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

Question

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. :(

Answer 1

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.

Answer 2

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!

Answer 3

for once Dr Bob is right lol. he is usually wrong

Answer 4

Dr Bob is almost always right, don't be so salty ashley

Answer 5

Hi,
Where did you get K to equal (H+)(HCO3-)/(H2CO3)?

Answer 6

General Ka expressions take the form Ka = [H+][A-] / [HA]. In this case A- is the anion HCO3- and HA is the acid H2CO3.