Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibrium equation
CO2(g) <-> CO2(aq) K=0.032
The acid dissociation constants listed in most standard reference texts for carbonic acid actually apply to dissolved CO2. For a CO2 partial pressure of 1.9x10^-4 atm in the atmosphere, what is the pH of water in equilibrium with the atmosphere?
What I did:
K=[CO2(aq)]/[CO2(g)]
0.032 = [x]/(1.9x10^-4)
x = 6.08x10^-6
-log(6.08x10^-6) = 5.22 = pH
It says my answer is wrong, and this is the hint:
This is the negative logarithm of the [H2CO3]. Since H2CO3 is a weak acid, [H2CO3]=/=[H+]. Use the acid dissociation constant of H2CO3 to determine the [H+]. Then use the [H+] to calculate the pH.
Please help!
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