Colligative Properties The reduction of the freezing point of a substance is an example of a colligative property: A property of a solvent that depends on the total number of solute particles present There are four colligative properties to consider: Vapor pressure reduction (Raoult’s Law) Freezing point depression Boiling point elevation Osmotic pressure

Colligative Properties – Vapor Pressure A solvent in a closed container reaches a state of dynamic equilibrium. The pressure exerted by the vapor in the headspace is referred to as the vapor pressure of the solvent. The addition of any nonvolatile solute (one with no measurable vapor pressure) to any solvent reduces the vapor pressure of the solvent.

Colligative Properties – Vapor Pressure Nonvolatile solutes reduce the ability of the surface solvent molecules to escape the liquid. Vapor pressure is lowered. The extent of vapor pressure lowering depends on the amount of solute. Raoult’s Law quantifies the amount of vapor pressure lowering observed.

Colligative Properties – Vapor Pressure Raoult’s Law: PA = X P° where PA = partial pressure of the solvent vapor above the solution X = mole fraction of the solvent P° = vapor pressure of the pure solvent

Colligative Properties – Vapor Pressure Example: The vapor pressure of water at 20oC is 17.5 torr. Calculate the vapor pressure of an aqueous solution prepared by adding 36.0 g of glucose (C6H12O6) to 14.4 g of water. Given: P°H2O= 17.5 torr mass solute = 36.0 g of glucose mass solvent = 14.4 g of water Find: PH2O Raoult’s Law: P = XP° Answer: 14.0 torr

Colligative Properties – Vapor Pressure Example: The vapor pressure of pure water at 110°C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.10 atm at the same temperature. What is the mole fraction of ethylene glycol in the solution? Both ethylene glycol and water are liquids. How do you know which one is the solvent and which one is the solute? The solvent is the one in greater amount

Colligative Properties – Vapor Pressure Given: P°H2O = 1070 torr Psoln = 1.10 atm Find: XEG Solution: Answer: XEG = 0.219 Raoult’s Law: P = XP° 836 = X(1070) XH2O = 0.78 760 torr 1.10 atm = 836 torr x XEG = 0.22 1 atm

Colligative Properties – BP Elevation The addition of a nonvolatile solute causes solutions to have higher boiling points than the pure solvent. Vapor pressure decreases with addition of non- volatile solute. Higher temperature is needed in order for vapor pressure to equal 1 atm.

Colligative Properties- BP Elevation The change in boiling point is proportional to the number of solute particles present and can be related to the molality of the solution: Tb = i Kb.m where Tb = boiling point elevation Kb = molal boiling point elevation constant m = molality of solution (mol/kg) i = van't Hoff factor (dissociation integer)

Colligative Properties - BP Elevation Example: Calculate the boiling point of 100 grams aqueous solution that is 20.0 % ethylene glycol (C2H6O2) by mass. Tb = i Kb  m Solute = Solvent = Kb (solvent) = ethylene glycol (20 grams) water (0.080 kilograms) 0.51 °C/m

Colligative Properties - BP Elevation since ethelyne glycol is molecular i = 1 Molality of solute: Tb = BP = 102.1oC 1 mol C2H6O2 20 g C2H6O2 x [(2·12.011)+(6·1.008) +(2·15.999)] g C2H6O2 0.322 mol C2H6O2 = 4.03 molal C2H6O2 0.080 kg H2O (1)(0.51) 4.03 = 2.05 C° BP = 102.05 °C

Colligative Properties - BP Elevation Example: The boiling point of an aqueous solution that is 1.0 m in NaCl is 101.02oC whereas the boiling point of an aqueous solution that is 1.0 m in glucose (C6H12O6) is 100.51oC. Explain why. NaCl dissociates into Na+ and Cl- i = 2 C6H12O6 does not dissociate (molecular) i = 1

Colligative Properties - BP Elevation Example: A solution containing 4.5 g of glycerol, a nonvolatile, nonelectrolyte, in 100.0 g of ethanol has a boiling point of 79.0oC. If the normal BP of ethanol is 78.4oC, calculate the molar mass of glycerol. Given: Tb = 79.0°C - 78.4°C = 0.6 C° mass solute = 4.5 g mass solvent = 100.0 g = 0.100 kg Kb = 1.2°C/m (Table) i = 1 (nonelectrolyte) Find: molar mass (g/mol) Tb = i Kb  m

Colligative Properties - BP Elevation Step 1: Calculate molality of solution Step 2: Calculate moles of solute present Step 3: Calculate molar mass Tb = i Kb  m 0.5 molal glycerol 0.6 = (1)(1.2)(m) 0.5 mols glycerol = 0.100 kg ethanol x 1 kg ethanol 0.05 mols glycerol 4.5 grams glycerol 90 g/mol = 0.05 mols glycerol

Colligative Properties - Freezing Pt Depression (it's the same – just different) The addition of a nonvolatile solute causes solutions to have lower freezing points than the pure solvent. Solid-liquid equilibrium line rises ~ vertically from the triple point, which is lower than that of pure solvent. Freezing point of the solution is lower than that of the pure solvent.

Colligative Properties - Freezing Pt Depression Tf = i Kf.m where Tf = freezing point depression Kf = molal freezing point depression constant m = molality of solution (mol/kg) i = van't Hoff factor (dissociation integer)

Colligative Properties - Freezing Pt Depression Example: Calculate the freezing point depression of a solution that contains 5.15 g of benzene (C6H6) dissolved in 50.0 g of CCl4. Given: mass solute = mass solvent = Kf solvent = Find: Tf 5.15 g C6H6 0.05 kg CCl4 Tf = i Kf  m 29.8 °C/m

Colligative Properties - Freezing Pt Depression since benzene is molecular i = 1 Molality of solution: Tf = 1 mol C6H6 5.15 g C6H6 x [(6·12.011)+(6·1.008)] g C6H6 0.071 mol C6H6 = 1.42 molal C6H6 0.050 kg CCl4 (1)(29.8)(1.42) = 42.3 C° FP = - 64.6 °C

Colligative Properties - Freezing Pt Depression Example: Which of the following will give the lowest freezing point when added to 1 kg of water: 1 mol of Co(C2H3O2)2, 2 mol KCl, or 3 mol of ethylene glycol (C2H6O2)? Explain why. KCl (largest i· m product) Co(C2H3O2)2 dissociates into Co2+ and 2 C2H3O2- i = 3 m = 1 molal KCl dissociates into K+ and Cl- i = 2 m = 2 molal C2H6O2 does not dissociate (molecular) i = 1 m = 3 molal

Colligative Properties - Osmosis Some substances form semipermeable membranes, allowing some smaller particles to pass through, but blocking other larger particles. In biological systems, most semipermeable membranes allow water to pass through, but solutes are not free to do so.

Colligative Properties - Osmosis Osmosis: the net movement of a solvent through a semipermeable membrane toward the solution with greater solute concentration. In osmosis, there is net movement of solvent from the area of lower solute concentration to the area of higher solute concentration. Movement of solvent from high solvent concentration to low solvent concentration

Colligative Properties - Osmosis Everyday examples of osmosis: A cucumber placed in brine solution loses water and becomes a pickle. A limp carrot placed in water becomes firm because water enters by osmosis. Eating large quantities of salty food causes retention of water and swelling of tissues (edema). There is a quantitative expression for osmotic pressure, but we're not going to use it . . . . . yet :)