Question

A solution of 0.108 M cysteine is titrated with 0.0540 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 of the titration.

Answer 1

Answer:

2.698

Explanation:

Cysteine + H+ <==> cysteineH+

at the equivalence point

cosidering 1 L of cysteine solution

volume of HNO3 to reach equivalence point = 0.108 M x 1 L/0.0540 M = 2.0 L

[CysteineH+] formed = (0.108 M x 1 L)/(1.0 + 2.0) L = 0.036 M

hydrolysis of salt

cysteineH+ + H_2O <==> cysteine + H_3O+

let x amount hydrolyzed

Ka1 = [cysteine][H3O+]/[cysteineH+]

pKa1 = -log[Ka1] = 1.70

Ka = 0.02

so,

0.02 = x^2/(0.036-x)

x^2 + 0.02x - 0.00072 = 0

x = 0.002

pH at equivalence point = -log(0.002) = 2.698