These are the species: [H2CO3], [HCO−3], [CO2−3], [H3O+], [OH−]
1 Answer

H2CO3 ionizes in two steps:
H2CO3 <—> H+ + HCO3 Ka = 4.3X10^7
HCO3 <—> H+ + CO32 Ka = 5.6X10^11
Using the first Ka:
Ka = [H+][HCO3]/[H2CO3] = 4.3X10^7
[H+] = [HCO3] = x Then,
x^2/0.160 = 4.3X10^7
x = [H+] = [HCO3] = 2.6X10^4 M
A small amount of the HCO3 will ionize by the second equation, but this will be very small relative to the first ionization. So,
5.6X10^11 = [H+][CO32]/[HCO3]
Since from the first ionization, [H+] = [HCO3], and because these will not change significantly because of the second ionization,
[CO32] = 5.6X10^11
Since [H+] = 2.6X10^4, [OH] = 1X10^14 / 2.6X10^4 = 3.8X10^11