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In partition chromatography, the change in the retention factor for a given anal

ID: 952203 • Letter: I

Question

In partition chromatography, the change in the retention factor for a given analyte when the composition of the mobile phase changes can be calculated using the following equations, For RPLC log(k_1/k_2) = (P_tot1 - P_tot2)/2 For NPLC log(k_1/k_2) = (P_tot2 - P_tot1)/2 where k_1 and k_2 are the retention factors for the analyte when two different mobile phases are used, which have solvent polarity indexes of P_tot1 and Ptot_2. A list of solvent polarity indexes for various liquids can be found here. When a 29.0% ethyl acetate/71.0% water mobile phase is used, an analyte elutes from a reversed-phase column with a retention factor of 14.7. For the same analyte, what will the retention factor be if the mobile phase is changed to 14.5% ethyl acetate/85.5% water? k = If instead you want the analyte to elute with a retention factor of 3.7, what mixture of ethyl acetate and water is needed? ethyl acetate water

Explanation / Answer

The relation between dielectric constand and mixtures is given by

The value of the dielectric constant for a mixture is obtained by multiplying the volume fraction of each solvent times its dielectric constant and summing.

Also the dielectric constant () of a compound is an index of its polarity.

So we can assume that polarity index of a mixture can be calculated in a similar way

Polarity index of Ethyl Acetate is 4.4 and that of Water is 10.2

so when we have 29% ethyl acetate and 71% water the overall polarity of the mobile phase is

0.29 x 4.4 + 0.71 x 10.2 = 8.52

14.5% ethyl acetat and 85.5% water will be

0.145 x 4.4 + 0.855 x 10.2 = 9.36

log(k1/k2) = (Ptot1-Ptot2)/2

log(14.7/k2) = (8.52-9.36)/2

log(14.7/k2) = -0.419

14.7/k2 = 10-0.419

14.7/k2 = 0.38

k2 = 14.7/0.38

k2 = 38.62

Is the retention factor in the new solvent.

log(k1/k2) = (Ptot1-Ptot2)/2

log(14.7/3.7) = (8.52-Ptot2)/2

log (3.97) = (8.52-Ptot2)/2

0.599 x 2 = (8.52-Ptot2)

1.19 =  (8.52-Ptot2)

Ptot2 = 8.52 - 1.19

Ptot2 = 7.32

so the polarity index of this solvent should be 7.32

x1 x 4.4 + 1-x1 x 10.2 = 7.32

4.4x1 + 10.2 -10.2x1 = 7.32

-5.8x1 = 7.32-10.2

-5.8x1 = -2.88

x1 = 2.88/5.8

x1 = 0.49

So we need a mixture have 49% ethyl acetat and 51% water.