Chemistry
posted by Chris on .
Calculate the concentrations of all species in a 1.41 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8.
Na^+ = 2.82M
SO3^2 = 1.40M
HSO3^ = 0.00047M
H2SO3 =
OH^ =
H^+ =
I have been able to calculate the first three, but do not know where to go from there to find the last three. Any help would be very much appreciated.

I agree with Na^+ and HSO3^. I would have used 1.41 M for SO3^2 since 1.410.00047 = essentially 1.41 M.
You must have written this equation to obtain the 0.00047.
SO3^2 + HOH ==> HSO3^ + OH^
So (HSO3^) = 0.00047 which makes OH^ = 0.00047 M.
For H^+, you know OH and
(H^+)(OH^) = Kw which gives you H^+.
Finally, for H2SO3, you have the second hydrolysis of HSO3^ as
HSO3^ + HOH ==> H2SO3 + OH^
Kb2 = (Kw/k2) = 1E14/1.4E2 and
1.4E2 = (H2SO3)(OH^)/(HSO3^).
However, in the first hydrolysis (kb1) you said (HSO3^) = (OH^) so Kb2 = (H2SO3) = about 7.1E13 M. 
YOLO

How did you get 0.00047?