Calculate the concentrations of all species in a 0.690 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8.

Answer 1

.......SO3^2- + H2O ==> HSO3^- + OH^-
I......0.690M...........0.........0
C.......-x..............x.........x
E.....0.690-x...........x..........x

Kb1 for SO3^2- = (Kw/ka2 for H2SO3) = (HSO3^-)(OH^-)/(SO3^2-)
Solve for x = (OH^-) = (HSO3^-)

Then...HSO3^- + OH^- ==> H2SO3 + OH^-
Kb2 = (Kw/Ka1) = (H2SO3)(OH^-)/(HSO3^-)
BUT (HSO3^-) = (OH^-) from the first hydrolysis; therefore,, (H2SO3) = (Kw/Ka1).
All of that gives you H2SO3, HSO3^-, SO3^2-, OH^- and you can use (H^+)(OH^-)= Kw = 1E-14 to calculate H+.