Question

Calculate the concentrations of all species in a 1.12 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4� 10�2 and Ka2 = 6.3� 10�8.

NA+ =

SO32- =

HSO3- =

H2SO3 =

OH- =

H+ =

Answer 1

1.12 M Na2SO3

Na2SO3 --------------------> 2 Na+ + SO3-2

[Na+] = 2 x 1.12 = 2.24 M

Ka1 = 1.4 x 10^-2 ,

Ka2 = 6.3 x 10^-8

Kw = 1.0 x 10^-14

Kb1 = Kw / Ka2 = 1.0 x 10^-14 / 6.3 x 10^-8 = 1.59 x10^-7

Kb2 = Kw / Ka1 = 1.0 x 10^-14 / 1.4 x 10^-2 = 7.14 x 10^-13

SO3^-2 + H2O -------------------> HSO3-   + OH-

1.12                                               0               0    ------------------> initial

1.12-x                                            x                x ---------------------> equilibrium

Kb1 = [HSO3-][OH-]/[SO3-2]

Kb1 = x^2 / 1.12-x

1.59 x 10^-7   = x^2 / 1.12-x

x = 4.22 x 10^-4

x = [OH-] = [HSO3-] = 4.22 x 10^-4 M

[OH-] = [HSO3-] = 4.22 x 10^-4 M

[SO3^-2] = 1.10 - x

[SO3^-2]      = 1.12 M

[H+] = Kw / [OH-] = 1.0 x 10^-14 / 4.22 x 10^-4

[H+] = 2.37 x 10^-11 M

H2SO3 = Kb2

[H2SO3 ] = 7.14 x10^-13 M

note : all bolded format are answers .