Question

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.

Answer 1

I agree with Na^+ and HSO3^-. I would have used 1.41 M for SO3^2- since 1.41-0.00047 = 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) = 1E-14/1.4E-2 and
1.4E-2 = (H2SO3)(OH^-)/(HSO3^-).
However, in the first hydrolysis (kb1) you said (HSO3^-) = (OH^-) so Kb2 = (H2SO3) = about 7.1E-13 M.

Answer 2

YOLO

Answer 3

How did you get 0.00047?