--%>

Explain the process of adsorption in solution.

The process of adsorption can occurs in solutions also. This implies that the solid surfaces can also adsorb solutes from solutions. Some clarifying examples are listed below:


(i) When an aqueous solution of ethanoic acid (CH3COOH) is shaken with charcoal. The concentration of acid in the solution is found to decrease because a part of acid is adsorbed by charcoal.

(ii) Litmus solution become colourless when it is shaken with charcoal which indicates that the dye of litmus solution is adsorbed by charcoal.

(iii) Mg(OH)2 is colorless but when it is precipitated in the presence of magneton reagent (a bluish dye) the precipitate acquire blue colour due to adsorption of dye on their surface.

It has also been observed that if we have a solution containing more than one solutes, then the solid adsorbents can adsorb certain solutes in preference to other solutes and solvents. The property is made use of in removing colouring matters from solutions of organic substances. For example, raw sugar solution is decolourised by animal charcoal.

Factors affecting adsorption from solutions

Adsorption from solutions by the solid adsorbents depends on the following factors.

(i) Nature of adsorbent and adsorbate

(ii) Temperature: the extent of adsorption, in general, decreases with rise in the surface area of adsorbent.

(iii) Surface area: the extent adsorption increase with the increase in the surface area of adsorbent.

(iv) Concentration: the effect of concentration on the extent of adsorption isotherm similar to Freundlich adsorption isotherm. The relationship between mass of solute adsorbed per gram (x/m) is given by the following relations:
                                                 
x/m = kC1/n              (n > 1)

Taking logarithm, equation becomes
                                               
log x/m = log k + (1/n) log C

A graph between log x/m and log C is a straight line, similar to graph.

Experimental verification of equation can be done by taking equal volume of acetic acid solutions of different concentrations in four different bottles fitted with stoppers. The contents should be well mixed. The concentration of acetic acid is found in each bottle. This concentration refers to equilibrium concentration (c). The difference of original concentration and equilibrium concentration gives the value of amount of acetic acid adsorbed by charcoal (x)log (x/m) values of different plotted against C

   Related Questions in Chemistry

  • Q : Thermodynamics 1 Lab Report I already

    I already did Materials and Methods section. I uploaded it with the instructions. Also, make sure to see Concept Questions and Thinking Ahead in the instructions that I uploaded. deadline is tomorow at 8 am here is the link to download all instructions because I couldn't attach all of t

  • Q : Calculation of molecular weight Provide

    Provide solution of this question. In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (mol. mass = 58) at 298K. The vapour pressure of the solution was found to be 192.5 mm Hg. The molecular weight of the solute is (vapour pressure of ace

  • Q : Neutralization of benzoic acid Choose

    Choose the right answer from following. How many grams of NaOH will be required to neutralize 12.2 grams of benzoic acid : (a) 40gms (b) 4gms (c)16gms (d)12.2gms

  • Q : Water under pressure problem-henry law

    Can someone help me in going through this problem. The statement “When 0.003 moles of a gas are dissolved in 900 gm of water under a pressure of 1 atm, 0.006 moles will be dissolved under the pressure of 2 atm", signfies: (a)

  • Q : How molecule-molecule collisions takes

    An extension of the kinetic molecular theory of gases recognizes that molecules have an appreciable size and deals with molecule-molecule collisions. We begin studies of elementary reactions by investigating the collisions b

  • Q : Problem on equilibrium constant Ethanol

    Ethanol is manufactured from carbon monoxide and hydrogen at 600 K and 20 bars according to the reaction2 C0(g) + 4 H2(g) ↔ C2H5OH(g) + H2O (g)The feed stream contains 60 mol% H2, 20 m

  • Q : Problem on mole fraction of glucose

    Provide solution of this question. While 1.80gm glucose dissolve in 90 of H2O , the mole fraction of glucose is: (a) 0.00399 (b) 0.00199 (c) 0.0199 (d) 0.998

  • Q : What are various structure based

    This classification of polymers is based upon how the monomeric units are linked together. Based on their structure, the polymers are classified as: 1. Linear polymers: these are the polymers in which monomeric units are linked together to form long straight c

  • Q : Difference in Mendeleevs table and

    Briefly describe the difference in the Mendeleev’s table and modern periodic table?

  • Q : Lowering of vapour pressure Help me to

    Help me to go through this problem. Lowering of vapour pressure is highest for: (a) urea (b) 0.1 M glucose (c) 0.1M MgSo4 (d) 0.1M BaCl2