1.
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)
0.28MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
2*0.28MÂ Â Â Â Â Â Â Â 0.28M
[Na^+]Â Â Â = 2*0.28M = 0.56M
[SO3^2-]Â Â = 0.28M
SO3^2- (aq) + H2O --------> HSO3^- (aq) + OH^- (aq)
IÂ Â Â Â Â Â Â
0.28Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
  0                      0
CÂ Â Â Â Â Â -
x                                      Â
    +x                      +x
EÂ Â Â Â 0.28-
x                                    Â
+x                     +x
kb2Â Â = Kw/Ka2
= 1*10^-14/6.3*10^-8 = 1.58*10^-7
Kb2Â Â Â = [HSO3^-][OH^-]/[SO3^2-]
1.58*10^-7 = x*x/0.28-x

2.
1.58*10^-7*(0.28-x) = x^2
x  = 0.00021
[OH^-]Â Â = x = 0.00021M
[HSO3^-]Â Â = x = 0.00021M
HSO3^- + H2O -------------> H2SO3 + OH^-
Kb1Â Â = Kw/Ka1
= 1*10^-14/1.4*10^-2Â Â = 7.14*10^-13
Kb1Â Â = [H2So3][OH^-]/[HSO3^-]
7.14*10^-13Â Â = [H2So3]* 0.00021/ 0.00021
7.14*10^-13M = [H2So3]