Question

What fraction of piperazine (perhydro-1,4-diazine) is in each of its three forms (H,A, HA, A2) at pH 7.65? The acid Kal 4.65 x 10 and K 1.86 x 10-10 dissociation constant values for piperazine aгe aHA

Answer 1

First dissociation equation is

H2A (aq) <----> HA- (aq) + H+ (aq)

given pH = 7.65 ,

hence [H+] = 10^ -pH = 10^ - 7.65 = 2.24 x 10^ -8 M

Ka1 = [HA-] [H+] / [H2A]

4.65 x 10^ -6 = [HA-] ( 2.24 x 10^ -8) / [H2A]

[HA-] / [H2A] = 208 ...........................(1)

now we have 2nd dissociation eq HA- (aq) <---> H+ (aq) + A2- (aq)

Ka2 = [H+] [A2-] / [HA-]

1.86 x 10^ -10 = (2.24 x 10^ -8) [A2-] / [HA-]

[A2-] / [HA] = 0.0083........................(2)

All HA- comes from H2A. Thus from equation (1) its clear that out of 209 H2A molecules 208 converted to HA-.

After dissociation [H2A] = 209-208 = 1 , [HA-] = 208 ,

Hence ratio of [HA-] / [H2A] after dissociation = 208 / 1 = 208.

Now from eq (2) All A2- comes from HA , thus when HA- is 208 value of [A2-] = 0.0083 x [HA-]

= 0.0083 x 208 = 1.73

Fraction of H2A = ( H2A left undissociated ) / ( initial H2A) = ( 1 / 209) = 0.005

Fraction of HA - = ( HA- formed / initial H2A) = ( 208 / 209) = 0.995

Fraction of A2- = ( A2- formed) / ( H2A initial) = ( 1.73 / 209) = 0.0083