Question

Can someone please help EXPLAIN this to me? I don't just want answers, I need EXPLANATIONS! I will mark it thumbs down if it is not explained to me and just given the answer. Thanks! Even if you could answer a few of them. Please, you'd be saving my grade. Just explain it.

Equivalnets of hydroxide 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 pH a) What is the one letter code for the amino acid that is being titrated? For all the following questions, use the letters above the titration to answer the question. Note that there may be multiple answers to some of the questions. Une noilsilloi 901) w510 di equoi ils to o ntz 199910 ebulai (b) This amino acid is present predominantly in a +2 net charge. olissinoi lo 1069 101 (c) The average net charge of the amino acid is +1.5.m arinismo noiho26 9VS OY (d) Half of the amino groups are ionized. So m (e) The pH is equal to the pka of the carboxyl group. on I goinistroa noitulo 6 susd voy (b i Hori od bluowierw 2. Hq (f) The pH is equal to the pka of the amino group. Piimsowierto sulsvigert 975 ir (9 (g) The amino acid has its maximum buffering capacity. (h) The average net charge of amino acid is zero. (i) The carboxyl group has been completely titrated. otely titrated volgd worla elitong Haer't isbiano (es (i) The amino acid is completely titrated. (1) The average net charge of the amino acid is -1. (m) Amino acid is present predominantly as a 50:50 mixture of the +1 and O charge states. (n) This is the isoelectric point. (o) This is the end of the titration. (p) These are the worst pH regions for buffering power.

Answer 1

The curve can be remade by putting pH on y-axis and Equivalent of hydroxide on x-axis as:

final equir. point Thund pka second equivalence point PH Dr secondoka Fiest equivalence oc point fiest pka a is 2.0 2.5 3.0 3.5 0.5 1.0 баш оон"

Now, as the curve has three vertical lines, it means it has three equivalence points (indicated on curve by C,E and G) each for a proton donor site and hence, amino acid is triprotic, i.e. it can donate three protons till it reaches its completely deprotonated stage.

If three proton donating sites are there, then there would be three K_a and hence three pK_avalues (indicated on curve by B, D and F).

The pKa values are identified by the region where pH remains almost constant.

At pH = pKa of a group, there is equal (50% each) concentration of both protonated and deprotonated form of that group.

As the carboxy terminal has lowest pKa and amino has highest pKa, so,

pKa	Associated group
~ 2.0 (B)	terminal carboxyl
~ 6.0 (D)	side chain
~ 10.0 (F)	terminal amino

As side chain pKa (D) is less than 6.0 so, it is Histidine.

Side chain of histidine has side chain amino group inside the imidazole ring and has pKa of slightly less than seven.

Now, at various pH, the charges on different groups and net charge on amino acid can be represented as:

pH range	Charge on terminal carboxyl	Charge on terminal amino	Charge on side chain amino	Overall charge
1.00-3.99	0	+1	+1	+2
4.00	-0.5 (pKa)	+1	+1	+1.5
4.01-5.99	-1	+1	+1	+1
6.00	-1	+1	+0.5 (pKa)	+0.5
6.01-9.99	-1	+1	0	0
10.00	-1	+0.5 (pKa)	0	-0.5
10.01-14.00	-1	0	0	-1

Isoelectric point (pI) : pH at which overall charge is zero.

So, pI = (6.01+9.99)/2 = 8.0

The answers are:

a) One letter code for Histidine : H

b) +2 net charge is at point : A

c) Average net charge is 1.5 : C

d) Half of amino groups ionized : E

e) pH equals pKa of carboxyl : B

f) pH equals pKa of amino : F

g) maximum buffering capacity : D

h) net charge zero : E

i) Carboxyl completely titrated (first equivalence point) : C

j) Amino completely titrated (third equivalence point) : G

l) Net charge -1 : G

m) +1 and 0 as 50:50 mixture : D

n) isoelectric point : E

o) End of titration : G

p) Worst regions for buffering : A, C, E, G.