Question

Estimate the ratio of [HSO₃^-]/[SO₃^2-] at equilibrium in a 1.0 M H₂SO₃ solution. (pK_a1 = 1.81; pK_a2 = 6.91)

Answer 1

The two reactions are:

H₂SO₃ + H₂O H₃O⁺ + HSO₃^-            pKa₁ = 1.81
HSO₃^- + H₂O H₃O⁺ + SO₃^2-             pKa₂ = 6.91

Ka₁ = anitlog (-pKa₁) = antilog (-1.81) = 1.55 10^-2

Ka₂ = antilog (-pKa₂) = antilog (-6.91) = 1.23 10^-7

Construct an ICE chart for the first hydrolysis step.

Molarity . . . . .H₂SO₃ + H₂O H₃O⁺ + HSO₃^-   
Initial . . . . . . ....1.0 .. . . . . . . . . .0 . . . . . 0
Change . . .. . . ..-x . . . . . . . . . . .x . . . . . x
Equilibrium . .    1.0-x . . . . . . . . . x . . . . . x

Ka₁ = [H₃O⁺][HSO₃^-] / [H₂SO₃] = (x)(x) / (1.0-x) = 1.55 x 10^-2

x² = (1.55 x 10^-2)(1.0-x)
x² + 0.0155x - 0.0155 = 0

Solving for x we get,

x = 0.117

[H₂SO₃] = 1.0 - x = 1.0 - 0.117 = 0.883 M
[H₃O⁺] = [HSO₃^-] = x = 0.117 M

Molarity . . . . .HSO₃^- + H₂O <==> H₃O⁺ + SO₃^2-
Initial . . . . . . .0.117 . . . . . . . . . . .0.117. . . .0
Change . . . . . . .-x . . . . . . . . . . . . .x . . . . ..x
Equilibrium . . 0.117-x . . . . . . . . .0.117+x . . x

Ka₂ = [H₃O⁺][SO₃^2-] / [HSO₃^-] = (0.117+x)(x) / (0.117-x) = 1.23 10^-7

Since Ka₂ is so small (10^-7), the x term will be small compared to 0.117 and we can drop it from 0.117-x and 0.117+x.

0.117x / 0.117 = x = 1.23 10^-7 = [SO₃^2-]

So, the ratio [HSO₃^-] / [SO₃^2-] = 0.117 / (1.23 10^-7) = 9.5 10⁵