Question

Calculate the concentrations of all species in a 0.560 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(aq) -------------> 2Na^+ (aq) + SO3^2- (aq)

0.56M                              2*0.56M       0.56M

[Na^+]    =     2*0.56M     = 1.12M

                SO3^2- (aq) + H2O(l) ----------------> HSO3^- (aq) + OH^-

I                0.56                                                    0                       0

C                  -x                                                    +x                      +x

E                0.56-x                                                  +x                   +x

                      Kb2   = Kw/Ka2

                                = 1*10^-14/6.3*10^-8   = 1.6*10^-7

                     Kb2   = [HSO3^-][OH^-]/[SO3^2-]

                    1.6*10^-7   = x*x/0.56-x

                     1.6*10^-7*(0.56-x) = x^2

                        x   = 0.0003

                    [HSO3^-]   =x    = 0.0003M

                     [OH^-]    =x       = 0.0003M

                      [SO3^2-]   = 0.56-x = 0.56-0.0003   = 0.56M

                      HSO3^- (aq) + H2O (l) ---------------> H2SO3(aq)   + OH^-(aq)

                  kb1   = kw/ka1

                             = 1*10^-14/1.4*10^-2    = 7.14*10^-13

                         Kb1    = [H2SO3][OH^-]/[HSO3^-]

                             7.14*10^-13 = [H2SO3]*0.0003/0.0003

                                [H2SO3]    = 7.14*10^-13M

                               [OH^-]   = 0.0003M

                              [H3O^+]   = Kw/[OH^-]

                                                 = 1*10^-14/0.0003 = 3.34*10^-11M