Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: AussieKenDoll on March 05, 2019, 01:50:42 PM
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Is this plotted graphs accord with Langmuir and Freundlich isotherms?
Please see the attached pic
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As the axes aren't labelled, it's impossible to say. Tell us what you did and what you have plotted.
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These graphs are for adsorption of acetic acid on charcoal!
On freundlich isotherm graph, x axis -log c ( c= equilibrium concentration) y axis - log(x/m) where x= adsorbed charcoal mass, m is the mass of adsorbent
On langmuir isotherm graph , x axis - c ( c= equilibrium concentration) y axis - c/(x/m)
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Have you tried fitting a Freundlich or Langmuir curve to the data? What do you expect the curves to look like?
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I am not sure if my graphs are correct because Freundlich isotherm gives me a negative intercept and a positive slope and Langmuir isotherm gives me a negative slope and a positive intercept which is unusual?
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Freundlich should give a positive slope, sign of intercept depends on sign of log k, could be negative or positive.
Langmuir (as you have plotted it) should give positive slope and positive intercept. Your plot isn't even a straight line.
I have concerns about your data (they might be fine, but I don't understand the procedure). Why is adsorbed mass not equal to initial mass - equilibrium mass? How are any of these related to consumed moles of NaOH? Is m exactly 1 in every case? Is c the initial concentration, or the equilibrium concentration after absorption? How was it measured?
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This is the procedure for calculation according to my lab guide
Calculate the initial mass of acetic acid in the first six flasks from the determined strength of acid.
Calculate the mass of acid in the solution in equilibrium with the adsorbent from the titration results.
Calculate the mass of acetic acid adsorbed in each case x= (I) -(II) and hence x/m.
Calculate the equilibrium concentration c in mass of acid per liter of solution from (II).
Plot log (x/m) against log c and test the applicability of the Freundlich isotherms. Deduce k and n.
log(x/m)=nlogc+logk
Plot c/(x/m) against c and the test the applicability of the Langmuir isotherm. Deduce alpha and beta
c/(x/m) =(1/α) + (β/α)c
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Why is adsorbed mass not equal to initial mass - equilibrium mass? How are any of these related to consumed moles of NaOH? Is m exactly 1 in every case? Is c the initial concentration, or the equilibrium concentration after absorption? How was it measured?
Should usually adsorbed mass equal to initial mass?
What NaOH react with?
This is the procedure for this experiment
Cleaned and dried 7 stoppered bottles and placed 1 g of charcoal (weighted accurately as possible) in each stoppered bottle.
Then, into 6 bottles, transferred 50.0, 40.0, 30.0, 20.0, 10.0 and 5 cm3 respectively using a 25 cm3 graduated pipette and added approximately 0.5 mol dm-3 acetic acid and made up the total volume in each to 50.0 cm3 with distilled water.
Added 50 cm3 of distilled water to the seventh stoppered bottle.
Thoroughly shaken the all 7 stoppered bottles and left it to adsorb for 30 minutes by periodically shaking all stoppered bottles.
Meanwhile standardize the given acetic acid solution using 0.10 mol dm-3 NaOH.
Filtered each solution into 7 beakers through a small filter papers rejecting the first 10 cm3 because the adsorption may have occurred on the filter paper.
Titrated two portions of 10 cm3 of each filtrate and two portions from the seventh bottle with standard 0.10 mol dm-3 NaOH solution.
Yes m= mass of adsorbent which is charcoal, we measured it as 1 gram for all bottles
C is equilibrium concentration c in mass of acid per liter of solution calculated from the mass of acid in the solution in equilibrium with the adsorbent from the titration results.
I have attached the excel file before please look at it again for the NaOH consumed
Corrected volume is because the charcoal we used on the experiment is basic so we corrected it for all the titration values
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I)Calculate the initial mass of acetic acid in the first six flasks from the determined strength of acid.
II)Calculate the mass of acid in the solution in equilibrium with the adsorbent from the titration results.
Calculate the mass of acetic acid adsorbed in each case x= (I) -(II) and hence x/m.
Calculate the equilibrium concentration c in mass of acid per liter of solution from (II).
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I'm sorry, I can't make any sense of your data.
Should usually adsorbed mass equal to initial mass?
No, initial mass minus equilibrium mass. Sorry if that wasn't clear. That is what you say you've done (x = (I) - (II) ). But your numbers don't agree with that. x is not equal to I - II.
The initial masses, if I understand your procedure correctly, ought to correspond to 5, 10, 20, 30, 40 and 50 mL AcOH, and should thus be in the ratio 1:2:4:6:8:10. But they are not; they are in the ratio 1:4:16:36:64:100. (Square of the correct ratios?)
Either your numbers are wrong, or you didn't follow the procedure you've written.
Please show, for at least one of your samples, the actual calculations you did to get those numbers.
Otherwise, with the numbers as they stand, they seem consistent with Freundlich but not Langmuir.
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With what does NaOH reacts when we titrate the filtrated solution? Is there remaining un-adsorbed HAc which react with NaOH or what?
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I corrected some of the calculation as you stated, this is the graphs and I got.
Please review
Please see the new excel spreadsheet from this link i uploaded to my onedrive
https://campuen-my.sharepoint.com/:x:/g/personal/mic815081_365office_site/EVarMQ1U8UJArYKHyP_UpusBqKhDBsnZYHtMjis_HGgMyQ?e=TX9dba
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Ah, I can see what you're doing now. There is still a mistake in column M - the volume you must multiply by is the solution volume of 50 ml, not the volume of NaOH consumed (G2 etc.) Correct this, and you'll get a reasonable straight line for Langmuir, not so good for Freundlich.
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Wow! This is the graphs i got!
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Thank you soo much for the *delete me* Now I can complete my lab report! <3