"pKw = pKa + pKb" formula explanation

David513

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Dec 25, 2010
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  1. Pre-Medical
    Hi guys,

    This may be a silly question, but I'm confused about the formula pKw=pKa+pKb (and I guess by extension Kw=Ka*Kb). My confusion came from an Examkrackers question asking "What is the pKb for HSO3-?" when given the pKa of H2SO3 (1.81) and pKa of HSO3- (6.91). The answer for the pKb of HSO3- is 12.19.

    My confusion arises from why the pKb for HSO3- is not 7.09. I thought...
    pKw=pKa+pKb
    14=6.91+pKb
    7.09=pKb of HSO3-

    What is the above pKb value if not the pKb of HSO3-? I don't understand, conceptually, how it makes sense that you would find the pKb of HSO3- by plugging in the pKa of H2SO3 as required...
    pKw=pKa+pKb
    14=1.81+pKb
    12.19=pKb of HSO3-

    Could someone kindly give me an explanation of how this makes sense? I'm missing the intuition behind this!

    Thank you very much in advance!
     

    NextStepTutor_3

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    Nov 6, 2014
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    1. Pre-Medical
      This is a great question! Most prep books don't properly explain this equation. The proper formula is NOT "pKa (species) + pKb (of same species) = pKw."

      Instead, it's "pKa (species) + pKb (conjugate base of that species) = pKw."

      In other words, if you're considering the pKb of HSO3-, you can't assume that the pKa of HSO3- and the pKb of HSO3- add up to 14. In fact, they almost definitely won't. Instead, you can either do pKa (H2SO3) + pKb (HSO3-) = pKw, OR you can do pKa (HSO3-) + pKb (SO3 2-) = pKw.

      Now, why is this true? I like to think of it this way. A strong acid (low pKa) will have a weak conjugate base (high pKb), and a weak acid (high pKa) will have a comparatively strong conjugate base (low pKb). The lower the pKa of the acid, the higher the pKb of its conjugate base. So, when we make sure to use an acid/conjugate pair in this equation, it makes perfect sense that the sum of the pKa of the acid and the pKb of the conjugate base would always be the same quantity (pKw).

      Also, it's a good habit to remember that pKb should always refer to the "base," or deprotonated version, of a species. With regard to the MCAT, you fell into a very tricky trap, because this question asked about the conjugate base of a polyprotic acid (HSO3-). It's really easy to think of HSO3- as either an acid or a base, so we can easily make the mistake of plugging it in as both species in the equation. Imagine if a different question had referenced H2SO4. Would we have plugged in pKa (H2SO4) + pKb (H2SO4) = pKw? Maybe, but it's much less likely, because we don't think of H2SO4 as a base at all, so it seems illogical to use its pKb. It's much more likely that we would have naturally done pKa (H2SO4) + pKb (HSO3-) = pKw.

      Anyway, hope that helps! Good luck :)
       
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      aldol16

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        This had already been more than adequately explained, but one thing most pre-meds need to understand is that equations are there to support concepts, not the other way around. In other words, you should think about what you're doing before you start putting numbers in.

        So, concisely, think about the process that's going on. It's asking you about Kb, so the reaction inherently yields hydroxide. The reaction in question then is HSO3- + H20 ----> H2SO3 + OH-. Kb thus equals [H2SO3][OH]/[HSO3]. Alright, now think about what you're given. Ka of H2SO3 refers to [HSO3-][H+]/[H2SO3] since it's an acid dissociation constant and likewise, Ka of HSO3- refers to [SO3][H+]/[HSO3-]. Now, you also know that [H+][OH-] = Kw.

        Think: is there a way I can manipulate these equilibria to eliminate all variables but one? Sure. Kb*Ka = [H2SO3][OH][HSO3-][H+]/[HSO3][H2SO3] = [OH-][H+] = Kw. Note which Ka I used here. It's the Ka of H2SO3. From here, it's basic manipulation of logs to realize that pKb of HSO3- plus pKa of H2SO3 equals Kw.

        So in sum, the important thing here is to not just go into it blindly applying equations but to think about why the equations work the way they do. That eliminates the need to memorize a lot of useless information and allows you to develop a methodical way to approach these problems.
         
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