Question

Suppose that the partition coefficient for an amine, B, is K = 3.0 and the acid dissociation constant for BH+ is Ka = 1.0 x 10−9. If 50.0 mL of 0.010 M aqueous amine is extracted with 100.0 mL of solvent two times, what percentage amine remains in aqueous solution if the pH of the aqueous phase is 9.00?

Answer 1

The partition co-efficient for the equilibrium

B (aqueous) <=====> B (organic)

can be written as

K = [B]₂/[B]₁ …….(1)

where 2 = organic phase and 1 = aqueous phase.

Let we have m moles of base B total in V₁ mL of water and is extracted with V₂ mL solvent. Let q be the fraction of B in the aqueous phase after the extraction. Therefore,

[B₁] = qm/V₁

and [B₂] = (1-q)m/V₂ …..(2)

The distribution co-efficient (an alternative form of partition co-efficient) is given as

D = [B]₂/([B]₁ + [BH⁺]) …..(3)

(the aqueous phase contains both the amine and its protonated form).

Divide both numerator and denominator of (3) by [B]₁ to obtain

D = ([B]₂/[B]₁)/([B]₁/[B]₁ + [BH⁺]/[B]₁) = K/(1 + [BH⁺]/[B]₁) ……(4)

Since B is a weak base that undergoes protonation in the aqueous phase, we can write (the proton dissociation reaction) as

BH⁺ <=====> B + H⁺

K_a = [B][H⁺]/[BH⁺]

====> [BH⁺]/[B] = [H⁺]/K_a …..(5)

Put (5) in (4) above to write

D = K/(1 + [H⁺]/K_a) = K.K_a/(K_a + [H⁺]) …..(6)

Given K = 3.0, K_a = 1.0*10^-9 and pH = 9.0;

Therefore, [H⁺] = 1.0*10^-9.

Plug in the values in (6) above and obtain

D = (3.0).)(1.0*10^-9)/[(1.0*10^-9) + (1.0*10^-9)] = (3.0).(1.0*10^-9)/(2).(1.0*10^-9) = 3/2 = 1.5

Now write for D in terms of q,m,V₁ and V₂ as

D = [(1-q)m/V₂]/(qm/V₁) = (1-q)V₁/qV₂

====> D.qV₂ = V₁ – qV₁

====> q.(V₁ + D.V₂) = V₁

====> q = V₁/V₁ + DV₂

Fraction remaining in aqueous phase (1) after two extractions is

q² = (V₁/V₁ + DV₂)².

Plug in values: V₁ = 50.00 mL; V₂ = 100.00 mL and D = 1.5.

Therefore, fraction remaining in aqueous phase = [50/50 + (1.5).(100)]² = [50/50 + 150]² = (50/200)² = (1/4)² = (0.25)² = 0.0625

Fraction of amine remaining in aqueous phase = (0.0625)*100 = 6.25 (ans).