(1) H2CO3 + H2O --> HCO3- + H3O+ [Ka = 4.5E-7]
(2) HCO3- + H2O --> CO32- + H3O+ [Ka = 4.7E-11]
Observe that the two Ka values differ from each other drastically. This will mean that in terms of pH, ONLY THE FIRST IONIZATION WILL AFFECT pH.
(1) H2CO3 + H2O --> HCO3- + H3O+ [Ka = 4.5E-7]
I.....0.165M..................0..........
C......-x......................+x........
E....0.165-x.................x...........
4.5E-7 = x^2/(0.165-x) <-- use the appoximation rule because Ka is very small
2.72E-4 = x
Thus, from equation (1), we see that the [H3O+] = 2.72E-4M, [H2CO3] = 0.165M, and [HCO3-]= 2.72E-4M. Note that THIS step will determine the pH of the polyprotic acid
To calculate the last species, just transfer these values the next equation
(2) HCO3- + H2O --> CO32- + H3O+ [Ka = 4.7E-11]
I....2.72E-4..................0..........
C
E
4.7E-11 = (x*2.72E-4)/2.72E-4
4.7E-11 = x = [CO32-]
To calculate the OH- from H3O+, just find the pH of H3O+, subtract it from 14 to find the pOH, and find the [OH-] from that.
