Question

Calculate the concentrations of all species in a 0.540 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8.

Answer 1

[Na2SO3] = 0.540 M

Na2SO3 dissociates in aqueous solution as:

Na2SO3           2Na+   +      SO3^2-

Hence,

[Na+] = 2 * 0.540

[Na+] = 1.08 M  

[SO3^2-] = 0.540 M    

Kb for So3^2- = Kw / Ka2

= 10^-14 / 6.3*10^-8

= 1.59*10^-7

Now,

                  SO3^-2     +     H2O              HSO3-     +     OH-
Initial            0.54                                            0                   0
Change         -x                                              +x                 +x
Final         0.54 - x                                           x                   x

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

1.59*10^-7 = x*x / (0.54 - x)

x = 2.92*10^-4

[HSO3-] = 2.92*10^-4 M

[OH-] = 2.92*10^-4 M

[SO3^2-] = 0.54 - 2.92*10^-4

[SO3^2-] = 0.540 M  

Kb of HSO3- = Kw / Ka1

= 10^-14 / 1.4*10^-2

= 7.14*10^-13

Again,

                      HSO3-     +        H2O                     H2SO3        +        OH-
Initial           2.92*10^-4                                                 0                         0
Change            -x                                                          +x                       +x
Final        2.92*10^-4 - x                                                x                         x

Kb = [H2SO3][OH-] / [HSO3-]

7.14*10^-13 = x*x / (2.92*10^-4 - x)

x = 1.44*10^-8

[H2SO3] = 1.44*10^-8 M

[OH-] = 1.44*10^-8 M

[HSO3-] = 2.92*10^-4 - 1.44*10^-8

[HSO3-] = 2.92*10^-4 M

[OH-] = 2.92*10^-4 + 1.44*10^-8

[OH-] = 2.92*10^-4 M

[H+] = 10^-14 / [OH-]

= 10^-14 /2.92*10^-4

[H+] = 3.42*10^-11 M