Question

A 0.942 M sample of carbonic acid, H2CO3, has a measured hydronium ion concentration of 6.36 ´ 10–4 M. Calculate the acid-ionization constant of carbonic acid. The equilibrium equation is:
H2CO3(aq) + H2O(l) <--><--> H3O+(aq) + HCO-3(aq)

Answer 1

Ka = (H3O^+)(HCO3^-)(H2CO3)

Set up an ICE chart, substitute into Ka expression, and solve for Ka.
Post your work if you get stuck.

Answer 2

what is an ICE chart?

Answer 3

ICE.
I = initial
C = change
E = equilibrium.

H2CO3 + H2O ==> H3O^+ + HCO3^-

Ka = (H3O^+)(HCO3^-)/(H2CO3).

initial:
H2CO3 = 0.942 M
(H3O^+) = 0
(HCO3^-) = 0

equilibrium:
(H3O^+) = 6.36 x 10^-4 M
(HCO3^-) = 6.36 x 10^-4 M
(H2CO3) = 0.942-6.36 x 10^-4 = 0.942

This solution assumes Ka is the only ionization constant for H2CO3; in reality, H2CO3 has a k1 and a k2 but we are ignoring k2. The error is small.