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  • Slide 1
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 1 Properties of Solutions Chapter 11
  • Slide 2
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 2 Overview Introduce student to solution composition and energy of solution formation. Factor affecting solubilities will be discussed: structure, pressure and temperature effects. Vapor pressure of solutions, boiling point, freezing point affected by solute addition. Colligative properties of electrolyte solutions and colloids.
  • Slide 3
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 3 A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount
  • Slide 4
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 4 A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature. A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature. Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate.
  • Slide 5
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 5 Solution Composition 1.Molarity (M) = 2.Mass (weight) percent = 3.Mole fraction ( A ) = 4.Molality (m) =
  • Slide 6
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 6 Concentration Units The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Percent by Mass % by mass = x 100% mass of solute mass of solute + mass of solvent = x 100% mass of solute mass of solution Mole Fraction (X) X A = moles of A sum of moles of all components
  • Slide 7
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 7 Concentration Units Continued M = moles of solute liters of solution Molarity (M) Molality (m) m = moles of solute mass of solvent (kg)
  • Slide 8
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 8 N = Number of equivalent liters of solution Normality (N) Acid-Base reaction: is the amount needed to accept one mole of H + Number of equivalent = mass Equivalent mass x 1 V Equivalent mass of H 2 SO 4 = MM 2 Equivalent mass of Ca(OH) 2 = MM 2
  • Slide 9
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 9 Redox reaction : is the amount needed to accept exactly one mole e - MnO 4 - + 5e - + 8H + Mn 2+ + 4H 2 O Equivalent mass of KMnO 4 = MM 5
  • Slide 10
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 10
  • Slide 11
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 11 What is the molality of a 5.86 M ethanol (C 2 H 5 OH) solution whose density is 0.927 g/mL? m =m = moles of solute mass of solvent (kg) M = moles of solute liters of solution Assume 1 L of solution: 5.86 moles ethanol = 270 g ethanol 927 g of solution (1000 mL x 0.927 g/mL) mass of solvent = mass of solution mass of solute = 927 g 270 g = 657 g = 0.657 kg m =m = moles of solute mass of solvent (kg) = 5.86 moles C 2 H 5 OH 0.657 kg solvent = 8.92 m
  • Slide 12
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 12 Steps in Solution Formation Step 1 -Expanding the solute (endothermic) Step 2 -Expanding the solvent (endothermic) Step 3 -Allowing the solute and solvent to interact to form a solution (exothermic) H soln = H step 1 + H step 2 + H step 3
  • Slide 13
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 13 The formation of a liquid solution can be divided into three steps: (1) expanding the solute, (2) expanding the solvent, and (3) combining the expanded solute and solvent to form the solution.
  • Slide 14
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 14 HH < 0Exothermic Solution will occur
  • Slide 15
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 15 The driving factor that favor a process of solution formation is an increase in disorder
  • Slide 16
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 16
  • Slide 17
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 17 Factor Affecting Solubility 1.Structure 2.Pressure 3.Temperature
  • Slide 18
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 18 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 CCl 4 in C 6 H 6 polar molecules are soluble in polar solvents C 2 H 5 OH in H 2 O ionic compounds are more soluble in polar solvents NaCl in H 2 O or NH 3 (l)
  • Slide 19
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 19 The molecular structures of (a) vitamin A (nonpolar, fat-soluble) and (b) vitamin C (polar, water-soluble). The circles in the structural formulas indicate polar bonds. Note that vitamin C contains far more polar bonds than vitamin A. Fat Soluble Hydrophobic Water Soluble Hydrophilic
  • Slide 20
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 20 Pressure Effects Pressure has little effect on solids and liquids. It increases the solubility of gases.
  • Slide 21
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 21 (a) A gaseous solute in equilibrium with a solution. (b) The piston is pushed in, increasing the pressure of the gas and number of gas molecules per unit volume. This causes an increase in the rate at which the gas enters the solution, so the concentration of dissolved gas increases. (c) The greater gas concentration in the solution causes an increase in the rate of escape. A new equilibrium is reached.
  • Slide 22
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 22
  • Slide 23
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 23 Henrys Law C = kP C = concentration of dissolved gas P = partial pressure of gaseous solute above the solution k = a constant The amount of a gas dissolved in a solution is directly proportional to the pressure of the gas above the solution.
  • Slide 24
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 24 Henrys Law Applied for Dilute solutions Gases that do not dissociate or react with solvent: O 2 /waterApplied HCl/waterIs not applied
  • Slide 25
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 25 Temperature Effect on Gases The solubilities of several gases in water as a function of temperature at a constant pressure of 1 atm of gas above the solution.
  • Slide 26
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 26 The solubilities of several solids as a function of temperature. Temperature Effect on Gases
  • Slide 27
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 27 The Vapor Pressure of Solutions Solutions have different physical properties from pure solvent. Solutions of nonvolatile solutes differ from solutions of volatile solvents.
  • Slide 28
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 28 An aqueous solution and pure water in a closed environment. (a) Initial stage. (b) After a period of time, the water is transferred to the solution. Vapor pressure of pure water is higher VP of solution is lower
  • Slide 29
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 29 Raoults Law P soln = solvent P solvent P soln = vapor pressure of the solution solvent = mole fraction of the solvent P solvent = vapor pressure of the pure solvent The presence of a nonvolatile solute lowers the vapor pressure of a solvent. solvent = Moles of solvents Moles of solvent + Moles of solute
  • Slide 30
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 30 Note Solutions that obey Raoults Law are called Ideal Solutions. Can be used to determine the molar mass of unknown. For ionic compounds you should multiply by the total number of ions per molecule e.g. Na 2 SO 4 n = 3 x Na 2 SO 4
  • Slide 31
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 31 For a solution that obeys Raoult's law, a plot of P soln versus x solvent gives a straight line.
  • Slide 32
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 32 Non-Ideal Solutions When a solution contains two volatile components, both contribute to the total vapor pressure.
  • Slide 33
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 33 P A = X A P A 0 P B = X B P B 0 P T = P A + P B P T = X A P A 0 + X B P B 0
  • Slide 34
  • Copyright2000 by Houghton Mifflin Company. All rights reserved. 34 (a) ideal liquid-liquid solution by Raoult's law. (b) This solution shows a positive deviation from Raoult's law. (c) This solution shows a negative deviation from Raoult's law.
  • Slide 35
  • Copyright200

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