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Unit 2: Solutions

Unit 2: Solutions. The Dissolution Process. Solutions are homogeneous mixtures of two or more substances which do not settle. Dissolving medium is called the solvent (usually the most abundant species present). Dissolved species are called the solute .

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Unit 2: Solutions

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  1. Unit 2: Solutions

  2. The Dissolution Process • Solutions are homogeneous mixtures of two or more substances which do not settle. • Dissolving medium is called the solvent (usually the most abundant species present). • Dissolved species are called the solute. • There are three states of matter (solid, liquid, and gas) which when mixed two at a time gives nine different kinds of mixtures. • Seven of the possibilities can be homogeneous. • Two of the possibilities must be heterogeneous.

  3. The Dissolution Process Seven Homogeneous Possibilities SoluteSolventExample • Solid Liquid salt water • Liquid Liquid mixed drinks • Gas Liquid carbonated beverages • Liquid Solid dental amalgams (liquid Hg dissolved in solid metals) • Solid Solid alloys • Gas Solid metal pipes • Gas Gas air Two Heterogeneous Possibilities • Solid Gas dust in air • Liquid Gas clouds, fog

  4. Spontaneity of the Dissolution Process • As an example of dissolution, let’s assume that the solvent is a liquid. • Two major factors affect dissolution of solutes: • Change of energy content or enthalpy of solution, Hsolution • If Hsolution is exothermic (< 0) dissolution is favored. • If Hsolution is endothermic (> 0) dissolution is not favored. Note: Dissolution is a physical process. Compare dissolving Na in H2O with dissolving NaCl in H2O

  5. Spontaneity of the Dissolution Process • Change in disorder (entropy), or randomness, of the solution Smixing • If Smixing increases (> 0) dissolution is favored. • If Smixing decreases (< 0) dissolution is not favored. • Thus the best conditions for dissolution are: • For the solution process to be exothermic. • Hsolution < 0 • For the solution to become more disordered. • Smixing > 0

  6. Spontaneity of the Dissolution Process • Disorder in mixing a solution is very common. • Smixing is almost always > 0. • What factors affect Hsolution? • There is a competition between several different attractions. • Solute-solute attractions such as ion-ion attraction, dipole-dipole, etc. • Breaking the solute-solute attraction requires an absorption of E. • Therefore weak solute-solute attractions favour solubility

  7. Spontaneity of the Dissolution Process • Solvent-solvent attractions such as hydrogen bonding in water. • This also requires an absorption of E. • Therefore weak solvent-solvent attractions favour solubility • Solvent-solute attractions, solvation, releases energy. • If solvation energy is greater than the sum of the solute-solute and solvent-solvent attractions, the dissolution is exothermic, Hsolution < 0. • If solvation energy is less than the sum of the solute-solute and solvent-solvent attractions, the dissolution is endothermic, Hsolution > 0. • Therefore strong solute-solvent attractions favour solubility

  8. Diagram 1 representing energy changes associated with a hypothetical dissolution process • The process depicted is exothermic • The amt of heat absorbed in steps a and b is less that the amt released in step c • If this were anendothermic process the heat of solution would behigherthan the original solvent + solute and the amt of energy absorbed in steps a & b would be greater than that released in step c

  9. Spontaneity of the Dissolution Process • Many solids do dissolve in liquids by an endothermic process • The reason for this is that the endothermicity can be outweighed by a large increase in disorder of the solute during the dissolution process

  10. Dissolution of Solids in Liquids Solute-Solute Attractions • The ability of a solid to go into solution strongly depends on its crystal lattice energy the strength of attractions among particles of the solid • The energy released (exothermic) when a mole of formula units of a solid is formed from its constituent ions (molecules or atoms for nonionic solids) in the gas phase is called the crystal lattice energy. • The crystal lattice energy is a measure of the attractive forces in a solid. • The crystal lattice energy increases as the charge density (magnitude of charge / radius) increases.

  11. Solvent-Solvent Attractions • Energy must be supplied to expand the solvent • If the solvent is water, then H-bonds must be disrupted

  12. Dissolution of Solids in Liquids Solute-Solvent Attractions • Solvation: The process by which solvent molecules interact with solute ions/molecules • If solvent is water then term ‘hydration’ is used. • Molar hydration energy: (= the sum of steps b + c in diagram 1) The energy change involved in the (exothermic) hydration of 1 mole of gaseous ions. • Hydration is usually highly exothermic for ionic or polar covalent compounds

  13. OH2 O H H OH2 H2O 2+ H O H H O H Ca Cl- H2O OH2 H H O OH2 Dissolution of Solids in Liquids Dissolution of CaCl2.

  14. Dissolution of Solids in Liquids • The overall heat of solution for a solid dissolving in a liquid is given by: ∆Hsolution = heat of solvation – crystal lattice energy • Non polar solids e.g. Naphthalene, C10H8 do not dissolve appreciably in polar solvents like water because the two do not attract each other significantly “Like dissolves like” means that polar solvents dissolve ionic and polar molecular solutes, and nonpolar solvents dissolve nonpolar molecular solutes.

  15. Dissolution of Solids in Liquids • Solute -solute attractions • crystal lattice energy for ionic solids • Solvent-solvent attractions • H-bonding for water • Solute-solvent attractions • Solvation or hydration energy • So then…Dissolution is a competition between:

  16. Hydration energy usually increases with increasing charge density (charge-to-size ratio) • But as the charge density increases for ionic solids the lattice energy increases more than the hydration energy. This makes dissolution of solids with highly charged ions e.g. AlF3, MgO too endothermic to be soluble in water Table 14-1, p. 505

  17. Dissolution of Solids in Liquids • Ammonium nitrate (NH4NO3) is an example of a salt that dissolves endothermically • Used as instant cold packs to treat sprains, etc p. 505

  18. Dissolution of Liquids in Liquids (Miscibility) • Most polar liquids are miscible in other polar liquids. • In general, liquids obey the “like dissolves like” rule. • Polar molecules are soluble in polar solvents. • Nonpolar molecules are soluble in nonpolar solvents. • For example, methanol, CH3OH, is very soluble in water: H-bonding in a solution of methanol and water

  19. Dissolution of Liquids in Liquids (Miscibility) • Nonpolar molecules essentially “slide” in between each other. • For example, carbon tetrachloride and benzene are very miscible Oil and water are immiscible

  20. Dissolution of Gases in Liquids • Polar gases are more soluble in water than nonpolar gases. • This is the “like dissolves like” rule in action. • Polar gases can hydrogen bond with water • Some polar gases enhance their solubility by reacting with water.

  21. Dissolution of Gases in Liquids

  22. A few nonpolar gases are soluble in water because they react with water. Because gases have very weak solute-solute interactions, gases dissolve in liquids in exothermic processes. Dissolution of Gases in Liquids

  23. Finely divided solids dissolve more rapidly than large crystals. Compare the dissolution of granulated sugar and sugar cubes in cold water. The reason is simple, look at a single cube of NaCl. The enormous increase in surface area helps the solid to dissolve faster. Breaks many smaller crystals up NaCl Rates of Dissolution and Saturation

  24. Rates of Dissolution and Saturation • Saturated solutions have established an equilbrium between dissolved and undissolved solutes • Example of saturated solutions include: • Some solids dissolved in liquids.

  25. Symbolically this equilibrium is written as: In an equilibrium reaction, the forward rate of reaction is equal to the reverse rate of reaction. Rates of Dissolution and Saturation

  26. Rates of Dissolution and Saturation • Supersaturated solutions have higher-than-saturated concentrations of dissolved solutes. • Usually prepared at high temperatures • Produce crystals rapidly if it is slightly disturbed or if it is “seeded” with a dust parrticle or a tiny crystal. Under such conditions enough solid crystalizes to leave the saturated solution.

  27. According to LeChatelier’s Principle when stress is applied to a system at equilibrium, the system responds in a way that best relieves the stress. Since saturated solutions are at equilibrium, LeChatelier’s principle applies to them. Possible stresses to chemical systems include: Heating or cooling the system. Changing the pressure of the system. Changing the concentrations of reactants or products. Effect of Temperature on Solubility

  28. What will be the effect of heating or cooling the water in which we wish to dissolve a solid? It depends on whether the dissolution is exo- or endothermic. For an exothermic dissolution, heat can be considered as a product. Warming the water will decrease solubility and cooling the water will increase the solubility. Predict the effect on an endothermic dissolution like this one. Effect of Temperature on Solubility

  29. Effect of Temperature on Solubility • For ionic solids that dissolve endothermically dissolution is enhanced by heating. • For ionic solids that dissolve exothermicallydissolution is enhanced by cooling. • Be sure you understand these trends.

  30. Effect of Temperature on Solubility • Example 1: The solubility of O2 in water decreases as temperature increases • Thermal pollution causes a decrease in the conc of dissolved oxygen in rivers, streams etc. • As a result, the water can no longer support marine life it ordinarily could. • Example 2:  • Calcium acetate, is more soluble in cold water than in hot water. When a cold concentrated solution is heated, solid calcium acetate precipitates.

  31. Effect of Pressure on Solubility • Pressure changes have little or no effect on solubility of liquids and solids in liquids. • Liquids and solids are not compressible. • Pressure changes have large effects on the solubility of gases in liquids. • Sudden pressure change is why carbonated drinks fizz when opened. • It is also the cause of several scuba diving related problems including the “bends”.

  32. Effect of Pressure on Solubility • The effect of pressure on the solubility of gases in liquids is described by Henry’s Law. • C gas = kPgas Where :- C= conc of gas of dissolved gas usually expressed as molarity P = pressure of the gas above the solution k = constant for a particular gas

  33. Effect of Pressure on Solubility Illustration of Henry’s Law: The solubility of a gas (red) that does not react completely with the solvent (yellow) increases with increases with increasing pressure of the gas above the solution.

  34. Molality and Mole Fraction • Recall, two important concentration units: • % by mass of solute: g Solute x 100 g solute + g solvent OR g Solute x 100 g solution

  35. Molality and Mole Fraction • Molarity • We must introduce two new concentration units in this chapter.

  36. Molality and Mole Fraction • Molality is a concentration unit based on the number of moles of solute per kilogram of solvent.

  37. Molality and Mole Fraction • What is the molality of a solution that contains 128g of CH3OH in 108g of water? • Plan: • Convert amt of solute (alcohol) mols • Express amt of solvent (water) in kg • Apply definition of molality  mols of solute / kg solvent ? Mols CH3OH = 128g x 1 mol = 4 mol 32.0g ? Kg water = 108g x 1 kg = 0.108kg 1000g ? mol/ kg = 4mol / 0.108 kg = 37.0 m

  38. Molality and Mole Fraction • Mole fraction is the number of moles of one component divided by the moles of all the components of the solution • Mole fraction is literally a fraction using moles of one component as the numerator and moles of all the components as the denominator. • In a two component solution, the mole fraction of one component, A, has the symbol XA.

  39. Molality and Mole Fraction • The mole fraction of component B is XB

  40. Molality and Mole Fraction • Example: What are the mole fractions of CH3OH and water in the previous example. (It contained 128g CH3OH and 108g water) You do it! • Mols of CH3OH  4.0 mol • Mols of water  6.0 mol • Therefore,X CH3OH = 4/(6 + 4) = 0.400 X H2O= 6.0/ (6+4) = 0.600

  41. Colligative Properties of Solutions • Colligative properties are properties of solutions that depend solely on the number of particles dissolved in the solution. • Colligative properties do not depend on the kinds of particles dissolved. • Colligative properties are a physical property of solutions.

  42. Colligative Properties of Solutions • Colligative properties depend on the number of dissolved particles rather than on their physical and chemical properties. • There are four common types of colligative properties: • Vapor pressure lowering • Freezing point depression • Boiling point elevation • Osmotic pressure • Vapor pressure lowering is the key to all four of the colligative properties.

  43. Lowering of Vapor Pressure and Raoult’s Law • Addition of a nonvolatile solute to a solution lowers the vapor pressure of the solution. • The effect is simply due to fewer solvent molecules at the solution’s surface. • The solute molecules occupy some of the spaces that would normally be occupied by solvent. • Raoult’s Law models this effect in ideal solutions.

  44. Lowering of Vapor Pressure and Raoult’s Law • Derivation of Raoult’s Law. Raoult’s Law: the vapour pressure of a solvent in an ideal solution is directly proportional to the mole fraction of the solvent in the solution.

  45. Lowering of Vapor Pressure and Raoult’s Law • Lowering of vapor pressure, Psolvent, is defined as:

  46. Lowering of Vapor Pressure and Raoult’s Law • Remember that the sum of the mole fractions must equal 1. • Thus Xsolvent + Xsolute = 1, which we can substitute into our expression.

  47. Lowering of Vapor Pressure and Raoult’s Law • This graph shows how the solution’s vapor pressure is changed by the mole fraction of the solute, which is Raoult’s law. An ideal solution of a solute in a volatile liquid: The vapour pressure exerted by the liquid (P0solvent) increase as its mole fraction (X solvent) in the solution increases

  48. Boiling Point Elevation • Addition of a nonvolatile solute to a solution raises the boiling point of the solution above that of the pure solvent. • This effect is because the solution’s vapor pressure is lowered as described by Raoult’s law. • The solution’s temperature must be raised to make the solution’s vapor pressure equal to the atmospheric pressure. • The amount that the temperature is elevated is determined by the number of moles of solute dissolved in the solution.

  49. Boiling Point Elevation • Boiling point elevation relationship is: • Kb is different for different solvents and does not depend on the solute • The units are oC / m NOTE: boiling point elevation (∆Tb )can also be thought of as:- ∆Tb = boiling pt of the solution – boiling pt of the pure solvent

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