- #1

skaai

- 16

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OK folks,

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...

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 25°C. We let the carbon dioxide come to equilibrium with the water so the maximum dissolves. What would the resultant pH be?

CO2 + 2H

Convert pKa to Ka: 10

so the lowest the pH can go in 1 liter of water in equilibrium with 100% CO

It sounds like a reasonable number, or did I totally mess something up somewhere?

thanks so much for any help on this!

*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 25°C. 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 25ºC 1atm is 1.45 gram/liter
- we have 1 liter of water

**What is the reaction?**CO2 + 2H

_{2}O → H_{2}CO_{3}+ H_{2}O → HCO_{3}^{-}+ H_{3}O^{+}- 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. CO

_{2}[itex]\frac{1 mol.}{44.01 g.}[/itex]= 0.032947 mol. CO_{2}**Calculate using a RICE table**(see the attached image table):Convert pKa to Ka: 10

^{-pKa}= Ka → 10^{-6.367}= KaKa (10

can we apply the "rule of 500?":^{-6.367}) = [itex]\frac{[A-][H+]}{[HA]}[/itex] = [itex]\frac{(x)(x)}{(0.0329-x)}[/itex]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{x²}{0.0329}[/itex]x²=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% CO

_{2}at 25ºC 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!