--%>

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 : Question associated to vapour pressure

    Choose the right answer from following. The vapour pressure lowering caused by the addition of 100 g of sucrose(molecular mass = 342) to 1000 g of water if the vapour pressure of pure water at 25degree C is 23.8 mm Hg: (a)1.25 mm Hg (b) 0.125 mm Hg (c) 1.15 mm H

  • Q : Concentration factor affected by

    Can someone please help me in getting through this problem. Which of the given concentration factor is affected by the change in temperature: (1) Molarity (2) Molality (3) Mole fraction (4) Weight fraction

  • Q : Mole fraction in vapours Choose the

    Choose the right answer from following. If two substances A and B have P0A P0B= 1:2 and have mole fraction in solution 1 : 2 then mole fraction of A in vapours: (a) 0.33 (b) 0.25 (c) 0.52 (d) 0.2

  • Q : What are the various types of drugs

    Drugs are broadly classified into following types depending on the purpose for which they are used. 1. Antipyretics

  • Q : Problem on convection coefficient An

    An experiment to determine the convection coefficient associated with airflow over the surface of a thick stainless steel casting involves insertion of thermocouples in the casting at distances of 10 mm and 20 mm from the surface.  When the experiment was perform

  • Q : Question on molality Provide solution

    Provide solution of this question. Which of the following concentration factor is affected by change in temperature : (a)Molarity (b) Molality (c)Mole fraction (d)Weight fraction

  • Q : Explain the polymers and its types.

    Polymers are the chief products of modern chemical industry which form the backbone of present society. Daily life without the discovery and varied applications of polymers would not have been easier and colourful. The materials made of polymers find multifarious uses and applications in all walk

  • Q : Molar mass of solute The boiling point

    The boiling point of benzene is 353.23 K. If 1.80 gm of a non-volatile solute was dissolved in 90 gm of benzene, the boiling point is increased to 354.11 K. Then the molar mass of the solute is: (a) 5.8g mol-1  (b)

  • Q : Calculate molarity of a solution

    Provide solution of this question. Molarity of a solution prepared by dissolving 75.5 g of pure KOH in 540 ml solution is: (a) 3.05 M (b) 1.35 M (c) 2.50 M (d) 4.50 M

  • Q : Problem based on molarity Select the

    Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal