We want to calculate the concentrations of all species in a 0.36 M Na2SO3 (sodium sulfite) solution.

Question

We want to calculate the concentrations of all species in a 0.36 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1=1.4×10−2 and Ka2=6.3×10−8. Let's start by first calculating the concentration of the spectator ion.

Calculate the concentrations of all species in a 0.870 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1=1.4×10^−2 and Kb2=6.3×10^−8.

Answer 1

For most calculations of this sort, because the Ka2 for H2SO3 (sulfurous acid) is more than 1000 times less than the Ka1, it can be ignored and considered to contribute a non significant amount of ionization. The following calculation assumes just that.

0.36 M Na₂SO₃ ==> 2Na⁺ + SO₃^2-

[Na+] = 2 x 0.36 M = 0.72 M

[SO₃^2-] = 0.36 M

Then, SO₃^2- hydrolyses as follows:

SO₃^2- + H₂O <==> H₂SO₃ and

H₂SO₃ ==> to H⁺ + HSO₃^- Ka1 = 1.4x10^-2

We can find [H⁺], [HSO₃^-] and [H₂SO₃] as follows:

1.4x10^-2 = (x)(x)/0.36 - x

x² = 7.06x10^-3 - 1.4x10^-2x

x = 0.077 M = [H⁺] = [HSO₃^-]

[H2SO3] = 0.36 M - 0.077 M = 0.28 M

Summary (ignoring 2nd ionization of H₂SO3 which would be HSO₃^- ==> H⁺ + SO₃^2-), we have..

[Na⁺] = 0.72 M

[H⁺] = 0.077 M

[HSO₃^-] = 0.077 M

[H₂SO₃] = 0.28 M

Do the other one the same way, except use Ka2 (not Kb2) as 6.3x10^-8. I'm guessing that's a typo).