Problem $1.$ The adsorption isotherms of two proteins in an ionic exchange resin are:

$q_A=\frac{k_1{C}_A}{k_2+C_A}$

$q_B=\frac{k_3{C}_B}{k_4+C_B}$

With

$k_1=0\mathrm{.}2mg$ of $\frac{A}{m}g$ of adsorbent

$k_2=0\mathrm{.}1mg$ of $\frac{A}{m}L$ of liquid

$k_3=0\mathrm{.}05mg$ of $\frac{B}{m}g$ of adsorbent

$k_4=0\mathrm{.}02mg$ of $\frac{A}{m}L$ of liquid

The two proteins are separated in a chromatography column filled with $3g$ of this resin. The dimensions

of the gel bed are: diameter: $2\mathrm{.}8cm,$ height: $25cm.$ If the volume of the mixture added is $50mL$ and the

porosity of the bed is $0\mathrm{.}35,$ determine:

$($a$)$ The position of each band in the column

$($b$)$ The ratios between the distances travelled by each of the proteins to the distance travelled by

the solvent when $C_A=0\mathrm{.}1m\frac{g}{m}L$ and $C_B=0\mathrm{.}05m\frac{g}{m}L.$