Question

A compound is known to have a free amino group with a pKa of 8.8, and one other ionizable group with a

pKa between 5 and 7. To 100 mL of a 0.2 M solution of this compound at pH 9.2 was added 40 mL of a solution

of 0.2 M hydrochloric acid. The pH changed to 6.2. Calculate the pKa of the second ionizable group.

Answer 1

the total dissociation of the compound can be represented by :

H2A <======> H+ + HA-

HA- <=======> H+ + A2-

Where the total compound concentration is represented by [H2A]+ [HA-] + [A2-]

100*0.2/1000=0.02 mol total compund

pH=8.2

HA- + A2- =0.02 mol

8.2=8.8+log(A2-/HA-)

Solving for HA- and A2-, we get

[HA]=0.016 mol

[A2-]=0.004 mol

Adding 40 ml of 0.2 M HCl

40*0.2/1000=0.008 mol H+

Since [A2-] is less than [H+], we dont have to add 0.008

pH=pKa + log ([A-]/[HA])

6.2=pKa+log(0.016/0.004)

6.2=pKa+log(4)

So, pKa=5.6