Question

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 M atm^-1

The acid dissociation constants listed in most standard reference texts for carbonic acid actually apply to dissolved CO2. For a CO2 partial pressure of 8.8×10-4 atm in the atmosphere, what is the pH of water in equilibrium with the atmosphere?

Answer 1

Solution: The given partial pressure = 8.8*10^-4 bar

atmospheric pressure = 0.9869 bar

so moles fraction of CO2 in air =8.8*10^-4/0.9826 =8.95*10^-4

Now for one mole of air so no of moles CO2 =8.95*10^-4

CO2 [g] <--------------> CO2[aq]

c-x x

given K =0.032= x/[c-x] {here c initial concentration of CO2}

so x= Kc/[1+K]= 0.032*8.8*10^-4/[1+0.032]

=2.72 *10^-5 M

This is the initial concentration of CO2 concentration in water

as Ka1 = [concentration of H+ ion]^2/x =4.46*10^-7

  [concentration of H+ ion]^2 =4.46*10^-7*2.72 *10^-5

=1.21*10^-11



gives [H+]=3.48*10^-6 M

so pH = -log 3.48*10^-6

=5.45