Question

What is the pH of a 0.611 M solution of sulfurous acid? (Ka1 = 1.7 x 10-2 and Ka2 = 6.2 x 10-8).

Answer 1

A pertinent step-by-step ionization equation for sulphurous acid is

H₂SO₃ + H₂O   HSO₃^-(aq) + H₃O⁺. ............ Ka1 = 1.7 * 10^-2.

HSO₃^-  + H₂O. SO₃^2-(aq) + H₃O⁺. ............ .Ka2 = 6.2 * 10^-8.

Second ionization seems to be giving negligible protons as inization constant value is so small.

Hence pH of the sulphurous acid solution will be decided on first ionization only.

ICE table for first ionization.

Initial concentration of H₂SO₃ = 0.611 M

Let at equilibrium "X" M H₂SO₃ ionizes hence,

H₂SO₃ + H₂O   HSO₃^-(aq) + H₃O⁺.

Initial conc. 0.611 - 0 0

Change -X - +X +X

At Eqm Conc. (0.611-X) - X X

Ka1 = [HSO₃^-][H₃O⁺] / [H₂SO₃] = 1.7 * 10^-2. = 0.017

At equilibrium we get above expression changed to,

(X)(X) / (0.611-X) = 0.017

X² = (0.611-X) * 0.017

X² = 0.611*0.017 - 0.017X

X² + 0.017X = 0.01

Let us solve this quadratic equation by perfect square method.

Add to both sides Third term = [1/2 * 0.017]² = 0.0001

X² + 0.017X + 0.0001 = 0.01 + 0.0001

(X + 0.0085)² = 0.0101

Taking square of both sides,

X + 0.0085 = + 0.1005 ............ (-ve square root would have given -ve value of X which is not acceptable)

X = 0.1005 - 0.0085

X = 0.092

Hence,

[H₃O+] = X = 0.092

Then

pH = -log[H₃O+] = -log(0.092) = 1.04

pH of 0.611 M Sulphurous acid is 1.04.

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