Na2SO3 ionizes into
2Na+ + SO3-2
SO3^2- + HOH ==> HSO3^- + OH^-
0.45
                         Â
0Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â
initial
-x                           Â
+x        Â
+x        Â
change
0.45-x                       Â
x         Â
x         Â
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
