Question

On a glutamic acid molecule, there are three hydrogens could dissociate in water. Each has a unique pka, they are pkai = 2.10, pKa2 = 4.07, pka3 = 9.47, respectively. This molecule is shown in the following figure. Please redraw the correct molecule in a solution with a pH at 11.5, by adding or removing the hydrogens from the functional groups. And indicate the overall charge of th pK1=2.10 1 [ Aud] [Base] pH = pka t lag 100 HO ОН NH2 pk.-9.47 pKg2=4.07

Answer 1

For the COOH group with pK_a1 = 2.10, we can write the following dissociation

R-COOH + R-COO + H+

Hence, the expression of K_a1 cn be written as

Kai = H+1R - C00-1 RCOOH

Given that the pH of the solution is 11.5, we can find the ratio of protonated and deprotonated state of our COOH group with pKa1 = 2.10 as follows:

Lets first take the negative logarithm on both side of the K_a1 expression.

[H+][R-C00- – log Kai = -log( [RCOOH] ([R - C00-] → pKai = -log[H+] - log RCOOH R-C00] 2.10 = pH – log [RCOOH log ( P 600) = pH – 2.10 = 11.5 – 2.10 = 9.4 /R-COO ) = 109.4 [RCOOH] ) [RC00-] = 109.4 x [RCOOH] [RC00-] >>>>[RCOOH

Hence, the COOH group with pK_a1 = 2.10 will be predominantly in the deprotonated state with charge -1.

Now,

for the COOH group with pK_a1 = 4.07, we can do the similar exercise above to find its state at pH 11.5

Hence,

/R-C00-1 pha2 = -log[ 108RCOOH) R-C00] = 4.07 = pH – log [RCOOH (R-C00] log ) = pH – 4.07 = 11.5 - 4.07 = 7.43 [RCOOH] ) ([R-C00-]) = 107.43 > (FRCOOH) = RC00-] = 107.43 x RCOOH [RCOO-] >>>>[RCOOH

Hence, every 1 molecule of the group in 10^7.43 molecules is in protonated state. Hence, here also this acid group stays in a deprotonated state with charge -1.

Now, coming to the amine group with pK_a3 = 9.47.

We can write the dissociation of NH³⁺ as follows

R- NH + R-NH2 + H+

Hence, the expression of K_a3 is

Ka3 = [R - NH2H+1 [RN H.)

Hence, taking negative logarithm on both sides

([R - NH][H+] – log Ka3 = -log [RNH3] ([R - NH] →pka3 = pH – log (RNH3] [R - NH2] = log | = 11.5 – 9.47 = 2.03 | [RNH3] [R - NH2] = 102.03 107 (RN1

Hence, less than 1 % of the NH₂ group is protonated. So we can assume that there is no charge contribution from the NH2 group as it is not protonated.

Hence, overall the two COOH groups are deprotonated and NH2 group is neutral. Hence, overall charge of the amino acid at pH 11.5 is -2.

Now, we can draw the predominant structure at pH 11.5 as follows:

NH2