For a solution of 3.3 M H2CO3 (Ka = 4.4 × 10-7), calculate:

Question

(a) [H+] ____?______M

(b) pH _____??______

(c) percent ionization ______?______%

Answer 1

I don't think it wise to write the Ka for H2cO3 as 4.4E-7 because it has TWO ionization constants. It is correct to write k1 = 4.4E-7.
...........H2CO3 ==> H^+ + HCO3^-
I.........3.3M.......0.......0
C.........-x.........x.......x
E.........3.3-x.......x......x

k1 = 4.4E-7 = (H^+)(HCO3^-)/(H2CO3)
Substitute into k1 exprsession and solve for x = (H^+) and convert that to pH.
% ion = [(H^+)/(H2CO3)]*100 = ?