--%>

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 : Statement of Henry law Determine the

    Determine the correct regarding Henry’s law: (1) The gas is in contact with the liquid must behave as an ideal gas (2) There must not be any chemical interaction among the gas and liquid (3) The pressure applied must be high (4) All of these.

  • Q : Net charge of a non-ionized atom

    Describe the net charge of a non-ionized atom?

  • Q : Calculating number of moles from

    Choose the right answer from following. If 0.50 mol of CaCl2 is mixed with 0.20 mol of Na3PO4, the maximum number of moles of Ca3 (PO2)2 which can be formed: (a) 0.70 (b) 0.50 (c) 0.20 (d) 0.10

  • Q : Significance of the organic chemistry

    Describe some of the significance of the organic chemistry in brief?

  • Q : Question based on maximum vapour

    Provide solution of this question. Which has maximum vapour pressure: (a) HI (b) HBr (c) HCl (d) HF

  • Q : Group Cations Explain how dissolving

    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

  • 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 : Molar concentration of hydrogen 20 g of

    20 g of hydrogen is present in 5 litre of vessel. Determine he molar concentration of hydrogen: (a) 4  (b) 1 (c) 3 (d) 2 Choose the right answer from above.

  • 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 : Formula of diesel Write a short note on

    Write a short note on the formula of diesel, petrol and also CNG?