Help with calculation of water pH when doubling pCO2

[JonasRoman]

Randy, can you help me also with this?

At i first glance i thought this question was simple..if i double pCO2 i will if water is in equilibrium also double CO2(aq), and double H2CO3...so far it is only constants ..and so far i am sure
PCO2xK=CO2(aq)...henleys law etc...
and
CO2(aq)-xK---H2CO3.. that K is around 0.13 % if i remember correct (never mind that is not important for my question.)
then next step of course
H2CO3---HCO3+ H

At my first thought i thinks that if i double PCO2 (and also CO2(aq) the as we assume equilibrium), then the H will double i think. And lg(2) is 0.3, so i think that if i double CO2 the ph will decrease with 0.3 unit.
BUT: then i consider it again---even the HCO3 will increase when adding CO2 driving the reaction to the right according to.
H2CO3---HCO3 + H....And:HCO3xH/CO2(aq) = K1
So HCO3 should with CO2 adding increase in same scale as H, for every HCO3 which is forming a H is forming...then i know that if i double CO2(aq), then the product (HCO3xH) doubles....so [H] increase should be √2..thus 1.4. and lg(1.4) is 0.15...so: if i double PCO2, shouldn't the ph decrease "only" with 0.15? (thus not 0.3...)...or??

Jonas Roman

 

[Randy Holmes-Farley Verified Community Expert]

Doubling CO2 in the range of pH of a reef tank will cause a pH change of about 0.3 pH units (but that is just an estimate and applies mostly at tank pH and lower), but in other pH ranges the answer can be different (like above the pKa of carbonate).

I show the equations here. You need to insert the values for the constants and solve for [H+].

http://www.advancedaquarist.com/2002/5/chemistry

Ac = (K1KHPCO2 / [H+]) + 2(K1K2KHPCO2 / [H+]^2)

changing to quadratic form...

Ac[H+]^2 - K1KHPCO2[H+] - 2K1K2KHPCO2 = 0

solving for it we get

[H+] = [K1KHPCO2 +/- [(K1KHPCO2)^2 +/- 8AcK1K2KHPCO2)]^(0.5)]2Ac


and a result in Figure 3 in the article


I just set up a spread sheet and here's what I get as the drop in pH

CO2.............................................pH
87.5 --> 175 ppm.......................8.57 --> 8.36 delta = 0.21
175 ---> 350 ppm....................8.36 --> 8.13 delta = 0.23
350 --> 700 ppm............... 8.13 ---> 7.87 delta = 0.26
700 ---> 1400 ppm..............7.87 --> 7.61 delta = 0.26
1400 --> 2800 ppm ..............7.61 --> 7.32 delta = 0.29
2800 --> 5600 ppm...................7.32 --> 7.03 delta = 0.29

 

[JonasRoman]

Thank you so much Randy i understand and i now remember that i tried to solve this equation to solve out H for a while. My try here is in a picture. I have not checked if was doing the second degree equation correct. (I am surgeon , not chemist or mathematics;-))Nevertheless i understanding the formula you post and the article in advance aquarist is excellent!
Thanks again.

 

[Randy Holmes-Farley Verified Community Expert]

FWIW, the formula for a quadratic always has two answers, but I only showed the "+" one in the calculation, the "-" answer gives the square root of a negative number so it not relevant to a real pH.

 

[JonasRoman]

I understanding

 

[JimWelsh]

Jonas, you may find this helpful: http://cdiac.ornl.gov/oceans/co2rprt.html

 

[JonasRoman]

Randy, i now thanks to this understand what i did wrong. I was right i think that it is the product hco3xh which double then h increase eith squareroot of 2 thus 1.4
But i forgot next sten so the product
Co3xh Will increase with 1.4 and thus the h Will increase there with square robot of 1.4
Total increase of h is then square of 1.4 and square root of 2..that is an increase if h with 2.583
Lg 2.583 is 0.22!!!thus pH decrease with 0.22
So it sema to be correct with your series
Am i right??

 

[JonasRoman]

Randy, i now thanks to this understand what i did wrong. I was right i think that it is the product hco3xh which double then h increase eith squareroot of 2 thus 1.4
But i forgot next sten so the product
Co3xh Will increase with 1.4 and thus the h Will increase there with square robot of 1.4
Total increase of h is then square of 1.4 and square root of 2..that is an increase if h with 2.583
Lg 2.583 is 0.22!!!thus pH decrease with 0.22
So it sema to be correct with your series
Am i right??

 

[Randy Holmes-Farley Verified Community Expert]

I'm not really able to follow your rationale to say if it is right or not.

Adding CO2 to lower pH keeps total alkalinity unchanged, but reduces CO3-- and increases HCO3--:

CO2 + H2O + CO3-- <--> H2CO3 + CO3-- <--> 2HCO3-

FWIW, my equation assumes carbonate alkalinity is unchanged. That is not true (it is total alkalinity that is unchanged, not carbonate alkalinity) and that error can becoem important at very high pH where borate becomes more of an error, but still smallish, and hydroxide becomes a big error when the pH gets really high (with low to no CO2).

 

[JonasRoman]

well, i understand you equations, just tried to do a simple one for estimate..it seems to work as i tried som figures...but i understand the errors in this. I also know the fact that adding CO2 will increase the KH and decrease other alkalinity-ion´s in samme order so the total alkalinity is preserved. Opposite when degassing..KH will be lower and other alkalinitycontributers will rise, so the total alk is same...that i know and you helped me with this a month ago. Thanks again

 

[Randy Holmes-Farley Verified Community Expert]

You're welcome

Happy Reefing.