Calculating the pH of a polyprotic acid?

H2SO3 -> HSO3- + H+ --> SO3-- + H+

ka1 = 1.3E-2

ka2 = 6.3E-8

If you have a 0.050 M H2SO3, 0.050 M HSO3-, and 0.050 M SO3-- in solution:

H2SO3 -> HSO3- + H+

[H2SO3] = 0.050-X

[HSO3-] = 0.050+x

ka = [HSO3-][H+]/[H2SO3]

1.3E-2 = (0.050+x)x / (0.050-x)

x = [H+] = 0.00902 M

[H2SO3] = 0.050 - 0.00902 = 0.04098 M

[HSO3-] = 0.050 + 0.00902 = 0.05902 M

HSO3- H+ + SO3-

[HSO3-] = 0.05902 - x

[SO3--] = 0.050 + x

[H+] = 0.00902 + x

ka = [H+][SO3-] / [HSO3-]

6.3E-8 = (0.00902+x)(0.050+x)/(0.05902-x)

x = We will just assume that x is so small, that it is has a negligible effect on the concentration and pH.

[SO3-] = 0.050 M

[H+] = 0.00902 M

[HSO3-] = 0.05902 M

[H2SO3] = 0.04098 M

pH = -log[H+] = -log[0.00902] = 2.04