PKa of fatty acids

  1. Aug 3, 2011 #1
    Can someone help me understand this statement:

    The pKa of most fatty acids is around 4.5, so most fatty acids exist in their anion form in the cellular environment.

    I know what a pKa is... it is -log(Ka), where Ka is the equilibrium constant of the reaction. In this case the Ka would be 3 x 10^(-5).

    Also I understand that the larger the Ka, the stronger the acid strength (the more it would dissociate).

    How do we know that a Ka of 3 x 10-5 is a large Ka?
  2. jcsd
  3. Aug 13, 2011 #2


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    You can see from the Henderson-Hasslebach equation that if the pH of the solution is equal to the pKa, exactly half of the fatty acid must be deprotonated (-log(1)=0). Work it out for yourself. What is the physiological pH? How many orders of magnitude more basic is it?
  4. Aug 13, 2011 #3
    The same way you know that a person is tall or short or fat or skinny. You compare to other known things.

    Its actually not that big of a number if you compare carboxylic acids to HCl or HBr for example which have Ka's ~106+.

    Its a pretty strong acid if you compare it to ammonia or methane for instance (with pKa's of ~39+).

    If you want the Biochemistry to make sense you should do what chemisttree suggested. Sit down with a pen, paper and table of pKa values and use the Henderson-Hasselbach equation to see what the behavior of the acid/base pair is like at physiological pH's (~7).

    Or you can think of it qualitatively as such; if the pH of solution > the pKa of the acid the species will tend to exist in its anionic/deprotonated form. If the pH of solution if < pKa of acid, then the protonated form will predominate. At pH=pKa the protonated=deprotonated. The extend to which the acid will dissociate can be calculated using the H-H equation but you can get a rough idea that at pH=10 there will be much more carboxylate than carboxylic acid present in solution.
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