Determine the concentration of CO32- ions in a 0.18 M H2CO3 solution. Carbonic acid is a diprotic acid whose Ka1 = 4.3 × 10-7 and Ka2 = 5.6 × 10-11.

Answer 1

(CO3^2-) = k2. The reasoning is this but it's a little long. I wish we had a board and I could talk.

...........H2CO3 ==> H^+ + HCO3^-
I..........0.18......0.......0
C...........-x.......x.......x
E.........0.18-x.....x.......x

k1 = 4.3E-7 = (x)(x)/(0.18-x) and solve for x = (H^+) = (HCO3^-)if you want an answer but that really isn't necessary. That answer is about 2.8E-4M. Then k2 takes over.
...........HCO3^- ==> H^+ + CO3^2-
I..........2.8E-4....2.8E-4...0
C..........-x.........x.......x
E.........2.8E-4-x..2.8E-4+x..x

so k2 = (H^+)(HCO3^-)/(HCO3^-). If you substitute the numbers you have
k2 = (2.8E-4+x)(x)/(2.8E-4-x)
if x is small the k2 = x = (CO3^2-)

As above, however, you really don't need to solve for H^+ and HCO3^-. Your reasoning tells you (H^+) = (HCO3^-) and since k2 = (H^+)(CO3^2-)/(HCO3^-),they cancel each other in the k2 equation and (CO3^2-) = k2

Answer 2

3.06