Question #152996

Albumin is being separated from Immunoglobin G (IgG) in a chromatography column
having a stationary phase fraction of 0.25. The mobile phase retention time is 10
min. The distribution coefficients for albumin and IgG are 0.1 and 1, respectively. If
the albumin peak has a characteristic width of 0.52 minutes, predict the selectivity
and resolution.

Expert's answer

the volume of the column is 39ml calculated using the given data

the mobile phase retention time can be calculated using equation

tM = VCεQ\frac{V_Cε}{Q}   ⟹  \implies 39(0.25)10\frac{39(0.25)}{10} = 0.975 min

the retention times can be calculated using equation

tR = tM {1+ ( 1−εε\frac{1-ε}{ε} ) K}


tR,alb = 0.975 { 1 + (1−0.250.25\frac{1-0.25}{0.25} ) 0.1 } = 1.25 min


tR,IgG = 0.975 { 1 + (1−0.250.25\frac{1-0.25}{0.25}) 1 } = 3.9 min


the number of theoretical plates in the column can be calculated from albumin retention time data using equation

N = 16(tgw\frac{t_g}{w} )2   ⟹  \implies 16 (1.270.52\frac{1.27}{0.52})2 = 95


the peak width of IgG can be calculated using equation

WIgG = tR.IgGN16\frac{t_{R.IgG}}{\sqrt\frac{N}{16}}   ⟹  \implies 3.99516\frac{3.9}{\sqrt\frac{95}{16}} = 1.6 min


the resolution of separation can be calculated using equation

R = tR2−tg10.5(w1+w2)\frac{t_{R2}-t_{g1}}{0.5(w_1+w_2)}   ⟹  \implies 3.9−1.270.5(1.6+0.52)\frac{3.9-1.27}{0.5(1.6+0.52)} = 2.58


the selectivity can be calculated using equation

α = K2K1\frac{K_2}{K_1} = tR2−tMtR1−tM\frac{t_{R2}-t_M}{t_{R1}-t_M}   ⟹  \implies 10.1\frac{1}{0.1} = 10


the height of the theoritical plate can be calculated using equation

H = lN\frac{l}{N} = 5095\frac{50}{95} = 0.526 cm


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