Question

A solution of 0.188 M cysteine is titrated with 0.0752 M HNO3. The pKa values for cysteine are 1.70, 8.36, and 10.74, corresponding to the carboxylic acid group, thiol group, and amino group, respectively. Calculate the pH at the equivalence point.

Answer 1

When HNO3 reacts with cystiene it reacts only with the amine group of cysteine because amino group only acts as base.

Example:

H+ + R-NH2 -----> R-NH3+

Buffer system forms when HNO3 added to cysteine

H+ ----> HNO3

R-NH2 ---> Cysteine

At equivalence point

moles of HNO3 = Moles of R-NH2

M1V1 = M2V2

0.188 / 0.0752 = Volume of HNO3

Volume of HNO3 = 2.5 L

H+ + R-NH2 -----> R-NH3+

Moles of Salt formed = 0.188*1 = 0.188 [ assumed 1 litre of cysteine]

[salt] = 0.188 / 1+2.5 = 0.05371

R-NH3+ -----> H+ + R-NH3+ [H+ from COOH group]

0.05371-x x x

Ka = x^2 / 0.05371-x

10^-1.7 = x^2/ 0.05371-x

x= 0.0242 M = [H+]

pH = -logx = 1.616