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Topic: Calculating the fraction of protonated side chain using Henderson-Hasselbalch  (Read 19787 times)

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Offline spruce_moose

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This was a problem from class and I'm having some trouble understanding how the fraction at the end of the problem is derived.

Ex.
Calculate the fraction of histidine that has its imidazole side chain protonated at pH 7.3. The pKa values for histidine are pK1=1.82, pK2=6.00(imidazole), and pK3=9.17.

pH=pKa + log[base/acid]
7.3 = 6.00 + log[base/acid]
1.3 = log[base/acid]
antilog(1.3)=[base/acid] = 20 = 2.0 X 10^1

And here is where I don't know what to do. The professor gave the ratio as 1 part protonated form in a total of 21 total parts histidine.
Where does that ratio come from?

Any assistance is much appreciated.
« Last Edit: August 28, 2009, 02:06:06 PM by spruce_moose »

Offline renge ishyo

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You got the answer correct; the ratio of [base]/[acid] is 20/1, because any number can be written as that number over 1. So you have 1 part acid and 20 parts base. Add them together and you get a total of 21 parts.

Hope this helped.

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