Question

An HPLC setup has an injection volume of 0.334 mL, a detector volume of 0.491 ml, and contains 27.0 cm of tubing, which has a diameter of 0.025 cm. The baseline width, wobs, of a band eluting from the column at a flow rate of 1.48 ml/min is observed to be 0.671 min. Assuming a diffusion coefficient of 6.8 x 10-9 m²/s, calculate the baseline width of the band, in minutes, if the column was the only source of broadening. Wcolumn =

Answer 1

given values

Dm = diffusion coefficient

= 6.8 X 10^-9 m² /s

where y =0.7 (consitent)

Van diameter equation

B (longitudinal diffusion)

B = 2y D_m

= 2 X 0,7 X 6.8 X 10^-9 m² /s

B = 9.52 X 10^-9 m² /s

linear velocity = L / t_R

L = length of column = 27 cm = 0.27 m

t_R = V_R / F_c = retention value / flow rate

t_R = 0.491 ml / 1.48 (ml/min)

t_R = 0.3317 min

t_R = 19.90 sec

linear velocity = L / t_R

= 0.27 / 19.90

=0.01356 m/s

van diameter equation (H)= longitudinal diffusion (B) / linear velocity

= (9.52 X 10^-9 m² /s) / 0.01356 m/s

= 7.02 X 10^-7 m

N = L / H

= 0.27 / (7.02 X 10^-7)

N = 3.85 X 10⁵  

but N= 16 X (t_R / σ)²

σ² = (16 X t_R²) / N

σ² = 16 X (0.3317)² / (3.85 X 10⁵ )

σ² = 4.57246 X 10^-6

baseline width of the band (σ) = 2.138 X 10^-3 min

=0.2138 min

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