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Calculating the pH of CO2 dissolved in 1l of water

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skaai
#1
Feb17-14, 07:51 AM
P: 16
OK folks,
first of all, let me thank those of you who have helped me with previous questions.... perhaps this is why I keep coming back!

This time, I'm just asking those of you who are better than I am at this if I did this right and thus, got the correct result. I think I did, but since this is not a homework problem, I have no way of checking my answer...

I set up my problem like this (see the attached diagram):

Imagine our system has a beaker with 1 liter of water and is enclosed in a constant-pressure (of 1 atm.) atmosphere of 100% carbon dioxide at 25C. We let the carbon dioxide come to equilibrium with the water so the maximum dissolves. What would the resultant pH be?

What do we know?
  • the pKa of carbonic acid is 6.367
  • the solubility of carbon dioxide in water at 25C 1atm is 1.45 gram/liter
  • we have 1 liter of water
What is the reaction?
CO2 + 2H2O → H2CO3 + H2O → HCO3- + H3O+
  • the number of moles of carbon dioxide that dissolve in water at equilibrium will equal the number of moles of carbonic acid in solution.
1.45 g. CO2[itex]\frac{1 mol.}{44.01 g.}[/itex]= 0.032947 mol. CO2
Calculate using a RICE table (see the attached image table):
Convert pKa to Ka: 10-pKa = Ka → 10-6.367 = Ka
Ka (10-6.367) = [itex]\frac{[A-][H+]}{[HA]}[/itex] = [itex]\frac{(x)(x)}{(0.0329-x)}[/itex]
can we apply the "rule of 500?":
Rule of 500: if [itex]\frac{[HA]}{Ka}[/itex]>500, ignore any nonzero changes in "x"
[itex]\frac{0.0329}{1x E-6.367}[/itex]=76,594 >> 500 so yes.
simplify Ka:
10-6.367=[itex]\frac{x}{0.0329}[/itex]
x=10-6.367(0.0329)=1.41317x10-8
x=√1.41317x10-8=1.18877x10-4=[H+]
pH = -log[H+]=-log(1.18877x10-4)=3.9249≈3.92
so the lowest the pH can go in 1 liter of water in equilibrium with 100% CO2 at 25C and 1 atmosphere pressure is 3.92.

It sounds like a reasonable number, or did I totally mess something up somewhere?
thanks so much for any help on this!
Attached Thumbnails
CO2 and Water.jpg   H2CO3 Dissociation RICE Table.png  
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Borek
#2
Feb17-14, 08:57 AM
Admin
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P: 23,574
Looks OK to me.

Note that actually we don't know what is the concentration of carbonic acid in water - we can measure solubility and pH, but that allows only to calculate equilibrium constant for the overall reaction

H2O + CO2 <-> H+ + HCO3-

but not for two separate ones:

H2O + CO2 <-> H2CO3

H2CO3 <-> H+ + HCO3-

We often do assume all carbon dioxide dissolved is present in the form of carbonic acid, that yields commonly known Ka1 value. As it yields a correct calculation results we simply don't care about reality.
skaai
#3
Feb18-14, 06:24 AM
P: 16
I see your point about carbonic acid... it's so mysterious, but all we can do is like with dark matter... infer its there from the effects...

thanks so much for the once-over!


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