Reef Chemistry Question of the Day #192 Cycling and Alkalinity

[120reefkeeper]

Here's my translation for you:

Now I understand this better than English .

 

[JimWelsh]

Now I understand this better than English .

*Sigh*

 

[SROPE]

Sea.

 

[mrsaltwatertank]

10.9 mg/L of NO3- is about 0.176 millimoles (NO3 is one N = atomic weight 14, plus three O = atomic weight 16, for a total of 62, so 10.9 mg/L divided by 62 = 0.176 millimoles) , so it will reduce alkalinity by 0.176 meq/L. Rounded, the answer is D.

Oh man. Makes me miss Chemistry.

 

[Bruce Burnett]

as a guess I have to say c not enough to make much change in alkalinity.

 

[john.m.cole3]

I used instant ocean regular to cycle my tank and at 35 ppt, after a diatom bloom, I had 10.2 dkh

A, is my answer

 

[reef_junkie]

[emoji15] is this alien language or English??

Its algeneese

 

[s2nhle]

I think my answer is E.

 

[beaslbob]

not able to do the math but as I remember ammonia to nitrite consumes an alk and nitrite to nitrate consumes an alk.

Both of which are returned when nitrate is reduced by plant or bacterial action.

 

[Myka]

10.9 mg/L of NO3- is about 0.176 millimoles (NO3 is one N = atomic weight 14, plus three O = atomic weight 16, for a total of 62, so 10.9 mg/L divided by 62 = 0.176 millimoles) , so it will reduce alkalinity by 0.176 meq/L. Rounded, the answer is D.

I think you're missing something Jim.

I don't know the chemistry equations to calculate the way you chemists are, but I do know that conversion from ammonia to nitrate consumes 7.14 mg/L alkalinity for every 1 mg/L ammonia from acid produced in the oxidation process. So there is 21.42 mg/L consumption from 9 dKH, or 21.42 ppm consumption from 161 ppm alkalinity. 161 - 21.42 = 139.58 ppm or 7.8 dKH or 2.8 meq/L.

So my answer is F.

Clearly I'm missing something too Jim.

 

[JimWelsh]

I think you're missing something Jim.

I don't know the chemistry equations to calculate the way you chemists are, but I do know that conversion from ammonia to nitrate consumes 7.14 mg/L alkalinity for every 1 mg/L ammonia from acid produced in the oxidation process. So there is 21.42 mg/L consumption from 9 dKH, or 21.42 ppm consumption from 161 ppm alkalinity. 161 - 21.42 = 139.58 ppm or 7.8 dKH or 2.8 meq/L.

So my answer is F.

Clearly I'm missing something too Jim.

I certainly could be missing something, Mindy. I've made bonehead mistakes in these quizzes before!

I did my best to answer with what I know without Googling anything. I made the following assumptions, based on how Randy asked the question: The "ammonia" in question is NH3, and not ammonium, or NH4+. That's because he specified that it was ammonium hydroxide, so I figured the ammonium NH4+ and the hydroxide OH- would be the same as NH3 and H2O for the purposes of the question. In terms of the alkalinity consumed, I figured that we were consuming one "molecule" of alkalinity, as it were, per molecule of ammonia since we were going from NH3 to NO3- (only one charge difference per molecule). If my assumptions are correct, then I'm confident in my math. I did contemplate that perhaps there were two charges involved in the conversion, as @beaslbob alluded to, but that would mean an answer somewhere between D and E, so that plus the other reasoning above caused me to go with the answer I gave.

After your reply, I did Google ammonia to nitrate conversion, and see the 7.14 number you are using cited in multiple references, and I agree with your math, based on that number. 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.

So, I'm sticking with my answer, but I'll allow that it might be "E" instead. No "F" appears to be necessary.

FWIW, I still stand by my diatribe about PPV vs. molarity and the conversion between the two, just because SOOOO many things discussed in this forum would be much easier if people would just understand how to do these conversions, even if I'm "missing something" about this particular calculation!

 

[Randy Holmes-Farley]

I can go through how to get the balanced reaction (aside from looking it up in my articles or elsewhere), but it is not a super easy DIY for nonchemists. For those interested in exactly how it is done, this link does it for the ammonia to nitrite conversion, and the process is the same for ammonia to nitrate:

http://ceae.colorado.edu/~silverst/cven5534/STOICHIOMETRY II nitrogen.pdf

However, knowing the fully balanced reaction is a good way (maybe the only generally accurate way) to determine if alkalinity is added or consumed, and by how much.

Adding ammonia and adding ammonium hydroxide are essentially the same, since ammonia plus water gives ammonium hydroxide. So we'll just think about ammonia.

NH3 + H2O --> NH4+ OH-

So the balanced reaction is

NH3 + 2O2 --> NO3- + H+ + H2O

It is the H+ produced that depletes alkalinity. One mole of alkalinity for each mole of ammonia.

It is actually the conversion of ammonia to nitrite that uses the 1 equivalent of alkalinity:

NH3 + 3/2 O2 --> NO2- + H+ + H2O

while the conversion of nitrite to nitrate does not alter alkalinity:

NO2- + ½ O2 --> NO3-

Jim's idea of just looking at the charge gave the right answer in this case because he knew that the other charged species in the reaction were ONLY H+ or OH- , both of which impact alkalinity. That wouldn't always be true, however, if other charged species that do not impact alk are produced or consumed. For example, suppose the oxidant is iron instead of oxygen. The balanced reaction can be written as:

8Fe+++ + NH3 + 3H2O --> 8Fe++ + NO3- + 9H+

In this case where iron is used, alk is depleted nine times as much as when oxygen is used, even though it is still one ammonia being converted into one nitrate.

Math coming up later (bet you can't wait )

 

[s2nhle]

yes, i can't wait.

 

[Bruce Burnett]

I will change my answer to D from C. I am not sure I would have been able to answer that question 45 years ago.

 

[Myka]

After your reply, I did Google ammonia to nitrate conversion, and see the 7.14 number you are using cited in multiple references, and I agree with your math, based on that number. 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.

Hi Jim. I did not think of that extra H in NH4 and I actually thought the conversion was using NH3 since the number 7.14 is referenced in related to toxic ammonia. Or at least that's what my brain remembers. Alzheimer's runs in the family though.... I think I need to go back and redo chemistry as I haven't done anything beyond Chem 10 in highschool, and I'm not following the balanced reactions anymore. I loved it in highschool, and I'm not sure why I didn't pursue it. I wasn't very focused at that time. I've been tossing around the idea of going to university, though I think there would be Profs younger than me.

 

[Bruce Burnett]

Hi Jim. I did not think of that extra H in NH4 and I actually thought the conversion was using NH3 since the number 7.14 is referenced in related to toxic ammonia. Or at least that's what my brain remembers. Alzheimer's runs in the family though.... I think I need to go back and redo chemistry as I haven't done anything beyond Chem 10 in highschool, and I'm not following the balanced reactions anymore. I loved it in highschool, and I'm not sure why I didn't pursue it. I wasn't very focused at that time. I've been tossing around the idea of going to university, though I think there would be Profs younger than me.

My problem high school chemistry and going back to school I think I would be older than anyone in a college or university

 

[KingBlingTX]

At first, I was thinking it would be a net zero effective. As Jim & Randy mentioned, the proton (H+) generated in the NO2 step lowers the alkalinity 1:1 ratio with the NH3 added. However, I kept thinking "what about the hydroxide ions we are adding in the first place?" That should raising the alkalinity (also at a 1:1 ratio with the NH3 addition if you just look at the stoichiometry of the equation. NH3 + H20 <-> NH4+ + OH-) After looking up the ionization constant for NH4OH, now I understand why that effect is ignored. Apparently, only 0.42% of the NH3 ionizes. Makes me wonder why it was ever called ammonium hyrdoxide to begin with! LOL

 

[Randy Holmes-Farley]

And the answer is... D. 8.5 dKH (3.0 meq/L)

So as folks have shown, the calculation is fairly easy once you know the balanced equation and know that each mole (a unit of measure based on numbers of ions or molecules rather than weight) of nitrate made (or ammonia consumed) depletes one mole of alkalinity (meaning it made one H+ for each nitrate formed).

So how much is 10.9 mg/L of nitrate produced in units of alkalinity?

Nitrate has a molecular weight of 62 grams per mole (14 for the N and 16 for each of the three oxygens) or 63 mg/mmole

So 10.9 mg/L / 62 mg/mmole = 0.176 mmole/L = 0.176 meq/L

That is the same as alkalinity in meq/L, so the drop in alk is 0.176 meq/L or 2.8 times that in units of dKH = 0.49 dKH ~0.5 dKH

So the answer is 9.0 dKH - 0.5 dKH = 8.5 dKH.

 

[Myka]

Thanks for the answer Randy. Could you please explain where I went wrong here:

I don't know the chemistry equations to calculate the way you chemists are, but I do know that conversion from ammonia to nitrate consumes 7.14 mg/L alkalinity for every 1 mg/L ammonia from acid produced in the oxidation process. So there is 21.42 mg/L consumption from 9 dKH, or 21.42 ppm consumption from 161 ppm alkalinity. 161 - 21.42 = 139.58 ppm or 7.8 dKH or 2.8 meq/L.

 

[Randy Holmes-Farley]

Thanks for the answer Randy. Could you please explain where I went wrong here:

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?