pKa of fatty acids


by gkangelexa
Tags: acids, fatty
gkangelexa
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#1
Aug3-11, 03:56 PM
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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?
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chemisttree
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Aug13-11, 11:50 AM
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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?
Yanick
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#3
Aug13-11, 02:53 PM
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Quote Quote by gkangelexa View Post
How do we know that a Ka of 3 x 10-5 is a large Ka?
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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