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Chemistry

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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.

  • Chemistry - ,

    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.

  • Chemistry - ,

    YOLO

  • Chemistry - ,

    How did you get 0.00047?

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