Colligative Properties of Solutions PDF

Colligative properties of solutions ❑ Colligative properties of solutions are those properties that depend on the number of particles of solute in a solution and not on their specific chemical nature. ❑ The word colligative derives from the Latin word “colligare” meaning to collect since these...

Colligative properties of solutions ❑ Colligative properties of solutions are those properties that depend on the number of particles of solute in a solution and not on their specific chemical nature. ❑ The word colligative derives from the Latin word “colligare” meaning to collect since these properties are determined by the number of particles in the entire “collection” not by what the particles are composed of. Colligative properties include the following: 1. Vapor pressure of solutions, 2. Elevation in boiling point 3. Depression in freezing point of solvents, and 4. Osmotic pressure One of the reasons for making a detailed study of colligative properties is that they can be used for experimental molecular weight determination. 1. Vapor pressure of solutions ❑ All liquids and solids give off vapor, consisting of molecules of the substance. ❑ If the substance is in an enclosed space, the pressure of the vapor will reach a maximum that depends only upon the nature of the substance and the temperature. ❑ The vapor is then saturated, and its pressure is the “saturated vapor pressure” or “vapor pressure” for short. ❑ Vapor pressure is independent of the amount of liquid present. ❑ A liquid boils when its vapor pressure is equal to the atmospheric pressure or other external pressure above the liquid. This is about vapor pressure in pure liquids, what would be the situation in solutions?  We are going to examine two situations: The effect of a non-volatile solute on the vapor pressure of a solvent The effect of a volatile solute on the vapor pressure of a solvent Case (1). The solute is non-volatile: ▪ When the solute is non-volatile e.g., glucose dissolved in water, then the vapor pressure of the solvent is lowered because some of the solvent surface will be occupied by the non-volatile solute particles. ▪ It is obvious that the extent to which the vapor pressure is lowered depends on the mole fraction of the solute. The vapor pressure of the solution (P) is given by Raoult’s law P = Po Nsolvent Where Po is the vapor pressure of the pure solvent, Nsolvent is the mole fraction of the solvent. Example What are the mole fractions of solute and solvent in a 1.00 m aqueous solution? Solution The molecular weight of water is 18.00. The number of moles in 1000g H2O are 1000/18 = 55.56 mol A 1.00 m aqueous solution contains 1.00 mol of solute. Total number of moles = 55.56 + 1.00 = 56.56 mol Example A solution containing 8 grams of solute dissolved in 100 g of acetone. The vapor pressure of such a solution was measured as 220.2 mm Hg at 25 C. Given the vapor pressure of pure acetone as 229 mm Hg at the same temperature, then calculate the molecular weight of the solute. (C = 12, O = 16, H = 1) Case (2). The solute is volatile: In many solutions e.g., benzene in carbon tetrachloride, the solute is volatile and has its own vapor pressure just like the solvent itself. At equilibrium between the vapor and solution, the total vapor pressure PT is given as the sum of the partial vapor pressure of solute (PA) and solvent (PB) PT = PA + PB  o PA ❑ Figure is a plot of the partial Vapour pressures PA and PB, and the Pressure PT PA total pressure PT as a function of o PB idea solution composition for a PB two-component mixture. B NA A NB Vapor pressure of an ideal two-component system  A boiling point diagram for a two-component mixture ❑Another way of presenting the behavior of two-component mixtures is boiling- point diagrams in which the boiling point of the mixture is drawn as a function of solution composition. ❑A splitting in these curves is always obtained since the vapor at any temperature will be richer in the more volatile component. ❑A boiling-point diagram for a two- component mixture is given in Figure ❑ On our boiling-point diagram we can indicate the composition of the vapor by the upper curve drawn in Figure (10-3). ❑ Here, points corresponding to the compositions of liquid and vapor in equilibrium can be obtained by drawing a horizontal line, called a tie line, between the curve for the liquid and that for the vapor. ❑ In Figure (10-3), a liquid of composition N1 boils at temperature T1 and gives a vapor having the composition N2 which is richer in the more volatile component (A). ❑ When this vapor is condensed and reheated, it boils at temperature T2 and gives a vapor with composition N3. ❑ Repetition of this process will produce fractions richer in A. ❑ This procedure is called fractional distillation and is widely used in many areas. Petroleum industry, for example, uses fractional distillation to separate crude oil into its various components, which include gasoline, kerosene, oils, and paraffin.  Boiling point diagram for mixtures of: (a) water-ethanol and (b) H2O-HCl. (a) (b) Temperature 100 o C 78.5 o C Ethanol Water H 2O HCl ❑ Very few mixtures behave ideally according to the above discussion. ❑ An example of an ideal mixture is benzene and carbon tetrachloride. ❑ Most mixtures show deviations from this ideal behavior due to difference in interactions between various molecules of solutes and solvents. These deviations are usually of two types, namely positive deviations, and negative deviations. ❑ An example of a binary mixture that shows a negative deviation in the boiling-point diagram is water-ethanol mixtures as shown in Figure (10-4a). ❑ A mixture of hydrochloric acid and water shows a negative deviation in the boiling-point diagram, as shown in Figure (10-4b). The terms positive or negative are based on vapor pressure-composition changes. ❑ A mixture of composition corresponding to a maximum or a minimum on the boiling point diagrams is called an azeotropic mixture. For such mixtures, the composition of liquid and vapor is the same. ❑ In other words, if we boil an azeotropic mixture, the vapor will have the same composition of the liquid, and the components of these mixtures cannot be separated by fractional distillation. ❑ For ethanol-water mixture, an azeotropic mixture is obtained of composition 95.57% ethanol. This mixture boils at 78 C and thus it is known that the alcohol obtained by distilling a water-alcohol mixture is not water-free and other methods must be used to make dry “absolute” alcohol. ❑ Hydrochloric acid forms a maximum boiling azeotropic mixture with water having the approximate composition 80 % HCl and 20 % H2O by mass, with a boiling point of 109 C.  Elevation of boiling point and depression in freezing point of solutions. TF = KF. m Vapour pressure Pure solvent TF is the freezing point depression. m is the Frozen P solution Solution molality, and KF is proportionality constants of units C m-1. We can think Temperature of their values as Tf To representing the freezing  Tf Depression in freezing point of solutions point depression for a 1 m solution.  Elevation of boiling point Tb = Kb. m In these equations Tb is the boiling point elevation,. m is the molality, and Kb is proportionality constants of units C m-1. We can think of their values as representing the boiling point elevation for a 1 m Elevation of boiling point solution. Substituting for molality (m) by solvent weight (w1) and solute weight (w2) and molecular weight (M), then  TF = KF. m  TF Example A neutral, nitrogenous compound, obtained from human urine, was recrystallized from ethanol. A solution prepared by dissolving 90 mg of the purified compound in 12 g of distilled water had a freezing point of 0.233 C lower than the freezing point of pure distilled water. Calculate the molecular weight of the compound. (Molal depression constant KF for water = 1.86 C) Example A solution containing 3.795 g sulphur in 100 g carbon disulphide having Tb= 0.36 °C. Given Kb for the CS2 solvent as 2.397°C, calculate the molecular formula of sulphur in CS2. Given the atomic weight of sulphur as 32.0. Example An antifreeze solution was prepared by mixing equal volumes of water and ethylene glycol, C2H4 (OH)2. Given the density of ethylene glycol as 1.113 g cm 3, calculate the freezing temperature of the mixture given KF for water as 1.855 C. Solution The molecular weight of ethylene glycol = 62 g mol-1. In the antifreeze solution, one liter of water is mixed with one liter of ethylene glycol. This is expressed in terms of weights as: 1000 g of water contains 1113 g of ethylene glycol. This means that depression in freezing point is 33.3 C below the freezing point of pure water i.e., the solution freezes at -33.3 C.  3. Osmotic pressure of solutions ❑ Osmosis (Greek for push) is the flow of solvent through a semipermeable membrane, i.e. a membrane that will permit the passage of the solvent but not of dissolved substances(*). ❑ There is a tendency for solutions separated by such a membrane to become equal in molecular concentration; thus, the solvent will flow from a weaker to a stronger solution, the solutions tending to become more nearly equal in concentration.  Osmotic pressure is equal to the pressure required to prevent osmosis when the solution is separated by a semi-permeable membrane from the pure solvent. The symbol of osmotic pressure is . The osmotic pressure of non-electrolytic solutions is related to temperature and volume by laws like gas laws. e.g., the Van’t Hoff equation:  V = n RT where  is the osmotic pressure, T is the absolute temperature V is the volume of solution w V= RT n is the number of moles of solute, R is the gas constant R = 8.314 k Pa lit mol-1 K-1= 8.314 J mol-1 K-1 M = 0.08 lit atm mol-1 K-1 Example A polymer solution contains 0.400 g in a liter of water. Given the osmotic pressure of this solution at 27 C as 0.499 k Pa (kilo Pascal), calculate the molecular weight of the polymer. Example A solution containing 20 g of hemoglobin dissolved in one liter of water. An apparatus as shown in Figure (10-9) was used to measure osmotic pressure. At equilibrium the difference in height h was 77.8 mm at 25 C. Calculate the molecular weight of hemoglobin. (1 atm = 760 mm Hg, R = 0.08 lit-atm mol-1 K-1) Semipermeable h membrane Solvent Solution An osmotic pressure apparatus  Osmolarity (Osm.) Osmolarity is a measure of the total concentration of solute particles in a solution. It is calculated using the formula: Osmolarity (Osm.)=Molarity (M)×Number of Particles Examples: 1- 1M Glucose Solution: 1.Glucose does not dissociate in solution. 2.Therefore, the number of particles is 1. 3.Osmolarity = 1 M×1=1 Osm. 2- 1M NaCl Solution: 1.NaCl dissociates into Na+ and Cl− ions, yielding 2 particles. 2.Osmolarity = 1 M×2=2 Osm. 3- Blood Plasma: 1.The osmolarity of blood plasma is typically 0.308 Osm. ❑ Both 0.308 M glucose and 0.154 M NaCl are isotonic solutions for RBCs. This isotonicity ensures that RBCs neither swell (in a hypotonic solution) nor shrink (in a hypertonic solution), maintaining their structural integrity and physiological function. ❑ 0.154 M Glucose Solution Osmolarity: 0.154 Osm. Compared to RBCs: Hypotonic (0.308 Osm) Effect: Water exits RBCs → Shrinking (crenation).  Because the osmotic pressure that can be developed between solutions of only slightly different concentrations is so great, it is very important that solutions added to the body intravenously do not significantly alter the osmotic pressure of the blood. If the blood fluids become too diluted, the osmotic pressure that develops within the blood cells can cause them to rupture. On the other hand, if the fluids are too concentrated, water will diffuse out of the cells, and they will shrink. They will no longer function properly. For this reason, care is taken to use a solution with the same osmotic pressure as the solution within the cell. Solutions that have the same osmotic pressure are called isotonic solutions.