Chapt12

Chapt12

• 1.
  Physical Properties ofSolutions Chapter 12 Dr. Ali Bumajdad
• 2.
  Chapter 12 Topics PhysicalProperties of Solutions • Solution, solute, solvent, unsaturated, saturated, supersaturated, crystallization •Solution Process •Concentrations (%mass, Mole fraction, molality, molarity, ppm) •Factors affecting Solubility (T and P) •Colligative properties of nonelectrolyte solution: 1)Vapour pressure lowering 2)B.P. Elevation 3) F.P. Depression 4) Osmotic Pressure (π ) •Using Colligative properties to determine molar mass •Colligative properties of electrolyte solution Dr. Ali Bumajdad
• 3.
  Solution, solute, solvent,unsaturated, saturated, supersaturated, crystallization •Solution: homogenous mixture of 2 or more substances •Solute: is(are) the substance(s) present in the smaller amount(s) •Solvent: is the substance present in the larger amount ∴Solution not necessarily liquid, it could be gas or solid
• 4.
  • Saturated solution:contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. • Unsaturated solution: contains less solute than the solvent has the capacity to dissolve at a specific temperature. • Supersaturated solution: contains more solute than is present in a saturated solution at a specific temperature (not stable). Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate • Crystallization: dissolved solute comes out of solution and forms crystals. Solute + solvent solvation crystallization Solution
• 5.
  Solution Process Three typesof interactions in the solution process: • solvent-solvent interaction (endothermic) • solute-solute interaction (endothermic) • solvent-solute interaction (endothermic or exothermic) ∆Hsoln>0, solution may form and maynot ∆Hsoln<0, solution favourable ∆Hsoln = ∆H1 + ∆H2 + ∆H3
• 6.
  When solute-solute orsolvent-solvent interaction are much more than solute-solvent interaction a solution will not form “like dissolves like” Two substances with similar intermolecular forces are likely to be soluble in each other. • non-polar molecules are soluble in non-polar solvents CCl4 in C6H6 • polar molecules are soluble in polar solvents C2H5OH in H2O • ionic compounds are more soluble in polar solvents NaCl in H2O or NH3 (l)
• 7.
  •Miscible: two liquidscompletely soluble in each other in all proportions •Solvation: ion or molecules surrounded by solvent molecules Q) Can we form a solution of octane (C8H18) in water? No because: - octane not polar while water is polar -In this case ∆H3 is small while ∆H1 and ∆H2 large +ve hence, ∆Hsoln is large positive Sa Ex. :Predict whether each of the following substances is more likely to dissolve in CCl4 or in Water: C7H16, Na2SO4, HCl, I2.
• 8.
  (a) Br2 inC6H6 (b) KCl in NH3 (c) CH2O in H2O
• 9.
  Concentration Units •Concentration ofa solution is the amount of solute present in a given quantity of solvent or solution. 1) Percent by Mass (1) mass of solute x 100% % by mass = mass of solute + mass of solvent mass of solute x 100% = mass of solution 2) Mole Fraction (X) moles of A XA = sum of moles of all components (2)
• 10.
  Concentration Units Continued 3)Molarity (M) M = moles of solute (3) liters of solution 4) Molality (m) m = moles of solute mass of solvent (kg) (4)
• 11.
  Concentration Type Mass Percentage Equation Equation MoleFraction Unit No unit (%) No unit Molarity mol/l or M Molality mol/kg or m 1) n= mass M.m 2) D= mass Vol
• 14.
  Sa. Ex.:A solutionis made by dissolving 13.5 g glucose (C6H12O6) in 0.100 kg of water. What is the mass percentage of solute in this solution? 11.9% . Ex. Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50. g of water. A commercial bleaching solution contains 3.62 mass% sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution? 2.91% 90.5g
• 15.
  Sa. Ex. a) Asolution is made by dissolving 4.35 g glucose (C6H12O6) in 25.0 ml of water. Calculate the molality of glucose in the solution. 0.966 m b) What is the molality of a solution made by dissolving 36.5 g of naphthalene (C10H8) in 425 g of toluene (C7H8) 0.670 m c) A solution of hydrochloric acid contains 36% HCl by mass (i) Calculate the mole fraction of HCl in the solution. (ii) Calculate the molality of HCl in the solution XHCl= 0.22 15 m
• 16.
  Q) What isthe molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL? moles of solute moles of solute m = M = mass of solvent (kg) liters of solution Assume 1 L of solution: (1) mass of solute: (n×Mm=mass) 5.86 moles ethanol contains 270 g ethanol (2) mass of solution: solution density ×solution volume 927 g of solution mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = 0.657 kg m = moles of solute mass of solvent (kg) = 5.86 moles C2H5OH 0.657 kg solvent = 8.92 m
• 17.
  Sa. Ex. Asolution contains 5.0 g of toluene (C7H8) and 225 g of benzene and has a density of 0.876 g/ml. Calculate the molarity of the solution. 0.21 M Sa. Ex. A solution containing equal masses of glycerol (C3H8O3, Mm=92gmol-1) and water has a density of 1.10 g/ml. Calculate: 10.9 m (a) The molality of glycerol 0.163 (b) The mole fraction of glycerol 5.97 M (c) The molarity of glycerol in the solution.
• 20.
  •Factors affecting Solubility(T and P, and nature of solute and solvent) (1) Temperature Like dissolve like a) Solid solubility and temperature Solubility increases with increasing temperature Solubility decreases with increasing temperature Usually s αT For solid
• 21.
  •Fractional crystallization: isthe separation of a mixture of substances into pure components on the basis of their differing solubilities. Suppose you have 90 g KNO3 contaminated with 10 g NaCl. Fractional crystallization: 1. Dissolve sample in 100 mL of water at 600C 2. Cool solution to 00C 3. All NaCl will stay in solution (s = 34.2g/100g) 4. 78 g of PURE KNO3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g
• 22.
  (1) Temperature (b) Gassolubility and temperature Solubility usually decreases with increasing temperature Usually s α1 T For gas
• 23.
  (2) Pressure a) Gassolubility and Pressure The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law). Henry’s law c = kP (5) (there are some exceptions when the gas react with water) c is the concentration (M) of the dissolved gas P is the pressure of the gas over the solution (vapor or partial pressure) k is a constant for each gas (mol/L•atm) that depends only on temperature Vapour Pressure: pressure of gas above the liquid low P high P low c high c
• 25.
  Sample Ex. :Calculatethe concentration of CO2 in soft drink (partial pressure of CO2 = 4.0 atm, and the Henery's law constant for CO 2 in water is 3.1 × 10-2 mol/L atm. c = kP (5) =0.12 mol/l
• 26.
  Colligative Properties ofNonelectrolyte Solutions •Colligative properties: properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. (e.g. adding salt and ethylene glycol reduce F.P.) 1) Vapor-Pressure Lowering Adding nonvolatile solute to a solvent always lower the vapour pressure 0 P1 = X1 P 1 (6) For nonvolatile solute (no measurable vapour pressure) Raoult’s law Where: P 1 = vapor pressure of solution after adding solute 0 P 1 = vapor pressure of pure solvent X = mole fraction of the solvent
• 27.
  If the solutioncontains only one solute: X1 = 1 – X2 0 P 1 = (1 - X2) P 1 P 1 - P1 = ∆P = X2 P 1 0 0 (7) X2 = mole fraction of the solute The decrease in vapour pressure (∆P) directly proportion to solute concentration (not to the nature of solute)
• 28.
  When both componentsare volatile:1) Ideal behaviour Ideal Solution 0 PA = XA P A 0 PB = XB P B PT = PA + PB 0 PT = XA P A + XB P 0 B
• 29.
  When both componentsare volatile: 2) Nonideal behaviour PT is greater than predicted by Raoults’s law PT is less than predicted by Raoults’s law Force Force Force < A-A & B-B A-B Force Force Force > A-A & B-B A-B
• 30.
  Fractional Distillation Apparatus Usedto separate two liquids from each other based on their boiling point
• 31.
  Vapour pressure ofwater at 30 °C = 31.82mmHg 0 P1 = X1 P 1 (6) Raoult’s Law P 1 - P1 = ∆P = X2 P 1 0 X1 = X2=1-X1 0 (2) n1 n1+n2 (7)
• 32.
  Sa. Ex. Glycerin(C3H8O3) is a nonvolatile nanelectrolyte with a density of 1.26 g/ml at 25 °C. Calculate the vapor pressure at 25 °C of a solution made by adding 50.0 ml of glycerin to 500.0 ml of water. The vapor pressure of pure water at 25 °C is 23.8 torr. 23.2 torr Sa. Ex. The vapor pressure of pure water at 110 °C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110 °C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glyco in the solution? 0.290
• 33.
  2) Boiling-Point Elevation Bythe addition of nanvolatile solute 0 ∆Tb = Tb – T b 0 T b is the boiling point of the pure solvent T b is the boiling point of the solution 0 Tb > T b ∆Tb > 0 ∆Tb = Kb m (8) m is the molality of the solution Tb0 Tb Kb is the molal boiling-point elevation constant (0C/m) for a given solvent
• 34.
  3) Freezing-Point Depression Bythe addition solute ∆Tf = T 0 – Tf f T 0 is the freezing point of the pure solvent T f is the freezing point of the solution f T 0 > Tf f ∆Tf > 0 ∆Tf = Kf m (9) m is the molality of the solution Tf Tf0 Kf is the molal freezing-point depression constant (0C/m) for a given solvent (depend only on solvent)
• 35.
  The decrease inTf upon addition of solute is due to the more energy needs to be removed to form order solution than to form order solvent (solution is more disorder) The solid that separate when a solution freeze is the pure solvent only that why the V.P. of solid is the same as that for pure liquid (see the phase diagram)
• 36.
  De-icing of airplanes
• 37.
  Q) What isthe freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g/mol.(Kf water = 1.86 0C/m) ∆Tf = Kf m m = (9) moles of solute mass of solvent (kg) = 478 g 62.01 g 3.202 kg solvent = 2.41 m ∆Tf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C ∆Tf = T 0 – Tf f Tf = T 0 – ∆Tf = 0.00 0C – 4.48 0C = -4.48 0C f
• 38.
  Sa. Ex. Automotiveantifreeze consists of ethylene glycol (C2H6O2), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25.0 mass% solution of ethylene glycol in water. (Kb water=0.52 and Kf water = 1.86 °C/m) ∆Tb = Kb m Tb = Tb0 +∆Tb 102.8 °C (8) ∆Tf = Kf m (9) Tf = Tf0 -∆Tb -10.0 °C Sa. Ex. Calculate the freezing point of a solution containing 0.600 kg of CHCl3 and 42.0 g of eucalyptol (C10H18O), a fragrant substance found in the leaves of eucalyptus trees.
• 39.
  4) Osmotic Pressure(π) •Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. •Semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. • Osmotic pressure (π ) is the pressure required to stop osmosis. Cucumber in concentrated brine lose water dilute more concentrated Pickle
• 40.
  High P Low P Osmotic pressure in unitatm π = MRT Colligative property = depend on concentration only M is the molarity of the solution R is the gas constant (0.0821 L atm/mol K) T is the temperature (in K) (10)
• 41.
  • When theconcentrations of the two solution are equal and, hence, have the same osmotic pressure, the solutions called isotonic •Isotonic: the two solutions are of equal concentration and, hence, they have the same osmotic pressure •When concentrations of the two solution are different, the lower concentration called hypotonic and the higher concentration called hypertonic •Hypotonic: the solution of lower concentration •Hypertonic: the solution of higher concentration
• 42.
  A cell inan: A cell in isotonic solution the two solution are equal in concentration A cell in hypotonic solution A cell in hypertonic solution the outer solution is less the outer solution is more concentrated than the concentrated than the solution in the cell solution in the cell
• 43.
  π = MRT (10) 1.23M
• 44.
  Sa. Ex.:The averageosmotic pressure of blood is 7.7 atm at 25 °C. what concentration of glucose (C6H12O6) will be isotonic with blood? 0.31 M Sa. Ex.:What is the osmotic pressure at 20 °C of a 0.0020 M sucrose (C12H22O11) solution? 0.048 atm
• 45.
  •Using Colligative Propertiesto determine molar mass B.P. Elevation ∆Tb = Kb m F.P. Depression (9) (8) ∆Tf = Kf m m = ∆Tb m = ∆Tf Kb Kf m= n m= mass of solvent in Kg m= mass of solvent in Kg mass in g m= M.m. (mass of solvent in Kg) M.m = mass in g n (11) m (mass of solvent in Kg) mass in g M.m. (mass of solvent in Kg) M.m = mass in g (11) m (mass of solvent in Kg)
• 46.
  Osmotic Pressure (10) π =MRT M= π RT M= n VL M= mass in g M.m. (VL) M.m = mass in g (12) M (VL) M.m = (mass in g) R T (VL) π (13)
• 47.
  Kf of benzene= 5.12 °C/m m = ∆Tf Kf M.m = mass in g (11) m (mass of solvent in Kg) 127 g/mol C10H8
• 48.
  1 atm =760 mmHg 6.51 ×104 g/mol
• 49.
  Sa. Ex.: Coamphor(C10H16O) melts 1t 179.8 °C, and it has a particularly large freezing-point-depression constant Kf = 40.0 °C/m. When 0.186 g of an organic substance of unknown molar mass is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7 °C. what is the molar mass of the solute?
• 50.
  Colligative Properties ofNonelectrolyte Solutions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. Vapor-Pressure Lowering 0 P1 = X1 P 1 (7) Boiling-Point Elevation ∆Tb = Kb m (8) Freezing-Point Depression ∆Tf = Kf m (9) π = MRT (10) Osmotic Pressure (π)
• 51.
  Colligative Properties ofElectrolyte Solutions 0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions •Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. 0.1 m NaCl solution van’t Hoff factor (i) = 0.2 m ions in solution actual number of particles in soln after dissociation number of formula units initially dissolved in soln i should be nonelectrolytes NaCl CaCl2 1 2 3
• 52.
  Colligative Properties ofElectrolyte Solutions Boiling-Point Elevation ∆Tb = i Kb m (8b) Freezing-Point Depression ∆Tf = i Kf m Osmotic Pressure (π) (9b) π = iMRT (10b)
• 53.
  π = iMRT
• 54.
  Sa. Ex.: Listthe following solutions in the order of their expected freezing points: 0.050 m CaCl2; 0.15 m NaCl; 0.10 m HCl; 0.050 m HC2H3O2; 0.10 m C12H22O11. Sa. Ex: Which of the following solutes will produce the largest increase in boiling point upon addition to 1 Kg of water: 1 mol Co(NO3)2, 2 mol of KCl, 3 mol of ethylene glycol (C2H6O2)?