--%>
Assignment Help - homework help

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 : Oxoacids of halogens Why oxidising

    Why oxidising character of oxoacids of halogens decreases as oxidation number increases?

    		•
  • Q : Calculation of concentration of the

    Choose the right answer from following. 200ml of a solution contains 5.85 dissolved sodium chloride. The concentration of the solution will be(Na= 23: cl = 35.5 ) (a) 1 molar (b) 2 molar (c) 0.5 molar (d) 0.25 molar

    		•
  • Q : Colligative property related question

    Select the right answer of the question. Which of the following is not a colligative property : (a) Osmotic pressure (b) Elevation in B.P (c) Vapour pressure (d) Depression in freezing point

    		•
  • Q : Which solution will have highest

    Which solution will have highest boiling point:(a) 1% solution of glucose in water  (b) 1% solution of sodium chloride in water  (c) 1% solution of zinc sulphate in water  (d) 1% solution of urea in waterAnswer: (b) Na

    		•
  • Q : Illustrations of the reversible reaction

    What are the various illustrations of the reversible reaction? Explain briefly?

    		•
  • Q : Molar mass of compound The freezing

    The freezing point of a solution having 4.8 g of a compound in 60 g of benzene is 4.48. Determine the molar mass of the compound (Kf = 5.1 Km-1) , (freezing point of  benzene = 5.5oC)          &n

    		•
  • Q : Isotonic Solutions Which one of the

    Which one of the following pairs of solutions can we expect to be isotonic at the same temperature:(i) 0.1M Urea and 0.1M Nacl  (ii) 0.1M Urea and 0.2M Mgcl2  (iii) 0.1M Nacl and 0.1M Na2SO4  (iv) 0.1M Ca(NO3<

    		•
  • Q : Reducible Representations The number of

    The number of times each irreducible representation occurs in a reducible representation can be calculated.Consider the C2v point group as described or Appendix C. you can see that (1) sum of

    		•
  • Q : Molarity based question Help me to

    Help me to solve this problem. 4.0 gm of NaOH are contained in one decilitre of solution. Its molarity would be: (a) 4 M (b)2 M (c)1 M (d)1.5 M

    		•
  • Q : Problem on molecular weight of solid

    The vapor pressure of pure benzene at a certain temperature is 200 mm Hg. At the same temperature the vapor pressure of a solution containing 2g of non-volatile non-electrolyte solid in 78g of benzene is 195 mm Hg. What is the molecular weight of solid: