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Solutions

14. Solutions. Endothermic Exothermic. Chapter Goals. The Dissolution Process Spontaneity of the Dissolution Process Dissolution of Solids in Liquids Dissolution of Liquids in Liquids (Miscibility) Dissolution of Gases in Liquids Rates of Dissolution and Saturation

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Solutions

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  1. 14 Solutions

  2. Endothermic Exothermic

  3. Chapter Goals The Dissolution Process Spontaneityof the Dissolution Process • Dissolution of Solids in Liquids • Dissolution of Liquids in Liquids (Miscibility) • Dissolution of Gases in Liquids • Rates of Dissolution and Saturation • Effect of Temperature on Solubility • Effect of Pressure on Solubility • Molality and Mole Fraction

  4. Chapter Goals Colligative Properties of Solutions • Lowering of Vapor Pressure and Raoult’s Law • Fractional Distillation • Boiling Point Elevation • Freezing Point Depression • Determination of Molecular Weight by Freezing Point Depression or Boiling Point Elevation • Colligative Properties and Dissociation of Electrolytes • Osmotic Pressure

  5. Chapter Goals Colloids • The Tyndall Effect • The Adsorption Phenomena • Hydrophilic and Hydrophobic Colloids

  6. The Dissolution Process溶解過程 • Solutions are homogeneous mixtures of two or more substances. • Dissolving medium is called the solvent. • 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.

  7. The Dissolution Process Seven Homogeneous Possibilities SoluteSolventExample • Solid Liquid salt water • Liquid Liquid mixed drinks • Gas Liquid carbonated beverages • Liquid Solid dental amalgams • Solid Solid alloys • Gas Solid metal pipes • Gas Gas air Two Heterogeneous Possibilities • Solid Gas dust in air • Liquid Gas clouds, fog

  8. Spontaneity of the Dissolution Process • Many solid do dissolve in liquids by endothermic processes. • A large increase in disorder of the solute during the dissolution process

  9. 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 • Change in disorder, or randomness, of the solution Smixing • If Smixingincreases (> 0) dissolution is favored • If Smixingdecreases (< 0) dissolution is not favored

  10. Spontaneity of the Dissolution Process • 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

  11. 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. • Weak solute-solute attractions favor solubility • Weak solvent-solvent attractions favor solubility • Strong solvent-solute attractions favor solubility • Solute-solute attractions such as ion-ion attraction, dipole-dipole, etc. • Breaking the solute-solute attraction requires an absorption of Energy.

  12. Dissolution of Solids in Liquids • 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.(always negative) Ionic solid: M+(g)+ X-(g) M+X-(s) + crystal lattice energy • The crystal lattice energy is a measure of the attractive forces in a solid. • These attractions are strong, a large amount of energy is released as the solid forms • The crystal lattice energy increases as the charge density increases.

  13. Dissolution of Solids in Liquids MX(s) + energy M+(g)+ X-(g) • This process can be considered the hypothetical first step in forming a solution of a solid in a liquid • It always endothermic • The smaller the magnitude of the crystal lattice energy, the more readily dissolution occurs

  14. Spontaneity of the Dissolution Process • Solvent-solvent attractions such as hydrogen bonding in water. • This also requires an absorption of Energy. • 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.

  15. Spontaneity of the Dissolution Process • Solvent-solute attractions, solvation, releases energy. • When the solvent is water, the more specific term is hydration • Hydration energy: the energy change involved in the hydration of one mole of gaseous ions Mn+(g)+ xH2O(l)M(OH2)xn++ energy (for cation) Xy-(g)+ rH2O(l)X(H2O)ry-+ energy (for anion) • Hydration is usually highly exothermicfor ionic or polar covalent compounds, because the polar water molecules interact very strongly with ions and polar molecules • Nonpolar solids do not dissolve appreciably in polar solvent, because they do not attract each other

  16. Spontaneity of the Dissolution Process 亂度增加

  17. Dissolution of Solids in Liquids • Dissolution is a competition between: • Solute -solute attractions • crystal lattice energy for ionic solids • Solvent-solvent attractions • H-bonding for water • Solute-solvent attractions • Solvation or hydration (the solvent is water) energy

  18. Dissolution of Solids in Liquids • Solvation is directed by the water to ion attractions as shown in these electrostatic potentials.

  19. Dissolution of Solids in Liquids • In an exothermic dissolution, energy is released when solute particles are dissolved. • This energy is called the energy of solvation or the hydration energy (if solvent is water). • Let’s look at the dissolution of CaCl2.

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

  21. Dissolution of Solids in Liquids • The energy absorbed when one mole of formula units becomes hydrated is the molar energy of hydration.

  22. 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

  23. 50ml H2O 50ml H2SO4 Sulfuric acid is always diluted by adding the acid slowly and carefully to water. Water should never be added to the acid 50ml H2SO4 +50ml H2O

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

  25. 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.

  26. 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

  27. 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

  28. Rates of Dissolution and Saturation • Saturated solutions have established an equilibrium between dissolved and undissolved solutes • Examples of saturated solutions include: • Air that has 100% humidity. • Some solids dissolved in liquids.

  29. 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

  30. Rates of Dissolution and Saturation • The solubilities of many solids increase at higher temperatureSupersaturatedsolutions have higher-than-saturated concentrations of dissolved solutes. • Metastable (temporarily stable)

  31. 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

  32. 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

  33. 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.

  34. Effect of Temperature on Solubility

  35. 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”.

  36. Effect of Pressure on Solubility • The effect of pressure on the solubility of gases in liquids is described by Henry’s Law. Cgas = kPgas Where Cgas =the concentration of gas k= is constant for a particular gas and solvent at a particular temperature Pgas= the pressure of the gas

  37. Effect of Pressure on Solubility

  38. Molality重量莫耳濃度and Mole Fraction莫耳分率 • In Chapter 3 we introduced two important concentration units. • % by mass of solute

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

  40. Molality and Mole Fraction • Molality is a concentration unit based on the number of moles of solute per kilogram of solvent. moles of solute m= kg of solvent In dilute aqueous solutions molarity and molality are nearly equal

  41. Example 14-1: Molality What is the molality of a solution that contain 128g of CH3OH in 108g of water? The molecule weight of CH3OH is 32 ? mole CH3OH=128/32=4 mole 4mole/108x10-3 kg =37m CH3OH Example 14-2: Molality How many grams of H2O must be used to dissolve 50.0g of sucrose to prepare a 1.25m solution of sucrose, C12H22O11? The molecule weight of C12H22O11is 342 1.25m=(50/342)/xkg x=0.117kg H2O

  42. Molality and Mole Fraction Example 14-1: Calculate the molarity and the molality of an aqueous solution that is 10.0% glucose, C6H12O6. The density of the solution is 1.04 g/mL. 10.0% glucose solution has several medical uses. 1 mol C6H12O6 = 180 g 10% means: 10g C6H12O6 in 90g H2O 10.0g C6H12O6 10.0g C6H12O6 =0.617 m C6H12O6 = m= ?mol C6H12O6 ?mol C6H12O6 180.0g C6H12O6 180.0g C6H12O6 90.0gx10-3 H2O L solution kg H2O (100g/1.04)x10-3 sol’n This is the concentration in molality = M= =0.587 M C6H12O6 This is the concentration in molarity

  43. Molality and Mole Fraction Example 14-2: Calculate the molality of a solution that contains 7.25 g of benzoic acid C6H5COOH, in 2.00 x 102 mL of benzene, C6H6. The density of benzene is 0.879 g/mL. 1 mol C6H5COOH = 122 g 2.0x102 ml x0.879 C6H6 7.25g C6H5COOH = ?mol C6H5COOH 122g C6H5COOH 1000 kg C6H6 =0.338 m C6H5COOH

  44. 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.

  45. Molality and Mole Fraction • The mole fraction of component B - XB

  46. Example 14-3: Mole Fraction What are the fractions of CH3OH and H2O in the solution described in Example 14-1? It contains 128g of CH3OH and 108g of H2O? ? mole CH3OH=128/32= 4mol ? Mole H2O=108/18= 6 mol XCH3OH= 4mol/(4mol+6mol) = 0.4 XH2O= 6mol/(4mol+6mol) = 0.6

  47. Molality and Mole Fraction • Example 14-3: What are the mole fractions of glucose and water in a 10.0% glucose solution (Example 14-1)? Let the solution equal 100g 10.0g of glucose, 90.0g of water ?mole C6H12O6 = 10.0g/180=0.0556 mol C6H12O6 ?mole H2O= 90.0g/18=5.0 mol H2O XH2O = = 0.989 5.0 mole H2O 0.0556 mole C6H12O6 5.0 mole H2O + 0.0556 mol C6H12O6 5.0 mole H2O + 0.0556 mol C6H12O6 = 0.011 XC6H12O6=

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