Is it 3, or 3.6? - The Planted Tank Forum
 
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post #1 of 7 (permalink) Old 11-15-2004, 02:11 AM Thread Starter
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Is it 3, or 3.6?

I've seen several posts and references for determining the CO2 level in tank water and there seems to be some discrepancy as to which multiplier to use. Is it 3.6*KH*10^(7-pH) or 3*KH*10^(7-pH)?

Thanks!

Paul

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post #2 of 7 (permalink) Old 11-15-2004, 02:44 AM
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The one being used is 3. There is actually a more theoretically correct equation though.

But all in all, the CO2 equations are not exact, but good estimates. KevinC did a post where you can accurately find correct CO2 levels but requires a ton of known water parameters.
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post #3 of 7 (permalink) Old 11-15-2004, 02:45 AM
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I'd say it's 3... mainly because I just like using the chart over at Chuck's Planted Aquarium Pages. That's the one he uses for the calculations - and most of his stuff has proved to be accurate enough for most hobbiests.

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post #4 of 7 (permalink) Old 11-15-2004, 02:47 AM Thread Starter
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I'll probably go with 3 then. I'm setting up a log in Excel for my tank and was wondering which number to use. Thanks for the info.

Although, maybe I should go with 3.3 to split the difference.... Hmmmmm....

Paul

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post #5 of 7 (permalink) Old 11-15-2004, 03:25 PM
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Is it 3 or 3,6 or … ?

pH = pKA – log( [A] / [B] )
pH = pKA – (log[A] – log[B])
pH = pKA – log[A] + log[B]
log[A] = pKA – pH + log[B]
[A] = 10^(pKA – pH + log[B])
A = M(A) · 10^(pKA – pH + log[B])
A = M(A) · 10^(pKA – pH) · [B]
A = M(A) · [B] · 10^(pKA – pH)

A = CO2(aq) ~ H2CO3
M(A) = M(CO2) = 44,01g/mol

[B] = KH
1dKH = 2 · (10mg CaO/L / M(CaO)) = 2 · 10mg/L / 56,08g/mol = 0,357mmol/L

pKA = 6,35

A = M(A) · [B] · 10^(pKA – pH)
CO2 = 44,01g/mol · 0,357mmol/L · dKH · 10^(6,35 – pH)
CO2 = 15,7mg/L · dKH · 10^(6,35 – pH)

Chucks equation
CO2 (in PPM) = 3 * KH * 10^( 7-pH )

looking at 10^(7-pH) → 10(6,35-pH) the differense is 100,65 = 4,47 but what about 3 → 15,7mg/L

15,7mg/L / 10^0,65 = 3,5

so the equation will be

CO2 = 3,5 * dKH * 10^(7-pH) ↔ CO2 = 15,7mg/L · dKH · 10^(6,35 – pH)
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post #6 of 7 (permalink) Old 11-15-2004, 03:59 PM
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Quote:
Originally Posted by Jacan

[big snip]

CO2 = 3,5 * dKH * 10^(7-pH) ↔ CO2 = 15,7mg/L · dKH · 10^(6,35 – pH)
Yep, that was the same equation we found a while ago. the one I was referring to as the theoretically correct one. But I guess it's up to Silent Running to decide.
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post #7 of 7 (permalink) Old 11-16-2004, 12:15 AM Thread Starter
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Man, talk about pressure, I don't want to have to decide . Thanks for the info and thanks for the equation jacan (even though I can't follow it ).

Paul

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My old 100 Gallon SeaClear established 11/7/2004. 4 GE Daylight Ultras on IceCap VHO ballasts. 3" Eco-complete substrate. Eheim 2028, pressurized CO2, external reactor. 18 watt Turbo Twist UV.

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