How is itpossible that the sorption capacity of your material is higher for the dye 1 than dye 2 and your removal rate is the opposite?

If initially you had the same concentration for both dyes, adsorption capacity and removal percentage should be proportionnal.

Reminding:

qe=(Ci -Cf)*V/m = (qm*b*Ce)/(1+b*Ce)

=> qm=qe*Constant where Constant= (1+C*Ce)/ (b*Ce)

and % Removal= (Ci-Cf)/ Ci

This stated: %Removal should be in the same order of your sorption capacities.

On the the other hand, you should expect a relationship between the physico-chemical properties of the adsorbant(pollutant) and the sorbant; this is why the sorbant must be selective to your pollutant.

Hope this answers your question

Recheck your formulas.

Yes, the relationship do exist....you can find that relationship and more on https://www.researchgate.net/publication/329238484_ADSORPTION_OF_MALACHITE_GREEN_DYE

You just want to share your paper. Your journal provide no answer to my question.

Thanks Iwuozor Ogemdi Kingsley.

Adsorption effectiveness as well as adsorption capacity depend on adsorbate properties. Molecular weight can be one of them. Comparing adsorption isotherms of different adsorbates you will also consider methodological aspects and equilibria conditions. Was it jar test or column experiment? What was the adsorbent-adsorbate contact type – simultaneous or separate? Were initial concentrations of both dyes equal? What about equilibrium concentrations of both dyes? Regards Marek

Marek Mołczan , thanks for your response. Actually, the initial concentrations of both dyes are equal at separate and simultaneous experiments.

I believe that it has something to do with the kind of material pores (meso or micro)

How is itpossible that the sorption capacity of your material is higher for the dye 1 than dye 2 and your removal rate is the opposite?

If initially you had the same concentration for both dyes, adsorption capacity and removal percentage should be proportionnal.

Reminding:

qe=(Ci -Cf)*V/m = (qm*b*Ce)/(1+b*Ce)

=> qm=qe*Constant where Constant= (1+C*Ce)/ (b*Ce)

and % Removal= (Ci-Cf)/ Ci

This stated: %Removal should be in the same order of your sorption capacities.

On the the other hand, you should expect a relationship between the physico-chemical properties of the adsorbant(pollutant) and the sorbant; this is why the sorbant must be selective to your pollutant.

Hope this answers your question

Recheck your formulas.

Rita Abi Hanna Really appreciate your contribution. Actually % removal is proportional to adsorption capacity. Fitting the adsorption data to Langmuir isotherm model give LANGMUIR maximum sorption capacity (Qmax) that is opposite.

Dear Abel,

While I am more familiar with gaseous adsorption than liquid phase adsorption, there are several commonalities which might help.

If we consider the Langmuir equation, the maximum sorption capacity can be regarded as the total amount of active sites where the probe can adsorb. If adsorption takes place using only Langmuir assumptions (point-like probe, one probe per adsorption site, no lateral interactions), then the molar mass of your dye shouldn't impact the capacity at equilibrium, although kinetics may be affected. On the other hand, depending on the system, these assumptions may not hold and you may get large probes using up more than one site, different interactions with sites, surface ordering etc.

However, in a porous material, maximum sorption capacity can also be interpreted as the total pore volume. Here, the molar mass of your adsorbate will be inversely proportional with the total capacity. Smaller probe -> more can fit in the pore, and vice versa with a larger probe. This relationship can be quantified by assuming that the probe is in a liquid-like state when adsorbed (which may not be the case!), and using the liquid density to calculate pore volume. This is commonly used in adsorption characterization (like the t-plot method).

In a liquid system like yours, there is likely competition between the dye and the solvent for the adsorption sites, which introduces more complexity but should not change the ideas in the previous paragraphs.

I hope this helps,

Paul

Hi Paul Iacomi

Thanks for details explanation, quite relevant and helpful.

Generally, the Langmuir adsorption capacity is higher as the molecular mass of dye is higher. To compare the adsorption capacity for 2 different dyes, express this in mmol/g units. In this way, will be more obvious which dye has higher affinity for adsorbent during of adsorption process.

Adsorption isotherm experiments usually take about 24 hours in order to give the opportunity to saturate all the active sites in the adsorbent. Since Langmuir isotherm deals with mono layer of adsorbate molecules it is reasonable that the adsorbent will have higher maximum adsorption capacity in the case of adsorbate of higher molecular mass than the case of lower molecular mass. However, the kinetic experiments to calculate the removal efficiency could take shorter times For this reason big molecules will not have enough time to diffuse in the micro pores as in the isotherm experiments. Consequently, the removal efficiency will have opposite behavior.

Laura Bulgariu Thank you for insightful answer.

Zakaria Al-Qodah really appreciate your view and response to the question.