The structure of tryptophan is shown at the right along with the pK values of th

ID: 981413 • Letter: T

Question

The structure of tryptophan is shown at the right along with the pK values of the ?-carboxyl and ?-amino groups. Tryptophan is a diprotic acid which can exist in three ionized forms: H2A , HA0, and A–. (HA0 consists primarily of the zwitterionic form with a negatively charged carboxyl group and positively charged amino group, but a very small amount of the completely neutral species does exist). Calculate the fraction (?) of each species present at pH 1.70

Here is the HINT. PLEASE DO NOT JUST ANSWER THE HINT. PLEASE ANSWER THE ACTUAL QUESTION IN THE FIRST PICTURE. Thank you in advance.

Explanation / Answer

If pH is lower than pka then the species is protonated.At pH=1.70 both –NH2 and –COOH with pka higher than the given pH(acidic conditions) will be in protonated state.

So the acid species will have net charge of +1 due to –NH3+ group .The species that exists in solution is H2A+

fraction of H2A+=1 (only there in solution)

fraction of HAo=0

fraction of A-=0

But, the dissociation of H2A+ takes place,with equilibrium constant K1

pk1=2.38=-logk1

k1=10^-2.38=0.00417

Let [HAo]=x considering the equilibrium condition,

H2A+ HAo +H+

Let

[H2A+]=x

H2A+ HAo +H+

k1=[H+][HAo]/[H2A+]

pH=1.7=-log[H+]

[H+]=10^1.70=0.02=y=[HAo]

ICE table

[H2A+]

[H+]

[HAo]

initial

0.02

change

-y

equilibrium

x-y

0.02+y

K1=(0.02+y)(0.02+y)/(x-y)

[ignore y with respect to x as y is very small comparatively due to very less dissociation]

0.00417=(0.02)(0.02)/X

Or, x=0.0004/0.0042=0.096=[H2A+]

[A-]=0

[a(H2A+)= [H2A+]/[ H2A+]+[HAo]+[A-]

[H2A+]=(0.096)/(0.096+0.02)=0.096/0.116=0.827

[HAo]=0.02/0.116=0.172