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Chapter 13 Properties of Mixtures

Chapter 13 Properties of Mixtures. • Two or more substances physically mixed together, but not chemically combined. • Nearly all solids, liquids, and gases that make up our world. Mixtures contain 10s to 1000s of different compounds and they are everywhere.

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Chapter 13 Properties of Mixtures

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  1. Chapter 13Properties of Mixtures • Two or more substances physically mixed together, but not chemically combined. • Nearly all solids, liquids, and gases that make up our world

  2. Mixtures contain 10s to 1000s of different compounds and they are everywhere. From the chemistry of a tiny bacterial cell……………

  3. …………….to the chemistry of a vast ocean. Solution • Small, suspended (dissolved) • Not visible (except if colored - dye) Colloids Colloids Suspension • Large, but not large enough to settle out. Remain suspended • Usually visible to naked eye • Settle out of solution (gravity) Interplay between gravity and intermolecular (van der Waals) forces.

  4. Section 13.1: Types of Solutions Definitions: solvent – the most abundant component of a given solution (Ex: H2O is the “universal solvent” – Chapter 12) solute – the component dissolved in the solution (Ex: Ca2+) Solubility (S) – the max amount of a specific solute that can dissolve in a fixed quantity of a particular solvent at a specified temperature and pressure Specific solutes: NaCl is more soluble in water than AgCl. • Quantitative description: SNaCl = 39.12 g/100 mL versus SAgCl = 0.0021 g/100 mL at 100 ºC • Qualitative description: dilute – contains relatively less solute concentrated – contains relatively more solute Particular solvent: NaCl is more soluble in water than in ethanol. Specified temperature: CaCO3 is more soluble at higher temperatures. Specified pressure: CaCO3 is more soluble at higher pressures.

  5. Section 13.1: Types of Intermolecular Forces in Solution (strength in kJ/mol) Hydration shell - # of H2O molecules surrounding ion depends on ion’s size (Li+ 4 H2O molecules Na+  6 H2O molecules) (ionic bond strength: 400 – 4000 kJ/mol) Sodium chloride (NaCl) All intermolecular forces discussed in Chap12 occur between solute and solvent particles • ion-dipole – the principle force involved in the • solubility of ionic compounds in water NaCl dissolves in water Dissolution: Ion-dipole attractive forces overcome ionic bonding forces.

  6. Section 13.1: Types of Intermolecular Forces in Solution O H H + + + + + + + + + + (2) hydrogen bonding – forces between atoms that have a H atom bonded to a small, highly electronegative atom with lone electron pairs N, O, and F all fit this profile. Bonding between H2O molecules involved H–O bonds. Bonding between H2O molecules and organic compounds with O and N in functional groups is primary mechanism for solubility of these compounds in cell fluids in body. Example: Amino acids. + -

  7. Section 13.1: Types of Intermolecular Forces in Solution (3) dipole-dipole forces – in absence of H bonding (a stronger dipole-dipole force), dipole-dipole forces in general account for solubility of polar solutes in polar, nonaqueous (not water) solvents.

  8. Section 13.1: Types of Intermolecular Forces in Solution Two types of charge-induced dipole forces: Rely on the polarizability of molecules. (4) ion-induced dipole forces – an ion’s charge distorts the e- cloud of a nearby nonpolar molecule Example: Binding of Fe2+ in hemoglobin to an O2 molecule in the blood stream

  9. Section 12.3: Trends in Polarizability Polarizability – the ease with which the e- cloud of a particle can be distorted Smaller atoms (ions) are less polarizable than larger ones  e-’s closer to the nucleus and, therefore, held more tightly Polarizability • Increases down a Group • Decreases from L  R • Cations less polarizable than their original atoms Anions are more polarizable than original atoms

  10. Section 13.1: Types of Intermolecular Forces in Solution O H H Example: Solubility of oxygen gas in water + + Fish can breathe! Two types of charge-induced dipole forces: Rely on the polarizability of molecules. (5) dipole-induced dipole forces – a polar molecule’s partial charge distorts the e- cloud of a nearby nonpolar molecule

  11. Section 13.1: Types of Intermolecular Forces in Solution Example: Cellular membrane structure (6) dispersion forces – contribute to the solubility of all solutes in all solvents – principle intermolecular force in solutions of nonpolar (hydrophobic) substances

  12. Section 13.1: Liquid Solutions • Oil does not dissolve in water because weak dipole-induced dipole forces cannot sub for strong hydrogen bonds between H2O molecules Like dissolves like: The forces created between solute and solvent must be comparable in strength to forces destroyed within both solute and solvent Instances of Insolubility (immiscibility): • Some salts are insoluble in polar substances (i.e. hexane) because weak ion-induce dipole forces cannot substitute for strong ionic bonds (ionic bond strength: 400 – 4000 kJ/mol) Sodium chloride (NaCl)

  13. Section 13.1: Solubility Series for a Liquid Solute in Two Different Solvents Solute: Alcohols – have a polar group (– OH) and a nonpolar chain ( – CH2CH3) Solvents: Water – polar molecules interact by H-bonding Hexane – nonpolar molecules inacteract by dispersion forces Predicting Solubility Behavior Solubility of alcohols change as length of nonpolar chain increases. Solute-Solvent Interactions: Water – H-bonding with polar group (– OH) of alcohol Hexane – dispersion forces with nonpolar chain ( – CH2CH3) of alcohol

  14. Section 13.1: Solubility Series for a Gas • Due to weak intermolecular forces, also not very soluble in water • Solubility correlates with boiling point, since strength of intermolecular forces (dispersion forces) govern both. • Molecular gases are nonpolar. • Nonpolar nature means intermolecular forces are weak and b.p. is low (does not take much energy to separate nonpolar gaseous atoms or molecules)

  15. Section 13.1: Other Types of Solutions (2) gas-solid solutions – gas dissolves in a solid and occupies the space between the closely packed particles Example: O2 dissolves in Cu wire  Cu2O Electrical conductivity of wire is reduced (3) solid-solid solutions – formed by melting solids, mixing them together, and allowing them to freeze Example: Metal alloys  Zn + Cu = Brass Rocks  molten rock (magma) made up of many elements, cools at surface of earth Focus is on liquid solutions (vast majority), but there are other types of solutions. • gas-gas solutions – gases are infinitely soluble in one another  intermolecular • forces between particles in a gas are all very weak

  16. Dissolution Section 13.3: Why Substances Dissolve G – Gibbs free energy  takes into account relative magnitudes of H and S • H – enthalpy – keeps track of the quantity of energy (Chapter 6) • S – entropy – keeps track of the distribution of energy in a system – energy becomes distributed more uniformly (more disorder) with time [Quantitative treatment in Chapter 20 – CHEM 163?]

  17. Section 13.3: Why Substances Dissolve • 3 events must occur for substances to dissolve: • Solute particles separate from one another • Solvent particles separate to make room for solute particles • Solute and solvent particles mix • (1)and (2) require energy  endothermic (less stable) • (3) Releases energy  exothermic (more stable) Total enthalpy change: heat of solution (∆Hsoln) = ∆Hsolute + ∆Hsolvent + ∆Hmix ∆Hsoln is negative  exothermic; ∆Hsoln is positive  endothermic

  18. Section 13.3: Why Substances Dissolve ∆Hhydr is always exothermic – breaking H bonds in water more than compensated for by forming strong ion-dipole forces ∆Hsoln = ∆Hsolute + ∆Hsolvent + ∆Hmix Simplify this equation by defining a new term: heat of hydration (∆Hhydr) = ∆Hsolvent + ∆Hmix This is done because ∆Hsolventand ∆Hmix are difficult to measure individually. New equation: ∆Hsoln = ∆Hsolute + ∆Hhydr Higher charge density (charge/volume) = more negative ∆Hhydr

  19. Section 13.3: Why Substances Dissolve ∆Hsolute – energy required to separate an ionic solute into gaseous ions (∆Hlattice) – always positive Highly Endothermic (Beaker is cold) Highly Exothermic (Beaker is hot) Overall: ∆Hsoln = ∆Hlattice + ∆Hhydr

  20. Section 13.3: Why Substances Dissolve S – entropy – the other factor that determines whether a solute dissolves in a solvent A solution usually has a higher entrophy than the pure solute and pure solvent. Ssoln > (Ssolute + Ssolvent) From everyday experience, we know the solutions form naturally, and pure solutes and solvents do not. Lots of energy is expended to purify solutions: water treatment, oil refineries, etc Solutions tends towards lower enthalpy and higher entropy.

  21. Section 13.4: Solubility as an Equilibrium Process Solute is constantly dissolving and recrystallizing. Equilibrium: Rate of dissolution = Rate of recrystallization  No ∆ concentration Dissolution: Rate of dissolution > Rate of recrystallization  Conc. increases Recrystallization: Rate of dissolution < Rate of recrystallization  Conc. decreases

  22. Section 13.4: Solubility as an Equilibrium Process Net dissolution Unsaturated – the solution contains less than max amount dissolved solute In equilibrium Saturated – the solution contains max amount of dissolved solute Net recrystallization Supersaturated – solution contains more than max amount dissolved solute More on supersaturated solutions: Figure 13.20 Formed by adding additional solute to an already saturated solution at a higher temp and then cooling solution to room temp. Add seed crystal or tap the container  excess solute crystallizes immediately Qualitative Description of How Much Solute is in Solution

  23. Section 13.4: Temperature Effects on Solubility Solid Solutes Most solids have a + ∆Hsoln because ∆Hlattice > ∆Hhydr ∆Hsoln = ∆Hlattice + ∆Hhydr (∆Hhydr = ∆Hsolvent + ∆Hmix) How does this explain why adding more heat results in more dissolution?

  24. Section 13.4: Temperature Effects on Solubility Gas Solutes The opposite is true for gases: Energy must be added to separate solids (heat helps), but gases are already separated (∆Hsolute ≈ 0) ∆Hhydr step is always exothermic (< 0) That means ∆Hsoln (∆Hsolute + ∆Hhydr) must be negative. Why is this? (Think about intermolecular forces.)

  25. Section 13.4: Temperature Effects on Solubility Gas Solutes Thermal pollution O2 less soluble and fish cannot breathe What are implications for ocean acidification?

  26. Section 13.4: Pressure Effects on Solubility Gas Solutes Henry’s Law: The quantitative relationship between gas pressure and solubility Sgas = kH x Pgas where Sgas solubility of a gas (mol/L) Pgas  pressure of gas (atm) kH  Henry’s law constant (mol/L atm) The partial pressure of CO2 in the atmosphere is 0.000314 atm. What is the concentration of CO2 in the ocean? Henry’s law constant for CO2 Is 2.3 x 102 mol/L atm.

  27. The Bends http://www.elmhurst.edu/~chm/vchembook/174temppres.html

  28. Section 13.5: Quantitative Ways of Expressing Concentration Molarity (M) = moles of solute / volume of solution Molality (m) = moles of solute / kg of solvent • Find the molarity and molality of a solution where 20 g of CaCO3 were dissolved in 2 L of water at 4 ºC (density of water = 1 kg/m3). Mass percent (% w/w) = [mass solute / (mass solute + mass solvent) ] x 100 Volume percent (%v/v) = [volume solute / volume solution] x 100 % w/v = [mass solute / solution volume] x 100 • Find the % w/w and the % w/v for the CaCO3 solution described above. • What is the % v/v for a solution where 70 mL of alcohol was added to 2 L of water?

  29. Section 13.6: Colligative Properties The presence of solute particles (ions, molecules) make the physical properties of a solution different from those of a pure solvent. It is the # of solute particles that matters, not their chemical identity. Colligative properties: • vapor pressure lowering • boiling point elevation • freezing point depression • osmotic pressure

  30. Section 13.6: Vapor Pressure Lowering Closed Container What happens when there are nonvolatile solute molecules present? The vapor pressure of a solution of a nonvolatile (does not vaporize) nonelectrolyte is always lower than the vapor pressure of the pure solvent. Review: vapor pressure - the pressure exerted by a vapor when it has reached equilibrium in a system that isclosed with respect to the vapor molecules

  31. Section 13.6: Vapor Pressure Lowering Entropy argument: Pure solvent – vaporizes because the vapor has a higher entropy than the liquid Solvent in a solution – already has a greater entropy than the pure solvents Therefore, less tendency to vaporize to increase entropy. Quantitatively: Raoult’s law: Psolvent = Xsolvent x Pºsolvent where Psolvent = vapor pressure of the solvent above the solution Xsolvent = mole fraction of solvent in the solution Pºsolvent = vapor pressure of the pure solvent Mole fraction = moles of solvent / (moles of solvent + moles of solute) *Applies only to ideal solutions (in other words, dilute solutions). - freshwater versus saltwater (in ocean or in your body)

  32. Section 13.6: Vapor Pressure Lowering How does the amount of solute affect the magnitude of the vapor pressure lowering (∆P)? If Xsolvent + Xsolute = 1 and Psolvent = Xsolvent x Pºsolvent then Pºsolvent - Psolvent = ∆P = Xsolute + Pºsolvent What is the vapor pressure lowering (∆P, in units of torr) when 10.0 mL of glycerol (C3H8O3) is added to 500.0 mL of water at 50 ºC? At this temperature, the vapor pressure of pure water is 92.5 torr and its density is 0.988 g/mL. The density of glycerol is 1.26 g/mL.

  33. Section 13.6: Boiling Point Elevation ∆Tb = Kbm where ∆Tb = boiling point elevation Kb = molal b.p. elev. constant m = molality of the solution A solution boils at a higher temperature than the pure solvent. – Why? Review: boiling point - the temperature at which the vapor pressure of a liquid is equal to the external pressure.

  34. Section 13.6: Freezing Point Depression ∆Tf = Kfm where ∆Tf = freezing point depression Kf = molal f.p. dep. constant m = molality of the solution

  35. Section 13.6: Boiling and Freezing Point Depression Problem You add 1.00 kg of C2H6O2 (antifreeze) to your car radiator, which contains 4450 g of water. What are the b.p. and f.p. of the solution?

  36. π = pressure difference at equilibrium (osmotic pressure) n = moles of solute V = solution volume R = universal gas constant (8.314 J/mol K) T = temperature (???) M = molarity of solution Section 13.6: Osmotic Pressure This colligative property appears when two solutions of different concentrations are separated by a semipermeable membrane (solvent passes, solute does not).

  37. Section 13.6: Vapor Pressure for Volatile Solutes….. …..or how to make Moonshine in Appalachia with this guy. How is the vapor pressure affected when the solute vaporizes too? Dalton’s Law of Partial Pressures: The total vapor pressure is the sum of the partial vapor pressures. Ptotal = Psolvent + Psolute = (Xsolvent x Pºsolvent) + (Xsolute + Pºsolute) (Raoult’s law) Example: A solution contains equal amounts (moles) of benzene and toluene Xbenzene = Xtoluene = 0.50; v.p.pure benzene = 95.1 torr, v.p.pure toluene = 28.4 torr How does vapor pressure change for both? Does the composition of the vapor change?

  38. Fractional distillation – a process used to separate a mixture of volatile components. At 20 ºC, the v.p. of pure ethyl alcohol is 9 kPa and the v.p. of pure water at this temperature is 2.4 kPa. The solution contains equal amounts of water and ethyl alcohol. What is the partial pressure of each liquid in the solution? What is the mole fraction of ethyl alcohol and water in the vapor?

  39. Section 13.6: Electrolyte Solutions Colligative properties: • vapor pressure lowering • boiling point elevation • freezing point depression • osmotic pressure Equations differ slightly when dealing with electrolyte solutions (non-volatile): i = particles in solution / solute added

  40. Optional Homework Problems Even numbered problems: 13.90 – 13.104

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