Question

In a strong acid solution, the amino acid histidine binds three protons. The acid dissociation constants numbered from the weakest acid dissociation are 6.92x10^-10, 1.00x10^-6 and 1.51x10^-2 at 298 K.

Calculate the concentrations of the four forms of histidine (His^-, HisH, HisH₂⁺ and HisH₃²⁺) in a 0.1 M solution of histidine at ph=7, assuming that these constants apply at the ionic strength of the solution.

Answer 1

His- +H3O+ HisH   ka1=6.92*10^-10, pKa1=-log ka=9.2

HisH +H3O+ HisH2⁺ ka2=1.00*10^-6 pka2=6

HisH2⁺ +H3O+ HisH3²⁺ ka2=1.51*10^-2 pka2=1.82

At pH=7 ,HisH(pka=6) and HisH2+ (pka=1.82) will only be deprotonated species.

using henderson-hasselbach equation,

pH=pka+log[base]/[acid]

1) [His-]=[HisH]=0.1M

2)pH=pka+log[HisH]/[HisH2+]

7=6-log (0.1)/[HisH2+]

1=log (0.1)/[HisH2+]

(0.1)/[HisH2+]=10^1

[HisH2+]=0.1/10=0.001M

2)pH=pka+log[HisH2+]/[HisH3+]

7=1.82 log(0.001)/[HisH3+]

solving,

0.001/[HisH3+]=10^5.18

[HisH2³⁺]=6.6*10^-9M