Question

1. An unknown molecule has a KP (ether/water) of 3.5 at room temperature.
   1. If a solution of 350 mg of the compound in 2.5 mL of water is extracted with 5.0 mL of ether, what is the mass of the compound in both the ether and water layers?
   2. Comment on whether it is more efficient to use a single large extraction or multiple smaller extractions.

Answer 1

a) Distribution coefficient (Kp)

= Concentration of A in ether/ Concentration of A in water

[ A is the compound]

when, 5.0 ml ether is used , suppose We gram of the compound pass into ether layer.

and Ww gram is left at water layer

hece, (We/5.0) ÷ (Ww/2.5) = Kp = 3.5

or, We / Ww = 3.5*2 = 7

or, We/( We +Ww) = 7/8

hence, (7*100)/8 = 87.5 % was extracted

or, We = 7*Ww.

now, total mass of the compound We + Ww = 350 mg

or, 7*Ww + Ww = 350 mg

or, Ww = 350/8 = 43.75 mg

and We = 350 - 43.75 = 306.25 mg

b) A general equation for successive extraction

Xn = Xo * ( Kp *V/Kp+ L)ⁿ (1)

where, Xn is the amount of solute left behind after nth extraction

Xo is intial quantity of substance,

L is volume of solvent (ether)

as the quantity in eq. 1 is less than 1.

greater the value of n , Xn will be smaller.

Hence, it is advantageous to use a given volume of the extracting solvent in small successive stages rather than in one whole at a time.