Question

Calculate the concentration of all species (H₂SO₃),(HSO₃),(H₃O⁺),(SO₃^-2),(OH-), pH and pOH in a 1.41M Na₂SO₃ (sodium sulfite) solution. The ionization constant for sulfurous acid are

H₂SO₃+H₂O->HSO₃^-+H₃O⁺   Ka1=1.4x10^-2

HSO₃^-+H₂O->SO₃^-2+H₃O⁺ Ka2=6.3x10^-8

Answer 1

We will have multiple equilibriums... I assume there is only salt and water

SO3^2-(aq) + H2O(l) <=> HSO3^- + OH^-

This is the hydrolysis equilibrium reaction that SO3^2- undergoes. This is called Kh or Kb. Kb is related to Ka2 through the relationship

Kb = Kw/Ka2 = (1.41 x 10^-14)/(6.8 x 10^-8) = 2.07 x 10^-7

Now that we have the value for Kb we can calculate the pH.

Kb = [HSO3^-][OH^-]/[SO3^2-]

Let x = [HSO3^-] = [OH^-] and 1.41 -x = [SO3^2-]

To solve this problem we will need to use the quadratic equation to be sure that using the estimate of [SO3^2-] = 1.41 will not cause an error in the calculation.

2.07 x 10^-7 = [x][x]/[1.41]

x =5.4*10^-4 = [OH^-]

Now from Kw = KaKb

[H^+][OH^-] = 1.00 x 10^-14

OH = 5.4*10^-4
[H^+] = (1.00 x 10^-14)/[OH^-] = (1.00 x 10^-14)/(5.4*10^-4) = 1.85 x 10^-11
pH = -log[1.85 x 10^-11 ] = 10.7

pOH = 14- pH= 14-10.7 = 3.3

Concentrations:

H2SO3 = 5.4*10^-4

HSO3- = 1.85 x 10^-11

H+ =1.85 x 10^-11

SO3-2 = 1.41 - 5.4*10^-4 = 1.41

OH- = 5.4*10^-4