Jarred Awesome
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Calculating dissolved Co2
in Organic Chemistry
Posted
Hello Everyone,
I’m by no means a chemist, but I have been trying to learn a little bit about it, for a project I’m working on.
I don’t know where else to ask, so I figured this forum might be a good place.
I have an aquarium, and I am trying to calculate how much co2 is dissolved in the tank.
If I understand properly, I can permanently put an upside down cup in the water, and use Henry’s Law to figure this out.
Attached you can find a drawing of my set up.
I can also use Antoine’s equation compensate for temperature
Here is what I came up with, but I don’t have enough knowledge to know if I’m doing this properly can someone see if I am doing this properly?
I took the temp of water, the applied Antoine Equation to get the Vapour Pressure in the cup. (Not sure if that’s what I am supposed to do)
I then took the PPM measured in the cup, and the result from the Antoines equation And applied it to this: PA=(ppm/PV)x10e6, in order to get the partial pressure.
I then applied the result into Henry’s law to get the result in M/L,
Then multiplied it by 44010 to get the readily as ppm
Here is the raw math.
co2 in cup = 195PPM
Water temp in aquarium = 27 C
Antoine Equation(with constants from water put in):
10^(a-b/(T+c))
10^(7.94917-1657.462/(27+227.02)
=19.887130064 mmHg
Ppm to Partial Pressure
Partial Pressure = PPM/10^6*Vapor Pressure
(195/10^6)*19.887130064
=0.039687554121637006 mmHg
Then Henry’s law
kh = 29.95
c = pa/kh
ppm = c*44010
Dissolved gas = Partial Pressure / Henry's constant
039687554121637006 / 29.95
=000129907122206828293 M/L
Then converting m/L into ppm
mmHg * Mollar Mass of given Gas
000129907122206828293 * 44010
Rounded that gives me 5.71ppm if co2 in the aquarium.
My hope is someone can tell me if my my math and implantation is correct?