Question

For 6.0*10^-2 M H2CO3, a weak diprotic acid, calculate the following values. Use ionization constants of H2CO3: Ka1=4.4*10^-7, Ka2=4.7*10^-11, as necessary.
PART A: [H3O^+] M
PART B: [HCO3^-] M
PART C: [CO3^2-] M

Answer 1

A.
........H2CO3 + H2O ==> H3O^+ + HCO3^-
I.......0.06..............0......0
C........-x...............x.......x
E......0.06-x..............x......x

Substitute the values from the ICE chart into ka1 and solve for x = (H3O^+) = (HCO3^-)

I've answered part B in part A answer.
C.
......HCO3^- + H2O ==> H3O^+ + CO3^2-
Note ka2 = (H3O^+)(CO3^2-)/(HCO3^-)
Write that out on a sheet of paper so you can see it. From part A you know (H3O^+) = (HCO3^-). Therefore, )H3O^+) in the numerator cancels (HCO3^-) in the denominator and (CO3^2-) = ka2.