Chemistry Ka1 and Ka2 question

Question

What is the pH of a 0.15-M solution of H2SO3?

Ka1=0.017, Ka2=6.4×10−8

Completely lost on this question. Don't understand how to solve it.

Answer 1

The easy way to solve it is to ignore Ka2 since it is several orders of magnitude (>5) smaller than Ka1 and so it will not contribute to the generation of H+ ions. The more difficult way is to include it as explained below.

Ignoring Ka2:

H₂SO₃ ==> H⁺ + HSO₃^-

Ka = 0.017 = [H⁺][HSO₃^-]/[H₂SO₃]

0.017 = (x)(x)/0.15 - x

x² = 0.00255 - 0.017x

x² + 0.017x - 0.00255 = 0

x = 0.0497 M = [H⁺]

pH = -log 0.0497

pH = 1.3

If you wanted to include the second dissociation, you would have the following:

HSO₃^- ===> H⁺ + SO₃^2-

0.0497.............0...........0.......Initial

-x...................+x...........+x........Change

0.0497-x.........x.............x........Equilibrium

Ka2 = 6.4x10^-8 =(x)(x)/0.0497-x

Solve for x which will be [H⁺] form the second ionization. Add this to [H⁺] from first ionization (0.0497) and take negative log. The answer will be as close to 1.3 as anything within experimental error. But that's the way to do it.