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Leucine is a type of amino acid. Let me write its chemical formula as $\ce{NH2-C5H10-COOH}$. On the wikipedia page, https://en.wikipedia.org/wiki/Leucine, there are two dissociation constants, one for $\ce{-COOH}\,$ given by $\mathrm{pK}_{a1}=2.36$, and one for $\ce{-NH3+}$ given by $\mathrm{pK}_{a2}=9.60$. But I also know that once $\ce{-COOH}\,$ loses its proton $\ce{H+}$ into the solution, the proton may combine with the $\ce{-NH2}\,$ group of the same molecule to form the zwitterion $\ce{NH3+-C5H10-COO-}$ so that the solution now has $4$ species with $4$ dissociation equilibria

$$\begin{array}{rclll} \ce{NH3+-C5H10-COOH}\!\!&\!\!\ce{<=>}\!\!&\!\!\ce{NH2-C5H10-COOH +H+},& \ &(1)\\ \ce{NH3+-C5H10-COOH}\!\!&\!\!\ce{<=>}\!\!&\!\!\ce{NH3+-C5H10-COO- +H+},& \ &(2)\\ \ce{NH2-C5H10-COOH}\!\!&\!\!\ce{<=>}\!\!&\!\!\ce{NH2-C5H10-COO- +H+},& \ &(3)\\ \ce{NH3+-C5H10-COO-}\!\!&\!\!\ce{<=>}\!\!&\!\!\ce{NH2-C5H10-COO- +H+}.& \ &(4) \end{array}$$

So there should be $4$ equilibrium constants. Which two do $K_{a1}$ and $K_{a2}$ refer to and how do I know the other two? Or is it that we can assume that $\ce{-COOH}\,$ and $\ce{-NH3+}$ dissociate independently so that the equilibrium constants of the $4$ equations are $K_{a2},K_{a1},K_{a1},K_{a2}$ respectively? I've also seen in some calculation that the zwitterion $\ce{NH3+-C5H10-COO-}$ and the original molecule $\ce{NH2-C5H10-COOH}$ are treated as the same species. But I don't quite understand why this is allowed. They clearly have different charge distributions and should be treated as distinguishable particles in entropy (or Gibbs free energy) calculations.

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  • $\begingroup$ related chemistry.stackexchange.com/questions/32645/… $\endgroup$ – Mithoron Jan 14 '18 at 23:12
  • $\begingroup$ Well, I guess 1 and 3 are "included" into effective pKa values or neglected as very minor. $\endgroup$ – Mithoron Jan 14 '18 at 23:14
  • $\begingroup$ @Mithoron, I think I finally figured this out after spending the whole afternoon on it. Let me post an answer so you can check if it's correct. $\endgroup$ – Zhuoran He Jan 14 '18 at 23:16
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Among the $4$ equilibria, only $3$ are linearly independent, because we have $(1)+(3)=(2)+(4)$. Let's then choose only three independent reactions

\begin{align} \ce{NH3+-C5H10-COOH}&\ce{<=> NH3+-C5H10-COO- + H+},\\ \ce{NH3+-C5H10-COO-}&\ce{<=> NH2-C5H10-COO- + H+},\\ \ce{NH3+-C5H10-COO-}&\ce{<=> NH2-C5H10-COOH}, \end{align}

as the basis equations, which are obtained from $(2)$, $(4)$, and $(1)-(2)$. The equilibrium constants of these equations are $K_{a1}$, $K_{a2}$, and $K_{a2}/K_{a1}=10^{-7.24}\ll 1$. Therefore most of the neutral amino acid molecules would exist in the zwitterionic form $\,\ce{NH3+-C5H10-COO-}$. Neglecting the last equilibrium, the equilibrium constants $K_{a1}$ and $K_{a2}$ are those of the $\ce{-COOH}\,$ and $\ce{-NH3+}$ groups in the zwitterions $\,\ce{NH3+-C5H10-COO-}$, which is much more abundant than the original molecules $\,\ce{NH2-C5H10-COOH}\,$ making the latter negligible.

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