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Properties of Solutions Chapter 13 BLB 11 th. Expectations:  g ↔ mol (using molar mass)  g ↔ mL (using density)  Other conversions: temp., pressure,

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Presentation on theme: "Properties of Solutions Chapter 13 BLB 11 th. Expectations:  g ↔ mol (using molar mass)  g ↔ mL (using density)  Other conversions: temp., pressure,"— Presentation transcript:

1 Properties of Solutions Chapter 13 BLB 11 th

2 Expectations:  g ↔ mol (using molar mass)  g ↔ mL (using density)  Other conversions: temp., pressure, etc.  Solve for any variable in a formula.  Distinguish between molecular and ionic compounds.  Convert between different concentration units.  Describe the properties of solutions.

3 13.1 The Solution Process  Solution – homogeneous mixture Solute – present in smaller quantity Solvent – present in larger quantity  Intermolecular forces are rearranged when a solute and solvent are mixed.

4 Making a Solution 1. Solute molecules separate (endothermic) 2. Solvent molecules separate (endothermic) 3. Formation of solute-solvent interactions (exothermic) ΔH soln = total energy ΔH soln – enthalpy change for the formation of a solution; exothermic – usually favorable; endothermic – usually unfavorable

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6 Will a solution form?  Solute-solvent interaction must be stronger or comparable to the separation of solute and solvent particles.  Intermolecular forces play a key role.  Entropy (disorder) is also a factor. Disorder is favorable. (2 nd law of thermodynamics) Solution formation increases entropy.  Dissolve vs. react (p )

7 Entropy in Solution Formation Ionic compound very ordered As the ionic compound dissolves, it becomes more disordered.

8 13.2 Saturated Solutions and Solubility  Saturated solution – solution is in equilibrium with undissolved solute. Solute + solvent ⇌ solution  Unsaturated – less solute than saturated  Supersaturated – more solute than saturated dissolution crystallization

9 A Saturated Solution A dynamic equilibrium – ions continually exchange between the solid and solution form.

10 13.3 Factors Affecting Solubility 1.Like dissolves like, i.e. same polarity. Polar solutes are soluble in polar solvents. Nonpolar solutes are soluble in nonpolar solvents. If two liquids: miscible or immiscible Examples: ✔ water + alcohol, NaCl + water, hexane + pentane ✘ water + hexane, NaCl + benzene, oil + water

11 Fat- and Water-Soluble Vitamins

12 13.3 Factors Affecting Solubility 2.Pressure Effects (for gases in any liquid solvent) Solubility increases as the partial pressure above the solution increases. Henry’s Law: S g = kP g S g – solubility of gas k – Henry’s Law constant; conc./pressure units P g – partial pressure of gas above solution

13 The partial pressure of O 2 in your lungs varies from 25 to 40 torr. What molarity of O 2 can dissolve in water at each pressure? The Henry’s Law constant for O 2 is 6.02 x M/torr.

14 13.3 Factors Affecting Solubility 3.Temperature Effects For solids: Solubility ↑ as temperature ↑ - usually.  If ΔH soln > 0 (endothermic)  If ΔH soln < 0 (exothermic) For gases: Solubility ↓ as temperature ↑ - always.  Kinetic energy plays a primary role.  Entropy is also a factor.

15 Ionic compounds

16 Gases (In liquids)

17 13.4 Ways of Expressing Concentration  Mass %, volume %, and ppm

18 13.4 Ways of Expressing Concentration, cont.  Mole fraction, molarity, and molality

19 44. A solution contains 80.5 g ascorbic acid (C 6 H 8 O 6 ) in 210 g water and has a density of 1.22 g/mL at 55°C. Calculate mass %, X, m, and M.

20 51. Commercial aqueous nitric acid has a density of 1.42 g/mL and is 16 M. Calculate mass % of HNO 3.

21 43. A sulfuric acid solution containing g of H 2 SO 4 per liter of solution has a density of g/cm 3. Calculate the mass %, mole fraction, molality, and molarity.

22 Concentration Problems  Practice!  See Figure 13.19, p. 545, for conversion map.  Several examples on pp

23 13.5 Colligative Properties  The addition of a solute to a pure solvent: 1. Lowers the vapor pressure 2. Lowers the freezing point 3. Raises the boiling point 4. Causes movement through a semipermeable membrane (osmosis)  Depends on the number of solute particles (moles), not the identity; more particles the greater the effect  Ionic compounds cause an even greater effect.

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25 1. Lowering the vapor pressure  Addition of solute blocks the solvent from evaporation.  More solute, less vapor, lower vapor pressure  Raoult’s Law (for a nonvolatile solute): P A = X A P A °P A – solvent v. p. over solution (P A < P A °) P A ° – pure solvent v. p. X A – mole fraction of solvent

26 1. Lowering the vapor pressure, cont.  When a volatile solute is added, both the solvent and solute contribute to the vapor pressure.  “Expanded” Raoult’s Law: P total = P A + P B = X A P A ° + X B P B °  If a solution obeys Raoult’s Law, it is an ideal solution.  Nonideal solutions have strong intermolecular interactions which lower the vapor pressure of the solution even further.

27 Vapor Pressure Lowering

28 62a. Calculate the vapor pressure above a solution of 32.5 g C 3 H 8 O 3 (glycerin-nonvolatile) in 125 g water at 343 K. The vapor pressure of water at 343 K is torr.

29 63. A solution is made from equal masses of water and ethanol (C 2 H 5 OH). Calculate the vapor pressure above the solution at 63.5°C. The vapor pressures of water and ethanol are 175 and 400. torr, respectively, at 63.5°C.

30 2. Boiling point elevation 3. Freezing point depression  Since a solution has a lower vapor pressure: A higher temperature is needed to boil solution A lower temperature is needed to freeze solution.  To calculate effect: b.p. ↑ΔT b = K b ·msolution − solvent f.p. ↓ ΔT f = K f ·msolvent − solution ΔT – difference between boiling or freezing points of the pure solvent and solution K – boiling or freezing pt. dep. constant (specific to solvent) m – molality

31 69a. Calculate the freezing and boiling points of a solution that is 0.40 m glucose in ethanol. For ethanol: f.p °C, b.p. 78.4°C, K f = 1.99 °C/m, K b = 1.22 °C/m

32 72. Calculate the molar mass of lauryl alcohol when 5.00 g of lauryl alcohol is dissolved in kg benzene (C 6 H 6 ). The freezing point of the solution is 4.1°C. For benzene: f.p. 5.5°C, K f = 5.12 °C/m

33 4. Osmosis Osmosis – movement of solvent molecules through a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration  Driving force – to dilute the higher concentration  Continues until: Equilibrium is reached between two solutions, or External pressure prevents further movement.

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35 Osmosis in red blood cells Hypertonic solution Hypotonic solution

36 4. Osmosis, cont. Osmotic pressure  = M R T  – osmotic pressure (atm) M – molarity R – L∙atom/mol∙K T – temperature (K) Good technique for measuring molar mass of large molecules like proteins

37 4. Osmosis, cont.  Applications: Kidney dialysis Intercellular transport  Reverse osmosis – apply external pressure to reverse the flow of solvent molecules Water purification – alternative to salt ion exchange Desalination – purification of salt water

38 78. A dilute aqueous solution of an organic compound is formed by dissolving 2.35 g in water to form L of solution. The resulting solution has an osmotic pressure of atm at 25°C. Calculate the molar mass of the compound.

39 13.6 Colloids  Colloid or colloidal dispersion Intermediate between a solution and a suspension Dispersing medium – analogous to solvent Dispersing phase – analogous to solute; typically large molecules with high molar masses Does not settle Tyndall effect – particles scatter light

40 Tyndall Effect

41 Types of Colloids

42 Surfactants  Change the surface properties so that two things that would not normally mix do Emulsifying agent Soap Detergent

43 Hydrophobic – water-fearing (nonpolar) Hydrophilic – water-loving (polar)

44 Action of soap on oil


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