Question

calculate the percentage of ionization of amobarbital (pka=8) at ph 2.0, 5.5, 8.0. What trend is seen?

Answer 1

Consider amobarbital as HA (weak acid) , Ka = 10^-8

HA H⁺ + A^-

Ka = [H⁺]×c/(c(1-)) = [H⁺] ×/(1-)

/(1-) = Ka/[H⁺]

At pH = 2.0 ,

/(1-) = 10^-8/10^-2 = 10^-6

= 10^-6/(1+10^-6) = 9.9999×10^-7

Percentage of ionization of amobarbital = 100 % = 9.9999×10^-5

= 1.0×10^-4 % (Answer)

At pH = 5.5 ,

/(1-) = 10^-8/10^-5.5 = 10^-2.5 = 3.16×10^-3

= 3.16×10^-3/(1+3.16×10^-3) = 3.15×10^-3

Percentage ionization of amobarbital = 3.15×10^-3 × 100%

= 0.315 % (Answer)

At pH = 8.0

/(1-) = 10^-8/10^-8 = 1

= 0.5

Percentage of ionization of amobarbital = 0.5×100%

= 50 % (Answer)

As the pH of solution is increased , the percentage of weak acid (amobarbital) increases because as pH increases solution becomes more basic and moves the equilibrium towards right .