--%>

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 : 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 : Difference among hcl gas and hcl acid

    What is the basic difference among hcl gas and hcl acid? Briefly state the difference?

  • Q : Explain vapour pressure of liquid

    Liquid solutions are obtained when the solvent is liquid. The solute can be a gas, liquid or a solid. In this section we will discuss the liquid solutions containing solid or liquid solutes. In such solutions the solute may or may not be volatile. We shall limit our d

  • Q : Unit of mole fraction Provide solution

    Provide solution of this question. Unit of mole fraction is: (a) Moles/litre (b) Moles/litre2 (c) Moles-litre (d) Dimensionless

  • Q : Lab question Explain how dissolving the

    Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid, establishes a buffer with a pH of approximately 5.

  • Q : Colligative property associated question

    Give me answer of this question. Which of the following is not a colligative property : (a)Optical activity (b)Elevation in boiling point (c)Osmotic pressure (d)Lowering of vapour pressure

  • Q : Base parachloroaniline is strong base

    parachloroaniline is strong base than paranitroaniline

  • Q : Explain Second Order Rate Equations.

    Integration of the second order rate equations also produces convenient expressions for dealing with concentration time results.A reaction is classified as second order if the rate of the reaction is proportional to the square of the concentration of one o

  • Q : Vapour pressure of the pure hydrocarbons

    Give me answer of this question. A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20°C are 440 mmHg for pentane and 120 mmHg for hexane. The mole fraction of pentane in the vapour phase would be: (a) 0.549 (b)

  • Q : Question relatede to calculate molarity

    Select the right answer of the question. What is molarity of a solution of HCl that contains 49% by weight of solute and whose specific gravity is 1.41 : (a) 15.25 (b) 16.75 (c) 18.92 (d) 20.08