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ICE 149: using pKa to find pKb
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Gilbert_5394
#1 Posted : Sunday, June 07, 2020 4:12:53 PM
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For Q149, they found the pkb for HSO3- through the pKa of H2SO3. How come we cant use the given pKa of HSO3- and do pKa+pKb = 14?
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INSTR_Katerina_102
#2 Posted : Sunday, June 07, 2020 5:36:10 PM
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Hi,

The pKa and pKb adding to 14 works for very specific scenarios.

Imagine the case of

H2A --> HA-.--> A^2-

There are two dissociations here. The first can be defined as H2A --> HA-, which has a pKa1 for the forward reaction and a pKb1 which dictates the reverse reaction (HA- --> H2A). These values of pKa1 and pKb1 will add to 14. Note that you need to add the pKa for H2A and the pKb for HA-, because you need the conjugate acid and base.

However, you also have pKa2 for (HA --> A^2-), and pKb2 for the reverse reaction. You can add pKa2 and pKb2 together to get 14 here, where pKa2 is for HA- and pKb2 for A^2-.

Note that because pKa and pKb are defined for very specific processes, you have to use the right one for the situation (ie the conjugate acid base pair). That is, pKb2 + pKa1 =/= 14 and pKb1 + pKa2 =/= 14.

Extending this logic to your example, you use H2SO3's pKa because it is the conjugate acid pair to HSO3-.
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