Question

A solution of 0.243 M cysteine is titrated with 0.0608 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

Let 100 ml of 0.243 M cysteine is titrated with 0.0608 M HNO3

at the equivalence point, 1 moles of HNO3 reacts wth neutral cysteine to form +vely charged NH3+ species, which dissociates as,

RNH3+ + H2O <==> RNH2 + H3O+

let x amount of NH3+ species has dissociated

Volume of HNO3 added = 0.243 M x 100 ml/0.0608 M = 399.7 ml

[RNH3+] formed by reaction with HNO3 = 0.243 M x 100 ml/499.7 ml = 0.05 M

pKa1 = 1.70

pKb = 14 - 1.70 = 12.3

Kb = 5.01 x 10^-13

Kb = 5.01 x 10^-13 = x^2/0.05

x = [H3O+] = 1.58 x 10^-7 M

pH at equivalence point = -log(H3O+) = 6.80