Question

A monoprotic acid with a Ka of 7.79E10-5 has a partition coefficient of 3.4 (favoring octanol) when distributed between water and octanol. Find the formal concentration of the acid in each phase when 100 mL of 0.10 M aqueous acid is extracted with 31 mL of octanol at (a) pH 4.00 and (b) pH 9.00.

Answer 1

Ka of acid = 7.79x10^-5

Partition coefficient, K_D = 3.4

Volume of octanol = 31 mL

Milimoles of acid = 100 * 0.10 = 10 mmoles

(a) at pH 4.00

[H +] = 10^-4 M

Distribution ratio, D = K_D * [H+] / [H+] + Ka

D = 3.4 x 10^-4 / 10^-4 +  7.79x10^-5

D = 3.4 x 10^-4 / 1.779 x 10^-4

D = 1.91

Fraction of acid remaining in aqueous phase -

Q(aq) = 100 / 1.91*31 + 100

Q(aq) = 0.62

Moles of acid in aqueous phase = 0.62 * 10

= 6.2

Concentration of acid in aqueous phase = 6.2 / 100 = 0.062 M

Moles of acid in organic phase = 10 - 6.2 = 3.8

Concentration of acid in organic phase = 3.8 / 31 = 0.122 M

(b) at pH 9.0

[H +] = 10^-9 M

Distribution ratio, D = K_D * [H+] / [H+] + Ka

D = 3.4 x 10^-9 / 10^-9 + 7.79x10^-5

D = 3.4 x 10^-9 / 7.7901 x 10^-5

D = 4.36x10^-5

Fraction of acid remaining in aqueous phase -

Q(aq) = 100 / 4.36x10^-5 *31 + 100

Q(aq) = 0.99

Moles of acid in aqueous phase = 0.99 * 10

= 9.9

Concentration of acid in aqueous phase = 9.9 / 100 = 0.099 M

Moles of acid in organic phase = 10 - 9.9 = 0.1

Concentration of acid in organic phase = 0.1 / 31 = 0.0032 M