0
$\begingroup$

I have done a biosorption study using a protein as a biosorbent and six different concentrations of $\ce{Pb}$ ions. I need to show the efficiency of the biosorbent, I was told to use the Langmuir isotherm to represent these results. Thus far, I have plotted a graph of $q$ vs. $C_\mathrm f$ using the formula

$$q = (C_\mathrm i-C_\mathrm f)\times\frac{V}{m}$$

I have also derived my $R$ value from the straight line obtained. How do I get to the Langmuir isotherm from here? What do I need to calculate and what I should plot to represent the data in Langmuir isotherm? Also, I am a biologist so I would appreciate step by step help as I am very confused with all the different equations for Langmuir.

$\endgroup$
3
  • $\begingroup$ Could you label your variables, so it's clear what $\mathrm{q}$, $\mathrm{C_i}$, $\mathrm{C_f}$, $\mathrm{V}$, $\mathrm{m}$ and $\mathrm{R}$ are? $\endgroup$
    – user41033
    Commented Mar 10, 2018 at 14:19
  • $\begingroup$ Sure, Ci -- initial concentration of Pb ions (ug/L) Cf- residual concentration of Pb ions (ug/L), V- volume of the reaction (L), and m - mass of biosorbent (g). $\endgroup$
    – user60108
    Commented Mar 10, 2018 at 14:23
  • $\begingroup$ So $\mathrm{q}$ is the amount of absorbed material per mass of biosorbent, is that right? $\endgroup$
    – user41033
    Commented Mar 10, 2018 at 15:00

1 Answer 1

2
$\begingroup$

Introduction

The Langmuir isotherm is a theoretical equation that describes the reversible adsorption of a non-interacting adsorbate on a surfaced that contains a fixed number of identical sites. Therefore, the main assumptions of the model are:

  • There is a maximum amount of adsorbate that can be fixed to the surface, $\mathrm{S_{max}}$. In other words, the adsorption can reach a saturation point.
  • All sites are equivalent, and there are no other factors (for instance roughness) that make sites more available or more binding than others.
  • Sites are not affected by the status of neighbouring sites; all sites are independent. So, neither the local nor the global degree of saturation impacts how likely it is for a single adsorbate particle to adsorb or desorb from a site.

With these assumptions, an expression can be found (it can be derived in terms of chemical equilibrium, chemical kinetics or statistical mechanics) that expresses the relationship between the adsorbed amount of material, $\mathrm{S}$, and the concentration of the adsorbate in the adjacent phase, $\mathrm{C}$:

$\mathrm{S = \frac{K_a C}{1 + K_a C}S_{max}}$

where $\mathrm{S}$ is the amount of adsorbed material, $\mathrm{S_{max}}$ is the maximum possible amount of adsorbed material (in the saturation point), $\mathrm{K_a}$ is the equilibrium constant (written in the direction of adsorption) that regulates the adsorption/desorption of an adsorbate particle on a single site, and $\mathrm{C}$ is the concentration of the adsorbate on the adjacent phase.

It is very common to make the equation intensive by expressing the amount of adsorbed material as a fraction of the maximum possible amount, called coverage ($\mathrm{\theta}$):

$\mathrm{\theta=\frac{S}{S_{max}}=\frac{K_a C}{1+K_a C}}$

Linearisation

The main disadvantage of that formulation of the Langmuir isotherm is that it's not linear, and fitting data to that expression is not straightforward. Consequently, there are various linearisation schemes that have been employed in the literature:

Hanes-Woolf: $\mathrm{\frac{C}{S}=\frac{1}{K_a S_{max}}+\frac{C}{S_{max}}}$

Lineweaver-Burke: $\mathrm{ \frac{1}{S} = \frac{1}{K_a S_{max}} \frac{1}{C} +\frac{1}{S_{max}} }$

Eadie-Hofstee: $\mathrm{S=S_{max}-\frac{1}{K_a}\frac{S}{C}}$

Scatchard: $\mathrm{\frac{S}{C}=K_a S_{max}-K_a S}$

All of them allow for a linear relationship involving $\mathrm{S}$ and $\mathrm{C}$ to be plotted. Although in principle all equations are equivalent to the formulation above and should lead to the same values, in practice they affect the fitting of real data differently.

In my experience, Lineweaver-Burke is the most commonly used as it introduces less bias in the estimation of the values as other linearisations, but it has the drawback of being less precise at low values of $\mathrm{S}$.

Lineweaver-Burke

So, to fit your data to a Lineweaver-Burke equation, you need to plot $\mathrm{1/S}$ (y axis) vs. $\mathrm{1/C}$ (x axis). Your points will then fit a linear equation of the type

$\mathrm{y=a+bx}$

where

$\mathrm{y=\frac{1}{S}}$

$\mathrm{x=\frac{1}{C}}$

$\mathrm{a=\frac{1}{S_{max}}}$

$\mathrm{b=\frac{1}{K_a S_{max}}}$

Your example

As far as I can tell, you express the adsorbed amount of material as a load $\mathrm{q}$ with units of micrograms of adsorbate per gram of biosorbent:

$\mathrm{q=\frac{n_{ads}}{m}=\frac{\Delta C·V}{m}=(C_f-C_i)\frac{V}{m}}$

so, in the formulation above, $\mathrm{q}$ corresponds to $\mathrm{S}$ and $\mathrm{C_f}$ corresponds to $\mathrm{C}$, and your Lineweaver-Burke expression is

$\mathrm{\frac{1}{q}=\frac{1}{K_a q_{max}}\frac{1}{C_f}+\frac{1}{q_{max}}}$

Plot $\mathrm{1/q}$ vs. $\mathrm{1/C_f}$ for your points and fit them to a linear equation. Your y-intercept corresponds to $\mathrm{1/q_{max}}$ and your slope corresponds to $\mathrm{1/K_a ·1/q_{max}}$. From there you can obtain the values for your equilibrium constant $\mathrm{K_a}$ and your saturation load $\mathrm{q_{max}}$.

$\endgroup$
2
  • 1
    $\begingroup$ "$K$ is the equilibrium constant that regulates the adsorption/desorption of an adsorbate particle on a single site" Correct me if I'm wrong, but isn't the $K$ for the adsorption and desorption processes different? $\endgroup$ Commented Mar 18, 2018 at 13:34
  • $\begingroup$ $k$ (the kinetic coefficient) for adsorption ($k_a$) and desorption ($k_d$) will in general be different, but $K$ (the kinetic pseudo-equilibrium constant) is precisely the quotient of these two kinetic coefficients. As with a true thermodynamic equilibrium constant, we can write it in both directions, but these are mathematically equivalent ($K_d = K_a^{-1}$ by definition). By convention, in adsorption chemistry, we always use the adsorption expression ($K_a=k_a/k_d$), so it often gets simplified to $K$. I've edited the answer to unambiguously identify that it's $K_a$. $\endgroup$
    – user41033
    Commented Mar 18, 2018 at 17:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.