Question

Suppose that the partition coefficient for an amine, B, is K = 2.9 and the acid dissociation constant of BH+ is Ka = 1.0 x 10-9. If 74 mL of 0.010 M aqueous amine are extracted with 137 mL of solvent, what will be the formal concentration of amine remaining in the aqueous phase at pH 7.0? (Round your answer to two sig. fig. and do NOT use scientific notation.)

Answer 1

BH+ <==> B + H+

Ka = [B][H+]/[BH+]

pH = 7 = -log[H+]

[H+] = 1 x 10^-7 M

distribution coefficient D,

D = K.Ka/(Ka + [H+])

with,

K = partition coefficient = 2.9

Ka = 1.0 x 10^-9

Feed value,

Distribution coefficient (D) = (2.9 x 1.0 x 10^-9)/(1.0 x 10^-9 + 1.0 x 10^-7) = 0.029

Fraction of amine remaining in aqueous phase = V1/(V1 + KV2)

with,

V1 = 74 ml

V2 = 137 ml

K = 0.029

we get,

fraction of amine in aqueous phase = 74/(74 + 0.029 x 137) = 0.95 or 94.90% left in Aq. phase at pH 7.0