For H2SO3 pK1 = 1.8 and pK2 = 7.2 and thus K1 = 1.58E-2 and K2 = 6.31E-8 For a given pH (get to this in a minute), it can be shown [H+]=10^-pH [OH-] = Kw / [H+] where Kw is the ion product of water (1E-14) [H2SO3] = [H+]^2 / f [HSO3-] = K1 [H+] / f [SO3--] = K1 K2 / f where f = [H+]^2 + K1[H+] = K1K2 Given that H2SO3 is a weak acid, lets assume that the pH is described by K1 or [H+] = sqrt(K1 C) where C is the acid conc (mol/l). This leads to a [H+] = 0.05 and a pH = 1.07 Using the [H+] and data above yields in mol/l [H+] 8.51E-02 [OH-] 1.17E-13 [H2SO3] 3.88E-01 [HSO3-] 7.22E-02 [SO3--] 5.35E-08 Note the above data describes the acid as 15% dissociated and the assumptiom that the pH of the solution is best described by K1 is not necessarily a good assumption. A better way is as follows :- The charge balance is this system is given by [H+] = [OH-] + [HSO3-] + 2*[SO3--] Set up all the above relationships in MS Excel and Goal Seek a pH that makes the above charge balance correct. Doing this yields a pH = 1.11 and the following concs (mol/l) [H+] 7.78E-02 [OH-] 1.28E-13 [H2SO3] 3.82E-01 [HSO3-] 7.78E-02 [SO3--] 6.31E-08 Be mindful that in all this calculation, ionic strength impacts (ie activities) have been ignored or activity coefficients have been taken as 1 Solution For H2SO3 pK1 = 1.8 and pK2 = 7.2 and thus K1 = 1.58E-2 and K2 = 6.31E-8 For a given pH (get to this in a minute), it can be shown [H+]=10^-pH [OH-] = Kw / [H+] where Kw is the ion product of water (1E-14) [H2SO3] = [H+]^2 / f [HSO3-] = K1 [H+] / f [SO3--] = K1 K2 / f where f = [H+]^2 + K1[H+] = K1K2 Given that H2SO3 is a weak acid, lets assume that the pH is described by K1 or [H+] = sqrt(K1 C) where C is the acid conc (mol/l). This leads to a [H+] = 0.05 and a pH = 1.07 Using the [H+] and data above yields in mol/l [H+] 8.51E-02 [OH-] 1.17E-13 [H2SO3] 3.88E-01 [HSO3-] 7.22E-02 [SO3--] 5.35E-08 Note the above data describes the acid as 15% dissociated and the assumptiom that the pH of the solution is best described by K1 is not necessarily a good assumption. A better way is as follows :- The charge balance is this system is given by [H+] = [OH-] + [HSO3-] + 2*[SO3--] Set up all the above relationships in MS Excel and Goal Seek a pH that makes the above charge balance correct. Doing this yields a pH = 1.11 and the following concs (mol/l) [H+] 7.78E-02 [OH-] 1.28E-13 [H2SO3] 3.82E-01 [HSO3-] 7.78E-02 [SO3--] 6.31E-08 Be mindful that in all this calculation, ionic strength impacts (ie activities) have been ignored or activity coefficients have been taken as 1.

1. For H2SO3 pK1 = 1.8 and pK2 = 7.2 and thus K1 = 1.58E-2 and K2 = 6.31E-8
For a given pH (get to this in a minute), it can be shown [H+]=10^-pH [OH-] = Kw / [H+]
where Kw is the ion product of water (1E-14) [H2SO3] = [H+]^2 / f [HSO3-] = K1 [H+] / f
[SO3--] = K1 K2 / f where f = [H+]^2 + K1[H+] = K1K2 Given that H2SO3 is a weak acid, lets
assume that the pH is described by K1 or [H+] = sqrt(K1 C) where C is the acid conc (mol/l).
This leads to a [H+] = 0.05 and a pH = 1.07 Using the [H+] and data above yields in mol/l [H+]
8.51E-02 [OH-] 1.17E-13 [H2SO3] 3.88E-01 [HSO3-] 7.22E-02 [SO3--] 5.35E-08 Note the
above data describes the acid as 15% dissociated and the assumptiom that the pH of the solution
is best described by K1 is not necessarily a good assumption. A better way is as follows :- The
charge balance is this system is given by [H+] = [OH-] + [HSO3-] + 2*[SO3--] Set up all the
above relationships in MS Excel and Goal Seek a pH that makes the above charge balance
correct. Doing this yields a pH = 1.11 and the following concs (mol/l) [H+] 7.78E-02 [OH-]
1.28E-13 [H2SO3] 3.82E-01 [HSO3-] 7.78E-02 [SO3--] 6.31E-08 Be mindful that in all this
calculation, ionic strength impacts (ie activities) have been ignored or activity coefficients have
been taken as 1
Solution
For H2SO3 pK1 = 1.8 and pK2 = 7.2 and thus K1 = 1.58E-2 and K2 = 6.31E-8
For a given pH (get to this in a minute), it can be shown [H+]=10^-pH [OH-] = Kw / [H+]
where Kw is the ion product of water (1E-14) [H2SO3] = [H+]^2 / f [HSO3-] = K1 [H+] / f
[SO3--] = K1 K2 / f where f = [H+]^2 + K1[H+] = K1K2 Given that H2SO3 is a weak acid, lets
assume that the pH is described by K1 or [H+] = sqrt(K1 C) where C is the acid conc (mol/l).
This leads to a [H+] = 0.05 and a pH = 1.07 Using the [H+] and data above yields in mol/l [H+]
8.51E-02 [OH-] 1.17E-13 [H2SO3] 3.88E-01 [HSO3-] 7.22E-02 [SO3--] 5.35E-08 Note the
above data describes the acid as 15% dissociated and the assumptiom that the pH of the solution
is best described by K1 is not necessarily a good assumption. A better way is as follows :- The
charge balance is this system is given by [H+] = [OH-] + [HSO3-] + 2*[SO3--] Set up all the
above relationships in MS Excel and Goal Seek a pH that makes the above charge balance
correct. Doing this yields a pH = 1.11 and the following concs (mol/l) [H+] 7.78E-02 [OH-]
1.28E-13 [H2SO3] 3.82E-01 [HSO3-] 7.78E-02 [SO3--] 6.31E-08 Be mindful that in all this
calculation, ionic strength impacts (ie activities) have been ignored or activity coefficients have
been taken as 1