Question

Calculate the concentrations of all species in a 0.390 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8.

Species:
Na+
SO3 2-
HSO3 -
H2SO3
OH-
H+

Thanks! The only thing I seem to be getting correct is the [SO3 2-]=.390

any help would be great, thanks!

Answer 1

Na2SO3 ==> 2Na^+ + SO3^2^-
0.390....2*0.390...0.390

Na+ ....see above
SO3 2-.....see above

Then hydrolysis of the sulfite ion
.....SO3^2- HOH ==> HSO3^- + OH^-
I...0.390.............0.......0
C....-x...............x.......x
E..0.390-x............x.......x

Kb for sulfite = (Kw/k2 for H2SO3) = (x)(x)/((0.390-x). Solve for x = (OH^-) = (HSO3^-).

HSO3 -..see above.


The HSO3^- can hydrolyze also.
........HSO3^- +HOH ==> H2SO3 + OH^-
I.......etc just like the sulfite to bisulfite. Kb for HSO3^- = Kw/k1 for H2SO3. Then Substitute and solve for x = (OH^-) = H2SO3.

H2SO3....from the hydrolysis just above.

OH-. The OH^^- is the sum of the OH from the first hydrolysis of sulife, the second hydrolysis of HSO3^- and the OH from pure water. One or more of these may be so small it(they) can be ignored.
H+ -- After you know OH, then (H^+) = Kw/(OH^-)

Answer 2

Okay, thank you! I will try this method!

Answer 3

I got it! I was using the incorrect Kb for each of the reactions! Thanks so much!!!