Adsorption of acetic acid on charcoal and the isotherm

Adsorption of acetic acid on charcoal and the isotherm

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 in 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.

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Download PDF of Adsorption of acetic acid on charcoal and validity of Freundlich’s adsorption isotherm and Langmuir’s adsorption isotherm

Freundlich’s adsorption isotherm:

Freundlich’s adsorption isotherm; represented by –

X/m = kc1/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

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Langmuir’s adsorption isotherm:

According to Langmuir for a uni-molecular layer, the following relation holds for adsorption,

x/m = k1k2c/(1+k1c)

where k1 and k2 are the constant.

The above equation can be written as-

c/(x/m) = 1/ k1k2 + c/k2

Thus a plot of c/(x/m) vs c gives a straight line with a slope of 1/k2 and an intercept of 1/ k1k2 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.

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Data :

Download PDF of Adsorption of acetic acid on charcoal as well as the validity of Freundlich’s adsorption isotherm and Langmuir’s adsorption isotherm

Download PDF of Adsorption of acetic acid on charcoal as well as the validity of Freundlich’s adsorption isotherm as well as Langmuir’s adsorption isotherm

Bottle No: 01

Activaated solution (mL) IBR FBR BR difference mean
10 0 a1 a1 let the mean be x1
10 a1 b1 a1-b1
10 b1 c1 b1-c1

Bottle No: 02

Activaated solution (mL) IBR FBR BR difference mean
10 0 a2 a2 let the mean be x2
10 a2 b2 a2-b2
10 b2 c2 b2-c2

Bottle No: 03

Activaated solution (mL) IBR FBR BR difference mean
10 0 a3 a3 let the mean be x3
10 a3 b3 a3-b3
10 b3 c3 b3-c3

Bottle No: 04

Activaated solution (mL) IBR FBR BR difference mean
10 0 a4 a4 let the mean be x4
10 a4 b4 a4-b4
10 b4 c4 b4-c4

Bottle No: 05

Activaated solution (mL) IBR FBR BR difference mean
10 0 a5 a5 let the mean be x5
10 a5 b5 a5-b5
10 b5 c5 b5-c5

Bottle No: 06

Activaated solution (mL) IBR FBR BR difference mean
10 0 a6 a6 let the mean be x6
10 a6 b6 a6-b6
10 b6 c6 b6-c6

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Calculation:

Bottle No Initial conc. of the acetic acid before adsorption (C0) Equlibrium conc. of the acetic acid before adsorption (Ce) Amount of acetic acid adsorped (C0-Ce)
01 50 x1 y1
02 40 x2 y2
03 30 x3 y3
04 25 x4 y4
05 20 x5 y5
06 10 x6 y6

Table for the prof of Freundlich’s adsorption isotherm

Bottle No. Amount of acetic acid adsorped (C0-Ce) = (x/m) log(x/m) Ce= C log C
01 y1 log y1 x1 log x1
02 y2 log y2 x2 log x2
03 y3 log y3 x3 log x3
04 y4 log y4 x4 log x4
05 y5 log y5 x5 log x5
06 y6 log y6 x6 log x6

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 slop 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. Equlibrium conc. of the acetic acid before adsorption (Ce) Amount of acetic acid adsorped (C0-Ce) = (x/m) Ce /(x/m)
01 x1 y1 x1/y1
02 x2 y2 x2/y2
03 x3 y3 x3/y3
04 x4 y4 x4/y4
05 x5 y5 x5/y5
06 x6 y6 x6/y6

From the Langmuir equation

Ce/(x/m) = 1/(k1k2) + c/k2

from the above graph we can see that the graph is a straight line and the slop is 1/k2 the intercept is 1/(k1k2). we can find the value of k2 from the slop. After that we can also find the k1 from the intersept and value of k2.

Concentration of aceticacid after adsorption:

  1. Initial concentration of acetic acid is = 0.50N
  2. Calculetion of the equlibrium concentration of acetic acid:
we know, VaSa = VbSb

Sa=VbSb/Va

here,

  • Va= Volume of Aceticacid
  • Vb= Volume of NaOH
  • Sa= Concentration of Aceticacid
  • Sb=Concentration of NaOH
Bottle No. Sa
01 S1
02 S2
03 S3
04 S4
05 S5
06 S6

Result:

  1. Validity of the Freundlich’s adsorption isotherm is proved from the above first plot. 
  2. Validity of the Langmuir adsorption isotherm is proved from the above second plot. 

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