Question

Calculate the concentrations of all species in a 0.580 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4

Answer 1

Na₂SO₃ + H₂O     2Na⁺ + HSO₃^- + OH^-

1 mole of Na₂SO₃ give 2 moles of Na⁺ ions and 1 mole of HSO₃^- & OH^- each.

Concentration of Na₂SO₃ = 0.580M

[Na⁺] = 2*0.580

= 1.16M

[HSO₃^- ] = 0.580M

[OH^-] = 0.580M

Now,

HSO₃^-   H⁺ + SO₃^2-    .............................Ka = 6.3X10^-8

Let,   [H⁺] = [SO₃^2-] =x

Ka = [H⁺][SO₃^2-] / [HSO₃^-]

6.3X10^-8 = x²/0.580

x² = 3.654X10^-8

x = 1.91X10^-4

  [H⁺] = [SO₃^2-] = 1.91X10^-4 M

H₂SO₃   2H⁺   + SO₃^2-

For H₂SO₃ , Ka = 1.4X10^-2

Ka = [H⁺]²[SO₃^2-] / [H₂SO₃]

1.4x10^-2 = (1.91X10^-4 )²(1.91X10^-4 )/ [H₂SO₃]

1.4x10^-2 = 6.96X10^-12 / [H₂SO₃]

[H₂SO₃] =  6.96X10^-12 / 1.4x10^-2

[H₂SO₃] = 4.97X10^-10 M