3
$\begingroup$

How did we derive the relation between the isoelectric pH and the dissociation constants for the amino and carboxyl groups on zwitterionic amino acids.

$$pI=\frac {pK_{a1}+pK_{a2}}{2}$$ where $$\ce{NH3^+-R2(COO^{-})<=>NH3^+-R2(COOH)-K_{a1}}$$ $$\ce{NH3^+-R2(COO^{-})<=>NH2-R2(COO^{-})-K_{a2}}$$

For more complicated cases (charged side groups) and their isoelectric pH, here is one site which just posts the equations without any derivation. Besides this, there is no reference I could find for this derivation.

$\endgroup$
1
$\begingroup$

Consider the equilibria in dilute aqueous solution:

$$\ce{H2A+ \rightleftharpoons HA + H+} \ \ \ \ \ \ \ K_{a_1}=\frac{[HA][H^+]}{[H_2A^+]}$$

$$\ce{HA \rightleftharpoons A^- + H+} \ \ \ \ \ \ \ K_{a_2}=\frac{[A^-][H^+]}{[HA]}$$

At the isoelectric point:

$$[H_2A^+]=[A^-]$$

Using the expressions for $K_{a_1}$ and $K_{a_2}$, we find:

$$\frac{{[HA]}[H^+]}{K_{a_1}}=\frac{K_{a_2}[HA]}{[H^+]}$$

$$[H^+]^2 = K_{a_1}K_{a_2}$$

$$pH_{isoelectric}=\frac{pK_{a_1}+pK_{a_2}}{2}$$

See also this section on Wikipedia.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.