Adsorption of acetic acid on charcoal and validity of Freundlich’s adsorption isotherm and Langmuir’s adsorption isotherm
Adsorption of acetic acid on charcoal is the topic of discus for today. Adhesion to a surface is called adsorption of atoms, ions, biomolecules, or gas, liquid, or dissolved solids molecules. The activated charcoal adsorbs acetic acid. The activated charcoal adsorbs the acids in their porous the meanwhile this process follows the adsorption isotherm of Freundlich as well as isotherm of Langmuir. Adsorption of acetic acid on activated charcoal is a very important topic to understand the validity of Freundlich’s adsorption isotherm as well as Langmuir’s adsorption isotherm.
The amount adsorbed by an adsorbent is dependent on pressure and temperature. Thus the amount (X) adsorbed is a function of pressure (P) and temperature (T), i.e.
X= f(P,T)
A plot of P vs X, keeping temperature constant is known as adsorption isotherm.
Freundlich’s adsorption isotherm:
Freundlich’s adsorption isotherm; represented by –
X/m = kc^{1/n }
where m= the amount of adsorbing material C is the equilibrium constant of adsorbate in the solution k is the constant depending on the nature of both adsorbate and adsorbant n is another constant depending on the nature of both adsorbates.
The value of 1/n is less than unity. On taking the logarithm of the equation,
log (x/m)= 1/n log c + log k —————————— (2)
If a plot of log (x/m) vs log c gives a straight line with a slope of 1/n and intercept of log k, Freundlich’s adsorption isotherm is valid
Langmuir’s adsorption isotherm:
According to Langmuir for a unimolecular layer, the following relation holds for adsorption,
x/m = k_{1}k_{2}c/(1+k_{1}c)
where k_{1} and k_{2} are the constant.
The above equation can be written as
c/(x/m) = 1/ k_{1}k_{2} + c/k_{2}
Thus a plot of c/(x/m) vs c gives a straight line with a slope of 1/k_{2} and an intercept of 1/ k_{1}k_{2} and shows the validity of Langmuir adsorption isotherm
Apparatus and chemicals:
Burette, pipette, Activated charcoal, 0.5N acetic acid, 0.1N NaOH, Stopped bottles, Shaking machine, etc.
Procedure:
 Prepare 0.5N acetic acid as well as 0.1N NaOH solution.
 Take 6 clean and dry stopped bottle and prepare the following solutions in the table
Bottle No  0.5N Acetic acid(mL)  Distilled water (mL)  Amount of charcoal (g) 
01.  50  0  1 
02.  40  10  1 
03.  30  20  1 
04.  25  25  1 
05.  20  30  1 
06.  10  40  1 
 Spotted each bottle after that shake for at least 1 hour.
 Filter the solution through a filter paper as well as collect the filtrate in the numbered bottle. reject the first 5 mL of each filtrate.
 Pipette out 10 mL of each filtrate and titrate with 0.1N NaOH solution.
 Note Down the temperature of the solution.
Data :
Bottle No: 01
Activated solution (mL)  IBR  FBR  BR difference  mean 
10  0  a_{1}  a_{1}  let the mean be x_{1} 
10  a_{1}  b_{1}  a_{1}b_{1}  
10  b_{1}  c_{1}  b_{1}c_{1} 
Bottle No: 02
Activated solution (mL)  IBR  FBR  BR difference  mean 
10  0  a_{2}  a_{2}  let the mean be x_{2} 
10  a_{2}  b_{2}  a_{2}b_{2}  
10  b_{2}  c_{2}  b_{2}c_{2} 
Bottle No: 03
Activated solution (mL)  IBR  FBR  BR difference  mean 
10  0  a_{3}  a_{3}  let the mean be x_{3} 
10  a_{3}  b_{3}  a_{3}b_{3}  
10  b_{3}  c_{3}  b_{3}c_{3} 
Bottle No: 04
Activated solution (mL)  IBR  FBR  BR difference  mean 
10  0  a_{4}  a_{4}  let the mean be x_{4} 
10  a_{4}  b_{4}  a_{4}b_{4}  
10  b_{4}  c_{4}  b_{4}c_{4} 
Bottle No: 05
Activated solution (mL)  IBR  FBR  BR difference  mean 
10  0  a_{5}  a_{5}  let the mean be x_{5} 
10  a_{5}  b_{5}  a_{5}b_{5}  
10  b_{5}  c_{5}  b_{5}c_{5} 
Bottle No: 06
ActivatedActivaated solution (mL)  IBR  FBR  BR difference  mean 
10  0  a_{6}  a_{6}  let the mean be x_{6} 
10  a_{6}  b_{6}  a_{6}b_{6}  
10  b_{6}  c_{6}  b_{6}c_{6} 
Calculation:
Bottle No  Initial conc. of the acetic acid before adsorption (C_{0})  Equilibrium conc. of the acetic acid before adsorption (C_{e})  Amount of acetic acid adsorped (C_{0}C_{e}) 
01  50  x_{1}  y_{1} 
02  40  x_{2}  y_{2} 
03  30  x_{3}  y_{3} 
04  25  x_{4}  y_{4} 
05  20  x_{5}  y_{5} 
06  10  x_{6}  y_{6} 
Table for the prof of Freundlich’s adsorption isotherm
Bottle No.  Amount of acetic acid adsorped (C_{0}C_{e}) = (x/m)  log(x/m)  C_{e}= C  log C 
01  y_{1}  log y_{1}  x_{1}  log x_{1} 
02  y_{2}  log y_{2}  x_{2}  log x_{2} 
03  y_{3}  log y_{3}  x_{3}  log x_{3} 
04  y_{4}  log y_{4}  x_{4}  log x_{4} 
05  y_{5}  log y_{5}  x_{5}  log x_{5} 
06  y_{6}  log y_{6}  x_{6}  log x_{6} 
Freundlich’s adsorption equation
log(x/m) = logk +1/n logC
from the above graph, we can see that the graph is a straight line and the slope is 1/n the intercept is log k. we can find the value of k from the intercept.
Table for the plot of Langmuir adsorption isotherm:
Bottle No.  Equilibrium conc. of the acetic acid before adsorption (C_{e})  Amount of acetic acid adsorped (C_{0}C_{e}) = (x/m)  C_{e }/(x/m) 
01  x_{1}  y_{1}  x_{1}/y_{1} 
02  x_{2}  y_{2}  x_{2}/y_{2} 
03  x_{3}  y_{3}  x_{3}/y_{3} 
04  x_{4}  y_{4}  x_{4}/y_{4} 
05  x_{5}  y_{5}  x_{5}/y_{5} 
06  x_{6}  y_{6}  x_{6}/y_{6} 
From the Langmuir equation
Ce/(x/m) = 1/(k_{1}k_{2}) + c/k_{2}
from the above graph, we can see that the graph is a straight line and the slope is 1/k_{2} the intercept is 1/(k_{1}k_{2}). we can find the value of k_{2} from the slop. After that, we can also find the k_{1} from the intercept and the value of k_{2}.
The concentration of acetic acid after adsorption:
 Initial concentration of acetic acid is = 0.50N
 Calculation of the equilibrium concentration of acetic acid:
we know, V_{a}S_{a} = V_{b}S_{b}
S_{a}=V_{b}S_{b}/V_{a} 
here,

Bottle No.  S_{a} 
01  S_{1} 
02  S_{2} 
03  S_{3} 
04  S_{4} 
05  S_{5} 
06  S_{6} 
Result:
 The validity of Freundlich’s adsorption isotherm is proved from the above first plot.
 The validity of the Langmuir adsorption isotherm is proved from the above second plot.
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Please how was the concentration before and after adsorption gotten?
nice one brother. Thanks a lot.
Please how was concentration before and after adsorption gotten