Question

Determine the pH of a 0.45M solution of Na₂SO₃.

Answer 1

Na₂SO₃ ionizes into 2Na⁺ + SO3^-2

SO3^2- + HOH ==> HSO3^- + OH^-
0.45                            0            0           initial
-x                             +x          +x          change
0.45-x                         x           x           equlibrium

Kb for SO3^2- = (Kw/K2 for H2SO3) = (HSO3^-)(OH^-)/(SO3^2-)

K2 = 1.0 * 10^-7

(1*10^-14/1*10^-7) = x^2/(0.45-x)

x = 0.000212

[OH-]= 0.00021

pOH = -log(0.00021) = 3.68

pH = 14- pOH = 14- 3.68 = 10.32