24.6: Biological Amines and the Henderson-Hasselbalch Equation
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Objectives
- identify the form that amine bases take within living cells.
- use the Henderson‑Hasselbalch equation to calculate the percentage of a base that is protonated in a solution, given the p K a value of the associated ion and the pH of the solution.
- explain why organic chemists write cellular amines in their protonated form and amino acids in their ammonium carboxylate form.
The Henderson-Hasselbalch equation is very useful relating the pKa of a buffered solution to the relative amounts of an acid and its conjugate base. In Section 20-3 , we used the Henderson-Hasselbalch equation to show that under physiological pH, carboxylic acids are almost completely dissociated into their carboxylate ions.
\[\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \left(\frac{\text { concentration of conjugate base }}{\text { concentration of weak acid }}\right) \tag{Henderson-Hasselbalch equation}\]
So, what does the side chain of a lysine amino acid residue look like if it is on the surface of a protein in an aqueous solution buffered pH 7.0? Is it protonated or deprotonated? The values in the Henderson-Hasselbalch can be used for an amine with the ammonium salt written as, HA = RNH 3 + , and the amine as being, A - = RNH 2 . With an approximate pK a of 10.8 for the protonated amine HA, it should be >99% protonated, in the positively-charged, ammonium form:
\[\begin{align*}7.0 &= 10.8 + \log \left(\dfrac{[\ce{RNH2}]}{[\ce{RNH3^{+}}]}\right) \\[4pt] \dfrac{[\ce{RNH2}]}{[\ce{RNH3^{+}}]} &= 1.6 \times 10^{-4}\end{align*}\]
So, \(\ce{[RNH3^{+}]} \gg \ce{[RNH2]}\) at this pH. Consequently, in an aqueous solution buffered at pH 7, carboxylic acid groups can be expected to be essentially 100% deprotonated and negatively charged (ie. in the carboxylate form), and amine groups essentially 100% protonated and positively charged (i.e., in the ammonium form).
Alcohols are fully protonated and neutral at pH 7, as are thiols. The imidazole group on the histidine side chain has a pK a near 7, and thus exists in physiological solutions as mixture of both protonated and deprotonated forms.
Exercise \(\PageIndex{1}\)
Would you expect an aromatic hetererocycle, pyrrole, to be protonated at pH = 7.3? Use the Henderson-Hasselbalch equation to determine your answer. pKa of protonated pyrrole is 0.4.
- Answer
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\[\begin{align*}7.3 &= 0.4 + \log \left(\dfrac{[\ce{RNH2}]}{[\ce{RNH3^{+}}]}\right) \\[4pt] \dfrac{[\ce{RNH2}]}{[\ce{RNH3^{+}}]} &= 7.9 \times 10^{6}\end{align*}\]
So, \(\ce{[RNH2]} \gg \ce{[RNH3^{+}]}\). Thus pyrrole would be almost completely unprotonated at pH = 7.