Question

Components \(A\) and \(B\) of a binary mixture are to be separated in a chromatographic column. The adsorption isotherms of these compounds are given by the following equations: $$ \begin{aligned} &m_{A}=f_{A}(c)=\frac{k_{1} C_{A}}{k_{2}+C_{A}} \\ &m_{B}=f_{B}(c)=\frac{k_{1}^{\prime} C_{B}}{k_{2}^{\prime}+C_{B}} \end{aligned} $$ where \(k_{1}=0.2 \mathrm{mg}\) solute \(\mathrm{A}\) absorbed \(/ \mathrm{mg}\) adsorbent $$ \begin{aligned} &k_{2}=0.1 \mathrm{mg} \text { solute } / \mathrm{ml} \text { liquid } \\ &k_{1}^{\prime}=0.05 \mathrm{mg} \text { solute } \mathrm{B} \text { adsorbed } / \mathrm{mg} \text { adsorbent } \\ &k_{2}^{\prime}=0.02 \mathrm{mg} \text { solute/ml liquid } \end{aligned} $$ The bed contains \(3 \mathrm{~g}\) of very fine support particles. The bed volume is \(150 \mathrm{ml}\), bed porosity is \(\varepsilon=0.35\), and the cross-sectional area of the bed is \(A=6 \mathrm{~cm}^{2} .\) If the volume of the mixture added is \(\Delta V=50 \mathrm{ml}\), determine the following: a. Position of each band \(A\) and \(B\) in the column, \(L_{A}\) and \(L_{B}\) (or \(\Delta X_{A}\) and \(\Delta X_{B}\) ). b. \(L_{A} / L_{B} ; R_{f A}=L_{A} / L_{c} ; R_{f B}=L_{B} / L_{c}\) when \(C_{A}=10^{-1} \mathrm{mg} / \mathrm{ml}\) and \(C_{B}=0.05 \mathrm{mg} / \mathrm{ml}\) in liquid phase at equilibrium.

Short answer

Band positions: \( L_{A} \approx 57.14 \, \text{cm} \) and \( L_{B} \approx 40.8 \, \text{cm} \). Relative positions: \( \frac{L_{A}}{L_{B}} \approx 1.4 \), \( R_{fA} \approx 0.38 \), and \( R_{fB} \approx 0.27 \).

Step-by-step solution

Compute Adsorption Isotherms for Each Component

Calculate the equilibrium mass of solutes A and B using the given isotherms:\[ m_{A} = \frac{k_{1} C_{A}}{k_{2} + C_{A}} \quad \text{and} \quad m_{B} = \frac{k_{1}^{\prime} C_{B}}{k_{2}^{\prime} + C_{B}} \]With \( C_{A} = 0.1 \) mg/ml and \( C_{B} = 0.05 \) mg/ml, substitute the values:\[ m_{A} = \frac{0.2 \, \text{mg/mg adsorbent} \times 0.1 \, \text{mg/ml}}{0.1 \, \text{mg/ml} + 0.1 \, \text{mg/ml}} = 0.1 \frac{\text{mg}}{\text{mg adsorbent}} \]\[ m_{B} = \frac{0.05 \, \text{mg/mg adsorbent} \times 0.05 \, \text{mg/ml}}{0.02 \, \text{mg/ml} + 0.05 \, \text{mg/ml}} = 0.0357 \frac{\text{mg}}{\text{mg adsorbent}} \]

Calculate Total Mass Adsorbed

Determine the total mass of adsorbate A and B in the column:\[ M_{A} = m_{A} \times \text{mass of adsorbent} = 0.1 \frac{\text{mg}}{\text{mg adsorbent}} \times 3000 \, \text{mg} = 300 \, \text{mg} \]\[ M_{B} = m_{B} \times \text{mass of adsorbent} = 0.0357 \frac{\text{mg}}{\text{mg adsorbent}} \times 3000 \, \text{mg} = 107.1 \, \text{mg} \]

Compute Porous Volume of the Column

Given the bed porosity (\( \, \varepsilon = 0.35 \)), determine the porous volume:\[ V_{p} = \varepsilon \times \text{bed volume} = 0.35 \times 150 \, \text{ml} = 52.5 \, \text{ml} \]

Determine Volumes Occupied by Compounds

Determine the volumes occupied by the compounds A and B:\[ V_{A} = M_{A} \div C_{A} = 300 \, \text{mg} \div 0.1 \, \text{mg/ml} = 3000 \, \text{ml} \]\[ V_{B} = M_{B} \div C_{B} = 107.1 \, \text{mg} \div 0.05 \, \text{mg/ml} = 2142 \, \text{ml} \]

Calculate Band Positions of A and B

Ratio of A and B volumes to porous volume gives the band positions (note: step simplified for the problem context):\[ L_{A} = \frac{3000}{52.5} = 57.14 \, \text{cm} \]\[ L_{B} = \frac{2142}{52.5} = 40.8 \, \text{cm} \]

Compute Relative Positions and Rf Values

The ratio of the band positions, and Rf values given the arbitrary characteristic column length \( L_{c} = 150 \, \text{ml} \):\[ \frac{L_{A}}{L_{B}} = \frac{57.14}{40.8} \approx 1.4 \]\[ R_{fA}=\frac{L_{A}}{L_{c}} = \frac{57.14}{150} \approx 0.38 \]\[ R_{fB}=\frac{L_{B}}{L_{c}} = \frac{40.8}{150} \approx 0.27 \]

Key concepts

adsorption isotherms

Adsorption isotherms describe how different solutes interact with an adsorbent. They graphically or mathematically represent the relationship between the amount of solute adsorbed onto the adsorbent and the solute concentration in liquid at equilibrium. For a binary mixture, components A and B follow specific isotherms.
Using the given formulas: \[ m_{A}=f_{A}(c)=\frac{k_{1} C_{A}}{k_{2}+C_{A}} \ m_{B}=f_{B}(c)=\frac{k_{1}^{\prime} C_{B}}{k_{2}^{\prime}+C_{B}} \] allows us to calculate the equilibrium mass of each solute adsorbed by substituting the respective constants and concentrations. Here, the equilibrium mass of solute A, denoted as \(m_{A}\), depends on the values \(k_{1}\), \(k_{2}\), and the concentration \(C_{A}\). Similarly, for solute B, \(m_{B}\) depends on \(k_{1}^{\prime}\), \(k_{2}^{\prime}\), and \(C_{B}\). This fundamental concept helps understand how solutes distribute between liquid and solid phases under equilibrium conditions.

binary mixture separation

Binary mixture separation in chromatography involves differentiating two components, A and B, as they move through a column. Each component interacts uniquely with the adsorbent, affecting its path and speed.
The isotherms determine how much of component A or B is adsorbed at given concentrations, impacting their separation. The differing adsorption properties lead to distinct band positions, allowing for effective separation. The varying speeds at which components travel result in different migration distances or bands.
In our example, constants like \(k_{1}\), \(k_{2}\), etc., and initial concentrations \(C_{A}\) and \(C_{B}\) are crucial. They define the behavior of each solute, helping to calculate the final band positions, denoted by \(L_A\) and \(L_B\). Understanding these properties is essential in achieving clear and effective separation of the binary mixture.

mass adsorbate calculation

Mass adsorbate calculation is key in determining the total mass of solute adsorbed on the column's adsorbent. This mass is calculated using the equilibrium mass from the isotherms and the total adsorbent mass.
For component A, using \(m_{A}\) from the isotherm and multiplying with the adsorbent mass (3g), we find: \[ M_{A} = m_{A} \times \text{mass of adsorbent} \ M_{B} = m_{B} \times \text{mass of adsorbent} \]
Inserting the numerical values, we identify the total adsorbed mass. This calculation tells how much solute is bound to the adsorbent within the column, assisting in understanding the separation efficiency and column performance. Component B is calculated similarly using its respective isotherm.

bed porosity

Bed porosity (\( \, \varepsilon\)) is the fraction of the column volume occupied by voids. It impacts the flow behavior of the eluent and solutes within the column.
Formulated as: \[ V_{p} = \varepsilon \times \text{bed volume} \] In our example, with a bed volume of 150 ml and porosity 0.35, the porous volume is: \[ V_{p} = 0.35 \times 150 = 52.5 \, \text{ml} \]
Understanding the porous volume is critical as it directly influences the space available for the mobile phase to move through the bed. This volume helps us compute the travel distance of each solute band within the column, vital for predicting band positions and ensuring efficient separation.

band position calculation

Band positions denote the spatial location of solutes within the column during separation. It relates to the total volumes in the stationary phase and mobile phase within the bed.
Using the total masses of adsorbed solutes, and porous volume, band positions \(L_A\) and \(L_B\) are calculated thus: \[ L_{A} = \frac{V_{A}}{V_{p}} \ L_{B} = \frac{V_{B}}{V_{p}} \] Inserting our given numbers: \[ L_{A} = \frac{3000}{52.5} \ L_{B} = \frac{2142}{52.5} \] yields the band positions for A and B. These ratios explain how far each solute is located along the column. Knowing these distances helps predict where concentrations peak, informing about the separation quality and resolution.

Rf values

The retention factor \(R_f\) values describe relative migration distance of solutes within the column compared to a reference position.
Calculated as: \[ R_{fA}=\frac{L_{A}}{L_{c}} \ R_{fB}=\frac{L_{B}}{L_{c}} \] Using a reference column length \(L_{c}\) of 150 ml: \[ R_{fA} = \frac{57.14}{150} \ R_fB = \frac{40.8}{150} \] gives the \(R_f\) values for solutes A and B.
These values, 0.38 for A and 0.27 for B, indicate how far each solute travels relative to \(L_{c}\). A higher \(R_f\) signifies a solute that moves faster and further. \(R_f\) values are essential for comparing different solutes and assessing chromatographic performance.