the Ka for sulfurous acid is?

Question

The pH of a 0.401 mol/L solution of sulfurous acid is 1.16 (used as an example, your value may be different). Based on this evidence, the Ka for sulfurous acid is a.bcd?

Answer 1

Actually, sulfurous acid (H₂SO₃) probably does not exist in aqueous solution, and if it did, it would have two different Ka values. One for the first dissociation of H₂SO₃ ==> H⁺ + HSO₃^- and a second dissociation of HSO₃^- ==> H⁺ + SO₃^2-.

Assuming one only considers the first dissociation (as it is much greater and contributes to the pH significantly more than the second), we have the following:

H₂SO₃ ==> H⁺ + HSO₃^- and Ka = [H⁺][SO3^-]/[H₂SO₃]

From pH = 1.16, we can find [H⁺] = 1x10^-1.16 = 6.9x10^-2 M

Ka = (6.9x10^-2)(6.9x10^-2)/0.401 - 6.9x10^-2

Ka = 4.76x10^-3/0.332

Ka = 1.4x10^-2