Part 3 – Put it all together. Calculate the concentrations of all species in a 0.280 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8. Solution Na2So3(aq) -------------> 2Na^+ (aq) + SO3^2- (aq) 0.28M                             2*0.28M        0.28M [Na^+]   = 2*0.28M = 0.56M [SO3^2-]  = 0.28M SO3^2- (aq) + H2O --------> HSO3^- (aq) + OH^- (aq) I       0.28                                        0                      0 C      -x                                           +x                      +x E    0.28-x                                     +x                     +x kb2  = Kw/Ka2 = 1*10^-14/6.3*10^-8 = 1.58*10^-7 Kb2   = [HSO3^-][OH^-]/[SO3^2-] 1.58*10^-7 = x*x/0.28-x 1.58*10^-7*(0.28-x) = x^2 x  = 0.00021 [OH^-]  = x = 0.00021M [HSO3^-]  = x = 0.00021M HSO3^- + H2O -------------> H2SO3 + OH^- Kb1  = Kw/Ka1 = 1*10^-14/1.4*10^-2  = 7.14*10^-13 Kb1  = [H2So3][OH^-]/[HSO3^-] 7.14*10^-13  = [H2So3]* 0.00021/ 0.00021 7.14*10^-13M = [H2So3] .