What is the structural formula for serine with a pH of 4?

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What is the structural formula for serine with a pH of 4?

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Answer 1 · 2018-07-22T06:16:20

Read over this, and then see if you can draw the structure. You should have the structure of serine memorized.

Well, serine has two pKas to consider; the carboxyl and the amine pKa's ( $2.21, 9.15$ $2.21,9.15$ ). As you have seen, the Henderson-Hasselbalch equation relates $pH$ $"pH"$ to ${pK}_{a}$ $"pK"_a$ :

$pH = {pK}_{a} + log (\frac{[A^{-}]}{[HA]})$ $"pH" = "pK"_a + log((["A"^-])/(["HA"]))$

Here we see that

if $[HA] < [A^{-}]$ $["HA"] < ["A"^(-)]$ , then $\frac{[A^{-}]}{[HA]} > 1$ $(["A"^-])/(["HA"]) > 1$ and $log (\frac{[A^{-}]}{[HA]}) > 0$ $log((["A"^-])/(["HA"])) > 0$ . Thus, $pH > {pK}_{a}$ $"pH" > "pK"_a$ and the solution is basic with respect to the proton whose ${pK}_{a}$ $"pK"_a$ is given.
if $[HA] > [A^{-}]$ $["HA"] > ["A"^(-)]$ , then $\frac{[A^{-}]}{[HA]} < 1$ $(["A"^-])/(["HA"]) < 1$ and $log (\frac{[A^{-}]}{[HA]}) < 0$ $log((["A"^-])/(["HA"])) < 0$ . Thus, $pH < {pK}_{a}$ $"pH" < "pK"_a$ and the solution is acidic with respect to the proton whose ${pK}_{a}$ $"pK"_a$ is given.

Clearly, since $pH = 4$ $"pH" = 4$ , and $2.21 < 4 < 9.15$ $2.21 < 4 < 9.15$ ,

The solution is basic with respect to the carboxyl group.
The solution is acidic with respect to the amine group.

Hence, the carboxyl group is DEprotonated, and the amine group is protonated. Their charges are thus $(-)$ $(-)$ and $(+)$ $(+)$ , respectively.

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What is the structural formula for serine with a pH of 4?

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