Question

4. A pharmaceutical company plans to use a liquid-liquid extraction to purify a drug that is a weak monoprotic base. The chemical has a pKb = 9.52 at 25oC. The KD for the neutral form of this chemical is 210 when using water and chloroform at 25oC.

a. What fraction of this drug will be extracted from a 15.0 mL aqueous sample into 10.0 mL of chloroform at pH = 9.00 and 25oC?

b. What fraction of this drug will be back extracted at 25oC from the 10.0 mL of chloroform and into a fresh 50.0 mL portion of water that is buffered at pH = 2.00?

c. What overall fraction of the drug will be isolated from the original sample and placed into the final 50.0 mL portion of water under these conditions?

Answer 1

liquid liquid extraction of weak base B is governed by following equilibria:

B_aq B_org K_D = 210

B_aq + H₂O HB⁺ + OH^- pK_b = 9.52 ,   pK_b = -log K_{b , therefore} K_b = 10^-9.5 = 3.16 * 10^-10

Distribution coeficient = K_D = [Borg ] \ [B_aq]

Distribution ratio i.e D =  [Borg ] _total\ [B_aq] _total =  [Borg ] \ [B_aq] + [ HB⁺]

K_b =  [HB⁺] [ OH^-] \  [B_aq ]

therefore D =  [B_org ] \{ [B_aq] + { K_b *  [B_aq ] \  [ OH^-] } }

=  [B_org ] \ [B_aq] { 1+ { K_b \  [ OH^-] } }

= K_D \  { 1+ { K_b \  [ OH^-] } }

D = { K_D *  [ OH^-] } \  { [ OH^-] + K_b }

a. at pH = 9 [ OH^-] = 1* 10^-5 M

  K_D = 210 , K_b = 3.16 * 10^-10 ( given)

D = { K_D *  [ OH^-] } \  {  [ OH^-] + K_b }

D = 210* 1* 10^-5 \ {1* 10^-5 + 3.16 * 10^-10 }

D = 209.99 (approx 210)

by formula , fraction of drug extracted in aqueous phase where V_aq =15 mL, V_org = 10 mL

q_{aq at pH = 9} = Vaq \ {D V_org + Vaq }  = 15\ { 210 *10 + 15} = 0.0071

thus, fraction of B remaining in organic phase q_org = 1-0.0071 = 0.9929

b. At pH = 2.00

[ OH^-] = 1* 10^-12M

  K_D = 210 ,  K_b = 3.16 * 10^-10 ( given)

D = { K_D *  [ OH^-] } \  {  [ OH^-] + K_b }

D = 210* 1* 10^-12 \ {1* 10^-12 + 3.16 * 10^-10 }

D = 0.66

by formula , fraction of drug extracted in aqueous phase where V_aq =50 mL, V_org = 10 mL

q_{aq at pH = 2} =  Vaq \ {D  V_org + Vaq }  = 50\ { 210 *10 + 50} = 0.023

therefore fraction of drug remaining in chloroform chloroform i.e q_{org at pH=2} = 1- 0.023= 0.977

c.

overall fraction of the drug will be isolated from the original sample and placed into the final 50.0 mL portion of water under these conditions = fraction of drug remaining in organic phase at pH=9 * fraction of drug remaining after second extraction at pH=2 in aqueous phase

=q_{org at pH=9 *} q_{aq at pH =2}

=0.9929 *0.023

=0.022