ADSORPTION ISOTHERM The process of adsorption involves the concentration or accumulation of gas, liquid or solid on the surface of liquid or solid with which it is in contact. Solid has the property of holding molecules at their surfaces, and this property is quite marked in the case of porous and finally divided materials. Adsorbent: material used to adsorb the gas, liquid or solid. e.g. kaolin, charcoal. Adsorbate: substance being adsorbed.

Adsorption isotherm: the relation ship between the amount of substance adsorbed and the amount existing in the bulk of the solution at a given temperature (at constant temperature). - The types of adsorption are generally recognized as physical or "Van der Waals" adsorption. and chemical adsorption or . "chemisorption".

The removal of the adsorbate from the adsorbent being known as desorption. A physically adsorbed gas may be desorbed from solid by increasing the temperature and reducing' the Pressure. - Chemical adsorption, in which the adsorbate is attached to the adsorbent by primary chemical bond, is irreversible.

The two main types of adsorption may be distinguished as follows: Chemical Physical 1-Heat of adsorption Low (4.kcal/mole) high (20kcal/mole) 2-Specificity Non-specific Specific 3-Reversibility Reversible Non-Reversible 4-Effect of temperature. Decrease with Increase increase or decrease the temperature . With increasing the temp. .

The amount of the solute adsorbed by a given quantity of adsorbent increase with increasing conc. of solution. Some time the layer of adsorbent is only one molecule deep, and further adsorption ceases with time. Why ? The study of adsorption of gases is concerned in such diverse applications as the removal of objectionable odors from rooms and food, the operation of gas masks, and the measurement of the dimensions of particles in a powder. - The principle of solid/ liquid adsorption are utilized in decoloring solutions, adsorption chromatography, detergency and wetting.

freundlich adsorption isotherm: Freundlich derived an equation for the adsorption of the dissolved solid on the surface of porous substances. The equation is: X α c m = K c where X = weight of material (adsorbate) in grams, adsorbed by m grams of adsorbent, C = the concentration of solute in g/100. n and k are constants. By taking the logarithm of the equation, we obtain: log x/m = log K + n log c According to this equation, a plot of logx/m versus logc, a ,straight line is obtained, and the constants may be evaluated. Intercept = log K. and the slope is n n n Slop = n Log x/m Log k Log c

Experimental work To study the adsorption of oxalic acid on kaolin & see the effect of increasing the concentration of oxalic acid on adsorption. Procedure: 1- Prepare 50 mls of the following conc. of oxalic acid 0.2, 0.4, 0.6, and 0.8 N from stock solution 1 N oxalic acid by dilution method. C1 V1 =C2 V2 2- Put 50 ml of each concentration & the stock solutions in 5 conical flask, 3- Introduce 2 g of kaolin into each flask. 4- Shake for 15 min. and set aside for. 15 min. to achieve equilibrium. 5- Filter, reject the first portion of the filtrate after washing the receiver with it. 6- Pipette 20 ml of the filtrate containing the non adsorbed oxalic acid (free) and titrate them N/2 NaOH using phenolphthalein as indicator.( E.P. is colorless to pink) 7- Calculate the amount adsorbed in each flask and plot Freundlich adsorption isotherm. Find the values of K & n.

Calculation:- Total amount of oxalic acid — free amount = adsorbed amount. During titration: 1 mole oxalic acid + 2 mole NaOH = product 1 M.Wt of Ox.A = 2 M.Wt of NaOH (M.Wt of Ox.A ) / 2 = 1 M.Wt of NaOH 1 eq. Wt. of Ox.A =1 eq. Wt. of NaOH 126/2 = 63 = 1000 ml of 1 N NaOH 63/1000 = 1 ml of 1 N NaOH (63/1000) x (1/2) = 1 ml of 1/2 N NaOH 1M1 of 1/2 N NaOH = 0.0315 g ox. acid E.P x 0.0315 = g ox. acid (free)/20 ml.

Calculate the amount of total ox.acid in each flask as follows:- Flask no.1:- if there is no kaolin present and we take 20 ml of solution and titrate with N/2 NaOH theoretically it take 40 ml NaOH 20 x 1 N = v x 1/2N V = 40 ml of NaOH we needed if there is no kaolin present in the flask. 40 x 0.0315 = gm ox.acid/20 ml (total) X = amount adsorbed = ( 40 x 0.0315) - (E.p 1 x 0.0315) = g/20 ml x = ( 40 x 0.0315) - (E.p1 x 0.0315) x 50/20 = 0.0315 (40 - E.p1) x 50/20 (g/50 ml) 0.0315 (40 — E.p1) 50 X/m = X = (40 - E.p1) x 0.0394 ( g/50 ml) 2 20

So, calculations for other flasks are the same for flask no.2 (0.8 N) x/m = ( 32 — E.p2) x 0.0394 for flask no.3 (0.6 N) x/m = ( 24 — E.p3) x0.0394 for flask no.4 (0.4 N) x/m = ( 16— E.p1) x0.0394 for flask no.5 (0.2N) : x/m = ( 8 — E.p5) x 0.0394

Calculation for C% in each flask is as follows:- 1- Flask no.1:- 1N means g.eq.wt/1000 m1 63 g in 1000 ml, so it is 6.3%. 2- Flask no.2:- 1 N 6.3% 0.8N ? So it is (6.3 x 0.8N)/IN = 5.04% For flask no.3, 4 & 5 the percent of conc. is 3.75%, 2.52% & 1.26%. - Tabulate the result as follows:- Flask no. E.p x/m logx/m C% logC% Plot logx/m versus log C%