Lasse
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That would equate to a total NH3-N concentration of 8.5 mg/l
What are you talking about?
un-ionized ammonia is NH3 or the gas NH3.
Ionized ammonia is NH₄⁺ - the ion from the gas NH3.
NH3-N is the N part of the gas NH3 or the N part of the un-ionized ammonia
NH₄⁺-N is the N part of the NH₄⁺ or the N part of the ionized ammonia.
"0.57 ppm of un-ionized ammonia" at a pH of 8.1, that would equate to a TOTAL ammonia level of 8.5 ppm - and that's unheard of!
Maybe you have not heard about it but they discover issues with the gills (sub lethal damage) at this concentration. Its not a LC50 figure
Its a strange question because in the methods the authors state - my boldwas your ammonia chloride dose that high?
0.23 ± 0.02, 0.57 ± 0.03, 0.97 ± 0.06, 1.05 ± 0.06, 1.22 ± 0.05 and 1.63 ± 0.14 mg/L NH3-N obtained by adding ammonium chloride
Apparently - they have add so much ammonium chloride it was needed to get 0.57 mg/L NH3-N. Note that 0.57 mg/L NH3-N is not with in the LC but they could show gill lesionsTotal ammonia nitrogen (TA-N) and nitrite were measured according to Grasshoff et al. (1999) and Aminot and Chaussepied (1983). Un-ionized ammonia (NH3) concentrations were calculated following TA-N, and temperature, salinity and pH values following the equations of Ostrensky et al. (1992), adapted from Whitfield (1974) and Bower and Bidwell (1978).
Once again
un-ionized ammonia is NH3 in British English even named ammoniac
Ionized ammonia is NH₄⁺ - in British English even named ammonium
Total ammonia is the sum of NH3 and NH₄⁺ or in British English ammoniac+ammonium
Is there a -N in the the formula - it means that the mg/l refers to only the N part of the subject. If you have NH3-N - you need to multiply with 17 (mol mass NH3) divided with 14 (mol mass N) = 1,21 if you want to express it as NH3
Most important is that their findings of LC50 is well in line with other similar investigations of salt water fish there the LC50 96h value can vary from around 0.5 up to 2-3 mg/L NH3-N for LC50 96 H (4 days)
Sincerely Lasse
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