Question

Calculate the concentration of bicarbonate ion in a 0.100 M solution of sodium carbonate. K_a1 and K_a2 of carbonic acid are 4.2 x 10^-7 and 4.8 x 10^-11, respectively.

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..........0.10......0.......0
C...........-x.......x.......x
E.........0.10-x.....x.......x

k1 = 4.2x10^-7 = [x] [x] / [0.10-x] {x<<<0.1 so x neglect}

and solve for x = (H⁺) = (HCO3^-)

x = 2.04x10^-4

Then k2 takes over.
...........HCO3^-    ==> H⁺ + CO₃^2-
I..........2.04x10^-4....2.04x10^-4......0
C..........-x.........x.......x
E.....(2.04x10^-4-x)......(2.04x10^-4+ x)......x

so k2 = (H⁺)(HCO₃^-)/(HCO₃^-). If you substitute the numbers you have
k2 = [2.04x10^-4+x] [x] / [2.04x10^-4 -x]

4.8x10^-11 = x {x<< 2.04x10^-4 so x neglect}

k2 = x = (CO₃^2-) = 4.8x10^-11