Question

Question 2

Tyrosine is a triprotic amino acid with pKa of 2.17, 9.19, and 10.47 The first proton is removed

from the carboxylic acid ( –COOH), the second from the protonated amine group (–NH3

+) . The

third, with a pKa of 10.47, is the phenolic proton (–OH) on the aromatic ring.

Tyrosine

a) What is the structure of the most protonated form of tyrosine in aqueous solution?

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b) Write the three acid dissociation equilibria with the correct charge (use Tyr as shorthand

for the part of the molecule excluding the acidic protons, i.e. Tyr(charge), HTyr(charge),

H2Tyr(charge), H3Tyr(charge)).

c) Suppose we titrate a 0.100 M solution of tyrosine hydrochloride salt (often written

Tyr!HCl) with a 0.100 M solution of a strong base (NaOH). Calculate pH at the

following points:

i. initially,

ii. half way to the first,

iii. at the first,

iv. half way to the second,

v. at the second, and

vi. half way to the third equivalence points

d) Using your results calculated in (c) sketch an accurate titration curve.

e) Based on your sketch, identify which equivalence point(s) can be used to identify end

point with an indicator dye.

f) For part (e), indicate what the pKa (or range) of indicator dye should be to minimize

determinate titration error.

g) Write the general expressions for the mass balance and the charge balance for part (c).

Answer 1

Tyrosine

a) Structure of fully protonated form given below

b) Dissociation equations

Ist dissociation : HTyr <==> Tyr + H+

IInd dissociation : Tyr <==> Tyr- + H+

IIIrd dissociation : Tyr- <==> Tyr^2-

c) Titration

i) initial

HTyr+ <==> Tyr + H+

0.0062 = x^2/0.1

x = [H+] = 0.025 M

pH = -log[H+] = 1.60

ii) halfway to the first equivalence point

pH = pKa1 = 2.20

iii) first equivalence point

pH = (pKa1+pka2)/2 = (2.20 + 9.21)/2 = 5.705

iv) half way to the second equivalence point

pH = pKa2 = 9.21

v) at the second equivalence point

pH = (pka2 + pka3)/2 = (9.21 + 10.46)/2 = 9.835

vi) half way to the third equivalence point

pH = pKa3 = 10.46

d) titration curve below

e) equivalence points to be used for indicator dye = 2.20, 9.21

f) A dye which does not have a pH range falling in the region of equivalence point of tyrosine would give and error in measurement.

g) [HTyr] = [Tyr] + [Tyr-] + [Tyr^2-] + [H3O+]