I understand that actual pKa of amino acid side chains is greatly influenced by the surrounding environment. I am trying to deeply understand the equilibrium between protonated and deprotonated form of amino acid side chain by taking Lys as an example. Lets imagine a protein having solvent exposed Lys residue. At physiological pH, theoretically this Lys residue on every protein is in largely protonated form (ofcourse some are deprotonated). Now say with some reagent I modify few of this particular Lys residue (using NHS ester etc.). Now does the remaining unmodified Lys residues undergo protonation/deprotonation to nullify the effect caused by the chemical modification and maintain the state of such equilibrium? Also, I am curious is it always impossible to achieve 100 % modification of this particular Lys residue due to the existence of equilibrium state?
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$\begingroup$ I want you to imagine the protons pressing on the radicals. When you increase the protons in solution, the pressure becomes greater. Some radicals will pick up protons before others can. Water also comes into play pulling the protons from the radicals so you have this sort of back and forth at the very small scale. Equilibrium is when these effects basically cancel out, the radicals lose some protons to water which they gain back the next moment. The system sensibly tends towards equilibrium if you think about it. Now try thinking about what happens if I decrease the amount of protons. $\endgroup$– user7777777Aug 1 at 22:59
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