Adsorption: Process, Isotherms, Adsorbents, and More

• Adsorption is a process of free moving solutes or gaseous molecules of a solution that come close and attach themselves onto the surface of a solid.
• Adsorption can be strong or weak, depending on the nature of forces between the solid surface (adsorbent) and the gas or dissolved solute (adsorbate).
• Desorption is the reverse of adsorption.
• Sorption is a combination of both adsorption and absorption.
• Adsorption involves the concentration or accumulation of gas, liquid, and solid on the surface of a liquid or solid which is in contact.
• Solids have the property of holding molecules at their surface; this occurs in the case of porous and finely divided materials.
• Adsorbent is the material used to adsorb gas, liquid, and solid, such as charcoal and kaolin.
• Adsorbate is the substance being adsorbed.
• Adsorption isotherm is the relationship between the amount of substance being adsorbed and the amount existing in the bulk of the solution at a constant temperature.

Types of Adsorption

• Physical or Van der Waals adsorption is associated with Van der Waals forces and is reversible. Physically adsorbed gas can be desorbed from solid by increasing temperature and decreasing pressure; the removal of adsorbate from adsorbent is known as desorption.
• Chemical or chemisorption, adsorbate is attached to the adsorbent by a primary chemical bond and is irreversible.

Applications

• Adsorption is used for the removal of objectionable odors from rooms and food.
• Adsorption is used in gas masks and the measurement of particle dimensions.
• Adsorption is used in decoloring solutions, detergents, and wetting processes.
• Adsorption is used in chromatography.

Factors Affecting Adsorption on a Solid

• Surface area
• Nature of the adsorbate
• Hydrogen ion concentration (pH) of the solution
• Temperature
• Mixed solutes
• Nature of the adsorbent

Freundlich Adsorption Isotherm

• An empirical equation was derived for the absorption of dissolved solids on the surface of porous substances.
• x/m = kp¹/n (Freundlich adsorption isotherm), where log x/m = log k + 1/n log P.
• x = weight of the gas adsorbed by m gm of the adsorbent at pressure p; x/m represents the amount of gas adsorbed by the adsorbents per gm (unit mass).
• k and n are constant at a particular temperature and for a particular adsorbent and adsorbate (gas).
• n is always greater than one, indicating that the amount of the gas adsorbed increases less rapidly than the pressure.
• According to the Freundlich equation, a plot of log x/m versus log P yields a straight line, and the constants k and 1/n can be determined from the intercept (log k) and slope (1/n).

Experimental Work:

• The aim of the experiment is to study the adsorption of oxalic acid on kaolin and see the effect of increasing the concentration of oxalic acid on adsorption.
• Materials and equipment include: oxalic acid, D.W., NaOH, kaolin, 1 N oxalic acid, 0.5 N NaOH, ph.ph indicator, volumetric flask (50 cc), conical flask, pipette (20 cc), filter paper, funnel, burette.

Procedure:

• Prepare 50ml of the following concentrations of oxalic acid (0.2, 0.4, 0.6, 0.8 N) from a stock solution of 1N oxalic acid.
• Place 50 ml of each concentration and stock solution in 5 conical flasks. Introduce 2gm of kaolin into each flask.
• Shake for 15 min and set aside for another 15 min to achieve equilibrium.
• Filter, rejecting the first portion of the filtrate after washing the receiver.
• Pipette 20 ml of the filtrate containing the non-adsorbed oxalic acid (free) and titrate them with 0.5 N NaOH using phenolphthalein indicator (end point change in color from colorless to pink).
• Calculate the amount adsorbed in each flask, plot Freundlich adsorption isotherm, and find the values of K and 1/n.

Calculation

• Total amount of oxalic acid - free amount = adsorbed amount.
• During titration, oxalic acid reacts with NaOH to form sodium oxalate.
• 1 mole of oxalic acid = 2 moles of NaOH
• 1 M.Wt ox. A = 2 * 1 M * 1L NaoH
• 126/2 = 1000 ml 1 N NaOH
• 63*0.5/1000 = 1ml 0.5 N NaOH
• 0.0315 gm ox.A= 1ml 0.5N NaOH
• Ε.Ρ.* 0.0315 = gm ox.A (free) /20 ml