Question

An open tubular column with a diameter of 193 um and a staionary phase thickness on the inner wall of 0.52 um passes unretained solute through in 61 s. A particular solute has a retention time of 453 s.

a) What is the partition coeffecient for this solute?

b) What fraction of time does this solute spend in the stationary phase?

Answer 1

Fraction of time spent in stationary phase = time spent in stationary phase/total time spent fraction of time spent in stationary phase

= (453 – 61 s)/453 s = 0.865 = 86.5%

k = (453 – 61 s)/61 s = 6.426

K = C_s/C_m = k(V_m/V_s)

V_m = V_cylinder = π(d² /4)L = π[(0.193 mm)² /4)L = 0.0293 mm² L

V_s = V_tube = Vouter cylinder – Vinner cylinder =

π[(0.193 mm)² /4]L – π{[(0.193 mm – 2(0.00052 mm)]² /4}L

Note: for a thin tube, V_tube = πtdL, where t = film thickness. This same equation also can be calculated from the first equation (V_s = Vtube = Vouter cylinder – Vinner cylinder ) since V_tube = π[(0.193 mm)² /4]L – π[(0.193 mm)² /4]L + π(0.193 mm)(0.00052 mm)L – π(0.00052 mm)² L.

The first two terms cancel out and since 0.193 mm >> 0.00052 mm, the last term is inconsequential. V_s = V_tube = π(0.193 mm)(0.00052 mm)L = 0.000315 mm² L

K = k(V_m/V_s) = 6.426(0.0293 mm² L/0.000315 mm² L) = 598 (note how L cancels out so the column length is not needed).

to get the same answer as the text , you must use the outer diameter not including the film (0.193 – 2*0.00052 mm) for the mobile phase volume (since the film is stationary phase, and thus not mobile phase) and the average tube diameter for the stationary phase volume or the more complete equation for tube volume (outer cylinder vol. – inner cylinder vol.). However, considering that the film thickness is only accurate to two sig. fig.s, the answer should only be given to 2 sig figs.