Reef Chemistry Question of the Day #192 Cycling and Alkalinity

[Myka]

1 mg/L ammonia is 0.0588 mM, so it will consume 0.0588 meq/L (= 2.94 ppm calcium carbonate equivalents).

So the initial assumption of 7 mg/L alkalinity per mg ammonia seems the problem.

Where did that come from?

Ok. It is a common number from water purification, but maybe it's just not the same thing...or maybe it's different in saltwater?? ...here are some random Google references:

http://www.weao.org/assets/docs/new-professionals/timconstantine-nitrification_and_removal.pdf (page 6)

http://waterfacts.net/Treatment/Activated_Sludge/Nitrification/nitrification.html (scroll halfway down the page to "Table 1: Oxygen Consumed During Nitrification")

http://www.environmentalleverage.com/Nitrification-balance.htm (third paragraph, just above "The Nitrification Process".)

 

[Randy Holmes-Farley]

So I looked at the first link, but it is talking about the conversion of ammonium to nitrate, not ammonia to nitrate. So that process uses twice as much as the question here because the first step in that analysis is to strip the ammonium to ammonia, releasing H+ and depleting alkalinity.

NH4+ → NH3 + H+

So for that process, I'd get twice what I got for our question of the day situation, or 0.118 meq/L, or 5.88 ppm calcium carbonate equivalents for 1 mg/L of ammonia.

But here's where it really gets odd...

The first link says:
Acid formed, consumes alkalinity, 7.14 g alkalinity per g NO3

The second link says:
Theoretically, 7.14 lb of alkalinity (measured as CaCO3) is consumed during the oxidation of 1 lb of ammonia (Table 1).
but then when you look at table 1 it is saying this value applies for "every pound of ammonia nitrogen removed"

The third one says:
7.14 lbs of M-alkalinity are destroyed per lb of ammonia-nitrogen oxidized


The second two seem to agree with each other, although the text sentence of the third one as written is quite misleading. The first one is totally different.

Anyway, both of the second two are talking about ammonia nitrogen, not ammonia. So we have to correct my numbers to compare to theirs.

Ammonia weighs 17 g/mole and nitrogen weighs 14 g/mole, we we have to multiply my result by 17/14 = 1.21 to get to mg/L calcium carbonate equivalents per mg/L of nitrogen in the ammonia from mg/L calcium carbonate equivalents per mg/L of ammonia.

So my 5.88 ppm calcium carbonate equivalents becomes 5.88 x 1.21 = 7.14 mg calcium carbonate equivalents per mg of ammonia (I dropped out the per liter since that was redundant) One can translate that to any units desired, such as pounds without changing the value. So my calculations, when corrected to this situation and units of measure, match.

The first link, however, is clearly wrong. It should read:
Acid formed, consumes alkalinity, 7.14 g alkalinity per g NH3-N, not NO3

 

[Myka]

So my calculations, when corrected to this situation and units of measure, match.

The first link, however, is clearly wrong. It should read:
Acid formed, consumes alkalinity, 7.14 g alkalinity per g NH3-N, not NO3

Ok, thanks Randy. So I guess I "just" overlooked measurements and types of nitrogen. I did think the second two links were different than the first one, but I couldn't put my finger on it. I liked it better though since it looked simpler.

 

[JimWelsh]

I did notice two things, though, that just may explain what you might be missing: First, they are describing the conversion from NH4+ to NO3-, and not NH3 to NO3-, so this might reduce the effect by a factor of two, and second, that 7.14 number is based on the amount of ammonia expressed as N, and not as NH4+ or NH3. One or both of these factors probably explains the discrepancy.

 

[Myka]

Preening? Haha! I did try the equation with Nitrogen, and my numbers were WAY off of Randy's numbers then! Hahahaha!

Ya, so I spent the morning looking at some different university courses I could take to broaden my horizons...gently feeling around the edges of Biochemistry maybe. I suppose by the time I got a PhD my student loans would so racked up I might have them paid off by retirement! "It's never too late." Right?

 

[kevantheman35]

While cycling large public aquariums 20-100k gallons this is one issue that we have always had to deal with. The cycle seems to stall out at nitrite for quite some time typically while we seemingly endlessly feed the system alk buffers.
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My question is that if we are using up alkalinity both during the nitrogen cycle, as well as it being taken up by corals in calcification, why do our systems typically use calcium and alkalinity in approximately the same ratios?

 

[Randy Holmes-Farley]

While cycling large public aquariums 20-100k gallons this is one issue that we have always had to deal with. The cycle seems to stall out at nitrite for quite some time typically while we seemingly endlessly feed the system alk buffers.
-------
My question is that if we are using up alkalinity both during the nitrogen cycle, as well as it being taken up by corals in calcification, why do our systems typically use calcium and alkalinity in approximately the same ratios?

It can mess with the ratio, and I discuss that and a few other processes here:

When Do Calcium and Alkalinity Demand Not Exactly Balance? by Randy Holmes-Farley - Reefkeeping.com
http://reefkeeping.com/issues/2004-12/rhf/index.htm

That said, we get ALL of the nitrogen cycle alkalinity loss back when the nitrate undergoes denitrification or incorporation into organisms such as algae or corals. It is the same process in reverse. The only time we really lose alkalinity permanently is when nitrate is actually rising in concentration. This is from the above article:

However, the news is not all bad. When this nitrate proceeds further along the nitrogen cycle, depleted alkalinity is returned in exactly the amount lost. For example, if the nitrate is allowed to be converted into N2 in a sand bed, one of the products is bicarbonate, as shown in equation 2 (below) for the breakdown of glucose and nitrate under typical anoxic conditions as might happen in a deep sand bed:

1. 
   (2) 4NO3- + 5/6 C6H12O6 (glucose) + 4H2O --> 2 N2 + 7H2O + 4HCO3- + CO2

In equation 2 we see that exactly one bicarbonate ion is produced for each nitrate ion consumed. Consequently, the alkalinity gain is 0.8 meq/L (2.3 dKH) for every 50 ppm of nitrate consumed.

Likewise, equation 3 (below) shows the uptake of nitrate and CO2 into macroalgae to form typical organic molecules:

1. 
   (3) 122 CO2 + 122 H2O + 16 NO3- --> C106H260O106N16 + 138 O2 + 16 HCO3-

Again, one bicarbonate ion is produced for each nitrate ion consumed.