Effect of solutes on the properties of the solvent Properties of pure solvents are affected when solutes are dissolved in the solvent. The presence of the solute affects the interactions between solvent molecules. COLLIGATIVE properties: properties of solvents which are affected by the concentration of the solute Example: boiling point, freezing point, vapor pressure
For dilute solutions, the extent to which these properties of the pure solvent are affected depends on the amount of solute added and not on the type of solute added. An ideal solution is a homogenous mixture of two or more components that mix without change in volume or flow of heat into or out of the system. In an ideal solution all molecules find themselves in an environment very much like that in the pure solute and solvent; most ideal solutions are with similar molecules.
Mass percentage = mass of component total mass of mixture X 100% total number of moles in mixture Mole fraction = moles of component molality (m) = moles of solute kilogram of solvent = mol kg -1 Concentrations of solutions molarity (M) = moles of solute liter of solvent = mol L -1
Effect of solute on vapor pressure Consider an ideal solution of a volatile solvent with a non- volatile solute. Since the solvent is volatile, but not the solute, the vapor pressure above the solution is due to the solvent only. The vapor pressure of the solution is LOWER than the vapor pressure of the pure solvent. Experiments indicate that, for ideal solutions, the vapor pressure of the solution is directly proportional to the mole fraction of the solvent in the solution. Animation
If P 1 is the vapor pressure of the solution, P 1 the vapor pressure of the pure solvent, and X 1 the mole fraction of the solvent in the solution,then: o P 1 = X 1 P 1 Raoult’s Law o
Most solutions obey Raoult’s law at low concentrations of solute. Deviations from Raoult’s law provides information on the nature of the solute-solvent interactions.
Relating the lowering of vapor pressure to the solute concentration The change in the vapor pressure of the solvent when a nonvolatile solute is added is: P 1 = P 1 - P o 1 P 1 = - X 2 P o 1 From Raoult’s Law P 1 = X 1 P o 1 - P o 1 = P o 1 (X 1 - 1) where X 1 is the mole fraction of the solvent Since X 1 + X 2 = 1 (X 2 is the mole fraction of the solute) Vapor pressure of solvent above a dilute solution is always lower than the pure solvent
Elevation of Boiling Point Since the equilibrium vapor pressure of the solvent is lowered when a solute is added, the boiling point of the solution is raised. The elevation in boiling point, T b is related to the concentration of the solute in solution as follows: T b = K b m where K b is a constant and m is the MOLALITY of the solution Note: K b depends on the solvent, not on the solute.
What happens if the solute dissociates in solution? Colligative properties depend on the number of solute molecules in solution T b = i K b mvan’t Hoff equation where i equals the number of particles released into solution per formula unit of solute i = 2 for NaCl; i = 3 for CaCl 2 Note: the van’t Hoff equation holds for ideal solutions and hence for strong electrolytes, T b calculated using this equation is typically an upper limit.
Depression of Freezing Point Addition of solute to a solvent causes its freezing point to decrease (explains why salt is used on icy roads, and a mixture of ethylene glycol and water works as “anti-freeze”) For dilute solutions of a non-dissociating solute: T f = - K f m The negative sign indicates a lowering of the freezing point K f is a constant that depends on the solvent and not on the solute. For dilute solutions of a dissociating solute: T f = - i K f m Animation
Water flows from the pure solvent side to the solution side, because there are more water molecules per unit volume in the pure water side. The net effect is that the solution becomes more diluted. OSMOSIS: If two solutions of different concentrations are separated by a semi-permeable membrane, the solvent flows from the side which has a higher concentration of solvent to the side with a lower concentration of solvent. Animation
The pressure that must be applied to the solution side to exactly stop the flow of solvent through the membrane, i.e. stop osmosis, is called the OSMOTIC PRESSURE ( ). = g d h h
In an ideal solution, the osmotic pressure obeys a relationship that looks like the ideal gas law: = n R T/V Since n/V is the molar concentration (c) = c R T For dissociating solutes: = i c R T Higher the concentration of the solute, higher is the osmotic pressure required to prevent osmosis.
Since depends on the concentration of the solute it is a colligative property. Measuring the osmotic pressure can be used to determine the molar mass of the solute. For example, measuring the osmotic pressure of a solution containing a protein can be used to estimate the molar mass of the protein. = n R T V = (mass) R T M V
Body fluid ~ 0.15 molal NaCl b) solution too concentrated c) solution too dilute
Normal osmosisReverse osmosis Purification of water - Desalination
Colligative properties of mixtures For an ideal solution of two volatile substances (A & B), if each component obeys Raoult’s Law: P A = X A P A o P B = X B P B o = (1-X A ) P B o The total vapor pressure of the mixture is determined by Dalton’s law P tot = P A + P B = X A P A o + (1-X A ) P B o
Separation of a mixture of two volatile compounds A mixture of two components with different vapor pressures can separated by FRACTIONAL DISTILLATION Because of the different vapor pressure of the components in the solution, the composition of the vapor above the solution is not the same as the solution itself. DEFINE VOLATILE GIVE EXAMPLES
Separation of a mixture of hexane (C 6 H 14 ) and heptane (C 7 H 14 ). Heptane and hexane form a nearly ideal mixture over the whole range of mole fractions. At 25 o C Vapor pressure of hexane, P 1 o = 0.198 atm Vapor pressure of heptane, P 2 o = 0.0600 atm Mixture contains 4.00 mol hexane and 6.00 mol heptane
From Raoult’s Law P hexane = P 1 = X 1 P o 1 = (0.400)(0.198 atm) = 0.0792 atm P heptane = P 2 = X 2 P o 2 = (0.600)(0.0600 atm) = 0.0360 atm From Dalton’s Law P tot = P 1 + P 2 = 0.1152 atm Mole fraction of hexane in the vapor X ’ 1 = 0.0792 atm 0.1152 atm Mole fraction of heptane in the vapor X ’ 2 = 0.0360 atm 0.1152 atm =0.688 =0.312
The vapor is enriched with the more volatile component (in this example, hexane) Fractional Distillation separates components in a mixture. Solution is boiled, and vapors are collected and condensed. The solution formed from the condensate is “richer” in the more volatile component. Boiling this solution and condensing the vapors enriches the resulting condensate with the more volatile component.