[tpowell]

A triprotic acid, H3A, has three ionization constants of Ka1 = 5.47 x10-4, Ka2 = 6.69 x10-7, Ka3 = 7.46 x10-12 for three steps, respectively. The following reaction represent the reactions that would take place during titration of this acid with a strong base. Using Kw and Ka values above, determine the equilibrium constant following three steps.


H3A    +    OH-   =    H2A-   +    H2O   :    K' =    
H2A-    +    OH-   =    HA2-   +    H2O   :    K'' =    
HA2-    +    OH-   =    A3-   +    H2O   :    K''' =    

I don't have the first clue as to how I would set this up, and my professor's instructions were extremely vague. Any ideas? The sooner the better! Thanks~

[Vidya]

write equation for K1

H3A     =    H2A-   +   H+


K1 = [H2A-][H+]/[H3A]


H2O      =   OH-   + H+
Kw  = [OH-][H+]
H3A    +    OH-   =    H2A-   +    H2O   :    K' =  [H2A-]/[H3A] [OH-]
k ' =  K1/Kw
similarly work for others