Asked by Sam

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibrium equation.

CO2(g) <----> CO2(aq)

The acid dissociation constants listed in most standard reference texts for carbonic acid actually apply to dissolved CO2. K=0.032 M/atm

For a CO2 partial pressure of 4.4×10-4 atm in the atmosphere, what is the pH of water in equilibrium with the atmosphere?

Note: Since carbonic acid is primarily dissolved CO2, the concentration of H2CO3 can be taken as equal to that of dissolved CO2.

My online homework is due tonight and I am hopelessly lost on how to solve this problem. :(
Answered by DrBob222
Try this.
C = pK
C = 4.4E-4*0.032 = about 1.41E-5

H2CO3 ==> H^+ + HCO3^-

k1 = (H^+)(HCO3^-)/(H2CO3)
(H^+)= (HCO3^-) = x
(H2CO3) = 1.41E-5
Solve for x = (H^+) and convert to pH.
Are you plugging this into a data base? If so let me know how it turns out? And show your work when you do.
Answered by Sam
Yes I have online homework which I use to submit my homework on. I calculated H2CO3 and got the same value as yours. then I solved for x and got 2.3 x 10^-6 M. Finally i plugged it into the ph formula to get a ph of 5.64. Thank you so much! It was correct!
Answered by Ashley
for once Dr Bob is right lol. he is usually wrong
Answered by keisha
Dr Bob is almost always right, don't be so salty ashley
Answered by Pauline
Hi,
Where did you get K to equal (H+)(HCO3-)/(H2CO3)?
Answered by Here's Something
General Ka expressions take the form Ka = [H+][A-] / [HA]. In this case A- is the anion HCO3- and HA is the acid H2CO3.
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