solutions and their behavior n.
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  2. CHAPTER OVERVIEW • This chapter examines homogeneous mixtures called solutions, which are made up of a solute and a solvent. • Concentrations of solutions can be expressed in a variety of units. • Properties of solutions that depend only on the number of solute particles and not their type are called colligative properties.

  3. 14.1 UNITS OF CONCENTRATION • Molarity, M, moles solute per liter solution • Molality, m, moles of solute per kilogram solvent • Mole fraction, XA, moles A divided by moles total • Weight percent (mass percent), wt.% A, (mass A divided by mass total) x 100%

  4. Molarity -vs- Molality Each flask contains 19.4 g of K2CrO4 Water to the 1.00 L mark Exactly 1.00 kg of water added

  5. UNITS OF CONCENTRATION • parts per million, ppm, is calculated like percent, but multiply by 106 • Remember that the mass of the solution equals the mass of the solute plus solvent. • Conversions between molarity and the other concentration units requires the density of the solution.

  6. 6 Solutions Why does a raw egg swell or shrink when placed in different solutions?

  7. 7 Some Definitions A solution is a HOMOGENEOUS mixture of 2 or more substances in a single phase. One constituent is usually regarded as the SOLVENT and the others as SOLUTES.

  8. 14.2 THE SOLUTION PROCESS • The key to understanding the solution process is intermolecular forces: solvent - solvent; solute - solute; solute - solvent.

  9. THE SOLUTION PROCESS • What prevents solubility is an energy barrier when the latter interaction is significantly weaker than the former interactions. • Like dissolves like is a general rule, but not and explanation of the solubility phenomenon.

  10. Liquids Dissolving in Liquids • Miscible liquids are soluble in all proportions. • Immiscible liquids do not mix, but form separate layers. • Isopropanol is miscible with water but gasoline is not. Explain why.

  11. SOLUTIONS • A saturated solution is one which has reached its equilibrium solubility at that temperature. • An unsaturated solution is one that has not reached its equilibrium solubility. • A supersaturated solution is one in which the equilibrium solubility has been temporarily exceeded.

  12. 12 Definitions Solutions can be classified asunsaturated or saturated. SUPERSATURATED SOLUTIONS contain more than is possible and are unstable. A saturated solution contains the maximum quantity of solute that dissolves at that temperature.

  13. 13 Energetics of the Solution Process If the enthalpy of formation of the solution is more negative than that of the solvent and solute, the enthalpy of solution is negative. The solution process is exothermic!

  14. Solids Dissolving in Liquids The same rules apply. • Compare the intermolecular forces. • I2 is quite soluble in CCl4, but not very soluble in water. Explain why?

  15. 15 Supersaturated Sodium Acetate • One application of a supersaturated solution is the sodium acetate “heat pack.” • Sodium acetate has an ENDOthermic heat of solution.

  16. Ionic Solutions • The heat of solution for ionic compounds is the sum of the lattice energy (+), bonds breaking, and the hydration energy (-), bonds forming. • It may be positive (endo) or negative (exo) depending on the relative magnitudes of these energies.

  17. DHsol’n can be calc’d using Hess' Law. 821 kJ/mol – 819 kJ/mol = +2 kJ/mol (endo)

  18. DHsol’n can be calc’d using Hess' Law.

  19. Ion Size also determines Solubility Remember Coulombs Law (charge n)(charge n) +- Force of Attraction = k 2 d

  20. Ionic Solutions • Temperature has a significant effect on solubility for salts and is consistent with Le Chatelier's principle.

  21. The heat of solution for many salts is positive, endothermic, as seen by the positive slope of the graph.

  22. 22 Supersaturated Sodium Acetate Sodium acetate has an ENDOthermic heat of solution. NaCH3CO2 (s) + heat ----> Na+(aq) + CH3CO2-(aq) Therefore, formation of solid sodium acetate from its ions is EXOTHERMIC. Na+(aq) + CH3CO2-(aq) ---> NaCH3CO2 (s) + heat

  23. Dissolving Gases • Gas solubility decreases with increasing temperature which means ΔHsolution < 0 , exothermic. • Gas solubilities increase with increasing pressure. • Write the general equation for the solubility of a gas showing that the process is exothermic and show how increasing the temperature decreases the solubility. Gas + Solvent ⇄ Solution + Heat

  24. 25 Colligative Properties • On adding a solute to a solvent, the properties of the solvent are modified. • Vapor pressure decreases • Melting point decreases • Boiling point increases • Osmosis is possible (osmotic pressure)

  25. 26 Colligative Properties • These changes are called COLLIGATIVE PROPERTIES. • They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles.

  26. 27 Concentration Units An IDEAL SOLUTION is one where the properties depend only on the concentration of solute. Need concentration units to tell us the number of solute particles per solvent particle. The unit “molarity” does not do this!

  27. 28 Concentration Units MOLE FRACTION, X For a mixture of A, B, and C WEIGHT %= grams solute per 100 g solution mol A X mol fraction A = = A mol A + mol B + mol C MOLALITY, m mol solute m of solute = kilograms solvent

  28. Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate mole fraction, molality, and weight % of glycol.

  29. 30 Calculating Concentrations Dissolve 62.1 g (1.00 mole) of ethylene glycol in 250. g of H2O. Calculate X, m, and % of glycol. 250. g H2O = 13.9 mole 1.00 mol glycol X = glycol 1.00 mol glycol + 13.9 mol H O 2 X glycol = 0.0672

  30. 31 Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate X, m, and % of glycol. Calculate molality 1.00 mol glycol conc (molality) = 4.00 molal = 0.250 kg H O 2 Calculate weight % 62.1 g % glycol = x 100% = 19.9% 62.1 g + 250. g

  31. 32 Dissolving Gases & Henry’s Law Gas solubility (M) = kH • Pgas kH for O2 = 1.66 x 10-6 M/mmHg When Pgas drops, solubility drops.

  32. 33 Lake Nyos, Cameroon

  33. COLLIGATIVE PROPERTIES • Changes in Vapor Pressure: Raoult's Law • The presence of a solute in the solvent lowers the vapor pressure of the solvent. Psolvent = XsolventPosolvent • If the solute is also volatile, a similar equation applies to the solute. Psolute = XsolutePosolute • The total pressure for the solution is given by: Ptotal = Psolvent + Psolute

  34. COLLIGATIVE PROPERTIES • If the solute is nonvolatile, the total pressure is just the pressure of the solvent and is lower than that of the pure solvent.   • Study examples and exercises.

  35. 36 Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution. H H—O surface H H—O H—O H—O H—O H H H

  36. 37 Understanding Colligative Properties To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution.

  37. Understanding Colligative Properties: Raoults’s Law VP of H2O over a solution depends on the number of H2O molecules per solute molecule. Psolventproportional toXsolvent OR Psolvent = Xsolvent • Posolvent VP of solvent over solution = (Mol frac solvent)•(VP pure solvent) RAOULT’S LAW 38

  38. 39 Raoult’s Law An ideal solution is one that obeys Raoult’s law. PA = XA • PoA Because mole fraction of solvent, XA, is always less than 1, then PA is always less than PoA. The vapor pressure of solvent over a solution is always LOWERED!

  39. 40 Raoult’s Law Assume the solution containing 62.1 g of glycol in 250. g of water is ideal. What is the vapor pressure of water over the solution at 30 oC? (The VP of pure H2O is 31.8 mm Hg; see App.) Solution Xglycol = 0.0672 and so Xwater = ? Because Xglycol + Xwater = 1 Xwater = 1.000 - 0.0672 = 0.9328 Pwater = Xwater • Powater = (0.9382)(31.8 mm Hg) Pwater = 29.7 mm Hg

  40. Raoult’s Law Or (see next slide): ΔPA = VP lowering = XBPoA VP lowering is proportional to mole fraction of the solute! For very dilute solutions, ΔPA = K•molalityB where K is a proportionality constant. This helps explain changes in melting and boiling points.

  41. See Exercise 14.6, p. 575

  42. 43 Changes in Freezing and Boiling Points of Solvent See Figure 14.13

  43. Boiling Point Elevation • If a solute is added to the pure solvent at its normal boiling point, the equilibrium vapor pressure will decrease and the liquid will no longer boil. • To reach the new boiling point the temperature must be increased, thus boiling point elevation. Δtbp = Kbpmsolute

  44. Figure 14.13

  45. 46 The boiling point of a solution is higher than that of the pure solvent.

  46. 47 Elevation of Boiling Point Elevation in BP = DtBP = KBP • m (where KBP is characteristic of solvent)

  47. 48 Change in Boiling Point Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the BP of the solution? KBP = +0.512 oC/molal for water (Table 14.4). Solution 1. Calculate solution molality = 4.00 m 2. DtBP = KBP • m DtBP = +0.512 oC/molal (4.00 molal) DtBP = +2.05 oC BP = 102.05 oC

  48. Freezing Point Depression • The freezing point is lowered by the presence of a solute since these particles cannot form the solid and some of them are occupying the low energy slots needed to form the solid solvent. Δtfp = Kfpmsolute • Note the K is a negative value.

  49. 50 Change in Freezing Point Pure water Ethylene glycol/water solution The freezing point of a solution is LOWER than that of the pure solvent. FP depression = ΔtFP = KFP•m