Oh, since someone bumped this thread, I'm adding some equations I derived:
The equation for percent dissociation of an acid as a function of pH:
Quote:
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%=100*(10-pKa)/(10-pH + 10-pKa)
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The equation for percent dissociation of a base as a function of pH:
Quote:
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%=100*(10pKa)/(10pH+10pKa)
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Remember that pka of a base is fourteen minus it's pkb (aka the pKa of it's conjugate acid).
Keep in mind some acids work by soaking up a OH
- ion, and some bases work by soaking up a H
+ ion, so the equation might relate to association, rather than dissociation.
This equation is useful when trying to figure out to what pH you need to get an aqueous solution of the compound to get it to precipitate or transfer into the nonpolar (this happens when the ionization is minimal). It's also useful when your target amine has both an acid and a base group, allowing you to calculate what the pH needs to be to ensure neither is ionized.
Morphine, for instance, has a tertiary amine with pKb 6.13, and a phenol with pKa 9.85. Plugging the pKa into the percent dissociation of an acid, and (14-pKb) into the ionization of an amine equation, set them equal, and solve for pH. We then see that at pH 8.86, both groups are only 9.3% ionized, meaning that %82.2 of the morphine is freebase and is available for solution in nonpolar solvent (though most nonpolar solvents will not dissolve much morphine).
Now, another set of equations, for those with weak acids and bases, who want to figure out pH as a function of concentration:
The equation for pH of a solution of a weak acid given Ka and concentration:
Quote:
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pH=-log((Ka(4(conc)+Ka))1/2-Ka) - log(2)
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The equation for pH of a solution of a weak base given Kb and concentration:
Quote:
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pH=14+log((Kb(4(conc)+Kb))1/2-Kb) - log(2)
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Oh, and be careful about what "log" you're using. In the above equations it's log base ten, aka: log
10
Also, in the second set of equations I use Ka and Kb, not pKa and pKb.