Please help with steps for 22-7 and 22-8! 22-7. Solute S has a partition coeffic
ID: 895786 • Letter: P
Question
Please help with steps for 22-7 and 22-8!
22-7. Solute S has a partition coefficient of 4.0 between water (phase 1) and chloroform (phase 2) in Equation 22-1. (a) Calculate the concentration of S in chloroform if [S(ag)l is 0.020 M. (b) If the volume of water is 80.0 mL and the volume of chlorofornm is 10.0 ml., find the quotient (mol S in chloroform)(mol S in water) 22-8. The solute in Problem 22-7 is initially dissolved in 80.0 mL of water. It is then extracted six times with 10.0-mL portions of chloro- form. Find the fraction of solute remaining in the aqueous phase. 561Explanation / Answer
22-7
(a) Partition coeff. = [S(chloroform)]/ [S(aq)]
therefore concentration of S in chloroform, [S(chloroform)] = partition coeff. x [S(aq)] = 4.0 x 0.020 M = 0.080 M
(b) [S(chloroform)] = mol S in chloroform / volume of chloroform
and [S(aq)] = mol S in water / volume of water
Now Partition coeff. = [S(chloroform)]/ [S(aq)]
= (mol S in chloroform / volume of chloroform)/ (mol S in water / volume of water)
or 4.0 = (mol S in chloroform / 10)/ (mol S in water / 80)
or 4.0 = (mol S in chloroform / mol S in water) x 8
hence mol S in chloroform / mol S in water = 4/ 8 = 0.5
22-8
Suppose n moles of S is dissolved in 80 mL of water
when extracted with 10 mL of chloroform, let nx mole remain in water
therefore number of moles extracted in 10 mL chloroform = n - nx
again as mol S in chloroform / mol S in water = 0.5 when volume of water is 80 mL and volume of chloroform is 10 mL
we have (n - nx) / nx = 0.5
or 1 - x = 0.5x
or x = 1/ 1.5 = 2/ 3
therefore number of moles remaining in water after one 10 mL extraction with chloroform = n x 2/3
so number of moles remaining in water after two10 mL extraction with chloroform = n x 2/3 x 2/3
so after six such extraction n x (2/3)6
so fraction remaining would be (2/3)6 = 0.088
22-1
Three extraction with 100 mL is better than one extraction with 300 mL