Question

I calculated the concentration in A to be .08 and the quotient to be .5 so you don't have to do that (feel free to double check me though)

Part A: If mysterious solute S has a partition coefficient of 4.0 between water and chloroform, calculate the concentration of S in chloroform if the concentration of S in water is 0.020 M. If the volume of water is 80.0 mL and the volume of chloroform is 10.0 mL, find the quotient (mol S in chloroform/ mol S inwater).

Part B)Continuing from the previous problem, assume that S is initially dissolved in 80 mL of water. If it is extracted six times with 10.0 mL of chloroform (each time), what is the fraction of S remaining in the aqueous phase?

Answer 1

A)
K = 4

K = [S]_C / [S]_W = 4

[S]_C / 0.02 M = 4

[S]_C = 0.08 M

mMol S in water = 0.02 M * 80 mL = 1.6

mMol S in chloroform = 0.08 M * 10 mL = 0.8

Quotient = 0.8 / 1.6 = 0.5

B)

Initial mMol of S in water, S₀ = 1.6 + 0.8 = 2.4

Let mMol of S remaining in water after 1^st extraction = x1

((S₀ – x1)/10) / (x1 /80 ) = 4

x1 = 2 S₀ /3

After second extraction, S₀ = x1

So, x2 = 2 x1 / 3

Thus, after sixth extraction,

x6 = S₀ (2/3)⁶

= 2.4 (2/3)⁶

= 0.21

Fraction of S remaining in aqueous phase

= x6 / S₀ = (2/3)⁶

= 0.088