Sulfurous acid (H2SO3 ) is a diprotic acid with Ka1=1.39×10−2 and Ka2=6.73×10−8 . Determine the pH of a 0.180 M sulfurous acid (H2SO3 ) solution.

Answer 1

pKa1 = 1.86

pKa2 = 7.17

These 2 values differ by more than 6 so the contribution of the 2nd dissociation to the total [+] will be about 1 millionth of the first and so can be neglected.

Ka1 = [H+][HSO₃^-]/[H₂SO₃]

1.39x10^-2 = (x)(x)/0.180 - x

x² = 0.00025 - 0.00139x

Use quadratic to solve for x. That will be the [H⁺]. Then take -log [H+] to get pH