Question

The following data apply to apply to a column for liquid chromatography:

                   Length of column: 25.7 cm

                   Flow rate:            0.313 mL/min

                   VM:                      1.37 mL

                   VS:                       0.164 mL

A chromatogram of species A and B provided the following data:

	Retention Time, min	Width of Peak Base (W), min
Non-retained	3.1	-
A	13.3	1.07
B	14.1	1.16

a)    the expected resolution, if the column length was doubled

b)    the expected capacity factors, if the column length was doubled

c)   the length of column necessary to give a resolution of 1.5

d)   the time required to separate A and B with a resolution of 1.5

Answer 1

a)    the expected resolution, if the column length was doubled

Ans: If the the column length was doubled then the expected resolution also increases by factor 1.44.

Resolution = 2(t_RC– t_RB)/(W_C+ W_B)

                  R = 2(14.1-13.3)/(1.16+1.07) = 0.72

As column length is doubled, Resolution is 0.72 X 2 = 1.44

b)    The Capacity Factor () = K_B/K_A Where, K is a Retention factor

         K_A = 13.3 - 3.1 / 3.1 = 3.3, and K_B = 14.1 - 3.1 / 3.1 = 3.5

Hence, = 2 X 3.5 / 3.3 = 2.12

c)   The length of column necessary to give a resolution of 1.5

Ans: R₁/R₂ = (L₁/L₂)^1/2

           0.72/1.5 = (25.7/L₂)^1/2

Therefore, L2 = 111.5

d)   The time required to separate A and B with a resolution of 1.5

        The time required to separate A from B is 14.1 min for 25.7 cm column . Now, if Column length is increases to          111.5, retention time will also increase

        t_R1/t_R2= L₁/L₂

14.1/t_R2 = 25.7/111.5 = 61 minutes.