Question

A weak base, B, (Kb=1.0 *10-7) equilibrates between water (phase 1) and benzene (phase2). a) Define the distribution coefficient for this system b) Explain the difference between the partition coefficient, K, and the distribution coefficient, D, for this system. c) Calculate D at a pH of 9 if K=5.0 d) Will D be greater at pH of 10 than 9? Explain.

Answer 1

a) A weak base is generally organic moiety so most the of the undissociated part will be in organic layer i.e.benzene.

so the distribution coefficient is given as D= [Base]_Benzene/ {[Base]_water +[base ion] }

where [Base]_Benzene= concentration of base in its molecular form inbenzene

[Base]_water = concentration of base in its molecular form in water

[base ion]water = concentration of base in its ionised form in water.

b) Partition coefficient (K) and distribution coefficient (D) is the all same except the former one is measure of lipophilicity where only molecular forms are involved, wherein in the latter case, the will be ionisation of moeity happens like in case of weak acid or weak base.
K= [Base]_Benzene/ {[Base]_water whereas D= [Base]_Benzene/ {[Base]_water +[base ion] }

c) pH= 9. so pOH= 5. PK_b= 7 so from the Henderson equation, one can easily find out the concentation of ionised and molecular form of base.

pOH= 1/2 PK_b - 1/2 log [base] ; 5= 3.5 - 0.5 log [base]; -3= log [base] ; [base]= 10^-3

pOH gives concentration of ionised base. [OH^_- ]= 10^-5 = [ionised base]

[Base] _water = 10^-3 M,  from K , partition coefficient [Base] _benzene = 5 x 10^-3 M

D= [Base]_Benzene/ {[Base]_water +[base ion] }

D= 5 x 10^-3/ (10^-3 + 10^-5 ) = 5/(1.01) = 4.95.

c) similarly, for pH= 10, pOH= 4 so [ionised base]= 10^-4 log[base] = 10^-1

from tha value of K=5 ; [base]_banzene =5 x 10^-1

therefore D= 5 x 10^-1 / (10^-1 + 10^-4 ) = 5/1.001 = 4.99.

D value at pH 10 will be greater than that of at pH 9.