Solute S has a partition coefficient of 4.0 between water (phase 1) and chloroform (phase 2) in Equation 23-1.

(a) Calculate the concentration of in chloroform if[S]aqis 0.020M.

(b) If the volume of water is 80.0 mL and the volume of chloroform is 10.0 mL, find the quotient localid="1654852656619" (molSinchloroform)(molSinwater).

Short Answer

Expert verified

The organic solvent is chloroform and aqueous is water, hence


Schloroform=K×SwaterSchloroform=4.0×0.02MSchloroform=0.08M

The equation for number of moles is n=c×V, hence

(molSincholoroform)(molSinwater)=Cchloroform×VchloroformCwater×VwaterCchloroform×VchloroformCwater×Vwater=0.08M×10mL0.02M×80mL=0.5

Step by step solution

01

of 3

  • In this task we have a solute S which has a partition coefficient of 4.0 between water (phase 1) and chloroform (phase 2) in equation 23 - 1 .
02

of 3

(a)

  • Here let us calculate the concentration of S in chloroform if Saq=0.02Mby using the following equation

K=SorgSaq=SohloroformSwater


  • Here, that organic solvent is chloroform and aqueous is water, hence Schloroform=K×SwaterSchloroform=4.0×0.02MSchloroform=0.08M
03

of 3

(b)

  • If the volume of water is 80mL and the volume of chloroform is 10mL, we will find the quotient (molSinchloroform)/(molSinwater):
  • Let us note that we are asked to find moles here and the equation for number of moln=c×V , hence
    (molSincholoroform)(molSinwater)=Cchloroform×VchloroformCwater×VwaterCchloroform×VchloroformCwater×Vwater=0.08M×10mL0.02M×80mL=0.5

Result:

  • a)The concentration of in chloroform if [S]aqis0.020Mis

Schloroform=K×SwaterSchloroform=4.0×0.02MSchloroform=0.08M

  • b)the quotient (molSinchloroform)(molSinwater)is (molSincholoroform)(molSinwater)=Cchloroform×VchloroformCwater×VwaterCchloroform×VchloroformCwater×Vwater=0.08M×10mL0.02M×80mL=0.5

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Most popular questions from this chapter

The distribution coefficient for extraction of a metal complex from aqueous to organic solvents is D=[totalmetal]oegf[totalmetal]aqlocalid="1654846486193" D=[totalmetal]oegf[totalmetal]aqGive physical reasons why localid="1654846489683" β and localid="1654846493841" Kαappear in the numerator of Equation 23-13,but localid="1654846498058" KLand localid="1654846502645" [H+]aqappear in the denominator.

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(c) Find the Relative retention for octane and nonane.

(d) Find the partition coefficient for Octane by assuming that the volume of the stationary phase equals half the volume of the mobile phase.

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(a) If broadening is mainly due to longitudinal diffusion, how should the flow rate be changed to improve the resolution?

(b) If broadening is mainly due to the finite equilibrium time, how should the flow rate be changed to improve the resolution?

(c) If broadening is mainly due to multiple flow paths, what effect will flow rate have on the resolution?

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