrate (NaNO3) as the source of nitrate. In that article, the test concentrations used for a similar species range from 5 to 160 mg NO3-N/L. This is where my understanding of chemistry is lacking. I'll try to illustrate my doubts with an example:
If I want to find the LC50 for a solution of 10 mg nitrate/L, what do I calculate exactly?
a) 10 mg of NaNO3
b) 13,71 mg of NaNO3 - corresponding to 10 mg of NO3-
c) 60,71 mg of NaNO3 - corresponding to 10 mg of N
My feeling is that b) is correct, but the notation NO3-N is throwing me off. Can anyone help out with this?
In my lab we're preparing a toxicity bioassay using nitrate (NO3-) as the test toxicant. The tests we're doing are range-finding tests, to help determine the lethal concentration (LC50) for that species with nitrate.
We are following the lead of another work (http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V74-4F02KWG-B&_user=10&_rdoc=1&_fmt=&_orig=search&_sort=d&view=c&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=172b82c36ff657803f003c89ddaca62d") which used sodium nitrate (NaNO3) as the source of nitrate. In that article, the test concentrations used for a similar species range from 5 to 160 mg NO3-N/L. This is where my understanding of chemistry is lacking. I'll try to illustrate my doubts with an example:
If I want to find the LC50 for a solution of 10 mg nitrate/L, what do I calculate exactly?
a) 10 mg of NaNO3
b) 13,71 mg of NaNO3 - corresponding to 10 mg of NO3-
c) 60,71 mg of NaNO3 - corresponding to 10 mg of N
My feeling is that b) is correct, but the notation NO3-N is throwing me off. Can anyone help out with this?
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