The PH-KH-CO2 equation completely wrong? UPDATE on 22nd post - The Planted Tank Forum
Old 08-26-2004, 02:54 AM Thread Starter
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The PH-KH-CO2 equation completely wrong? UPDATE on 22nd post

THIS POST IS LONG. BUT PLEASE, ALL CHEM BUFFS, STICK TO IT.
Four days ago I sat down to figure out the equation that governs all pH-KH-CO2 tables. I feel pretty comfortable with chemistry, so it seemed like a relatively simple task. But thats one nothing started to make sense. Since then, I've been searching, reading, searching...and more reading...and think I have come to a conclusion. Except, my conclusion is so outrageous I believe I must have messed up somewhere! To all other chem buffs, tell me if I'm right or wrong.

First I'm going to leave out all the chemical equations, gas and solubility laws, equilibrium constants/equations...everything that explains why we can find CO2 from KH and pH. If you understand what I'm going to go through then you you know about all that. I'll begin with the current origins of the equation we use to calculate CO2, the Henderson-Hasselbach equation.

pH = pKa + log([A-]/[HA])
Knowing the pKa of a certain weak acid, and the molar concentrations of the weak acid and its conjugate base (a buffer exists), we can find pH. In our case, A- is HCO3 and HA is H2CO3. H2CO3 by convention can be replaced by CO2. the pKa of H2CO3 is 6.37. So equation becomes (first line) and can be rearranged to find [CO2]:

pH = 6.37 + log([HCO3]/[CO2])
pH = 6.37 + log[HCO3] - log[CO2]
log[CO2] = log[HCO3] + 6.37 - pH
[CO2] = [HCO3]*10^(6.37-pH)

The last equation to calculate [CO2] is 100% correct. We could use this for our purposes, but it's impractical for the average aquarist to use foreign molar concentration written in scientific notation. It's better to just plug in our well known unit of dKH and get our answer. Thats what this link written by George Booth on APD provides for us, and is the current equation that we are using now.

THIS IS WHERE THE ASSUMED ERROR BEGINS
To make the equation aquarist-friendly, we need to first find a conversion factor that interchanges dKH to molarity HCO3. According to the website 1 dKH = 2.92E-4M HCO3*. The author did as so:

dKH*17.8 = 'HCO3'mg/L
'HCO3'mg/L * (mol/61020mg) = 2.92E-4M HCO3

This is NOT correct. The author uses the common 17.8 factor to change dKH into mg/L, but doesn't realize that he has mg/L CaCO3. He assumes the units are mg/L HCO3 so proceeds in converting the mg into mol, by the molar mass of HCO3. Obviously as you know, using the molar mass of HCO3 is not going to convert mg/L CaCO3 into molarity HCO3. First we need to change mg/L CaCO3 into mg/L HCO3, then proceed into molarity. Before making the correction, I will continue the author's process:

[CO2] = [HCO3]*10^(6.37-pH)
[CO2] = 2.92E-4*dKH*10^(6.37-pH)

Right now the answer is [CO2]. We need mg/L CO2.

[CO2]*44010(**) = 2.92E-4*dKH*10^(6.37-pH)
CO2 mg/L = 12.838*dKH*10^(6.37-pH)

This is the current equation that we use for our CO2 charts. However, there was an error in finding it. So it must be wrong. THIS IS WHERE I NEED SOMEONE TO TELL ME THERE WAS NO ERROR AND I WAS WRONG. . I'll go ahead and show how to find the 'assumed' correct equation.

dKH*17.8 = CaCO3 mg/L
CaCO3 mg/L * (2*61.02/100.09) = HCO3 mg/L
HCO3 mg/L * (mol/61020) = 3.56E-4M HCO3

Therefore the only difference between my equation and the current one: 1dKH = 3.56E-4M HCO3 instead of 1dKH = 2.92E-4M HCO3. Completing the new equation...

[CO2] = [HCO3]*10^(6.37-pH)
[CO2] = 3.56E-4*dKH*10^(6.37-pH)
[CO2]*44010 = 3.56E-4*dKH*10^(6.37-pH)
CO2 mg/L = 15.65*dKH*10^(6.37-pH)

Comparing and testing (KH=5, pH=6.8 ) both equations side-by-side:

Original - CO2 mg/L = 12.838*dKH*10^(6.37-pH)
Correct? - CO2 mg/L = 15.65*dKH*10^(6.37-pH)
Original - CO2 = 23.85 mg/L
Correct? - CO2 = 29.10 mg/L

Just this example shows an 18% error. So now comes the scary part, am I wrong????. I believe I'm going to find out very fast I am because saying the current, long used CO2 charts are completely wrong is outrageous. At the same time, the author of the current equation made a very obvious error, so I do not know which equation to trust!

*Corrected. The author should have used 61020 mg/mol, not 61. This isn't the source of error however.
**Corrected. The author should have used 44010 mg/mol, not 44. This isn't the source of error however.

Last edited by Rolo; 04-23-2015 at 08:35 PM. Reason: edit
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Old 08-26-2004, 03:35 AM
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When you guys figure this out, let me know!

John P.
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Old 08-26-2004, 03:52 AM
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You're putting us on........right?

Len

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Old 08-26-2004, 04:23 AM Thread Starter
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Quote:
Originally Posted by djlen
You're putting us on........right?

Len
No, not at all! Except yeah, doesn't it sounds ridicules that the CO2 calculations we've been using might have been wrong this whole time? It took me four days of re-studying, researching, calculating, recheaking...making sure of myself that I MIGHT be right. But I'm still waiting for the first person to point out my very simple/most careless mistake so you guys can laugh at me.

I wonder how many people actually 'achieved' reading that post...
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Old 08-26-2004, 08:25 AM
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I want to know too.

However, let's say that Chris is right . So new charts/calcs are produced. All that would happen would be that, according to the new calculations, it is 12-15mg/l of CO2 that is the right amount for planted tanks and not 20-25.

In other words, what we thought was 20-25mg/l of CO2 was actually 12-15mg/l. Same thing in the end vis a vis the plants no? The "optimum" would just go down to the 12-15 level.

Or is my logic off here?
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Old 08-26-2004, 01:05 PM
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This is good - after posting something on this in another thread, I realized I made the same error you did below:

Quote:
Originally Posted by Rolo
dKH*17.8 = CaCO3 mg/L
CaCO3 mg/L * (61.02/100.09) = HCO3 mg/L
HCO3 mg/L * (mol/61020) = 1.78E-4M HCO3
The problem above is that 1mole of CaCO3 neutralizes TWO moles of acid, so TWICE as much as HCO3- can neutralize. Therefore, you must also multiply by 2 when converting to HCO3 mg/L.

Quote:
Originally Posted by Rolo
Therefore the only difference between my equation and the current one: 1dKH = 1.78E-4M HCO3 instead of 1dKH = 2.92E-4M HCO3. Completing the new equation...

[CO2] = [HCO3]*10^(6.37-pH)
[CO2] = 1.78E-4*dKH*10^(6.37-pH)
[CO2]*44010 = 1.78E-4*dKH*10^(6.37-pH)
CO2 mg/L = 7.83*dKH*10^(6.37-pH)

Comparing and testing (KH=5, pH=6.8 ) both equations side-by-side:

Original - CO2 mg/L = 12.838*dKH*10^(6.37-pH)
Correct? - CO2 mg/L = 7.83*dKH*10^(6.37-pH)
Original - CO2 = 23.85 mg/L
Correct? - CO2 = 14.55 mg/L
Now the above "Correct" CO2 (mg/L) is actually 15.66*dkH*10^(6.37-pH)

and "correct" from your example is actually 29.10mg/L. This means the original equation gives only an 18% error (still not great - and since the chemistry is wrong anyway, who cares!).

BUT - let me add to the confusion. There is a little thing we deal with in upper-level chemistry courses called "activity" - this represents the effect of waters of hydration (the ionic atmosphere) on the ion. It depends on both the ionic strength of the solution (total dissolved ions) and the size of the ion being considered. The bicarbonate ion has an activity coefficient between 0.82 and 0.96 depending on the ionic strength - this value is multiplied by the concentration to get the activity. Since we assumed that [X] = Activity(X) for everything above, the CO2 concentration is merely an estimate anyway.

Fun stuff!

Kevin

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Old 08-26-2004, 05:36 PM Thread Starter
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Thanks Kevin for the help!

I was thinking to myself the same exact thing when converting CaCO3 into HCO3, would it be 1 or 2 mol HCO3. But when using 2 mol of H+, aren't we assuming CaCO3 is directly taking part in the reaction. I was always under the understanding CaCO3 was merely just a representation; like for instance, that we could even use Cl2 gas as a measure of hardness if we wanted. And isn't the net balanced equation, where the product is HCO3:

CaCO3 + H+ <---> HCO3- + Ca++

So you would need 2 mol of CaCO3 to make 2 mol HCO3, meaning a 1:1 ratio, right? Or am I just completely off? lol. Yeah this is very fun stuff.

Now the original equation has only an error of 18% (actually as you go lower in CO2 concentration the error is less, but higher the error is more). However, I still believe that original equation should not be followed, if really in fact it is incorrect. It makes no sense in using something we know is wrong - and especially wrong in making such an elementary error. In light of the activity subject, I think this proves all the more we need to use the correct equation. Since we are only acquiring an estimate, compounding error on top of error just makes the whole pH-KH-CO2 relation useless.

I'm not ready yet just to take the quotes off our "correct" equation and make it official. Having only two people now agree isn't enough to completely erase all the old CO2 charts.
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Old 08-26-2004, 06:25 PM Thread Starter
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Yes laith your logic is correct.

Just a simple question. For anyone whos been into planted tanks long enough, when did the pH-KH-CO2 relationship begin to show up. Do you know of the source it probably first originated from?
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Old 08-26-2004, 06:43 PM
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Rolo:

I was going to quote some of your message, but then realized I would be quoting the whole thing - kind of pointless. Anyway, here's the deal:

1. alkalinity/buffering capacity/carbonate hardness actually is meant to be measured as the acid equivalents needed to neutralize to the carbonic acid endpoint (thus pH 4.5). Since carbonate ion requires two protons to become carbonic acid and bicarbonate ion only requires one, it is a 2:1 ratio. Expressing this as ppm CaCO3 (or dKh) is weird, but it is the convention.

2. I had sort of ASSUMED (know what that means), that the chart was the result of measurements of CO2, pH, and kH concs in a lab - this is very do-able with fairly simple (for a lab) equipment. Once this discussion came up it reminded me that I have seen a way to make your own CO2 sensor using a pH meter. The pH meter must be able to show the mV reading (not just the pH), but it is easy to construct and easy to calibrate. No, I'm not volunteering to do this experiment (I teach this for a living - don't want to do too much of it at home), just thought it might be of interest to some of you:

Quote:
Originally Posted by InfoTrac
Journal of Chemical Education, Sept 1999 v76 i9 p1253(3)
CO(sub.2) - potentiometric determination and electrode construction, a hands-on approach. (carbon dioxide)(Statistical Data Included) Santiago Kocmur; Eduardo Corton; Liliana Haim; Guillermo Locascio; Lydia Galagosky.
Abstract: A student-used hands-on approach for potentiometric determination and electrode construction for carbon dioxide is discussed. To improve learning students build the sensitive device to detect CO(sub.2) and solve real-life problems, measuring the compound in a carbonated beverage, as do bottlers of soft drinks, mineral waters and some beers. In a post-lab analysis session, students discuss biological rationale for importance of monitoring gas content in yeast cultures in different growing conditions.

Kevin

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Old 08-26-2004, 07:05 PM Thread Starter
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Quote:
Originally Posted by KevinC
Rolo:

I was going to quote some of your message, but then realized I would be quoting the whole thing - kind of pointless. Anyway, here's the deal:

1. alkalinity/buffering capacity/carbonate hardness actually is meant to be measured as the acid equivalents needed to neutralize to the carbonic acid endpoint (thus pH 4.5). Since carbonate ion requires two protons to become carbonic acid and bicarbonate ion only requires one, it is a 2:1 ratio. Expressing this as ppm CaCO3 (or dKh) is weird, but it is the convention.
This makes perfect sense to me now. I made the corrections in the first post.

Quote:
Originally Posted by KevinC
2. I had sort of ASSUMED (know what that means), that the chart was the result of measurements of CO2, pH, and kH concs in a lab - this is very do-able with fairly simple (for a lab) equipment. Once this discussion came up it reminded me that I have seen a way to make your own CO2 sensor using a pH meter. The pH meter must be able to show the mV reading (not just the pH), but it is easy to construct and easy to calibrate. No, I'm not volunteering to do this experiment (I teach this for a living - don't want to do too much of it at home), just thought it might be of interest to some of you:

Kevin
Yes this does interest me. So where can I find the insturctions/equipment?
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Old 08-26-2004, 07:28 PM
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Quote:
Originally Posted by Rolo
Yes this does interest me. So where can I find the insturctions/equipment?
What I posted is a reference to an article in the Journal of Chemical Education. Your nearby college/university library will either have hard copies of this journal or will be able to get you a copy of the article through Interlibrary Loan. I tried to find my copy here in my office, but it is not in the file folder it should be in. As I remember, you take a glass pH electrode and wrap a piece of porous material (like a nylon filter membrane) around it. There is a small amount of liquid between the membrane and the electrode. On exposure to a solution containing CO2, CO2 crosses the membrane, decreasing the pH of the internal solution. The response in mV is linearly related to the CO2 concentration. The more CO2, the higher the mV reading. I had thought about doing this for one of my courses, but haven't yet, so I don't have firsthand experience with how well it works.

As far as the equipment, you MIGHT be able to order directly from places like Fisher Scientific (fishersci.com) for a pH meter that measures mV. Be prepared - we're talking the \$400 to \$4000 area for a meter+pH electrode (that lets you see the mV reading). Maybe your local college/high school would let you borrow one.
Kevin

Kevin

72g bowfront planted, CO2, 4x - T5HO, Eheim 2213 and 2217, 2 angels, pristella tetras, blue tetras, betta, albino bristlenose pleco, albino cories. Sword, vals, hygros, ludwigias, java moss and fern, anubias

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Old 08-26-2004, 09:43 PM
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Why not contact high school and community college chemistry professors and see if they could possibly have students do a lab over it...maybe an extra credit project.

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Old 08-26-2004, 09:56 PM
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Would be nice to see the CO2 table of original calculation & new calculation side-by-side...

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Old 08-26-2004, 09:59 PM
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rolo i will throw in \$0.02. although i really don't know if this is correct. i'm wondering if there may be some sort of solubility co-effecient missing from one of the equations. when you are talking about converting caco3 to hco3, i'm wondering if there should be a correction factor since caco3 is not 100% soluble.

please don't be too hard on me if i'm way off base! i haven't done chemistry in a few years.

consider for example the normal ph = 7.4, normal pco2 = 40 and normal hco3 = 24. pka of carbonic acid is 6.1. plug in the #'s and it won't work unless you multiply the pco2 by 0.03.

you are certainly to be commended for your efforts to work this out!
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Old 08-26-2004, 10:20 PM Thread Starter
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Jart-

I have no reason to be hard on you. We are all learning this together! When we speak about are hardness in CaCO3, our test kits really didn't measure CaCO3 and is not actually what we need. The only unit of hardess that is important is [HCO3]. CaCO3 by convention is the common standard that we express our hardness in. For instance, say something cost \$10USD (HCO3), but we simply express our answer into \$15 Canadian (CaCO3). If you can understand the poor example, I think you got it. Kevin, I might have explained this pretty bad, so mabey you can fill in the holes?

Bigpow -

I'm going try to contact Chuck (the one with the fert calculators) and see if he could put in the new equation. I'll see about a side-side comparison too.

Kevin, Malkore -

I didn't know I was asking for at least \$400 to do this experiment, so I'm can't say just yet I'll voulneteer for the experiment. But I'm on really good grounds with my highschool chemistry teacher, so as soon as I back in september I'll see what I can do. An experiment however shouldn't be essential, as the math alone from the henderson-hasselbach equation provides only provides a good estimation of CO2.
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