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
