Calculating Protonation of Lysine at pH 9.5

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leopard
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How can I calculate how many percent of the epsilon amino groups in lysine are protonated at pH 9.5?

The pKa values are 2.2, 9.0 and 10.5. I have calculated the pI: 9.75. The epsilon amino is mostly protonated at this value, and also at pH 9.5 I guess.
 
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From the definition of acid dissociation constant:

[tex]K_a = \frac {[H^+][A^-]} {[HA]}[/tex]

you get

[tex]\frac {[A^-]} {[HA]} = \frac {K_a} {[H^+]}[/tex]
 
Ka = 10.5 gives a really big number for [A-]/[HA]
 
pH = 9.5 --> [H+] = 3.16E-10

[A-]/[HA] = 10.5/3.16E-10 = 3.32E10
 
I think your deduction is true after all if Ka = 10.5.



But the premise is false: Ka = 10.5 is not what is given - rather pKa = 10.5 !
 
leopard said:
Ka = 10.5 gives a really big number for [A-]/[HA]

Do you know what is a difference between Ka and pKa?

Besides, such large numbers happen quite often, pH scale is logarithmic and covers 14 orders of magnitude, not without a reason.
 
epenguin said:
I think your deduction is true after all if Ka = 10.5.



But the premise is false: Ka = 10.5 is not what is given - rather pKa = 10.5 !

Lol, of course. Then I get Ka = 3.1E10, so that [A-]/[HA] = 1E20. An even larger number. How can I use this to find what percentage has been protonated?
 
You will get there faster if you develop a combination of understanding, intuition and rules of thumb, so you will at least have an idea what answer is reasonable and therefore be able to correct your mistakes which then feeds back to better understanding etc.

At the pH equal to the pK the substance is half protonated. Just one pH unit away, the protons have increased or decreased by a factor 10. You wouldn't expect this to have changed [A-] by a factor like 10E20.

In fact one unit below or above the pK [A-]/[HA] has by the equation given by Borek become 10 or 0.1 . Then e.g. A is not exactly 10% or 90% protonated, but it is easy to work out exactly how much.

Another tip, when students come to enzyme kinetics they often do not realize that they are dealing with much the same equations, at least there is considerable overlap, so that a Michaelis saturation curve has exactly the same form as a pH titration curve if you plot v against log .
 
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