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Solutions

Solutions. Chapter 14. Types of Solution. A solution is a homogeneous mixture of two or more substances. Solvent – medium in which the solute dissolves Usually a liquid Usually the most abundant species 10 grams of H 2 O(l) in 25 grams of CH 3 OH(l) Solute – substance that is dissolved

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Solutions

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  1. Solutions Chapter 14

  2. Types of Solution • A solution is a homogeneous mixture of two or more substances. • Solvent – medium in which the solute dissolves • Usually a liquid • Usually the most abundant species • 10 grams of H2O(l) in 25 grams of CH3OH(l) • Solute – substance that is dissolved Discuss (review) homogeneous and heterogeneous mixtures

  3. Spontaneity of the Dissolution Process • Dissolving a substance in a solvent (usually a liquid) • Dissolving with a reaction • 2Na(s) + 2H2O(l)  2Na+(aq) + 2OH-(aq) + H2(g) • Dissolving without a reaction We will focus on the latter type of dissolution.

  4. Spontaneity of the Dissolution Process • Two major factors determine the dissolution of solutes • Change in energy, Hsolution • An _____ process (a decrease in energy) favors dissolution. • An _____ process (an increase in energy) does not favor dissolution. • Change in disorder or randomness, Smixing • ______ in disorder favors dissolution. • ______ in disorder does not favor dissolution. • When a substance dissolves the disorder of the system almost always increases.

  5. Spontaneity of the Dissolution Process • Heat of solution, Hsolution, primarily depends on the strength of ____________ between solute and solvent particles. A negative value indicates that heat is _____ favoring dissolution. The larger the magnitude of the value, the more dissolution is favored. • Factors affecting Hsolution • Solute-solute attractions • Weak attractions favor solubility (why?) • Solvent-solvent attractions • Weak attractions favor solubility (why?) • Solvent-solute attractions • Strong attractions favor solubility (why?)

  6. Spontaneity of the Dissolution Process Input of energy is required to separate solute-solute (step a) and solvent-solvent (step b) attractions. Energy is released due to solute-solvent attractions (step c). If the amount of energy released in step c is greater than the energy absorbed in steps a and b the process is exothermic and favored for dissolution. What if it is less?

  7. Spontaneity of the Dissolution Process • Many solids, however, will dissolve in liquids by endothermic processes. The increase in disorder of the system upon mixing (magnitude of Smixing) is enough to outweigh the endothermic process. • The solute particles go from highly ordered in a crystalline solid to highly disordered in solution. • Most dissolving processes involve an overall increase in disorder. • Discuss mixing of gases

  8. Dissolution of Solids in Liquids • The crystal lattice energy is the energy change accompanying the formation of one mole of formula units in the crystalline state from constituent particles in the gaseous state. • The crystal lattice energy is always ______ (i.e. the process is exothermic). • M+(g) + X-(g)  MX(s) + energy • The amount of energy released increases with the strength of attraction between particles (e.g. ions in the solid) • Energy released increases with increasing charge density. Why?

  9. Dissolution of Solids in Liquids • Solute-solute attractive interactions present in the solid must be overcome for the solid to dissolve in the liquid. • This can be related to the crystal lattice energy • MX(s) + energy  M+(g) + X-(g) • If the magnitude of the crystal lattice energy is small only a small amount of energy is needed to start the dissolution process. • This can be thought of as step a in Figure 14-1. • Is this step exothermic or endothermic?

  10. Dissolution of Solids in Liquids • Energy is also required to break up or expand the solvent molecules (i.e. solvent-solvent interactions). This can be fairly large if the solvent is water. Why? • This can be thought of as step b in Figure 14-1. • Energy is released, however, due to attractive solute-solvent interactions. • This can be thought of as step c in Figure 14-1. • Solvation is the process by which solvent molecules interact and surround solute particles or ions. • Hydration refers to when the solvent molecules are water.

  11. Dissolution of Solids in Liquids • Solvation energy is the energy change involved in the solvation of one mole of gaseous ions. • Process is almost always exothermic. • Equivalent to the sum of steps b and c in Figure 14-1. • Hydration energy is the energy change when water is the solvent. Hydration is highly exothermic for ionic or polar covalent compounds. Why? Hsolution = (solvation energy) – (crystal lattice energy) If Hsolution is negative the dissolution process is exothermic.

  12. Dissolution of Solids in Liquids • A nonpolar solid such as naphthalene, C10H8, does not dissolve in polar solvents such as H2O. It does, however, dissolve in nonpolar solvents. Why? • “like dissolves like” • Let’s look at the dissolving process of a soluble salt such as NaCl (Figure 14-2). • The individual ions become solvated (hydrated if the solvent is water) due to electrostatic interactions between the ions and water molecules. • Most cations (e.g. Na+) are surrounded by 4 to 9 H2O molecules.

  13. Dissolution of Solids in Liquids • An increasing charge density (charge/radius ratio) increases the hydration energy (heat of hydration). Table 14-1

  14. Dissolution of Solids • An increasing charge density, however, also increases the magnitude of the crystal lattice energy. • For many salts that possess low-charge species (e.g. NaCl) the hydration energy and crystal lattice energy nearly cancel each other. • Demo: The dissolution of NH4NO3 is endothermic. What can you say about the heat of solvation and crystal lattice energy? The dissolution of Ca(CH3COO)2 is exothermic. • With large charge densities the magnitude of the crystal lattice energy increases more than the hydration energy. • This is the reason why the dissolution process for many solids that contain highly charged ions (e.g. Cr2O3) is endothermic. Many are insoluble. Look at equation.

  15. Dissolution of Liquids in Liquids • Miscibility is the ability of one liquid to dissolve in another. If two liquids are miscible one liquid completely dissolves in the other. • Types of attractive forces to consider when determining miscibility • Solute-solute attractions (step a) • Solvent-solvent attractions (step b) • Solute-solvent attractions (step c) When will two liquids be the most miscible?

  16. Dissolution of Liquids in Liquids • ‘Like dissolves like’ determines miscibility • Polar liquids tend to dissolve in other polar liquids • H2O and CH3OH Discuss in terms of the types of interactions. What are some other liquids that will dissolve in water?

  17. Dissolution of Liquids in Liquids • Will hexane, C6H14, dissolve in water, H2O? Why or why not? Explain in terms of types of interactions. • Will hexane dissolve in gasoline which is nonpolar? Why or why not? What are the strengths of attractive interactions?

  18. Dissolution of Gases in Liquids • ‘Like dissolves like’ holds fairly well for gases dissolving in liquids. • Polar gases dissolve in water • Polar gases can also react with water

  19. Dissolution of Liquids in Liquids • Nonpolar gases (e.g. O2) dissolves to a limited extent in H2O due to dispersion forces • Dissolved O2 is responsible for keeping fish alive • CO2, a nonpolar gas, dissolves in water appreciable because it reacts with H2O CO2(g) + H2O(l) H2CO3(aq) H2CO3(aq) H+(aq) + HCO3-(aq) HCO3-(aq) H+(aq) + CO32-(aq) What happens to the acidity of water when CO2(g) is dissolved ? Because gases have such weak solute-solute attractions, gases dissolve in liquids exothermically.

  20. Rates of Dissolution and Saturation • Rate of dissolution of a solid increases if the size of the solid particles is decreased (e.g. ground to a powder). Why? • Sugar cubes versus granulated sugar • The amount of solid in a liquid will increase until the rate of dissolution equals the rate of crystallization. • The opposing processes are in dynamic equilibrium. The solution is saturated. It is holding all the solute it can at a given temperature. DEMO: NaCl crystals increasing in size in a saturated solution.

  21. Rates of Dissolution and Saturation • Saturated solutions have an established equilibrium between dissolved and undissolved particles. • NaCl(s) Na+(aq) + Cl-(aq) • The forward and reverse rates are equal. • Supersaturated solutions contain higher-than-saturated concentrations of solute. How is this possible? • A supersaturated solution is in a ‘metastable’ state. There needs to be mechanism to start the crystallization. • Hot packs, sodium acetate in H2O

  22. Effect of Temperature on Solubility • LeChatelier’s Principle states that a chemical system responds in a way that best relieves the stress or disturbance. • Exothermic dissolution • Endothermic dissolution

  23. Effect of Temperature on Solubility • What will happen to the solubility if the temperature is changed? The system will respond according to LeChatelier’s Principle. • Adding heat to an exothermic process ______ dissolution. Why? • Adding heat to an endothermic process _____ dissolution. Why? • For most solids dissolving in liquids the process is endothermic. • Most gases dissolve in liquids by exothermic processes. What does this mean if the temperature is increased? Demo: Dissolve NaCH3COO. Is this process endothermic or exothermic? What will happen if the temp. is increased? Cool the mixture. What type of solution results?

  24. Effect of Temperature on Solubility • Increasing the temperature increases the solubility of most solids in H2O. Are these processes exothermic or endothermic? • Na2SO4 is the only exothermic process

  25. Effect of Pressure on Solubility • Changing the pressure has no appreciable effect on the solubilities of solids or liquids in liquids. • Pressure changes have large effects on the solubilities of gases in liquids. • Carbonated beverages • Scuba divers get the ‘bends’

  26. Effect of Pressure on Solubility • Henry’s Law expresses that the concentration of dissolved gas is directly related to the pressure of the gas above the solution. Pgas = kCgas • Pgas = pressure of the gas above the sollution • k is a constant for a particular gas and solvent at a specific T • Cgas = the concentration of dissolved gas (molarity)

  27. Molality and Mole Fraction • Previously, we covered molarity and weight precent to express concentration. What are they? • Molality – number of moles of solute per kilogram of solvent • What is the molality of a solution that contains 142 grams of CH3OH in 315 grams of water?

  28. Molality and Mole Fraction • Calculate the molality and the molarity 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 • Calculate the molality of a solution that contains 7.25 g of benzoic acid C6H5COOH, in 200 mL of benzene, C6H6. The density of benzene is 0.879 g/mL. 1 mol C6H5COOH = 122 g

  29. Molality and Mole Fraction • Mole fraction is the number of moles of one component per moles of all the components. • The sum of the mole fractions (XA + XB) = 1 What are the mole fraction for methanol and water in the previous problem? What are the mole fractions of glucose and water in a 10.0% glucose solution?

  30. Colligative Properties of Solutions • Colligative properties depend solely on the number of particles dissolved in the solution and not the kinds of particles dissolved. • Physical property of solutions • Four kinds that will be discussed • Vapor pressure lowering • Freezing point depression • Boiling point elevation • Osmotic pressure

  31. Lowering of Vapor Pressure and Raoult’s Law • Addition of a nonvolatile solute to a solution lowers the vapor pressure of the solution • Fewer solvent molecules present at the surface since some solute molecules occupy the space. • As a result, molecules evaporate at a slower rate • Raoult’s Law describes this effect in ideal solutions • Where Xsolvent represents the mole fraction, is the vapor pressure of the pure solvent, and Psolvent is the vapor pressure of the solvent in the solution. • Figure 14-9

  32. Lowering of Vapor Pressure and Raoult’s Law • The change in vapor pressure of the solvent in solution can be expressed in terms of the solute mole fraction. Where Psolvent is the change in vapor pressure of the solvent in the solution. This relationship assumes ideal solutions and that the solute is nonvolatile (i.e. has not vapor pressure). This is Raoult’s Law.

  33. Lowering of Vapor Pressure and Raoult’s Law • Determine the vapor pressure of a solution, at 25C, that is made by dissolving 5.00 grams of sucrose, C6H12O6, in 15.0 grams of water. The vapor pressure of pure water at 25C is 23.8 torr (Appendix E). • Determine the vapor pressure of a 5.25 molal aqueous sucrose solution at 47C.

  34. Determining the Vapor Pressure of a Two-Component System • Both components are considered as volatile and contribute to the total vapor pressure. • A solution of hexane and heptane • Each component behaves as if it were pure. Therefore, the vapor pressure would be a sum of its components. Ptotal = PA + PB From Raoult’s Law Therefore,

  35. Determining the Vapor Pressure when a Two-Component System • The left-hand side corresponds to pure B (XB=1) and the right-hand side to pure A (XA=1). Which is more volatile? • Notice that the black line is always equal to the sum of PA and PB.

  36. Determining the Vapor Pressure when a Two-Component System • At 45C, the vapor pressure of pure heptane is 112 torr and the vapor pressure of pure octane is 36 torr. The solution contains two moles of heptane and three moles of octane. Calculate the vapor pressure of each component (heptane and octane) and the total vapor pressure above the solution. What is the composition (in mole fractions) of the vapor above the solution? • The volume above the solution is equal to 1.50 L. Assuming that the gases behave ideally calculate the amount of heptane and octane in the gas phase. Also, do Example 14-5 for practice.

  37. Boiling Point Elevation • The boiling point of a liquid is the temperature at which its _____ ______ equals the external pressure. It was stated with Raoult’s law that the addition of a nonvolatile solute decreases the vapor pressure. Therefore, a _____ temperature must be acquired to cause the liquid to boil (vapor pressure equals atmospheric pressure). • The amount of boiling point elevation depends on the number of moles of solute particles dissolved in the solvent.

  38. Boiling Point Elevation • The boiling point change can be expressed as Tb = Kbm • Tb is the change in boiling point (add) • Kb is the molal boiling point constant (Table 14-2) • Kb corresponds to the change in boiling point by a one-molal solution • m is the molality of the solution What is the normal boiling point of a 2.50 m glucose, C6H12O6, solution?

  39. Boiling Point Elevation • Do this later. Predict the boiling point elevation if 33.0 grams of NaCl is added to 100 grams of water? Note: Calculate moles of solute particles not moles of solute.

  40. Freezing Point Depression • Addition of a nonelectrolyte lowers the freezing point according to the following expression Tf=Kfm • Tf is the change in the freezing point (subtract) • Kf is the the molal freezing point depression constant (Table 14-2) • m is the molality of the solution Upon freezing, the solvent solidifies as a pure substance. Solute molecules make it more difficult for the solvent molecules to come together and freeze. A lower temperature must be acquired to freeze the solvent.

  41. Freezing Point Depression • Calculate the freezing point of a 2.50 m aqueous glucose solution. • CaCl2(s) is commonly added to melt snow on roads at lower temperatures. What is the freezing point when 20.0 grams of CaCl2 is dissolved in 100 grams of H2O at 20C? For comparative purposes, calculate the lowering of the freezing point when 20.0 grams of NaCl is dissolved in 100 grams of H2O at the same temperature. • Antifreeze (HOCH2CH2OH) addition to water is another example of freezing point depression.

  42. Changes in Boiling and Freezing Point Temperatures • The relationships for calculating the change in freezing and boiling points are very similar. • It is essentially the same effect. The difference is in the ‘size’ or magnitude of the effect which is indicated by the constants, Kf and Kb.

  43. Determination of Molecular Weight from Freezing Point Depression or Boiling Point Elevation • These colligative properties (especially freezing point depression) can be used to determine molecular weight of a solute. • Kf and the amount of solvent in kilograms is usually known. The freezing point lowering was measured. Therefore, the molality can be determined, and the moles of solute can be calculated. The molecular weight can then be determined.

  44. Determination of Molecular Weight from Freezing Point Depression or Boiling Point Elevation • A 37.0 g sample of a new covalent compound, a nonelectrolyte, was dissolved in 200 g of water. The resulting solution froze at -5.580C. What is the molecular weight of the compound? • Either ethylene glycol (C2H6O2) or propylene glycol (C3H8O2) can be used to make an antifreeze solution. A 15.0 gram sample of either ethylene glycol or propylene glycol dissolved in 50.0 grams of water lowers the freezing point to –7.33C. Which glycol is added to the solution? The solutions behave ideally at these concentrations.

  45. Colligative Properties and Dissociation of Electrolytes • Since a colligative property only depends on the ‘number’ of solute particles in a given mass, electrolytes have a larger effect on boiling point elevation and freezing point depression. • How many moles of solute particles will be produced from one mole of sucrose (C6H12O6)? How many from 1 mole of MgCl2? • As a result, the boiling point elevation and freezing point depression are larger for a given molar quantity of a typical electrolyte.

  46. Colligative Properties and Dissociation of Electrolytes • Solute particles, however, are not randomly distributed in an ionic solution. This causes the boiling point elevation and freezing point depression to be not as large. • Ionic solutions behave nonideally. The ions start to undergo association, and the number of solute particles decreases. The effective molality, therefore, is reduced. • Especially noticeable at higher concentrations

  47. Colligative Properties and Dissociation of Electrolytes • Due to association in ionic solutions, the dissociation (or ionization) is reduced. The extent of dissociation is measured by the van’t Hoff factor, i. • i has an ideal value of 2 for NaCl • What is the ideal value for i with CaCl2? Look at Table 14-3 at values for i and notice that the actual values of i are closer to the ideal values of i at lower concentrations.

  48. Colligative Properties and Dissociation of Electrolytes • The freezing point of 0.0100 m NaCl solution is -0.0360oC. Calculate the van’t Hoff factor and apparent percent dissociation of NaCl in this aqueous solution. • A 0.0500 m acetic acid solution freezes at -0.09480C. Calculate the percent ionization of CH3COOH in this solution.

  49. Osmostic Pressure • Osmosis is the net flow of solvent between two solutions separated by a semipermeable membrane. • Solvent will travel from lower concentration solutions to higher concentration solutions • Semipermeable membranges • Skin, Saran wrap, and cells

  50. Osmotic Pressure Semipermeable membrane The solvent molecules will pass through the semipermeable membrane into the more concentrated solution. The sugar cannot pass through the membrane. The solvent passes through the semipermeable membrane into the more concentrated solution at a faster rate.

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