Chapter 13Sec. 2Solutions and Their Colligative Properties

Solutions have them but pure substances do not have them.Colligative Properties -Properties of solutions which depend on the number of solute particles in the solution and not the nature of the solute.Three Colligative Properties1)Vapor pressure lowering2)Boiling point elevation3)Freezing point depression

II.Units of Concentration Molarity and % mass are not useful in colligative properties because the exact amount of solvent is unknown.Only molality concentration units reflect the number of solute particles per solvent molecules and are useful with colligative properties.

III.Colligative Properties The vapor pressure of the solution is lowered because the solute particles at the liquid/vapor boundary block the solvent particle from jumping into the vapor state.A. Vapor Pressure LoweringVapor pressure the pressure above a liquid created by the liquid releasing gas molecules as it evaporates. The pressure is directly related to temperature.

Dynamic Equilibrium When the rate of freezing is the same as the rate of melting, the amount of ice and the amount of water won't change on average (although there are short-term fluctuations at the surface of the ice). The ice and water are said to be in dynamic equilibrium with each other. The balance between freezing and melting processes can easily be upset. If the ice/water mixture is cooled, the molecules move slower. The slower-moving molecules are more easily captured by the ice, and freezing occurs at a greater rate than melting. Molecules in Motion Flash plug-in

IV.Colligative Properties For nonvolatile, nonelectrolyte solvents the change in boiling point (DTbp) is:B. Boiling Point ElevationDTbp = KbpmsoluteKbp =boiling point elevation constantmsolute =molality of soluteThe Boiling pointCurve is shifted to the right. Raising the boiling Pt.

IV.Colligative Properties For nonvolatile, nonelectrolyte solvents the change in freezing point (DTfp) is:C. Freezing Point Depressionmsolute =molality of soluteThe Freezing point curve shifts to the left. Lowering the F.P. of the solvent.

Practice Problems1. Determine the freezing point of a solution of 60.0 g of glucose, C6H12O6, dissolved in 80.0 g of water. m= mol/kg m = 60.0g / 180g/mol / 0.080kg = 4.167m Tfp = (-1.86C/m) (4.167 m)

ans: -7.75C

Practice Problems2. What is the boiling point of a solution of 645 g of urea, CON2H4, dissolved in 980. g of water? . m= mol/kg m = 645g / 60g/mol / .980 kg = 10.97m Tbp = (0.512C/m) (10.97 m) Tbp = 5.62CBP = 100 + 5.62C = 105.6C

IV.Colligative Properties The change in VP, BP or FP is greater than expected for electrolyte (ionic salt) solutions.D. Colligative Properties of Solutions Containing IonsPredicted BP elevation of an aqueous 0.100 m NaCl solution DTbp = Kbp msoluteDTbp, calculated = (0.512 C/m)(0.100 m) = 0.0512 CActual BP elevation of an aqueous 0.100 m NaCl solution DTbp, measured = 0.09470 C (Almost double the DTbp calculated)Colligative properties depend on the total number of solute particles in solution. Ionic compounds form ions in solution so the total number of solute particles in solution is equal to the total ions in solution.

IV.Colligative Properties D. Colligative Properties of Solutions Containing IonsNaCl(s) Na+(aq) + Cl-(aq) 0.100 mTHE MULTIPLIER -vant Hoff factor (i) We will assume 100% ionizationSo for electrolytes or ionic solutions:0.100 m0.100 m0.200 m total

IV.Colligative Properties D. Colligative Properties of Solutions Containing IonsNaCl(s) Na+(aq) + Cl-(aq) Predicting vant Hoff factors1 particle + 1 particle = 2 particlesipredicted = 2 Na2SO4(s) 2 Na+(aq) + SO42-(aq) 2 particles + 1 particle = 3 particlesipredicted = 3

Practice Problems3. What is the expected boiling point of a brine solution containing 30.00 g of KBr dissolved in 100.00 g of water? m= mol/kg KBr two ions , i = 2 m = 30.00g / 119.0g/mol / .100 kg = 2.52 m Tbp = 2 (0.512C/m) (2.52 m) = 2.6C B.P. of water = 100.0C + 2.6Cans: 102.6C

Practice Problems4. What is the expected boiling point of a CaCl2 solution containing 385 g of CaCl2 dissolved in 1.230 kg of water? m= mol/kg CaCl2 three ions , i = 3 m = 385 g / 111.1 g/mol / 1.230 kg = 2.82 m Tbp = 3 (0.512C/m) (2.82 m) = 4.3C B.P. of water = 100.0C + 4.3Cans: 104.3C

IV.Colligative Properties Colligative properties can be used to determine the molar mass of a solute when it is dissolved in a solvent of appreciable vapor pressure and a known Kbp or Kfp.E. Colligative Properties and Molar Mass DeterminationExample:Butylated hydroxyanisole (BHA) is used as an antioxidant in margarine and other fats and oils; it prevents oxidation and prolongs the shelf life of food. What is the molar mass of BHA if 0.640 g of the compound, dissolved in 25.0 g of chloroform, produces a solution whose boiling point is 62.22 C. (Chloroform BP = 61.70 C, Kbp = 3.63 C/m)

Practice Problems5. A solution of 0.827 g of an unknown non-electrolyte compound in 2.500 g of water has a freezing point of -10.18C. Calculate the molar mass of the compound.

Molar mass = g/mol solve for moles thru m

-10.18 = (-1.86C/m) (m) ; m = 5.47 mol/kg m = 5.47 mol/kg = mol/0.0025kg mol =.01367Molar mass = 0.827g /.01367 molans: 60.5 g/mol