Calculate the pH of 0.45 M K2CO3

Question

What are the concentrations of HCO3- and H2CO3 in the solution? H2CO3 is a diprotic acid with Ka1 = 4.3 x 10-7 and Ka2 = 5.6 x 10-11.

Answer 1

Looking at the equilibrium of CO3^2 in H2O, you have...

CO3^2-(aq) + H2O(l) <=> HCO3^-(aq) + OH^-(aq)

The CO3^2- acts as a base, so we use Kb in the expression Kb = [HCO3^-][OH^-]/[CO3^2-] to solve for [OH^-]

Kb = Kw/Ka = 1x10^-14/5.6x10^-11 = 1.79x10^-4

1.79x10^-4 = (x)(x)/0.45-x and neglecting x in denominator so as to avoid the quadratic...

x^2 = 8.1x10^-5

x = 9x10^-3 M = [OH-] = [HCO3^-]

pOH = -log 9x10^-3 = 2.05

pH = 14 - pOH = 11.95

The original question asked for the pH.  If you need [HCO3^-]  and [H2CO3], the [HCO3^-] is already calculated above.

For [H2CO3] use H2CO3 ==> H^+ + HCO3^-

Ka = 4.3x10^-7 = [H^+][HCO3^-]/[H2CO3] and solve for [H2CO3]