Question

The partition coefficient, K, for compound A between toluene and water is 7.25. What is the concentration of compound A remaining in the aqueous phase after 40.0 mL of 0.205 M A is extracted with each quantity of toluene indicated? one 35.0 mL portion: [A]= two 17.5 mL portions: (A)2 = five 7.00 mL portions: [A]g = ten 3.50 mL portions: [A] 20 =

Answer 1

Fraction of solute remaining in aqueous phase after n extractions is calculated by using formula,

q ⁿ = ( V ₁ / V ₁ + K V ₂ ) ⁿ

Where V ₁ is volume of aqueous phase , K is partition coefficient , V ₂ is volume of organic phase & n is number of extractions.

PART 1

Fraction of A remaining in aqueous phase after 1 extraction = ( V ₁ / V ₁ + K V ₂ )

Fraction of A remaining in aqueous phase after 1 extraction = ( 40.0 / 40.0 + 7.25 (35.0 ) ) = 0.136 = 13.6 %

The concentration of A in the aqueous phase is 13.6 % of 0.205 M = 0.02788 M = 0.0279 M

PART 2

Fraction of A remaining in aqueous phase after 2 extraction = ( V ₁ / V ₁ + K V ₂ ) ²

Fraction of A remaining in aqueous phase after 2 extraction = ( 40.0 / 40.0 + 7.25 (17.5 ) ) ²

Fraction of A remaining in aqueous phase after 2 extraction = 0.0574 = 5.74 %

The concentration of A in the aqueous phase is 5.74 % of 0.205 M = 0.0118 M

PART 3

Fraction of A remaining in aqueous phase after 5 extraction = ( V ₁ / V ₁ + K V ₂ ) 5

Fraction of A remaining in aqueous phase after 5 extraction = ( 40.0 / 40.0 + 7.25 (7.00 ) ) ⁵

Fraction of A remaining in aqueous phase after 5 extraction = 0.0166 = 1.66 %

The concentration of A in the aqueous phase is 1.66 % of 0.205 M = 0.00340 M

PART 4

Fraction of A remaining in aqueous phase after 10 extraction = ( V ₁ / V ₁ + K V ₂ ) 10

Fraction of A remaining in aqueous phase after 10 extraction = ( 40.0 / 40.0 + 7.25 (3.50 ) ) ¹⁰

Fraction of A remaining in aqueous phase after 10 extraction = 0.00735 = 0.735 %

The concentration of A in the aqueous phase is 0.735 % of 0.205 M = 0.00151 M